Textbook on Differential Equations & Integral Transforms [2 ed.]

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a's n h s TEXT BOOK on Kri

D i fferential E quations & I ntegral T ransforms (For B.A. and B.Sc. IInd year students of All Colleges affiliated to universities in Uttar Pradesh)

As per U.P. UNIFIED Syllabus (w.e.f. 2012-2013)

By A. R. Vasishtha Retd. Head, Dep’t. of Mathematics Meerut College, Meerut (U.P.)

KRISHNA Prakashan Media (P) Ltd. KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India

Jai Shri Radhey Shyam

Dedicated to

Lord

Krishna Authors & Publishers

Preface This book on DIFFERENTIAL EQUATIONS & INTEGRAL TRANSFORMS has been specially written according to the latest Unified Syllabus to meet the requirements of the B.A. and B.Sc. Part-II Students of all Universities in Uttar Pradesh. The subject matter has been discussed in such a simple way that the students will find no difficulty to understand it. The proofs of various theorems and examples have been given with minute details. Each chapter of this book contains complete theory and a fairly large number of solved examples. Sufficient problems have also been selected from various university examination papers. At the end of each chapter an exercise containing objective questions has been given. We have tried our best to keep the book free from misprints. The authors shall be grateful to the readers who point out errors and omissions which, inspite of all care, might have been there. The authors, in general, hope that the present book will be warmly received by the students and teachers. We shall indeed be very thankful to our colleagues for their recommending this book to their students. The authors wish to express their thanks to Mr. S.K. Rastogi, Managing Director, Mr. Sugam Rastogi, Executive Director, Mrs. Kanupriya Rastogi Director and entire team of KRISHNA Prakashan Media (P) Ltd., Meerut for bringing out this book in the present nice form. The authors will feel amply rewarded if the book serves the purpose for which it is meant. Suggestions for the improvement of the book are always welcome.

Preface to the Revised Edition The authors feel great pleasure in presenting the thoroughly revised edition of the book DIFFERENTIAL EQUATIONS & INTEGRAL TRANSFORMS and wish to record thanks to the teachers and students for their warm reception to the previous edition. The present edition has been specially designed, made up-to-date and well organised in a systematic order according to the latest syllabus. The authors have always endeavoured to keep the text update in the best interests of the students community- a gesture which the authors hope would be appreciated by the students and teachers alike. Suggestions for the improvement of the book will be thankfully received.

— Authors

Syllabus D i fferential E quations & I ntegral T ransforms U.P. UNIFIED (w.e.f. 2012-13)

B.A./B.Sc. Paper-II

M.M. : 33 / 65

Differential Equations Unit-1: Formation of a differential equation (D.E.), Degree, order and solution of a D.E., Equations of first order and first degree : Separation of variables method, Solution of homogeneous equations, linear equations and exact equations, Linear differential equations with constant coefficients, Homogeneous linear differential equations, Unit-2: Differential equations of the first order but not of the first degree, Clairaut's equations and singular solutions, Orthogonal trajectories, Simultaneous linear differential equations with constant coefficients, Linear differential equations of the second order (including the method of variation of parameters), Unit-3: Series solutions of second order differential equations, Legendre and Bessel functions (Pn and Jn only) and their properties. Order, degree and formation of partial differential equations, Partial differential equations of the first order, Lagrange's equations, Charpit's general method, Linear partial differential equations with constant coefficients. Unit-4(i): Partial differential equations of the second order, Monge's method.

Integral Transforms Unit-4(ii): The concept of transform, Integral transforms and kernel, Linearity property of transforms, Laplace transform, Inverse Laplace transform, Convolution theorem, Applications of Laplace transform to solve ordinary differential equations. Unit-5: Fourier transforms (finite and infinite), Fourier integral, Applications of Fourier transform to boundary value problems, Fourier series.

B rief C ontents Dedication.........................................................................(v) Preface ...........................................................................(vi) Syllabus ........................................................................(vii) Brief Contents ...............................................................(viii) Section-A: Differential Equation .............................D-01—D-398 1. Differential Equations of First Order and First Degree...................................D-03—D-46 2. Differential Equations of the First Order but not of the First Degree...........D-47—D-72 3. Orthogonal Trajectories...........................................................................................D-73—D-82 4. Linear Differential Equations with Constant Coefficients............................D-83—D-118 5. Homogeneous Linear Differential Equations..................................................D-119—D-130 6. Ordinary Simultaneous Differential Equations..............................................D-131—D-152 7. Linear Equations of Second Order with Variable Coefficients.................D-153—D-194 8. Partial Differential Equations of the First Order..........................................D-195—D-246 9. Linear Partial Differential Equations of Second and Higher Order .................................. with Constant Coefficients................................................................................D-247—D-286 10. Partial Differential Equations of Second Order with Variable ........................................... Coefficients...........................................................................................................D-287—D-306 11. Monge's Method...................................................................................................D-307—D-324 12. Series Solutions of Differential Equations....................................................D-325—D-348 13. Legendre's Functions.........................................................................................D-349—D-376 14. Bessel’s Functions..............................................................................................D-377—D-398

Section-B: Integral Transforms ...............................T-01—T-192 1. The Laplace Transform...............................................................................................T-03—T-42 2.

The Inverse Laplace Transform...............................................................................T-43—T-76

3.

Applications of Laplace Transform.......................................................................T-77—T-104

4.

Fourier Transforms.................................................................................................T-105—T-134

5.

Finite Fourier Transforms......................................................................................T-135—T-148

6. Applications of Fourier Transforms in Initial and Boundary ............................................. Value Problems.........................................................................................................T-149—T-168 7.

Fourier Series............................................................................................................T-169—T-192

SECTION

A DIFFERENTIAL EQUATIONS C hapters 1.

Differential Equations of First Order And First Degree

2.

Differential Equations of The First Order But Not of The First Degree

1. Orthogonal Trajectories 3.

4.

5.

Linear Differential Equations With Constant Coefficients

Homogeneous Linear Differential Equations

6.

Ordinary Simultaneous Differential Equations

7.

Linear Equations of Second Order With Variable Coefficients

8.

Partial Differential Equations of The First Order

9.

Linear Partial Differential Equation of Second and Higher Order with Constant Coefficients

10.

Partial Differential Equations of Second Order with Variable Coefficients

11. Monge's Method

12. Series Solutions of Differential Equations

13. Legendre's Functions 14. Bessel's Functions

D-3

1 D ifferential E quations of F irst O rder and F irst D egree

1.1 Definitions differential equation is an equation containing the dependent and independent variables and different derivatives of the dependent variables w.r.t. one or more independent variables.

A

The order of a differential equation is the order of the highest derivative (or differential coefficient) occurring in the equation. (Lucknow 2007; Meerut 09B, 10B; Bundelkhand 10)

The degree of a differential equation is the degree of the highest derivative (or diff. coeff.) which occurs in it, after the differential equation has been rationalized (i. e., made free from radicals and fractions so far as derivatives are concerned). A differential equation is called ordinary, if the unknown function depends on only one argument (independent variable). (Lucknow 2007; Meerut 09B, 10B; Bundelkhand 10)

A differential equation is said to be partial if there are two or more independent variables. A differential equation is said to be linear if the dependent variable, say, ‘ y ' and all its derivatives occur in the first degree, otherwise it is non-linear.

D-4

A function y = f ( x) is called a Solution (or the primitive) of a differential equation if, when substituted into the equation, it reduces the equation to an identity and the process of finding all the solutions is called integrating (or solving) the differential equation. General solution:

(Lucknow 2007; Gorakhpur 09)

A solution of a differential equation, containing independent arbitrary constants equal in number to the order of the differential equation is called its general solution. Particular solution:

(Lucknow 2007)

A solution obtained by giving particular values to the arbitrary constants in the general solution is called a particular solution or particular integral. Arbitrary Constants: The solution of a differential equation may contain as many arbitrary constants as is the order of the differential equation i. e., the solution of an nth order differential equation may contain n arbitrary constants.

Example 1: Find the differential equation of the family of curves y = Ae x + ( B / e x ), for different

values of A and B. Solution: We have y = Ae x + Be − x .

…(1)

To obtain the required differential equation the constants A and B are to be eliminated with the help of the given equation (1) and the two equations obtained by differentiating (1) once and twice. Thus differentiating (1), we get dy = Ae x − Be − x . dx

…(2)

Now differentiating (2), we get d2 y dx2

= Ae x + Be − x .

Eliminating A and B between (1), (2) and (3), we obtain

…(3) d2 y dx2

= y, which is the

required differential equation. Example 2: Find the differential equation of all circles of radius a, or By the elimination of the

constants h and k, find the differential equation of which ( x − h)2 + ( y − k )2 = a2 , is a solution. Solution: The equation of all circles of radius a is given by

( x − h)2 + ( y − k )2 = a2 , h and k being parameters (i. e., arbitrary constants). Differentiating (1), we get

…(1)

D-5

2 ( x − h) + 2 ( y − k )

dy = 0. dx

…(2)

Differentiating (2), we get 1 + ( y − k)

d2 y dx2

2

 dy  +   = 0.  dx 

…(3)

From (2) and (3), we obtain ( x − h) = − ( y − k ) (dy / dx) and ( y − k ) = −

[1 + (dy / dx)2 ] . d2 y / dx2

Substituting these values in (1) and simplifying, we obtain  1 + 

3

2

2  2   dy   2 d y    = a  2  ,  dx    dx  

which is the required differential equation.

Comprehensive Exercise 1 1.

Find the differential equation of the family of curves y = Ae2 x + Be −2 x , for different values of A and B.

2.

(Bundelkhand 2003)

Find the differential equation corresponding to y = ae2 x + be −3 x + ce x , where a, b, c are arbitrary constants.

3.

Find the differential equation of the family of curves y = e x ( A cos x + B sin x). where A and B are arbitrary constants.

4.

By eliminating the constants a and b obtain the differential equation of which xy = ae x + be − x + x2 is a solution. (Purvanchal 2014)

5.

Find the differential equation corresponding to the family of curves y = c ( x − c )2 , where c is an arbitrary constant.

6.

Show that Ax2 + By2 = 1 is the solution of  d2 y  dy  2  dy = 0. xy 2 +    − y  dx   dx  dx 

7.

Show that v = (a / r) + B is a solution of d2 v dr2

+

2 dv = 0. r dr

D-6

A nswers 1 1.

d2 y 2

dx 2

3. 5.

d y dx2

2.

=4y −2

dy +2y =0 dx

8 y2 = 4 xy

dy  dy  −  dx  dx 

d3 y 3

dx

4. x

−7

d2 y dx2

dy +6y =0 dx

+2

dy − xy + x2 − 2 = 0 dx

3

1.2 Differential Equations of First Order and First Degree The differential equations of first order and of first degree can always be written in the form M + N ( dy / dx) = 0

or M dx + N dy = 0,

where M and N are some functions of x and y or are constants. Here the differential equation being of 1st order, its general solution (or primitive) will contain only one arbitrary constant. All differential equations of the first order cannot be always solved. However they can be solved by suitable methods if they belong to any one of the following standard forms : (a)

Variables separable,

(b) (c)

Homogeneous equations, dy ax + by + c Equations of the form = , dx a1 x + b1 y + c1

(d)

Linear equations,

(e)

Equations reducible to the linear form,

(f)

Exact differential equations,

(g)

Equations reducible to exact form,

(h)

Equations which by suitable change of variable can be transformed to any of the above forms.

1.2.1 Variables Separable If a differential equation of the first order and of the first degree is of the form f1 ( x) dx = f2 ( y) dy,

…(1)

where f1 ( x) is a function of x only and f2 ( y) is a function of y only, then we say that the variables are separable in the differential equation. In such equations it is possible to get dx and all the terms involving x on one side and dy along with all the terms involving y on the other side.

D-7

To solve such a differential equation integrate the two sides and add an arbitrary constant of integration to any one of the two sides. Thus, integrating both the sides of (1), we get its solution as

∫ f1 ( x) dx = ∫ f2 ( y) dy + c, where c is an arbitrary constant. The arbitrary constant can be chosen in any form suitable for the answer, i. e., we can replace it by log c, tan−1 c, sin c , e c , etc.

Example 3: Solve (1 + x2 ) dy = (1 + y2 ) dx

(Meerut 2003) 2

2

Solution: The given equation can be written as dy /(1 + y ) = dx /(1 + x ), in which the

variables have been separated. ∴ integrating, we get tan−1 y = tan−1 x + tan−1 c, where c is an arbitrary constant or

tan−1 y − tan−1 x = tan−1 c

or

tan−1

or

( y − x) = c (1 + yx), which is the required solution.

y−x = tan−1 c, 1 + yx

or

y−x =c 1 + yx

Example 4: Solve (1 + e x ) y dy = ( y + 1) e x dx. Solution: Here

or ∴

y ex dy = dx, (the variables being separated) y +1 1+ ex  1  ex dx. 1 −  dy = y + 1 1+ ex 

integrating, we get y − log ( y + 1) = log (1 + e x ) + log c , (c being an arbitrary constant)

or

y = log [c ( y + 1) (1 + e x )]

or

c ( y + 1) (1 + e x ) = e y, which is the required solution.

Example 5: Solve dy / dx = ( x + y)2 .

(Avadh 2011)

Solution: Here the variables are not separable but some suitable substitution will

reduce the differential equation to a form in which the variables are separable. Here we put x + y = v. Differentiating both sides w.r.t., ‘x ’, we have dy dv dy dv or 1+ = = − 1. dx dx dx dx By these substitutions the given equation reduces to dv dv or or − 1 = v2 , = v2 + 1, dx dx

dv v2 + 1

= dx.

D-8

Thus the variables being separated and so integrating, we get dv or tan−1 v = x + c = dx + c , 2 v +1

∫

∫

tan−1 ( x + y) = x + c,

or or

[∵ v = x + y]

x + y = tan ( x + c ) as the required solution where c is an arbitrary constant.

Example 6: Solve dy / dx = sin ( x + y) + cos ( x + y). Solution: Let x + y = v. Then differentiating, we have

1+

dy dv , = dx dx

dy dv = − 1. dx dx

or

Substituting these values in the given equation, we get dv dv or − 1 = sin v + cos v, = 1 + sin v + cos v dx dx dv or = dx, separating the variables (1 + cos v) + sin v dv

or

= dx 1 1 1 v + 2 sin v cos v 2 2 2 1 Now integrating, we get log (1 + tan v) = x + c 2 1 or log {1 + tan ( x + y)} = x + c, 2 2 cos2

or

1 1 sec2 v 2 2 dv = dx. 1 1 + tan v 2

which is the required solution containing an arbitrary constant c.

Comprehensive Exercise 2 Solve the following differential equations : 1. (i) (1 + x) y dx + (1 − y) x dy = 0 . (ii) (1 − x2 ) (1 − y) dx = xy (1 + y) dy. 2. (i)

x2 ( y + 1) dx + y2 ( x − 1) dy = 0 .

dy  dy    (ii)  y − x  = a  y2 +  ⋅   dx  dx 

(Purvanchal 2006; Avadh 10; Gorakhpur 08, 11)

2

2

3. (i) sec x tan y dx + sec y tan x dy = 0. (ii) dy / dx = e

x− y

2 −y

+x e

(Agra 2006; Meerut 09B)

. (Agra 2005; Avadh 07; Meerut 09; Purvanchal 10, 11; Rohilkhand 10; Bundelkhand 04)

y

y

4. (i) (e + 1) cos x dx + e sin x dy = 0. x

x

(ii) 3 e tan y dx + (1 − e ) sec

2

y dy = 0 .

(Rohilkhand 2010) (Meerut 2008; Bundelkhand 04)

D-9

5. (i)

(ds / dx) + x2 = x2 e3 s .

(ii) (dy / dx) tan y = sin ( x + y) + sin ( x − y). 6. (i)

log (dy / dx) = ax + by.

(ii) xy (dy / dx) = (1 + y2 ) (1 + x + x2 ) /(1 + x2 ). 7. (i)

If dy / dx = e x + y and it is given that for x = 1, y = 1, find y when x = − 1.

(ii) ( x − y)2 (dy / dx) = a2 . 8. (i)

cos ( x + y) dy = dx. (Lucknow 2005; Gorakhpur 09; Purvanchal 10; Avadh 14)

(ii) cos ( x + y) dx = dy. 9. (i)

sin

−1

(Gorakhpur 2005)

(dy / dx) = x + y.

(ii) (dy / dx) + 1 = e x + y. 10. (i)

dy / dx = (4 x + y + 1)2 .

(Lucknow 2006)

 x + y − a  dy  x + y + a  ⋅  (ii)  =  x + y − b  dx  x + y + b  11. ( y − x) (dy / dx) = a2 .

(Meerut 2006B)

A nswers 2 1. (i)

xy = ce y− x

(ii) log{ x (1 − y)2 } = 2. (i) (ii)

1 2 1 2 x − y −2y + c 2 2

1 2 ( x + y2 ) + ( x − y) + log { c ( x − 1)( y + 1)} = 0 2 y = c (a + x) (1 − ay) 1 3 x +c 3

3. (i)

tan x tan y = c

(ii)

e y = ex +

4. (i)

sin x (e y + 1) = c

(ii)

tan y = c (1 − e x )3

5. (i)

(e3 s − 1) = c1 e(3 s + x ), where c1 = e3 c

6. (i) (ii)

3

− (1/ b) e

− by

(ii) sec y = c − 2 cos x

+c

1 log (1 + y2 ) = log x + tan−1 x + log c 2

7. (i)

y = −1

8. (i)

y = c + tan

9. (i)

= (1/ a) e

ax

1 ( x + y) 2

1 [− 2 /( x + c )] = 1 + tan ( x + y)] 2

(ii)

y+c=

(ii)

tan

 x − y − a a log   2  x − y + a

1 ( x + y) = x + c 2

(ii) ( x + c ) e( x + y) + 1 = 0

D-10

10. (i)

4 x + y + 1 = 2 tan (2 x + k )

(ii) (b − a) log {( x + y)2 − ab} = 2( x − y + c ) 11. y + a2 log (a2 + x − y) = c

1.2.2 Homogeneous Equations Definition: A differential equation of the form

dy f ( x, y) , where f1 ( x, y) and = 1 dx f2 ( x, y)

f2 ( x, y) are homogeneous functions of x and y of the same degree, is called a homogeneous equations. (Bundelkhand 2005) To solve such an equation, put dy dv y = vx ; whence =v+ x ⋅ dx dx These substitutions transform the given equation into an equation of the form dv dv v+ x = f (v) i. e., x = f (v) − v. dx dx The variables are now separable. Separating the variables and integrating, we get dv = log x + c , where c is an arbitrary constant. f (v) − v

∫

Now replacing v by ( y / x) after integration, we get the required solution. Working Rule. To get the solution of a homogeneous differential equation proceed as follows : dy dv (i) Put y = vx and . =v+ x dx dx (ii)

The equation thus obtained will be of the form in which the variables are separable. Separate the variables and integrate.

(iii) After integration replace v by y / x and get the required solution.

Example 7: Solve ( x2 − y2 ) dx + 2 xy dy = 0 .

(Gorakhpur 2008)

Solution: The given equation can be written as

dy y2 − x2 = dx 2 xy

…(1)

This is a homogeneous differential equation as each term in the N r and Dr of R.H.S. is of the second degree in x and y. dy dv putting y = vx and consequently in (1), we get ∴ =v+ x dx dx v+ x

dv v2 x2 − x2 v2 − 1 = = dx 2v 2 vx2

D-11

v2 − 1 − 2 v2 −1 − v2 dv v2 − 1 = −v= = dx 2v 2v 2v 2 v dv dx , in which the variables have been separated. =− x v2 + 1

or

x

or ∴

integrating, we get log (v2 + 1) = − log x + log c

or

log {(v2 + 1) ⋅ x} = log c

x (v2 + 1) = c

or

x {( y / x)2 + 1} = c

or

( y2 + x2 ) = cx, which is the required solution.

or

[putting v = y / x]

Example 8: Solve y2 dx + ( xy + x2 ) dy = 0 .

(Avadh 2010)

Solution: The given equation can be written as

dy y2 , which is homogeneous. =− dx xy + x2 ∴

putting y = vx , this equation reduces to dv v2 x2 v2  = − v + x  = −  dx  v +1 x . vx + x2

or

x

v2 + v2 + v 2 v2 + v dv v2 , =− −v=− =− dx v +1 v +1 v +1

in which the variables are separable. (v + 1) dx (v + 1) dv ∴ − = = dv . x (2 v2 + v) v (2 v + 1) Integrating, we get − log x + log c = Now let ∴

v +1

∫ v (2v + 1) dv ,

where c is an arbitrary constant.

v +1 A B . Then A = 1, B = − 1. = + v (2 v + 1) v 2 v + 1 v +1

1

1 

1

∫ v (2v + 1) dv = ∫  v − 2v + 1 dv = log v − 2 log (2v + 1) .

Hence the solution is − log x + log c = log v −

1 log (2 v + 1) 2

or

1 log (2 v + 1) = log v + log x − log c 2

or

log (2 v + 1)1 /2 = log (vx / c )

or

{2 ( y / x) + 1}1 /2 = y / c

or

(2 y + x) / x = y2 / c 2

[∵ v = y / x] or

c 2 (2 y + x) = xy2 .

D-12 Example 9: Solve x2 y dx − ( x3 + y3 ) dy = 0 .

(Gorakhpur 2007; Agra 08)

Solution: The given equation can be written as

dy x2 y . = 3 dx x + y3

…(1)

This is a homogeneous equation as each term in the N r and Dr on the R.H.S. is of the same degree (3rd degree). dy dv putting y = vx and in (1), we get ∴ =v+ x dx dx v+ x or or ∴

x

x2 . vx dv v = 3 = dx x + v3 x3 1 + v3

dv v v4 = −v=− 3 dx 1 + v 1 + v3

or

(1 + v3 ) dv dx =− x v4

dx 1 1 = − 4 dv − dv . x v v integrating, we get log x + log c = 1/(3 v3 ) − log v

or

log (cxv) = 1/(3 v3 )

or

3 log (cxv) = 1/ v3

or

c 3 x3 v3 = e1 / v

or

3

y3 = ke x

3

3

/ y

or

log (cxv)3 = 1/ v3

or

c 3 y3 = e x

3

/ y3

[∵ v = y / x]

, which is the required solution.

Example 10: Solve ( x3 − 3 xy2 ) dx = ( y3 − 3 x2 y) dy.

(Meerut 2007)

Solution: The given equation can be written as

dy x3 − 3 xy2 , = 3 dx y − 3 x2 y

…(1)

which is homogeneous. ∴

putting y = vx and v+ x

dy dv in (1), we get =v+ x dx dx

x3 − 3 x . v2 x2 1 − 3 v2 dv = 3 3 = 3 2 dx v x − 3 x . vx v − 3 v

1 − 3 v2 − v4 + 3 v2 1 − v4 dv 1 − 3 v2 = 3 −v= = 3 3 dx v − 3 v v − 3v v − 3v

or

x

or

dx v3 − 3 v = dv = x 1 − v4

 1 1 2v  + − 2   dv , 2 1 2 1 ( v ) ( v ) + − v + 1  (by partial fractions)

∴

integrating, we get log x + log c =

1 1 log (v + 1) + log (v − 1) − log (v2 + 1) 2 2

D-13

or

log (cx) = log [{(v + 1)1 /2 (v − 1)1 /2 } /(v2 + 1)]

or

cx (v2 + 1) = (v2 − 1)1 /2

or

c 2 x2 {( y2 / x2 ) + 1}2 = {( y2 / x2 ) − 1},

or

2

2

or

c 2 x2 (v2 + 1)2 = v2 − 1 [∵ v = y / x]

2

2 2

2

 

y y   ( y dx + x dy) = y  sin  ( x dy − y dx).  x x

c ( y + x ) = ( y − x ), which is the required solution.

Example 11: Solve x  cos

Solution: The given equation can be written as

y  dy  y   dy    x cos   y + x  = y sin   x − y    x  dx  x   dx or or

dy  y y y y 2  x cos − y sin  = − y sin − xy cos dx  x x x x dy y { y sin ( y / x) + x cos ( y / x)} = dx x{ y sin ( y / x) − x cos ( y / x)} x

…(1)

dy dv in (1), we get =v+ x dx dx dv vx (vx sin v + x cos v) v+ x = dx x (vx sin v − x cos v)

Now putting y = vx and

dv v (v sin v + cos v) = dx (v sin v − cos v)

or

v+ x

or

x

or

 − v sin v + cos v 2 dx  dv = , separating the variables. − v cos v x  

∴

dv v2 sin v + v cos v − v2 sin v + v cos v , variables separable = dx v sin v − cos v

integrating, we get − log (v cos v) = 2 log x + log c

or

log {1/(v cos v)} = log (c x2 )

or

c x2 = 1/(v cos v)

or

c x2 v cos v = 1

or

c x2 . ( y / x)cos ( y / x) = 1

or

cxy cos ( y / x) = 1, which is the required solution.

Comprehensive Exercise 3 Solve the following differential equations : 1. x + y (dy / dx) = 2 y.

(Gorakhpur 2005)

2.

(Gorakhpur 2010)

y − x (dy / dx) = x + y (dy / dx).

3. ( x2 + y2 ) dx − 2 xy dy = 0 .

(Avadh 2006; Bundelkhand 03)

D-14

4. ( x2 + y2 ) (dy / dx) = xy. 2

(Kanpur 2002)

2

2

5. dy / dx = y /( xy − x ).

6.

x dy + y ( x + y) dx = 0 .

7. x ( x − y) dy = y ( x + y) dx.

8.

x

dy y2 + = y. dx x

9. x dy − y dx = √ ( x2 + y2 ) dx. 10. (1 + e

x/ y

) dx + e

x/ y

(Meerut 2008; Lucknow 10)

{1 − ( x / y)} dy = 0 .

dy y y = + tan . dx x x  dy  13. x   = y + x e y / x .  dx 

(Meerut 2006)

dy y = y − x cos2 . dx x dy  y  y 14. x cos   = y cos   − x.  x  x dx

11.

12. x

A nswers 3 1. log ( y − x) = c + x /( y − x)

2.

1 log ( x2 + y2 ) + tan−1 ( y / x) = log c 2

3. x2 − y2 = cx

4.

cy = e x

5.

y = ke y / x

7. c 2 xy = e − x / 9.

2

/(2 y2 )

6. ( y + 2 x) = c 2 x2 y y

8.

y + √ ( y2 + x2 ) = cx2

10.

11. sin ( y / x) = cx

cx = e x /

y

x + ye x /

y

=c

12.

tan ( y / x) = log (c / x)  y 14. sin   + log x = c  x

13. e − y / x + log x = c

1.2.3 Equations Reducible To Homogeneous Form A differential equation of the form dy ax + by + c a b , where = ≠ dx a1 x + b1 y + c1 a1 b1 can be reduced to homogeneous form by taking new variables X and Y such that x = X + h and y = Y + k , where h and k are constants to be so chosen as to make the given equation homogeneous. With the above substitutions we get dx = dX and dy = dY , so that dy / dx = dY / dX . Hence the given equation becomes a ( X + h) + b (Y + k ) + c aX + bY + (ah + bk + c ) dY . = = dX a1 ( X + h) + b1 (Y + k ) + c1 a1 X + b1Y + (a1h + b1k + c1) Now choose h and k such that ah + bk + c = 0,

and

a1h + b1k + c1 = 0 .

D-15

Then the differential equation becomes aX + bY dY , which is homogeneous. = dX a1 X + b1Y Now this equation can be solved as in 1.2.2 substituting Y = vX . Finally by replacing X by ( x − h) and Y by ( y − k ) we shall get the solution in original variables x and y. If, however, a / a1 = b / b1 = m (say) then the differential equation becomes of the form dy m (a1 x + b1 y) + c . = dx a1 x + b1 y + c1 To solve such a differential equation put v = a1 x + b1 y, get rid of y and then the transformed equation will be such that the variables are separable.

Example 12: Solve Solution: Here

dy y − x +1 . = dx y + x +5

a b i. e., the coefficients of x and y in the numerator and ≠ a1 b1

denominator of the expression for (dy / dx) are not proportional. Such equations can be reduced to homogeneous form by taking new variables X and Y such that x = X + h and y = Y + k . where h and k are constants to be taken at our choice. With these substitutions the given equation reduces to Y + k − X − h + 1 Y − X + (k − h + 1) dY . …(1) = = dX Y + k + X + h + 5 Y + X + (k + h + 5) Choose h and k such that k − h + 1= 0

and

k + h + 5 = 0.

Solving these, we get k = − 3, h = − 2. With these values of h and k the equation (1) becomes dY Y − X . = dX Y + X

…(2)

Now (2) is a homogeneous equation, so putting Y = vX , we have v+ X or or

v +1 v2 + 1 2v v2 + 1

dv v − 1 = , dX v + 1 dv = − dv +

dX X 2

v2 + 1

or or dv = −

X

− 1 − v2 dv v − 1 = −v= dX v + 1 v +1 v

v2 + 1

dv +

2 dX . X

Integrating, we have log (v2 + 1) + 2 tan−1 v = − 2 log X + c

dv v2 + 1

=−

dX X

D-16

or

log {(v2 + 1) X 2 } = − 2 tan−1 v + c

or

log (Y 2 + X 2 ) = − 2 tan−1 (Y / X ) + c , 2

or

2

log [( y + 3) + ( x + 2) ] + 2 tan

−1

[∵ v = Y / X ]

{( y + 3) / ( x + 2)} = c ,

[∵ Y = y − k = y + 3, X = x − h = x + 2] which is the required solution. dy 2 x + 2 y − 2 Example 13: Solve = ⋅ dx 3x + y − 5

(Lucknow 2005)

Solution: Here (a / a1) ≠ (b / b1), therefore putting x = X + h and y = Y + k the given

equation reduces to dY 2 ( X + h) + 2 (Y + k ) − 2 2 X + 2Y + (2 h + 2 k − 2) . = = dX 3 ( X + h) + (Y + k ) − 5 3 X + Y + (3 h + k − 5)

…(1)

Now choose h and k such that 2h + 2k − 2 = 0

and

3h + k − 5 = 0.

Solving these, we get h = 2 and k = − 1. With these values of h and k, the equation (1) becomes dY 2 X + 2Y = ⋅ dX 3X + Y This is a homogeneous equation so putting Y = vX , we get dv 2 X + 2 vX 2 + 2 v v+ X = = dX 3 X + vX 3+v or or

or

2 + 2 v − 3 v − v2 2 − v − v2 dv 2 + 2 v = −v= = dX 3+v 3+v 3+v 3+v 3+v dX − = 2 dv = dv X ( v + 2) (v − 1) v + v−2 X

 4 1  = −  dv , by resolving into partial fractions 3 ( v − 1 ) 3 ( v + 2)  dX  1 4  3 = − dv. X v + 2 v − 1 

Integrating, we get 3 log X = log (v + 2) − 4 log (v − 1) + log c or log { X 3 (v − 1)4 } = log { c (v + 2)} or

X 3 [(Y / X ) − 1]4 = c {(Y / X ) + 2},

[∵ v = Y / X ]

4

or

(Y − X ) = c (Y + 2 X )

or

{( y + 1) − ( x − 2)}4 = c {( y + 1) + 2( x − 2)}

or

( y − x + 3)4 = c (2 x + y − 3), which is the required solution.

Example 14: Solve (2 x + y + 1) dx + (4 x + 2 y − 1) dy = 0 . Solution: We have

dy 2x + y + 1 . =− dx 4x + 2 y − 1

(Lucknow 2011; Kanpur 12)

D-17

Here ∴

a b . = a1 b1 let 2 x + y = v so that 2 + (dy / dx) = dv / dx.

With these substitutions the given equation reduces to v +1 dv , −2= − dx 2v − 1 or

v + 1 3v − 3 dv =2− = dx 2v − 1 2v − 1

or

3 dx =

∴

2v − 1 2 (v − 1) + 1 1   dv = dv = 2 +  dv.  v −1 v −1 v − 1

integrating, we have 3 x + c = 2 v + log (v − 1)

or or

3 x + c = 2 (2 x + y) + log (2 x + y − 1),

[∵ v = 2 x + y]

x + 2 y + log (2 x + y − 1) = c , which is the required solution.

Comprehensive Exercise 4 Solve the following differential equations : 1. (2 x + y − 3) dy = ( x + 2 y − 3) dx.(Bundelkhand 2010; Lucknow 09; Rohilkhand 11) 2. dy / dx = (2 x − y + 1) /( x + 2 y − 3). 3. (2 x + 3 y − 5) (dy / dx) + (3 x + 2 y − 5) = 0 . 4. ( x − y − 2) dx + ( x − 2 y − 3) dy = 0. 5. ( x + y − 1) dy = ( x + y) dx. 6. ( x + y) (dx − dy) = dx + dy.

(Gorakhpur 2008)

7. ( x − y − 2) dx = (2 x − 2 y − 3) dy. dy x − y +3 8. . = dx 2 x − 2 y + 5 9. (2 x + 4 y + 1) dy = (2 y + x + 1) dx. 10. (2 x − 2 y + 5) dy − ( x − y + 3) dx = 0 .

A nswers 4 1. ( x + y − 2) = c 2 ( x − y)3 2. (5 y − 7)2 + (5 x − 1) (5 y − 7) − (5 x − 1)2 = c 3. 3 x2 + 4 xy + 3 y2 − 10 x − 10 y = c

(Kashi 2011)

D-18

4. log { c 2 (2Y 2 − X 2 )} = (1/ √ 2) log {(Y √ 2 − X ) /(Y √ 2 + X )}, where X = x − 1 and Y = y + 1. 5. 2 ( y − x) − log (2 x + 2 y − 1) = c

6. ( y − x) + log ( x + y) = c

7. log ( x − y − 1) = x − 2 y − c

8.

x − 2 y + log ( x − y + 2) = c

9. 4 (2 y − x) − log (4 x + 8 y + 3) = c 10. x − 2 y + log ( x − y + 2) = c

1.2.4. Linear Differential Equations Definition: A differential equation is said to be linear when the dependent variable y and all its derivatives occur in the first degree only and are not multiplied together. An equation of the form dy + Py = Q, dx where P and Q are functions of x only is called a linear differential equation of the first order with y as the dependent variable. To solve such an equation, multiply both the sides by e ∫ P dx . The equation then becomes e∫

P dx

d dx

or

(dy / dx) + Pye ∫ P dx = Qe ∫ P dx

{ ye∫ P dx} = Qe∫ P dx .

Integrating both sides w.r.t., ‘x’, we get ye ∫ P dx =

∫ Qe∫

P dx

dx + c ,

which is the solution of the differential equation. Note 1: The factor e ∫ P dx , on multiplying by which the L.H.S. of the differential

equation becomes a differential coefficient of some function of x and y, is called an integrating factor (I.F.) of the differential equation. (Agra 2007) Note 2: Sometimes a given differential equation becomes linear if we take y as the

independent variable and x as the dependent variable. dx Then it is of the form + Px = Q, dy where P and Q are some functions of y only. The integrating factor (I.F.) in this case is e ∫ P dy and the solution is x ⋅ e ∫ P dy =

∫ Q ⋅ e∫

P dy

dy + c .

Working Rule :

(i)

Write the given equation in the form dy dx or + Py = Q + Px = Q as the case may be. dx dy

(ii) Find the integrating factor e ∫ P dx

or

e ∫ P dy.

D-19

(iii) The solution of the differential equation is either

∫ {Q ⋅ (I. F. )} dx + c x ⋅ (I. F. ) = { Q ⋅ (I. F. )} dy + c as the case may be. ∫ y ⋅ (I. F. ) =

or

2 dy + 2 xy = e − x . dx

Example 15: Solve

(Meerut 2001, 03, 10B)

Solution: The given differential equation is

2 dy + 2 xy = e − x , which is linear with y as dx

the dependent variable. 2

Here P = 2 x, and Q = e − x . We have

∫

P dx =

1

2

∫ 2 x dx = 2 ⋅ 2 x

Therefore I.F. = e ∫

P dx

2

= ex .

Hence the solution is y. (I.F.) = 2

or

y ⋅ ex =

or

y ⋅ ex

2

= x2 .

− x2

∫ (e . e = dx + c ∫

∫ {Q ⋅ (I. F. )} dx + c, where c is an arbitrary constant

x2

) dx + c or

2

ye x = x + c .

Example 16: Solve ( x2 − 1) (dy / dx) + 2 xy = 1.

(Meerut 2010)

Solution: The given equation can be written as

dy 2x 1 + y= 2 dx x2 − 1 x −1 Here P =

2x 2

x −1

, and so

∫

dy   From dx + Py = Q 2 x dx

2

∫ x2 − 1 = log ( x

P dx =

∴

integrating factor (I.F.) = e ∫

∴

the solution is y. ( x2 − 1) =

or

y ( x2 − 1) =

P dx

= e log ( x 1

2

− 1) 2

∫ ( x2 − 1) ⋅ ( x

− 1).

= ( x2 − 1).

− 1) dx + c ,

 1  ∵ Q = 2  ( x − 1) 

∫ dx + c = x + c.

Hence y ( x2 − 1) = x + c is the required solution. Example 17: Solve x (dy / dx) + 2 y = x2 log x. Solution: The given equation can be written as

dy 2 + y = x log x, dx x

dy   + Py = Q  from   dx

D-20

Here P = (2 / x) and so ∴

I.F. = e ∫

∴

the solution is

∫

P dx

y (I. F. ) =

P dx = = e log

2

∫ (2 / x) dx = 2 log x = log x .

x2

= x2 .

∫ Q ⋅ (I. F. ) dx + c

yx2 =

2

[∵ Q = x log x]

∫ ( x log x) x dx + c, yx2 = x3 log x dx + c ∫

i. e., or

x4 − 4

1 x4 dx + c , ⋅ x 4 1 1 1 or yx2 = x4 log x − ⋅ ( x4 ) + c 4 4 4 1 Hence 4 yx2 = x4 log x − x4 + k is the required solution. 4 yx2 = (log x) ⋅

or

∫

Example 18: Solve cos2 x (dy / dx) + y = tan x.

[integrating by parts]

(Meerut 2003, 13; Avadh 05)

dy Solution: The given equation can be written as + sec2 x . y = tan x sec2 x, which is dx linear with y as the dependent variable. Here P = sec2 x, and Q = tan x sec2 x. We have

∫ P dx = ∫ sec

2

x dx = tan x.

∴

I. F. = e ∫ P dx = e tan x .

∴

the solution is y . e tan

or

ye tan

x

x

=

∫ tan x sec

2

x . e tan

ye tan

x

dx + c

t

2

∫ t e dt + c, putting tan x = t and sec x dx = dt = [t e t − e t dt] + c , integrating by parts ∫ =

= t e t − e t + c = tan x e tan or

x

x

− e tan

x

+c

= e tan x (tan x − 1) + c .

Example 19: Solve (1 + y2 ) dx = (tan−1 y − x) dy. (Agra 2005; Avadh 07; Rohilkhand 10; Bundelkhand 03, 04; Meerut 13B)

Solution: The given equation can be written as

dx tan−1 y − x = dy 1 + y2

or

tan−1 y dx 1 . + ⋅x = 2 dy 1 + y 1 + y2

This is linear equation with y as independent variable and x as dependent variable, Here P = 1/(1 + y2 ).

D-21 2

∫ P dy = ∫ {1/(1 + y )} dy = tan

∴

Hence I.F. = e ∫

P dy

= e tan

−1

x e tan

−1

y

tan−1 y

=

∫

=

∫te

1 + y2 t

y.

y.

Now the solution is x . (I. F. ) = i. e.,

−1

∫ {Q . (I. F. ) } dy + c

⋅ e tan

−1

y

 tan−1 y   ∵ Q =  1 + y2  

dy + c ,

dt + c , putting tan−1 y = t and dy /(1 + y2 ) = dt

= [t e t − 1 . e t dt] + c , integrating by parts

∫

t

= t e − e t + c = (t − 1) e t + c or

x . e tan

−1

y

= (tan−1 y − 1) . e tan

−1

y

+ c.

Example 20: Solve (dy / dx) + 2 y tan x = sin x, given that y = 0 when x = π / 3. Solution: We have

dy + 2 tan x . y = sin x, which is linear with y as dependent variable. dx

Here P = 2 tan x and so

∫ P dx = ∫ 2 tan x dx = 2 log sec x = log sec

sec2 x

∴

I.F. = e ∫ P dx = e log

∴

the solution is y . (I. F. ) =

x.

= sec2 x.

∫ [Q . (I. F. )] dx + c

i. e.,

y sec2 x =

or

y sec2 x = sec x + c .

∫

2

sin x . sec2 x dx + c =

Now it is given that y = 0 at x =

∫

tan x sec x dx + c …(1)

1 π. 3

Therefore, from (1) 0 . sec2

1 1 π = sec π + c 3 3

or

c = − sec

1 π = − 2. 3

Hence from (1), y sec2 x = sec x − 2 is the required solution.

Comprehensive Exercise 5 Solve the following differential equations : 1. (dy / dx) + (1/ x) y = x n.

2. x2 (dy / dx) + y = 1.

3. x (dy / dx) + y = x2 + 3 x + 2

(Avadh 2010)

4. (sin x) (dy / dx) + 3 y = cos x.

(Meerut 2004)

D-22

5. ( x2 + 1)(dy / dx) + 2 xy = 4 x2 .

(Agra 2008)

2

6. (1 + x ) (dy / dx) + 2 xy = cos x. 2

2

(Meerut 2009) 3

7. x ( x + 1) (dy / dx) = y (1 − x ) + x log x.

(Meerut 2007B)

8. (dy / dx) + (2 y / x) = sin x. 9. (dy / dx) + ( y / x) = sin x2 10. ( y sin x − 1) dx + cos x dy = 0.

(Bundelkhand 2008)

11. sin 2 x (dy / dx) − y = tan x.

(Agra 2007)

12. (1 + x) (dy / dx) − xy = 1 − x. 13. ( x + 2 y3 ) dy = y dx.

(Rohilkhand 2006; Avadh 11)

14. ( x + y + 1) (dy / dx) = 1. 15. (1 + y2 ) dx + ( x − e tan

−1

(Gorakhpur 2006; Rohilkhand 08) y

) dy = 0 .

(Gorakhpur 2007; Lucknow 09; Purvanchal 11; Bundelkhand 04)

A nswers 5 1. xy = x n + 2 / (n + 2) + c

2. y = 1 + ce1 / x

3. 6 xy = 2 x3 + 9 x2 + 12 x + c 1 1 1 1 y tan3 ( x) = 2 tan ( x) − x − tan3 ( x) + c 2 2 3 2 4 5. y (1 + x2 ) = x3 + c 6. y (1 + x2 ) = sin x + c 3 1 1 7. y ( x2 + 1) = cx + x3 log x − x3 2 4 1 8. yx2 = − x2 cos x + 2 x sin x + 2 cos x + c 9. yx = − cos x2 + c 2 10. y sec x = tan x + c 11. y = tan x + c √ (tan x) 4.

12. y (1 + x) = x + ce x

13. x = y3 + cy

14. x = ce y − y − 2

15. xe tan

−1

y

=

1 2 tan −1 e 2

y

+c

1.2.5. Equations Reducible to the Linear Form Some equations can be reduced to the linear form by making suitable substitutions, and hence can be solved easily. (a) Bernoulli’s Equation:

(Gorakhpur 2008)

A differential equation of the form dy + Py = Q y n, dx where P and Q are functions of x alone is called Bernoulli’s equation.

D- 23

To solve this equation dividing both sides by y n, we get y −n

dy + Py − n + 1 = Q dx [Note that the R.H.S. is now a function of x alone.]

Now putting y − n +1 = v, so that (1 − n) y − n

dy dv , the equation (1) transforms to = dx dx

1 dv + Pv = Q 1 − n dx dv + (1 − n) P . v = (1 − n) Q, dx

or

which is a linear differential equation with v as dependent variable and can be solved by the method discussed in 1.2.4. (b) Equations of the form dy + Pφ ( y) = Q f ( y), dx where P and Q are functions of x alone. To reduce such an equation to the linear form, dividing both sides by f ( y) and thus getting rid of f ( y) from the R.H.S., we have dy φ ( y) 1 ⋅ + P⋅ = Q. f ( y) dx f ( y) Now we try the substitution

φ ( y) = v⋅ f ( y)

dv d  φ ( y)  1 dy , =   = K⋅ dx dx  f ( y) f ( y) dx

If

where K is some constant, the equation reduces to the form

1 dv + Pv = Q, or K dx

dv + K Pv = KQ, which is a linear differential equation, with v as the dependent dx variable.

Example 21: Solve

dy 1 + y = x2 y6 . dx x

Solution: The given equation on dividing out by y6 , becomes

1 dy 1 1 + ⋅ = x2 . y6 dx x y5 Put

1 y5

= v, so that

…(1)

D-24

−

5 dy dv = y6 dx dx

or

1 dy 1 dv . =− 6 dx 5 dx y

With these substitutions the equation (1) becomes 1 dv 1 dv 5 or − + ⋅ v = x2 − v = − 5 x2 . 5 dx x dx x This is a linear equation with v as the dependent variable. Here P = − 5 / x and Q = − 5 x2 . We have

∫ P dx = ∫ (− 5 / x) dx = −5 log x = log x ∴

I.F. = e ∫ P dx = e log (1 / x

∴

the solution is

5

)

−5

= log (1/ x5 ).

= 1/ x5 .

∫ {Q ⋅ (I. F. )} dx + c v ⋅ (1/ x5 ) = − 5 x2 ⋅ (1/ x5 ) dx + c = −5 x −3 dx + c ∫ ∫

v ⋅ (I. F. ) = i. e.,

(1/ y5 ) ⋅ (1/ x5 ) =

i. e.,

5 −2 x + c. 2

[Note that v = 1/ y5 ]

5 ⋅ (1/ x2 ) + c is the required solution. 2 dy  x  Example 22: Solve +  y = x √ y. dx 1 − x2  Hence 1/( x5 y5 ) =

Solution: Dividing both sides of the given equation by 2 y1 /2 , we have

1 dy 1  x  1 +   √ y = x. 2 2 √ y dx 2 1 − x2  Now put y1 /2 = v, so that

1 −1 / 2 y (dy / dx) = (dv / dx). 2

With these substitutions the equation (1) becomes dv 1  x  1 +   ⋅ v = x, 2   dx 2 1 − x 2 which is linear with v as the dependent variable. 1 1 Here P = { x /(1 − x2 )} and Q = x. 2 2 1 x 1 − 2x We have P dx = dx = − dx 2 1 − x2 4 1 − x2

∫

∫

=−

∫

1 log (1 − x2 ) = log (1 − x2 )−1 /4 = log{1/(1 − x2 )1 /4 }. 4

∴

I.F. = e ∫ P dx = e log{1 /(1− x

∴

the solution is

2 1 /4

v /(1 − x2 )1 /4 =

)

}

= 1/(1 − x2 )1 /4 .

1 { x /(1 − x2 )1 /4 } dx + c 2

∫

…(1)

D- 25

1 4

or

v /(1 − x2 )1 /4 = −

or

√ y /(1 − x2 )1 /4 = −

or

√ y /(1 − x2 )1 /4

∫t

−1 / 4

dt,

putting (1 − x2 ) = t so that − 2 x dx = dt

1 4 3 /4 ⋅ t + c, 4 3 1 = − (1 − x2 )3 /4 + c . 3

[∵ v = √ y] [∵ t = (1 − x2 )]

Example 23: Solve x (dy / dx) + y = y2 log x. (Bundelkhand 2006; Gorakhpur 10; Purvanchal 07, 08, 10; Kanpur 06)

Solution: The given equation can be written as

or

dy 1 + y= dx x 1 dy 1 + ⋅ y2 dx x

y2 log x, dividing throughout by x x 1 log x , = y x

…(1)

dividing throughout by y2 . Now put 1/ y = v, so that (− 1/ y2 )(dy / dx) = (dv / dx). With these substitutions equation (1) becomes log x dv 1 − + v= dx x x log x dv 1 or . − ⋅v= − dx x x This is linear with v as the dependent variable. Here P = − 1/ x and Q = − (log x) / x. ∴

I.F. = e ∫ P dx = e ∫ (−1 / x) dx = e − log

x

= e log (1 / x) = (1/ x).

Hence the solution is

∫ {− (log x) / x} ⋅ (1/ x) dx + c v / x = − (1/ x2 ) log x dx + c ∫ = − [(− 1/ x) log x − (1/ x)(−1/ x) dx] + c, ∫

v (1/ x) = or

integrating by parts taking 1/ x2 as the second function = (1/ x) log x − or

2

∫ (1/ x ) dx + c = (1/ x) log x + (1/ x) + c

1/( xy) = (1/ x)(1 + log x) + c ,

[∵ v = 1/ y]

or

1 = y (1 + log x) + cxy. dy y y Example 24: Solve + log y = 2 (log y)2 . dx x x Solution: Dividing both sides of the given equation by y (log y)2 , we have

dy 1 1 1 + ⋅ = 2 y (log y) dx x log y x 1

2

…(1)

D-26

Now put 1/ log y = v, so that − {1/(log y)2 } ⋅ (1/ y) ⋅ (dy / dx) = dv / dx. With these substitutions the equation (1) becomes dv v 1 dv 1 1 or − + =+ 2 − ⋅v= − 2 . dx x dx x x x This is linear with v as the dependent variable. and Q = − 1/ x2 ,

Here P = − 1/ x ∴

I.F. = e ∫ P dx = e − ∫ (1 / x)dx = e − log

∴

the solution is v / x =

or or

x

= e log(1 / x) = 1/ x.

2

∫ (−1/ x ).(1/ x)dx + c (1/ log y)(1/ x) = − x −3 dx + c , ∫

[∵ v = 1/ log y]

1 1 = + c. x log y 2 x2 dy 1 e y . + = dx x x2

Example 25: Solve

Solution: On dividing out by e y, the given equation becomes

e− y

dy 1 1 + e− y ⋅ = 2 . dx x x

…(1)

Now put e − y = v, so that − e − y (dy / dx) = (dv / dx). With these substitutions equation (1) becomes dv 1 1 dv 1 1 or − + ⋅v= 2 − ⋅v= − 2 . dx x dx x x x This is linear with v as the dependent variable. Here P = − 1/ x and Q = − 1/ x2 . ∴

I.F. = e ∫ P dx = e ∫ −(1 / x) dx = e − log

∴

the solution is v. (I.F.) = 1 =− x

1

x

= e log (1 / x) = 1/ x.

∫ {Q . (I. F. )} dx + c 1

1

∫ x2 ⋅ x dx + c = 2 x2 + c

i. e.,

v.

or

e − y . (1/ x) = 1/(2 x2 ) + c ,

or

e − y(1/ x) = (1 + 2 cx2 ) /(2 x2 ),

(∵ v = e − y) or

2 x = (2 cx2 + 1) e y.

Comprehensive Exercise 6 Solve the following differential equations : 1. (dy / dx) + ( y / x) = y2 .

(Bundelkhand 2007) 1 /3

2. (dy / dx) + { y /( x − 1)} = xy

.

(Meerut 2005B)

D- 27

3. ( x3 y2 + xy) dx = dy. 3

(Kanpur 2002)

3

4. dy / dx = x y − xy.

(Gorakhpur 2005; Meerut 13B) 3

2

5. 3 (dy / dx) + 2 y /( x + 1) = x / y . 6.

y (2 xy + e x ) dx − e x dy = 0 . 1 / x3

2

7. ( xy − e

8. dy / dx = e

(Kanpur 2001)

2

) dx − x y dy = 0 .

x− y

x

(Lucknow 2007)

y

(e − e ).

9. (dy / dx) − y tan x = − y2 sec x.

(Meerut 2009B)

10. cos x dy = (sin x − y) y dx. 3

(Rohilkhand 2007) 2

11. (dy / dx) + x sin 2 y = x cos 2

y.

(Gorakhpur 2007, 11; Purvanchal 08; Avadh 14)

2

12. (dy / dx) + y cot x = y sin x. 13. 2 (dy / dx) − y sec x = y3 tan x. 14. (1 − x2 ) 15. x

dy + xy = xy2 . dx

(Purvanchal 2006)

dy + y log y = xye x . dx 2

(Bundelkhand 2005) 3

3

16. Solve xy (dy / dx) − 2 y = 2 x , given that y = 1 when x = 1. 17. Solve

dy dφ dφ + y = φ ( x) , where φ is some function of x only. dx dx dx

A nswers 6 1. xy (c − log x) = 1 2.

y2 /3 ( x − 1)2 /3 =

2 3 x ( x − 1)5 /2 − ( x − 1)8 /3 + c 5 20 1 2

− x 1 2 x ) − cy e 2 2 1 2 1 y3 ( x + 1)2 = x6 + x5 + x4 + c 6 5 4

2

3. 1 = 2 y (1 −

4. (1/ y2 ) = ce x + x2 + 1

5.

6. e x + y (c + x2 ) = 0

3

7. 3 y2 = 2 x2 e1 / x + 3 cx2

8. e y = e x − 1 + ce −(e

9. sec x = y (tan x − c )

x

)

10. sec x = y (tan x − c )

11. 2 tan y = ( x2 − 1) + 2 ce − x

2

12. (1/ y) = (cos x − c ) sin x

2

13. − (sec x + tan x) = y (sec x + tan x − x + c ) 14. 1/ y n−1 = 2 sin x − {2 /(1 − n)} + ce(n − 1)sin x 15. x log y = e x ( x − 1) + c 16. y3 = − 2 x3 + 3 x6

17. ye φ =

∫φe

φ

dφ + c

D-28

1.2.6. Exact Differential Equations

(Lucknow 2006)

A differential equation is said to be exact if it can be obtained from its primitive (solution) directly by differentiation, without involving any subsequent process of multiplication, elimination, etc. Thus the differential equation of the form M dx + N dy = 0, where M and N are some functions of x and y, is exact if it can be obtained directly by differentiating an equation of the form u = c , where u is some function of x and y and c is an arbitrary constant. Theorem: The necessary and sufficient condition for the ordinary differential equation M (Lucknow 2006, 09) dx + N dy = 0 to be exact is that ∂M / ∂y = ∂N / ∂x . The condition is necessary: Suppose the differential equation …(1)

M dx + N dy = 0

is exact. Let the primitive of (1) be u = c , where u is some function of x and y and c is an arbitrary constant. Since u is a function of x and y, therefore from partial differentiation, we have ∂u ∂u du = dx + dy . ∂x ∂y Now

u = c ⇒ du = 0 ⇒

∂u ∂u dx + dy = 0 . ∂x ∂y

…(2)

Since the equation (1) is exact, therefore equation (2) must be identical with (1), therefore, we have ∂u …(3) =M ∂x ∂u and …(4) = N. ∂y Differentiating (3) and (4) partially with respect to y and x respectively, we obtain ∂2 u ∂M ∂2 u ∂N . = , = ∂y ∂x ∂y ∂x ∂y ∂x ∂2 u ∂2 u ∂M ∂N . = . Therefore = ∂y ∂x ∂x ∂y ∂y ∂x

But

Hence the condition is necessary. The condition is sufficient: We have to show that if ∂M / ∂y = ∂N / ∂x, then M dx + N dy = 0 must be an exact differential equation. Let

∫ M dx = P , then

∂P ∂2 P ∂M ∂N , by hypothesis. = M, so that = = ∂x ∂y ∂x ∂y ∂x Now

∂N ∂2 P ∂N ∂2 P ∂N ∂  ∂P    = ⇒ = ⇒ = ∂x ∂x ∂y ∂x ∂x ∂y ∂x ∂x  ∂y 

⇒

N =

∂P + f ( y), where f ( y) is some function of y alone. ∂y

D- 29

Now putting M =

∂P ∂P and N = + f ( y) in M dx + N dy, we have ∂x ∂y

M dx + N dy = =

 ∂P  ∂P dx +  + f ( y) dy ∂x  ∂y 

∂P ∂P dx + dy + f ( y) dy = dP + f ( y) dy ∂x ∂y

= d [ P + F ( y)], where d [ F ( y)] = f ( y) dy = an exact differential of some function of x and y. Therefore the differential equation M dx + N dy = 0 is exact and hence the condition is sufficient. Working rule for solving an exact differential equation:

(Meerut 2008)

To solve a differential equation of the form M dx + N dy = 0, first ascertain with the help of the condition ∂M / ∂y = ∂N / ∂x whether the equation is exact or not. If the equation is exact, then (i)

Integrate M with respect to x treating y as constant.

(ii)

Integrate w.r.t. y only those terms of N which do not contain x.

(iii) Equate the sum of these two integrals [found in (i) and (ii)] to an arbitrary constant and thus get the required solution. Thus if the differential equation M dx + N dy = 0 is exact, its solution is

∫

M dx

+

treating y as constant

∫

N dy

=c

take only those terms in N which do not contain x.

Example 26: Solve (ax + hy + g) dx + (hx + by + f ) dy = 0.

(Meerut 2003)

Solution: Here M = ax + hy + g and N = hx + by + f .

We have

∂M ∂N = h and = h . Thus we observe that ∂M / ∂y = ∂N / ∂x and therefore the ∂y ∂x

given differential equation is exact. Hence its solution is

∫

M dx

+

treating y as a constant

∫

N dy

taking only those terms in N which do not contain x.

i. e.,

∫ (ax + hy + g) dx treating y as a constant

+

∫ (by + f ) dy = c

=c

D-30

i. e.,

1 2 1 ax + hxy + gx + by2 + fy = c 2 2

i. e.,

ax2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0 ,

where c is an arbitrary constant. We have replaced − 2c by c. Example 27: Solve (4 x + 3 y + 1) dx + (3 x + 2 y + 1) dy = 0 . (Purvanchal 2009; Meerut 11)

Solution: Here M = 4 x + 3 y + 1 and N = 3 x + 2 y + 1 .

We have

∂M ∂N ∂M ∂N , and hence the given equation is exact. = = 3 and = 3 i. e., ∂y ∂x ∂y ∂x

Integrating M , i. e., 4 x + 3 y + 1 with respect to x, treating y as a constant, we obtain 1 4 ⋅ x2 + 3 xy + x i. e., 2 x2 + 3 xy + x . 2 Again, the only terms in N which do not contain x are 2 y + 1 . Integrating (2 y + 1) with respect to y, we obtain 1 2 ⋅ y2 + y i. e., y2 + y. 2 Hence the solution of the given differential equation is 2 x2 + 3 xy + x + y2 + y = c , where c is an arbitrary constant 2 x2 + 3 xy + y2 + x + y = c .

i. e.,

Example 28: Solve ( x2 − ay) dx − (ax − y2 ) dy = 0 . Solution: Here M = x2 − ay and N = − (ax − y2 ).

We have ∂M ∂N ∂M ∂N = − a and = − a i. e., = ∂y ∂x ∂y ∂x and hence the given equation is exact. Now

∫

M dx (regarding y as a constant) =

and

∫

or or

− ay) dx =

1 3 x − ayx 3

…(1)

N dy (taking in N only those terms which do not contain x) =

∴

2

∫ (x ∫y

2

dy =

1 3 y 3

the required solution is, (1) + (2) = c (an arbitrary constant) 1 3 1 3 x − ayx + y = c, 3 3 x3 + y3 − 3 axy = 3 c .

…(2)

D- 31 Example 29: Solve x dx + y dy +

x dy − y dx x2 + y2

=0 . (Bundelkhand 2007)

Solution: The given equation can be written as

   y  x dx + y +   dy = 0 . x − 2  x + y2  x2 + y2    y x . and N = y + 2 M=x− 2 2 x + y2 x + y

Here We have

1 ⋅ ( x2 + y2 ) − y ⋅ 2 y x2 − y2 y2 − x2 ∂M , =0 − = = ∂y ( x2 + y2 )2 ( x2 + y2 )2 (x2 + y2 )2 1 ⋅ ( x2 + y2 ) − x ⋅ 2 x y2 − x2 ∂N . =0 + = ∂x ( x2 + y2 )2 ( x2 + y2 )2

and

Thus ∂M / ∂y = ∂N / ∂x and hence the given equation is exact. Therefore its solution is 

y



∫  x − x2 + y2  dx regarding y as a constant

or

y2 x2 1 x − y ⋅ tan−1 + =c 2 y y 2

or

x2 − 2 tan−1 ( x / y) + y2 = 2 c = k .

∫

+

y dy

=c

taking only those terms in N which do not contain x

Comprehensive Exercise 7 Solve the following differential equations : 1.

dy / dx = (2 x − y) /( x + 2 y − 5).

2.

(2 ax + by) y dx + (ax + 2 by) x dy = 0 .

3.

(1 + 4 xy + 2 y2 ) dx + (1 + 4 xy + 2 x2 ) dy = 0 .

4.

x dx + y dy = a2

5.

[1 + e x / y] dx + e x / y[1 − ( x / y)] dy = 0 .

6.

y

(Bundelkhand 2004)

x dy − y dx

⋅ x2 + y2(Agra 2006; Meerut 08; Gorakhpur 11; Bundelkhand 11) y

(e + 1) cos x dx + e sin x dy = 0 . 2x

7.

(sin x cos y + e ) dx + (cos x sin y + tan y) dy = 0 .

8.

( y sin 2 x) dx − (1 + y2 + cos2 x) dy = 0 .

9.

[ y {1 + (1 / x)} + cos y] dx + [ x + log x − x sin y] dy = 0 .

(Meerut 2004B, 06) (Agra 2006)

D-32

A nswers 7 1.

x2 − y2 − xy + 5 y = c .

2. ayx2 + bxy2 = c

3.

x + 2 x2 y + 2 xy2 + y = c

4. x2 + y2 + 2 a2 tan−1 ( x / y) = c

5.

x + y ex /

6. (e y + 1) sin x = c

7. 9.

y

=c

1 2x e − cos x cos y + log sec y = c 2 y ( x + log x) + x cos y = c

8. y cos 2 x + 2 y +

2 3 y =c 3

1.3 Integrating Factors (Lucknow 07)

A differential equation which is not exact can sometimes be made exact by multiplying by some suitable function of x and y. Such a function is called an integrating factor (I.F.) of the equation. Methods used to find the integrating factors: Method 1: Integrating factor found by inspection: Sometimes integrating factors are found by inspection. The students should remember the following exact differentials. These help us in finding the integrating factors.  x y dx − x dy (i) d ( xy) = x dy + y dx , (ii) d   = , y2  y  y  x dy − y dx , (iii) d   =  x x2

(iv)

 x2  2 yx dx − x2 dy , d   = y2  y

 y2  2 xy dy − y2 dx  = (v) d  ,  x2  x 

(vi)

 y dx − x dy x , d  tan−1  = y x2 + y2  

y  x dy − y dx  (vii) d  tan−1  = ,  x x2 + y2

(viii)

y  x dy − y dx  (ix) d  log  = ,  x xy

(x)

(xi) d {log √ ( x2 + y2 )} =

x dx + y dy x2 + y2

.

Example 30: Solve (1 + xy) y dx + (1 − xy) x dy = 0 . Solution: The given equation can be written as

 y dx − x dy x , d  log  = y xy   ex  ye x dx − e x dy , d   = y2  y

D- 33

( y dx + x dy) + xy2 dx − x2 y dy = 0 , or

d ( yx) + xy2 dx − x2 y dy = 0 .

[from (i) of article (1.3)]

Dividing out by x2 y2 , we have d ( yx) x2 y2

+

1 1 dx − dy = 0 . x y

…(1)

Now putting xy = v in (1), we have (1/ v2 ) du + (1/ x) dx − (1/ y) dy = 0 . Now integrating each term, we get − (1/ v) + log x − log y = c , where c is a constant (∵ v = xy)

or

−(1/ xy) + log x − log y = c,

or

log ( x / y) = c + (1/ xy), is the required solution.

Example 31: Solve y dx − x dy + (1 + x2 ) dx + x2 sin y dy = 0 .

(Lucknow 2005)

Solution: Dividing each term by x2 , the given equation becomes

y dx − x dy 2

or or

 1  +  2 + 1 dx + sin y dy = 0 x 

x x dy − y dx

1  +  2 + 1 dx + sin y dy = 0 x2 x   y  1  − d   +  2 + 1 dx + sin y dy = 0 , [Refer article 1.3, result (iii)]   x x −

Integrating each term, we have y 1 − − + x − cos y = − c , x x

or

y 1 + − x + cos y = c . x x

Comprehensive Exercise 8 Solve the following differential equations : 1. x dy − y dx = xy2 dx .

2. x dy − y dx + 2 x3 dx = 0 .

3. ( y2 e x + 2 xy) dx − x2 dy = 0 . 4.

y (axy + e x ) dx − e x dy = 0 .

5.

y (2 x2 y + e x ) dx − (e x + y3 ) dy = 0 .

6.

2

(Gorakhpur 2008)

(Gorakhpur 2009; Avadh 09)

2

y sin 2 x dx = (1 + y + cos x) dy .

7. x dy − y dx = ( x2 + y2 ) dx . 8.

x dx + y dy + ( x2 + y2 ) dy = 0 .

9. ( x2 + y2 + a2 ) y dy + ( x2 + y2 − a2 ) x dx = 0 .

(Rohilkhand 2006, 07)

D-34

A nswers 8 1.

yx2 + 2 x = 2 cy .

2. y + x3 = cx . 1 2 ax + (e x / y) = c . 2 1 3 6. − y cos2 x = y + y +c. 3

3. e x + ( x2 / y) = c . 5.

4.

2 3 1 2 x − y + (e x / y) = c . 3 2

7. tan−1 ( y / x) = x + c .

8. x2 + y2 = e c − 2 y.

9. ( x2 + y2 )2 + 2 a2 ( y2 − x2 ) = c . Method 2: If the equation M dx + N dy = 0 is of the form [ f ( x, y)] y dx + [ F ( x, y)] x dy = 0, and

Mx − Ny ≠ 0 , then 1/( Mx − Ny) is an integrating factor.

Note: If Mx − Ny = 0 , then Mx = Ny

or

M/ y = N/x

i. e., the differential equation M dx + N dy = 0 reduces to y dx + x dy = 0 whose solution is xy = c . Example 32: Solve ( x3 y3 + x2 y2 + xy + 1) y dx + ( x3 y3 − x2 y2 − xy + 1) x dy = 0 . Solution: Here

Mx − Ny = ( x3 y3 + x2 y2 + xy + 1) y ⋅ x − ( x3 y3 − x2 y2 − xy + 1) xy = 2 ( x3 y3 + x2 y2 ) = 2 x2 y2 ( xy + 1). ∴

I.F. = 1 / {2 x2 y2 ( xy + 1)}.

The given equation may be written as [ x2 y2 ( xy + 1) + ( xy + 1)] y dx + [( x3 y3 + 1) − xy ( xy + 1)] x dy = 0 . Now multiplying by I.F. this becomes [ x2 y2 ( xy + 1) + ( xy + 1)] y dx 2 x2 y2 ( xy + 1)

+

[( xy + 1){( x2 y2 − xy + 1) − xy}] x dy 2 x2 y2 ( xy + 1)

or

 x2 y2 + 1 ( x2 y2 − 2 xy + 1)  2 2  y dx +   x dy = 0 x2 y2  x y   

or

( y dx + x dy) +

or

d ( xy) +

d ( xy) 2

x y

2

y dx + x dy 2

x y −

2

2 dy = 0 . y

Now integrating each term, we get xy + (−1/ xy) − 2 log y = c .

−

2 x2 y x2 y

dy = 0

=0

D- 35

Comprehensive Exercise 9 Solve the following differential equations : 1.

( xy2 + 2 x2 y3 ) dx + ( x2 y − x3 y2 ) dy = 0 .

2.

( x2 y2 + xy + 1) y dx + ( x2 y2 − xy + 1) x dy = 0 .

3.

( xy sin xy + cos xy) y dx + ( xy sin xy − cos xy) x dy = 0 .

A nswers 9 1.

x2 = cy e1 /( xy)

2. xy + log x − log y − 1/( xy) = c

3. x sec ( xy) = cy

Method 3: When Mx + Ny ≠ 0 and the equation is homogeneous, then the integrating factor of M dx + N dy = 0, is 1/( Mx + Ny) . Example 33: Solve x2 y dx − ( x3 + y3 ) dy = 0 . Solution: Here Mx + Ny = x3 y − x3 y − y4 = − y4 ≠ 0

and the equation is homogeneous. ∴

I.F. = 1/( Mx + Ny) = −1/ y4 .

Multiplying the given equation by the I.F., we get −

y

3

dx +

x3 + y3 y

4

or

dy x2 x3 = 3 dx − 4 dy y y y

integrating each term, we get log y =

Method 4: If

1 x3 + log c , 3 y3

or

y = ce x

3

/3 y3

.

1  ∂M ∂N   is a function of x alone, say f ( x), then the integrating  − N  ∂y ∂x 

factor for M dx + N dy = 0 is e ∫ Example 34: Solve ( y +

f ( x) dx

.

1 3 1 2 1 y + x ) dx + ( x + xy2 ) dy = 0 . 3 2 4

1 3 1 2 1 y + x and N = ( x + xy2 ) . 3 2 4 ∂M ∂N 1 = 1 + y2 and = (1 + y2 ) . ∂y ∂x 4

Solution: Here M = y +

∴

dy = 0,

dy 1  x3  . = d y 3  y3 

or ∴

x2

(Lucknow 2007)

D-36

∴

= ∴

 ∂M ∂N  4 1  =  {(1 + y2 ) − (1 + y2 )} − 4 ∂x  x (1 + y2 )  ∂y

1 N

4

3 3 (1 + y2 ) = , which is a function of x alone, say f ( x). 4 x (1 + y ) 4 2

I.F. = e ∫

⋅

f ( x) dx

= e ∫ (3 / x) dx = e3 log

x

= e log

x3

= x3 .

Multiplying the given diff. equation by the I.F. x3 , we get 1 1 1 ( x3 y + x3 y3 + x5 ) dx + ( x4 + x4 y2 ) dy = 0 , 3 2 4 which is an exact diff. equation and its solution is

∫ M dx (treating y as a constant) + ∫ N dy (taking in N terms without x) = c i. e.,

x4 y x4 y3 x6 + + = c, 4 12 12

or

3 x4 y + x4 y3 + x6 = 12 c = k .

Method 5: If

1  ∂N ∂M    is a function of y alone, say f ( y), then the integrating − M  ∂x ∂y 

factor for M dx + N dy = 0 is e ∫

f ( y) dy

.

Example 35: Solve ( xy3 + y) dx + 2 ( x2 y2 + x + y4 ) dy = 0 . 3

2

2

(Lucknow 2006)

4

Solution: Here M = xy + y and N = 2 ( x y + x + y ).

∴

∂M ∂N = 3 xy2 + 1 and = 2 (2 xy2 + 1) . ∂y ∂x

∴

1 M

 ∂N ∂M  1 1   = − {4 xy2 + 2 − 3 xy2 − 1} = (xy2 + 1) ∂y  xy3 + y y ( xy2 + 1)  ∂x =

∴

I.F. = e ∫

f ( y) dy

= e ∫ (1 /

1 , which is a function of y alone, say f ( y). y

y) dy

= e log

y

= y.

Multiplying the given equation with the I.F. y, we have ( xy4 + y2 ) dx + 2 ( x2 y3 + xy + y5 ) dy = 0 , which is an exact differential equation and its solution is

∫ M dx (treating y as a constant) + N dy (taking in N only those terms which do not contain x) ∫ = c, (a constant) 4

2

i. e.,

∫ ( xy

or

1 2 4 1 6 x y + xy2 + y =c. 2 3

+ y ) dx ( y constant) +

∫2 y

5

dy = c

D- 37

Comprehensive Exercise 10 Solve the following differential equations : 1.

( x2 y − 2 xy2 ) dx − ( x3 − 3 x2 y) dy = 0 .

2.

( x2 + y2 ) dx − 2 xy dy = 0 . 2

4

3

(Lucknow 2011) (Gorakhpur 2008)

3

2

3.

(3 x y + 2 xy) dx + (2 x y − x ) dy = 0 .

4.

( xy2 − x2 ) dx + (3 x2 y2 + x2 y − 2 x3 + y2 ) dy = 0 .

A nswers 10 2. x2 − y2 = cx

1.

( x / y) − 2 log x + 3 log y = c

3.

x3 y2 + ( x2 / y) = c

4.

1 1 1 1 1 e6 y[ x2 ( y2 − x) + ( y2 − y+ )] = 0 2 3 6 18 108

Method 6: If the equation is of the form x a y b (my dx + nx dy) + x c y d ( py dx + qx dy) = 0 ,

…(A)

where a, b, m, n, c , d, p, q are constants, then the integrating factor is x h y k , where h, k can be obtained by applying the condition that after multiplication by x h y k , the equation (A) must become exact. Example 36: Solve ( y2 + 2 x2 y) dx + (2 x3 − xy) dy = 0 . Solution: The given equation can be written as

y ( y dx − x dy) + 2 x2 ( y dx + x dy) = 0 . This equation is of the form (A) , as mentioned above. So let the possible integrating factor be x h y k . Multiplying the given equation by the proposed I.F. x h y k , we have ( x h y k + 2 + 2 x h + 2 y k +1) dx + (2 x h + 3 y k − x h +1 y k +1) dy = 0. Now and

h k +1

(∂M / ∂y) = (k + 2) x y

(∂N / ∂x) = 2 (h + 3) x

+ 2 (k + 1) x h+2

h+2

y

k

k

y − (h + 1) x h y k + 1.

If the equation (1) is exact, then (∂M / ∂y) = (∂N / ∂x). ∴

equating the coefficients of x h y k +1 and x h + 2 y k on both sides, we get

…(1)

D-38

and

k + 2 = − h −1

i. e.,

h+ k +3=0

2k + 2 = 2h + 6

i. e.,

h− k +2=0.

Solving these, we get h = − 5 / 2 and k = − 1/ 2. ∴

the integrating factor is x h y k = x −5 /2 y − 1 /2 .

With these values of h and k the equation (1) becomes ( x −5 /2 y3 /2 + 2 x −1 /2 y1 /2 ) dx + (2 x1 /2 y −1 /2 − x −3 /2 y1 /2 ) dy = 0 . This differential equation is of the form M dx + N dy = 0 and is exact as can be seen by verifying the condition ∂M / ∂y = ∂N / ∂x . Hence its solution is

∫ M dx (treating y as a constant) + N dy (taking in N only those terms which do not contain x) ∫ = c, (a constant) i. e.,

−

Hence −

2 −3 / 2 3 / 2 x y + 4 x1 /2 y1 /2 + 0 = c . 3

2 −3 / 2 3 / 2 x y + 4 x1 /2 y1 /2 = c is the required solution. 3

(Gorakhpur 2007)

Comprehensive Exercise 11 Solve the following differential equations : 1.

(2 y dx + 3 x dy) + 2 xy (3 y dx + 4 x dy) = 0 .

2.

(3 x + 2 y2 ) y dx + 2 x (2 x + 3 y2 ) dy = 0 .

3.

x (3 y dx + 2 x dy) + 8 y4 ( y dx + 3 x dy) = 0 .

(Lucknow 2010)

A nswers 11 1.

x2 y3 + 2 x3 y4 = c

2. x3 y4 + x2 y6 = c

3. x3 y2 + 4 y6 x2 = c

1.4 Change of Variables In some cases a suitable substitution (change of variables) reduces a given differential equation to one or the other of the forms already discussed and hence the equation can be solved.

D- 39

Example 37: Solve

 a2 − x2 − y2  x dx + y dy  . =  2 2  x dy − y dx  x + y  (Rohilkhand 2005; Purvanchal 09; Avadh 09; Bundelkhand 03)

Solution: Put x = r cos θ and y = r sin θ . Then

and

x2 + y2 = r2 ,

…(1)

y / x = tan θ.

…(2)

Differentiating (1), we get or

2 x dx + 2 y dy = 2 r dr Differentiating (2), we get x dy − y dx x2 or

x dx + y dy = r dr .

= sec2 θ dθ,

x dy − y dx = x2 sec2 θ dθ = r2 cos2 θ sec2 θ dθ = r2 dθ .

Substituting these values in the given differential equation, we get  a2 − r2  √ (a2 − r2 )   . =  r2  = r r2 dθ   r dr

∴

dr = √ (a2 − r2 ) dθ

dr

or

2

√ (a − r2 )

= dθ , [separating the variables]

Integrating, we get sin−1(r / a) = θ + c , or or

r = a sin (θ + c )

√ ( x2 + y2 ) = a sin {tan−1( y / x) + c } . [∵ r = √ ( x2 + y2 ) and θ = tan−1 ( y / x)]

1.5 Geometrical Problems Remember the following formulae of the differential calculus: Length of the tangent = y √ {1 + (dx / dy)2 } . Length of the subtangent = y (dx / dy) = y / (dy / dx). Length of the normal = y √ {1 + (dy / dx)2 }. Length of the subnormal = y (dy / dx). Intercept made by the tangent on the x-axis is x − y (dx / dy), and on the y-axis is y − x (dy / dx) . Angle φ between the radius vector and the tangent is given by tan φ = r (d θ / dr),

or

cot φ = (1/ r)(dr / dθ) .

D-40

Length of the polar subtangent = r2 (dθ / dr). Length of the polar subnormal = (dr / dθ) . Radius of curvature ρ = [1 + (dy / dx)2 ]3 /2 / (d2 y / dx2 ).

Example 38: Find the curves in which the polar subnormal is of constant length. Solution: Let the constant length of the polar subnormal be a.

But the length of the polar subnormal = dr / dθ . ∴

as given in the question, dr / dθ = a .

This is the differential equation of the required curves. To solve, put it into the form dr = a dθ , ∴

[separating the variables]

integrating, we get r = aθ + c , where c is an arbitrary constant.

This is the polar equation of the required family of curves. Example 39: Find the curves for which the sum of the reciprocals of the radius vector and the polar

subtangent is constant. Solution: The polar subtangent = r2 (dθ / dr).

∴

as given in the question,

1 1 dr + = λ , where λ is a constant. r r2 dθ

This is the differential equation of the required curves. This equation may be written as dr 1  = r2  λ −  = r (λr − 1)  dθ r or ∴

dθ =

 λ dr = − r (rλ − 1)  λr − 1

1  dr , r

[separating the variables]

integrating, we have θ + c = log (λr − 1) − log r , where c is an arbitrary constant

or

θ + c = log {(λr − 1) / r},

or

(λr − 1) / r = e θ + c .

Hence λr − 1 = re θ + c is the equation of the required family of curves. Example 40: Find the curves in which the cartesian subtangent varies as the abscissa. Solution: The cartesian subtangent is y /(dy / dx).

∴

as given in the question y /(dy / dx) = λx, where λ is a constant

or

λ dx = λx dy,

∴

integrating, we have log x = λ log y + log c , where c is an arbitrary constant

or

Hence log x = log ( y λ . c )

(dx / x) = λ (dy / y), separating the variables.

or

x = cy λ is the required family of curves.

D- 41 Example 41: Show that the curve in which the angle between the tangent and the radius vector at

every point is half of the vectorial angle is a cardioid. 1 θ, where θ is the vectorial angle. 2 1 tan φ = tan θ . But tan φ = r (dθ / dr). 2 1 1 or r (dθ / dr) = tan θ , (dr / r) = (cot θ) dθ , 2 2

Solution: As given, φ =

∴ ∴

in which the variables have been separated. 1 integrating, we have log r = 2 log (sin θ) + log c , ∴ 2 where c is an arbitrary constant 1 1 or or log r = log (sin2 θ) + log c , log (r / c ) = log (sin2 θ) 2 2 1 1 1 or (r / c ) = sin2 θ = (1 − cos θ) , or r = c (1 − cos θ) 2 2 2 1 or r = a (1 − cos θ), where a = c . This is a cardioid. 2 Example 42: Find the curve for which the tangent at each point makes a constant angle α with the

radius vector. Solution: As given φ = α (a constant)

or ∴

tan φ = tan α . But tan φ = r (dθ / dr) . r (dθ / dr) = tan α ,

or

(dr / r) = (cot α) dθ .

Now the variables have been separated. Therefore integrating, we have log r = θ cot α + log c , where c is an arbitrary constant or

log r − log c = θ cot α,

or

log (r / c ) = θ cot α

or

r / c = e θ cot α . Hence r = ce θ cot α is the required curve.

Example 43: Show that the curve in which the slope of the tangent at any point equals the ratio of

the abscissa to the ordinate of the point is a rectangular hyperbola. Solution: The slope of the tangent at any point ( x, y) = tan ψ = dy / dx .

dy abscissa x = = dx ordinate y

∴

as given,

or

y dy = x dx , in which the variables have been separated. 1 1 integrating, we have y2 = x2 + c , where c is an arbitrary constant. 2 2

∴

Hence y2 − x2 = 2 c = a2 (say) is the required curve which is the equation of a rectangular hyperbola.

D-42 Example 44: Find the family of curves whose tangent forms an angle π / 4 with the hyperbola

xy = c . Solution: Let m1 and m2 be the gradients of the tangents of the required family of curves

and the given hyperbola respectively. Since the angle between these tangents is given 1 to be π , therefore 4 m − m2 m − m2 tan (π / 4) = 1 , or 1= 1 1 + m1m2 1 + m1m2 or

1 + m1m2 = m1 − m2

or

m1 = (1 + m2 ) / (1 − m2 ) .

…(1)

Now m1 = gradient of the tangent of the required family of curves = dy / dx and

m2 = (dy / dx) for hyperbola xy = c .

∴

m2 =

d dx

c c   =− 2 ,  x x

c  ∵ y = x 

Hence from (1), we have dy 1 + (− c / x2 ) x2 − c  2c = = = 1 − 2 dx 1 − (− c / x2 ) x2 + c  x +

∴

  2c  or dy = 1 − 2   c x + 

  dx, c 

in which the variables have been separated. 2c  x  integrating, we have y = x − tan−1   + c1,  √ c √c

where c1 is an arbitrary constant. This is the required family of curves.

Comprehensive Exercise 12 1.

By the substitution y2 = v − x reduce the equation y3 (dy / dx) + x + y2 = 0 to the homogeneous form and hence solve the equation.

2.

Find the curve in which the polar subtangent is constant.

3.

Find the equation of the family of curves for which the sum of the reciprocals of the radius vector and the polar subnormal is constant.

4.

Find the curve in which the cartesian sub-normal is equal to the abscissa.

5.

Find the equation of the curve for which the cartesian subtangent varies as the reciprocal of the square of the abscissa.

6.

Show that the parabola is the only curve in which the sub-normal is constant.

7.

Show that the curve for which the normal at every point passes through a fixed point is a circle.

8.

Find the equation of the curve in which the perpendicular from the origin on any tangent is equal to the abscissa of the point of contact.

D- 43

9.

The normal PN to a curve meets the x-axis in N. If the distance of N from the origin is twice the abscissa of P, prove that the curve is a rectangular hyperbola.

10. The tangent of any point P of a curve meets the x-axis in Q. If Q is on the positive side of the origin O and OP = OQ , show that the family of curves having this property are parabolas whose common axis is the x-axis.

A nswers 12 1. tan−1 {( y2 + x) / x} − 2. r (θ + c ) + a = 0 4. y2 − x2 = c 2 6.

1 log { x2 + ( y2 + x)2 } = log c 2 3. θ = λr − log r + c 1 5. λ log y = x3 + c 3

y2 = 2 λ { x + (c / 2 λ )}

8.

y2 + x2 = cx

O bjective T ype Q uestions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

The order of the differential equation d4 y dx4 (a) 3 (c) 4

2.

The integrating factor of the differential equation cos2 x

2

(c) cos x

4.

(Rohilkhand 2005)

(b) 6 (d) 2

(a) sec2 x

3.

2

 d3 y  d2 y dy + 6 y = 0 is − 3  3  + 4 2 − 5 dx dx  dx 

dy + y = tan x is dx

(b) e tan x (d) tan x

dy + y cot x = 2 cos x is dx (a) sin x (b) log sin x (c) cot x (d) cos x The differential equation M dx + N dy = 0 , where M and N are functions of x and y, is exact if (Avadh 2005) ∂M ∂N ∂M ∂N (a) (b) = = ∂y ∂x ∂x ∂y

The integrating factor of the differential equation

(c) M + N = 0

(d) M = N

D-44

5.

The complete solution of the differential equation ( x2 − ay) dx − (ax − y2 ) dy = 0 is

6.

(a) x3 + y3 + 3 axy = c

(b) x3 + y3 − 3 axy = 4

(c) x3 + y3 − 3 axy = c

(d) x3 + y3 − 3 axy = 0

The integrating factor of differential equation (a) e ∫ P dy

(b) e ∫ P dx

(d) e ∫ − P dx dy The differential equation ( x2 + y2 + a2 ) y + x ( x2 + y2 − a2 ) = 0 is dx (c) e

7.

dx + P ( y) x = Q( y) is dy (Bundelkhand 2007)

∫ Q dx

(Rohilkhand 2007)

(a) variables separable

(b) linear

(c) Homogeneous

(d) Exact

Fill in the Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

The order of the differential equation d3 y 3

dx

−2

d2 y 2

dx

+3

dy + 2 y = 0 is …… . dx 3

2.

dy  dy  The degree of the differential equation   − 4 xy + 8 y2 = 0 is …… .  dx  dx

3.

  d2 y  The order and degree of the differential equation  2  − 1 +  dx  

2

…… and …… 4.

3

2  dy      = 0 is  dx   

(Rohilkhand 2006)

A differential equation of the form

dy f ( x, y) = 1 , where f1 ( x, y) and f2 ( x, y) dx f2 ( x, y)

are homogeneous functions of x and y of the same degree, is called a …… . 5.

To solve the differential equation x2 y dx − ( x3 + y3 ) dy = 0 , we put…… .

6.

To solve the differential equation

7.

dy 2x + y + 1 = , we put …… . dx 4 x + 2 y − 1

A differential equation is said to be …… when the dependent variable y and all its derivatives occur in the first degree only and are not multiplied together. dy 8. The integrating factor of the differential equation + Py = Q , where P and Q dx are functions of x only, is …… . dy 9. The integrating factor of the differential equation ( x2 − 1) + 2 xy = 1 is …… . dx dy 10. The integrating factor of the differential equation + 2 y tan x = sin x is …… . dx

D- 45

11. The ordinary differential equation M dx + N dy = 0 , where M and N are functions of x and y, is exact if and only if …… . (Meerut 2003; Bundelkhand 08; Rohilkhand 10)

12. A differential equation which is not exact can sometimes be made exact by multiplying by some suitable function of x and y. Such a function is called an …… of the equation. 13. The differential equation

tan−1 y dx 1 + ⋅ x = dy 1 + y2 1 + y2

is linear with …… as dependent variable. 14. If the differential equation M dx + N dy = 0 is homogeneous and if Mx + Ny ≠ 0 , then the integrating factor of this differential equation is …… . (Bundelkhand 2007) 2

15. The differential equation ( x + y) dy = a dx is linear with …… as independent variable. 16.

A solution of differential equation of three order has …… constants. (Agra 2007)

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

The differential equation y2 dx + ( xy + x2 ) dy = 0 is homogeneous.

2.

The order of the differential equation

3.

The degree of the differential equation

4. 5.

6.

7.

d3 y dx3

d2 y dx2

2

 dy  − 7   + 6 y = 0 is 2.  dx  2

 dy  + 4   − 5 y = 0 is 2.  dx 

dy + y = 1 is linear with y as dependent variable. dx dy The integrating factor of the differential equation + y sec x = tan x is dx (sec x + tan x). dy The differential equation ( x + y + 1) = 1 is linear with y as dependent dx variable. The differential equation x2

The differential equation (4 x + 3 y + 1) dx + (3 x + 2 y + 1) dy = 0 is exact. dy y 8. The integrating factor of the differential equation + = sin x2 is log x. dx x dy 1 9. The differential equation + y = x2 y6 is linear with y as dependent dx x variable. dy 1 10. The differential equation x + y = x2 y4 can be made linear by putting 3 = v . dx y

D-46

11. A differential equation of the form

dy + Py = Qy n, dx

where P and Q are functions of x alone, is called Bernoulli’s equation. (Meerut 2003)

12. If the differential equation M dx + N dy = 0 is exact, its solution is (Meerut 2003)

∫

M dx

∫

+

treating y as constant

N dy

c

=

taking only those terms in N which do not contain x

A nswers Multiple Choice Questions 1. 6.

(c) (a)

2. (b) 7. (d)

3. (a)

4. (a)

5. (c)

Fill in the Blank(s) 1. 3 5. y = vx 9. x2 − 1 13.

x

2. 3 6. 2 x + y = v 10. sec2 x 14.

3. 2, 2 7. linear ∂M ∂N 11. = ∂y ∂x

1 15. Mx + Ny

y

4. homogeneous equation 8. e ∫ P dx 12.

integrating factor

16. 3

True or False 1. 6. 11.

T F T

2. 7. 12.

F T T

3. 8.

F F

4. 9.

T F

5. 10.

T T

¨

D-47

2 D ifferential E quations of T he F irst O rder B ut N ot of T he F irst D egree

2.1 Introduction n the present chapter we shall discuss the solutions of differential equations which are of the first order but are of degree higher than one. Such differential equations will contain only the first differential coefficient dy / dx but it will occur in a degree higher than one. It is usual to denote dy / dx by p. The general form of such a differential equation is then

I

pn + A1 pn − 1 + A2 pn − 2 + … + An − 1 p + An = 0 , where A1 , A2 , … , An are some functions of x and y. Now we shall consider the various methods of solving the differential equations of the above type.

2.2 Equations Solvable for p Suppose a differential equation of first order and of degree n can be solved for p i. e., it can be resolved into n linear factors in p of the type { p − f1( x, y)} { p − f2 ( x, y)}...{ p − f n ( x, y)} = 0 .

D-48

We can equate each factor to zero and the resulting differential equations of the first order and first degree can be solved. Let their solutions be φ1 ( x, y, c1) = 0 , φ2 ( x, y, c2 ) = 0 , ..., φn( x, y, c n) = 0 , where c1, c2 , … , c n are arbitrary constants. There is no loss of generality if we replace the arbitrary constants c1, c2 , … , c n by a single arbitrary constant c because in any of the above n solutions c is free to take any real value. Thus the n solutions of the given differential equation are φ1 ( x, y, c ) = 0 , φ2 ( x, y, c ) = 0 , … , φn ( x, y, c ) = 0 . Combining the above equations, we get a single composite solution as φ1 ( x, y, c ) φ2 ( x, y, c ) … φn ( x, y, c ) = 0 .

Example 1: Solve p2 − 7 p + 12 = 0 .

(Bundelkhand 2008; Kanpur 04)

Solution: Resolving into linear factors, the given differential equation can be written as

( p − 3) ( p − 4) = 0 . Its component equations are p = 3, p = 4 . Solving the differential equation p = 3 i. e., dy / dx = 3, we get y = 3 x + c . Also the solution of the differential equation p = 4 is y = 4 x + c . So the solutions of the given differential equation are y = 3 x + c , y = 4 x + c . The single combined solution is ( y − 3 x − c) ( y − 4 x − c) = 0 . Example 2: Solve p2 + 2 py cot x = y2 .

(Gorakhpur 2005; Rohilkhand 09; Kanpur 08; Bundelkhand 04)

Solution: The given differential equation is

p2 + 2 py cot x − y2 = 0 . Solving for p , we get p=

dy − 2 y cot x ± √ (4 y2 cot2 x + 4 y2 ) = dx 2

= − y cot x ± y cosec x = y (− cot x ± cosec x) . Thus the component equations are dy = y (− cot x + cosec x) , dx dy and = − y (cot x + cosec x) . dx In each of the above differential equations, the variables are separable. From (1), separating the variables, we have dy = (− cot x + cosec x) dx . y

…(1) …(2)

D-49

Integrating, we get 1 x 2 1  1 1     tan 2 x   sin 2 x / cos 2 x   y log   = log   = log  1 1  c  sin x   2 sin x cos x     2 2 

log y − log c = − log sin x + log tan

or

     1 1 = log  = log   1 2 + 1 cos   2 cos x  2  y / c = 1/(1 + cos x)

∴

or

 . x …(3)

y = c /(1 + cos x).

From (2), separating the variables, we have dy = − (cot x + cosec x) dx . y Integrating, we get log y − log c = − (log sin x + log tan

1 x) 2

1 x)} 2 1 1 1 1 1 = − log {2 sin x cos x ⋅ (sin x /cos x)} = − log (2 sin2 x) 2 2 2 2 2

or

log ( y / c ) = − log{(sin x)(tan

= − log (1 − cos x) = log (1 − cos x)−1 = log {1/(1 − cos x)} . or

y / c = 1/(1 − cos x)

∴

y = c /(1 − cos x).

…(4)

Thus the solutions of the given differential equation are given by (3) and (4). The single combined solution is     c c  =0. y−  y− 1 + cos x   1 − cos x   Example 3: Solve ( p − xy) ( p − x2 ) ( p − y2 ) = 0 . Solution: Equating each factor to zero, the component equations are

and

p − x2 = 0 ,

…(1)

p − xy = 0 , p − y2 = 0 .

From (1), p = xy

or

…(2) …(3)

dy / dx = xy

or

(1/ y) dy = x dx.

Integrating, we get log y = or

y/c = 2

From (2), p = x

Integrating, we get

or

1 2 x + log c 2

1 2 x e2 2

dy / dx = x

or

log ( y / c ) =

or

y=

or

2

1 2 x 2 . ce

dy = x dx .

1 2 x 2 …(4)

D-50

1 3 1 x + c 3 3

y= From (3), p = y2

dy / dx = y2

or

or

3 y − x3 = c .

or

(1/ y2 ) dy = dx.

…(5)

Integrating, we have − (1 / y ) = x + c

or

xy + cy + 1 = 0 .

…(6)

Now (4), (5) and (6) are the solutions of the given differential equation. The single combined solution is 1 x2    y − ce2  (3 y − x3 − c ) ( xy + cy + 1) = 0 .  

Example 4: Solve p ( p − y) = x ( x + y) . Solution: The given differential equation can be written as

p2 − py − ( x2 + xy) = 0 . Solving for p, we get p=

dy y ± √ ( y2 + 4 x2 − 4 xy) y ± √ {( y + 2 x)2 } y ± ( y + 2 x) . = = = dx 2 2 2

Thus the component equations are and

dy / dx = { y + ( y + 2 x)} / 2 = y + x,

…(1)

dy / dx = ( y − y − 2 x) / 2 = − x .

…(2)

From (1), we have (dy / dx) − y = x , which is a linear differential equation with y as the dependent variable. Integrating factor is e∫

− dx

= e− x .

Therefore its solution is ye − x =

∫ xe

−x

dx + c

or

ye − x = − xe − x −

or

ye − x = − xe − x − e − x + c

or

ye − x = − e − x ( x + 1) + c

or

y = − ( x + 1) + ce x

∫ (− e

−x

) dx + c , integrating by parts

or

y + x + 1 − ce x = 0 .

…(3)

or

2 y + x2 − c = 0 .

…(4)

From (2), we have dy = − x dx . Integrating, we get y=−

1 2 1 x + c 2 2

Thus (3) and (4) are the required solutions of the differential equation. The single combined solution is ( y + x + 1 − ce x ) (2 y + x2 − c ) = 0 .

D-51

 y4  y y2 Example 5: Solve 1 − y2 + 2  p2 − 2 p+ 2 =0 . x x  x  Solution: Multiplying throughout by x2 , the given differential equation becomes

( x2 − x2 y2 + y4 ) p2 − 2 xyp + y2 = 0 or

x2 p2 − 2 xyp + y2 = p2 ( x2 y2 − y4 )

or

( xp − y)2 = p2 y2 ( x2 − y2 )

∴

xp − y = ± py √ ( x2 − y2 )

or

1 dx x ± y √ ( x2 − y2 ) . = = p dy y

p [ x ± y √ ( x2 − y2 )] = y

or

Putting x = vy so that (dx / dy) = v + y (dv / dy), we have v+ y

dv vy ± √ (v2 y2 − y2 ) = = v ± y √ (v2 − 1) dy y

dv = ± y √ (v2 − 1) dy

or

y

or

dv = ± √ (v2 − 1) dy

or

dy = ± [1/ √ (v2 − 1)] dv, separating the variables.

Integrating, we get y + c = ± cosh −1 v or

y + c = ± cosh −1( x / y)

or

cosh −1 ( x / y) = ± ( y + c )

or

x / y = cosh { ± ( y + c )} .

Since cosh (− x) = cosh x , therefore, we have x = y cosh ( y + c ) and x = y cosh ( y + c ), as the required solutions. The single combined solution is [ x − y cosh ( y + c )]2 = 0 .

Comprehensive Exercise 1 Solve the following differential equations : 1. 2. 3.

p2 − 5 p + 6 = 0 . 2

(Rohilkhand 2006; Kanpur 02)

2 2

y + xyp − x p = 0 . 2

2

2

xyp − ( x + y ) p + xy = 0 . 2

2

(Meerut 2007B) (Gorakhpur 2011)

2

4.

xyp + p (3 x − 2 y ) − 6 xp = 0 .

5.

x (dy / dx)2 + ( y − x)(dy / dx) − y = 0 .

(Meerut 2006B; Bundelkhand 09, 10)

D-52

6.

y (dy / dx)2 + ( x − y)(dy / dx) − x = 0 .

(Meerut 2005B)

A nswers 1 1.

( y − 2 x − c) ( y − 3 x − c) = 0

2.

( y2 − c x1+√5 )( y2 − c x1−√5 ) = 0

3.

( x2 − y2 − c ) ( y − cx) = 0

4.

( y − cx2 ) ( y2 + 3 x2 − c ) = 0

5.

( y − x − c ) ( xy − c ) = 0

6.

( y − x − c ) ( x2 + y2 − c 2 ) = 0

2.3 Equations Solvable for y Suppose the given differential equation is solvable for y. Then it can be put in the form …(1)

y = f ( x, p). Differentiating (1) w.r.t. x and denoting dy / dx by p, we obtain dp  p = φ  x, p,  ,  dx 

…(2)

which is a differential equation in two variables x and p. Suppose it is possible to solve the differential equation (2). Let its solution be …(3)

F ( x, p, c ) = 0, where c is the arbitrary constant.

Eliminating p between (1) and (3), we get the required solution of (1) in the form ψ ( x, y, c ) = 0 . If it is not easily practicable to eliminate p between (1) and (3), we may solve (1) and (3) to get x and y in terms of p and c in the form x = f1 ( p, c ), y = f2 ( p, c ) ,

…(4)

which give us the required solution of (1) in the form of parametric equations, the parameter being p. Special case: Equations that do not contain x: In this case the equation has the form f ( y, p) = 0 . If it is solvable for p, it will give p = φ ( y)

i. e.,

dy / dx = φ ( y) ,

which can be easily solved by separating the variables. If it is solvable for y, it will give y = ψ ( p), which can be solved by the method just explained in article 2.3.

D-53

Example 6: Solve y + px = p2 x4 . (Rohilkhand 2006; Avadh 08; Agra 08; Purvanchal 11; Lucknow 11)

Solution: Solving for y, the given differential equation can be written as

y = − px + p2 x4 .

…(1)

Differentiating (1) w.r.t. x and denoting dy / dx by p, we get dp dp p= − p− x + 2 p x4 + 4 p2 x3 dx dx dp or 2 p − 4 p2 x3 + x (1 − 2 x3 p) = 0 dx dp or 2 p (1 − 2 x3 p) + x (1 − 2 x3 p) = 0 dx dp  or (1 − 2 x3 p) 2 p + x  = 0 .  dx  dp = 0, dx

∴

2p + x

and

1 − 2 x3 p = 0 .

From (2), dp / dx = −2 p / x

or

…(2)

dp / p = −(2 / x) dx .

Integrating, we get log p = − 2 log x + log c or

log px2 = log c

or

px2 = c

or

p = c / x2 .

Substituting this value of p in (1), we get y = − x (c / x2 ) + (c 2 / x4 ). x4 or

y = − (c / x) + c 2

or

xy = c 2 x − c , as the required solution of (1).

Note: If we eliminate p between (1) and (3), we get

y = − (1/ 2 x3 ) x + (1/ 4 x6 ). x4 = − (1/ 2 x2 ) + (1/ 4 x2 ) = −1/ 4 x2 , which is also a solution of (1) because it satisfies (1). This solution itself does not contain any arbitrary constant. Also it cannot be obtained from the general solution xy = c 2 x − c by giving any particular value to c. Such a solution is called the singular solution and we shall discuss it in details later on. Example 7: Solve y = 2 px − p2 .

(Meerut 2001, 04B; Kanpur 07)

Solution: The given differential equation is

y = 2 px − p2 . Differentiating (1) w.r.t. x and denoting dy / dx by p, we get dp dp p = 2p + 2x − 2p dx dx

…(1)

D-54

dp ( x − p) = 0 dx

or

p+2

or

dx 2 + x = 2, dp p

or

p

dx + 2x = 2p dp …(2)

which is a linear differential equation with x as dependent variable and p as the independent variable. 2

∴

integrating factor = e ∫ (2 / p) dp = e2 log p = e log p = p2 .

∴

the solution of (2) is xp2 =

or

x=

2

2

3

∫ 2 p dp + c = 3 p

+c

2 p + cp−2 3

…(3)

Here it is not easily practicable to eliminate p between (1) and (3). So putting the value 2 of x from (3) in (1), we get y = 2 p ( p + cp−2 ) − p2 3 1 2 −1 or …(4) y = p + 2 cp . 3 The equations (3) and (4), which express x and y in terms of a parameter p, constitute the required solution of (1). 2

Example 8: Solve y − x = x

dy  dy  +  . dx  dx 

(Meerut 2008)

Solution: Denoting dy / dx by p, the given differential equation can be written as

y − x = xp + p2 . Solving it for y, we get y = x (1 + p) + p2 . Differentiating (1) w.r.t. x and writing p for dy / dx, we get dp dp p = 1+ p + x + 2p dx dx dp dp or 1+ x + 2p =0 dx dx dx or + x = − 2 p, dp

…(1)

…(2)

which is a linear differential equation. Here the I.F. = e ∫ 1 dp = e p . ∴

the solution of (2) is x ep =

∫ − 2 pe

p

dp + c

or

x e p = −2 ( pe p − e p ) + c , integrating by parts

or

x e p = − 2 e p ( p − 1) + c

or

x = − 2 ( p − 1) + ce − p .

…(3)

D-55

Substituting this value of x in (1), we get y = (1 + p) { − 2 ( p − 1) + ce − p } + p2 y = c (1 + p) e − p + 2 − p2 .

or

…(4)

The equations (3) and (4), which express x and y in terms of a parameter p, constitute the required solution of (1). Example 9: Solve y = a √ (1 + p2 ). Solution: The given differential equation is

y = a √ (1 + p2 ) .

…(1)

Differentiating (1) w.r.t. x and writing p for dy / dx, we get a dp 1 or = dx. p = a ⋅ (1 + p2 )−1 /2 ⋅ 2 p (dp / dx) 2 √ (1 + p2 ) Integrating, we get x = a log { p + √ (1 + p2 )} + c .

…(2)

Now from (1), y2 = a2 + a2 p2 Putting

or

a2 p2 = y2 − a2 or p = √ ( y2 − a2 ) / a .

√ (1 + p2 ) = y / a and p = √ ( y2 − a2 ) / a in (2), we get  √ ( y2 − a2 ) y  x = a log  + +c a a  

or

x = a log { y + √ ( y2 − a2 )} − a log a + c

or

x = a log { y + √ ( y2 − a2 )} + c ,

writing c for − a log a + c because c is an arbitrary constant. Hence the required solution is x = log { y + √ ( y2 − a2 )} + c .

Comprehensive Exercise 2 Solve the following differential equations : 1. 2.

y = 2 px + p4 x2 .

(Meerut 2006; Avadh 09)

3

4 p + 3 xp = y .

3. x2 + p2 x = yp . 4.

y − 2 x p = f ( xp2 ) .

5.

y = x { p + √ (1 + p2 )} .

6.

y = sin p − p cos p.

7.

(Meerut 2007) (Kanpur 2009; Rohilkhand 09) (Gorakhpur 2006) 3

2

By differentiating with respect to x the equation p + xp = y, obtain its general solution in the form x = f ( p), y = φ ( p) .

D-56

A nswers 2 1. 2. 3.

( y − c 2 )2 = 4 cx 12 2 3 p + cp−3 /2 , y = − p3 + 3 cp−1 /2 7 7 1 3 1 1 x = − p + cp1 /2 , y = (− p2 + cp1 /2 )2 / p + p2 (− p2 + cp1 /2 ) 3 3 3 x=−

4.

y = 2 c √ x + f (c 2 )

5.

x3 + y2 − 2 xc = 0

6.

x = c − cos p , y = sin p − cos p 1 (2 c + 3 p2 − 2 p3 ) 2 cp2 + 2 p3 − p4 2 , = y x= ( p − 1)2 2 ( p − 1)2

7.

2.4 Equations Solvable for x Suppose the given differential equation is solvable for x. Then it can be put in the form …(1)

x = f ( y, p) . Differentiating (1) w.r.t. y and writing 1 / p for dx / dy, we get  dp 1 = φ  y, p,  , p dy  

…(2)

which is a differential equation in two variables y and p. Suppose it is possible to solve the differential equation (2). Let its solution be …(3)

F ( y, p, c ) = 0, where c is the arbitrary constant.

Eliminating p between (1) and (3), we get the required solution of (1) in the form ψ ( x, y, c ) = 0 . In case it is not easily practicable to eliminate p between (1) and (3), we may solve (1) and (3) to get x and y in terms of p and c in the form x = f1( p, c ), y = f2 ( p, c ),

…(4)

which give us the required solution of (1) in the form of parametric equations, the parameter being p. Special Case: Equations that do not contain y:

In this case the equation has the form f ( x, p) = 0 . If it is solvable for p, it will give p = φ ( x)

i. e.,

dy / dx = φ ( x) , which can be easily integrated.

D-57

If it is solvable for x, it will give x = ψ ( p), which can be solved by the method just explained in article 2.4.

Example 10: Solve y = 2 px + y2 p3 .

(Purvanchal 2007, 10; Lucknow 10; Avadh 07)

Solution: Solving for x, the given differential equation can be written as

2 px = y − y2 p3 or

x=

y y2 p2 . − 2p 2

…(1)

Differentiating (1) w.r.t. y and writing 1/p for dx / dy, we obtain y dp 2 yp2 y2 dp 1 1 = − 2 − − ⋅2p p 2 p 2 p dy 2 2 dy or

 dp  y 1 =0 + yp2 +  2 + py2  2p  dy 2 p

or

 dp  1   1 =0 p  2 + py + y  2 + py  dy 2 p  2 p

or

  1 dp    2 p2 + py  p + y dy  = 0 .  

∴

p+ y

The equation From p + y or

1 2 p2

dp =0 dy

or

1 2 p2

+ py = 0 .

+ py = 0 will give us the singular solution of (1).

dp dp p = 0, we have =− dy dy y (1/ p)dp = −(1/ y) dy.

Integrating, we get log p = − log y + log c or

py = c

or

log py = log c

or p = c / y.

Substituting this value of p in the given differential equation, we get y = 2 x ⋅ (c / y) + y2 (c 3 / y3 ) or

or

y = 2 cx / y + c 3 / y

y2 = 2 cx + c 3 , which is the required solution.

Example 11: Solve the differential equation

p3 − 4 xyp + 8 y2 = 0 . Solution: Solving the given differential equation for x, we get

D-58

4 xyp = p3 + 8 y2

or

x=

p2 2 y + . 4y p

…(1)

Differentiating (1) w.r.t. y and writing 1/ p for dx / dy, we get p2 2 p dp 2 2 y dp 1 =− + + − 2 p 4 y dy p p2 dy 4y  p 2 y p2 1  =  − − 2 2 2 y p p  4y 

or

dp dy

or

dp  p 2 y p  − 2  =  dy  2 y p  2y

or

dp p = , dy 2 y

or

(2 / p) dp = (1/ y) dy, separating the variables.

 p 2 y    2 y − p2   

cancelling the common factor on either side which corresponds to the singular solution of (1)

Integrating, we get 2 log p = log y + log c or

log p2 = log cy

or

p2 = cy .

…(2)

Now we shall eliminate p between the given differential equation and the equation (2). From the given differential equation, we have 8 y2 = p (4 xy − p2 ) or

8 y2 = (cy)1 /2 (4 xy − cy), substituting for p from (2)

or

8 y2 = c1 /2 y3 /2 (4 x − c )

or

8 y1 /2 = c1 /2 (4 x − c )

or

64 y = c (4 x − c )2

or

64 y = 16 c ( x −

1 1 c ( x − c )2 4 4

or

or

y=

1 2 c) , 4

y = c ( x − c ) , writing c for

1 c. 4

Hence the required solution is y = c ( x − c )2 . Example 12: Solve ap2 + py − x = 0 .

(Rohilkhand 2005; Meerut 13B)

Solution: Solving the given differential equation for x, we get

x = py + ap2 .

…(1)

Differentiating (1) w.r.t. y and writing 1/ p for dx / dy, we get dp dp 1 = p+ y + 2 ap p dy dy

or

1 − p2 dp dp = y + 2 ap p dy dy

or

1 − p2 dy − y = 2 ap, multiplying both sides by dy / dp p dp

or

dy p 2 ap2 , + 2 y=− 2 dp p − 1 p −1

which is a linear differential equation.

…(2)

D-59 2

−1)}dp

Here the I.F. = e ∫ {p /( p ∴

1 log ( p2 −1)

= e2

= ( p2 − 1)1 /2 .

the solution of (2) is y ( p2 − 1 )1 /2 =

∫

− 2 ap2 p2 − 1

= − 2a

∫

( p2 − 1)1 /2 dp + c

( p2 − 1) + 1 √ ( p2 − 1)

dp + c

  1 = − 2 a √ ( p2 − 1) +  dp + c 2 √ ( p − 1)  1 1 = − 2 a [ p √ ( p2 − 1) − cosh −1 p + cosh −1 p] + c 2 2

∫

= − ap √ ( p2 − 1) − a cosh −1 p + c or

c − a cosh −1 p

y=

√ ( p2 − 1)

…(3)

− ap.

Substituting this value of y in (1), we get   c − a cosh −1 p x = p  − ap + ap2 2   √ ( p − 1) or

x=

p (c − a cosh −1 p) √ ( p2 − 1)

…(4)

.

The equations (3) and (4) constitute the parametric equations of the required solution.

Comprehensive Exercise 3 Solve the following differential equations : 1.

y2 log y = xyp + p2 .

2.

2

(Agra 2007; Purvanchal 08)

p y + 2 px − y = 0 . 2

4

(Meerut 2005; Purvanchal 06; Lucknow 08)

3.

4 ( xp + yp) = y .

4. (2 x − b) p = y − ayp2 .

5.

xp3 = a + bp .

6.

7.

x + p / √ (1 + p2 ) = a .

p = tan { x − p /(1 + p2 )} .

A nswers 3 1.

log y = cx + c 2

2.

y2 − 2 cx + c 2 = 0

3.

y = 4 c ( xyc + 1)

4.

ac 2 + (2 x − b) c − y2 = 0

D-60

5.

x = (a / p3 ) + (b / p2 ), y = (3 a / 2 p2 ) + (2 b / p) + c

6.

x = p /(1 + p2 ) + tan−1 p, y = c − 1/(1 + p2 )

7.

( x − a)2 + ( y + c )2 = 1

2.5 Clairaut’s Equation (Meerut 2004; Lucknow 08)

(a) The differential equation y = px + f ( p) ,

…(1)

is known as Clairaut’s equation. Here f ( p) is some function of p only. This differential equation is very important and the students should note its form carefully. To solve this differential equation, we shall obviously adopt the method explained in article 2.3. Thus differentiating (1) w.r.t. x and writing p for dy / dx, we get dp dp p= p+ x + f ′ ( p) dx dx dp or { x + f ( p)} = 0 . dx dp …(2) ∴ =0. dx or …(3) x + f ′ ( p) = 0 . From (2), we have p = constant = c, (say).

…(4)

Eliminating p between (1) and (4), we obtain y = cx + f (c ) ,

…(5)

which is the required general solution of (1). If we eliminate p between (1) and (3), we get the singular solution of (1). Remember: To obtain the general solution of a differential equation in Clairaut’s form

simply replace p by c. (b) To solve the differential equation y = x f1 ( p) + f2 ( p).

…(1)

This differential equation is not in Clairaut’s form. However it can be solved by the method we adopted in solving Clairaut’s equation. Thus differentiating (1) w.r.t. x and writing p for dy / dx, we get dp dp p = f1( p) + x f1 ′ ( p) + f2 ′ ( p) dx dx dp dp or p − f1( p) = x f1 ′ ( p) + f2 ′ ( p) dx dx dx or [ p − f1( p)] − x f1 ′ ( p) = f2 ′ ( p) dp or

f ′ ( p) f ′ ( p) dx + 1 ⋅x= 2 dp f1( p) − p p − f1( p)

…(2)

D-61

which is a linear differential equation with x as the dependent variable and p as the independent variable. Let the solution of (2) be …(3)

φ ( x, p, c ) = 0 . Then eliminating p between (1) and (3), we get the required solution.

Example 13: Solve y = x (dy / dx) + (dy / dx)2 .

(Meerut 2011)

Solution: Denoting dy / dx by p, the given differential equation is y = px + p2 , which is

in Clairaut’s form. So replacing p by the arbitrary constant c, the required solution is y = cx + c 2 . Example 14: Solve p = log ( px − y).

(Gorakhpur 2007; Lucknow 09)

Solution: The given differential equation can be written as px − y = e p or y = px − e p ,

which is in Clairaut’s form. So replacing p by the arbitrary constant c in the given differential equation, the required solution is c = log (cx − y) . Example 15: Solve sin px cos y = cos px sin y + p. (Meerut 2003; Rohilkhand 06, 08, 10; Lucknow 07)

Solution: The given differential equation is

or

sin px cos y − cos px sin y = p sin ( px − y) = p

or

px − y = sin−1 p

or

y = px − sin−1 p, which is in Clairaut’s form.

So changing p to the arbitrary constant c, the required solution is y = cx − sin−1 c . Example 16: Solve p2 ( x2 − a2 ) − 2 pxy + y2 − b2 = 0 . Solution: The given differential equation can be written as

p2 x2 − 2 pxy + y2 = p2 a2 + b2 or

or

( y − px)2 = b2 + a2 p2

y = px ± √ (b2 + a2 p2 ),

each of which is in Clairaut’s form. Hence the required solution is ( y − cx)2 = b2 + a2 c 2 . Example 17: Solve y = 2 px + pn. Solution: The given differential equation is

y = 2 px + pn. Differentiating (1) w.r.t. x and writing p for dy / dx, we get

…(1)

D-62

p = 2 p + 2 x (dp / dx) + npn −1(dp / dx) or

p + 2 x (dp / dx) = − npn −1(dp / dx)

or

p (dx / dp) + 2 x = − npn −1, multiplying both sides by dx / dp

or

dx 2 + x = − npn − 2 , dp p

…(2)

which is a linear differential equation. 2

Hence the I.F. = e ∫ (2 / p) dp = e2 log p = e log p = p2 . ∴

the solution of (2) is xp2 = −

or

xp2 = − n

or

x = cp−2 − { n /(n + 1)} pn −1.

∫

n −2 2

∫np

p dp + c

pn dp + c = − n pn +1 /(n + 1) + c …(3)

Substituting this value of x in (1), we get y = 2 p [cp−2 − { n / (n − 1)} pn −1] + pn y = 2 cp−1 + pn −

or

n −1 n 2n n p = 2 cp−1 − p. n +1 n +1

…(4)

The equations (3) and (4), which express x and y in terms of a parameter p, constitute the required solution.

2.6 Equations Reducible to Clairaut’s Form Some differential equations by suitable change of variables may be reduced to Clairaut’s form. Example 18: Use the transformation x2 = u, y2 = v to solve

( px − y) ( py + x) = h2 p .

(Agra 2006; Lucknow 08, 10)

Solution: The given differential equation can be written as

p2 xy + p ( x2 − y2 − h2 ) − xy = 0 . 2

…(1)

2

Putting x = u and y = v, we have 2 x dx = du and 2 y dy = dv . ∴

y dy dv = x dx du

or

dy x dv = dx y du

or

p=

x dv . y du

Putting this value of p in the differential equation (1), we get 2

xy

x2  dv  x dv 2 + ( x − y2 − h2 ) − xy = 0 2  du y du y  2  dv  2  2 2 2 dv − y2  = 0 x   + (x − y − h ) du   du 

or

x y

or

x2 (dv / du)2 + ( x2 − y2 − h2 ) (dv / du) − y2 = 0

D-63

u (dv / du)2 + (u − v − h2 )(dv / du) − v = 0 ,

or

[putting u for x2 and v for y2 ] or

u P2 + (u − v − h2 ) P − v = 0 , where P = dv / du

or

u P ( P + 1) − v ( P + 1) − h2 P = 0

or

v ( P + 1) = uP ( P + 1) − h2 P

or

v = uP − h2 P /( P + 1), which is in Clairaut’s form y = px + f ( p) .

∴

replacing P by the arbitrary constant c, we get the required solution as v = uc − h2 c /(c + 1) or y2 = cx2 − h2 c / (c + 1)

Comprehensive Exercise 4 Solve the following differential equations : 1.

y = px + a / p.

2.

y = px + ap (1 − p) .

3.

(Rohilkhand 2006, 07; Agra 05; Gorakhpur 08, 11; Purvanchal 06) (Agra 2006)

3

y = px + p − p .

(Kanpur 2010)

4. ( y − px) ( p − 1) = p.

(Agra 2005; Gorakhpur 06; Lucknow 09, 11)

2

5. ( x − a) p + ( x − y) p − y = 0 . 2

2

2

2

(Agra 2007)

2

6. p ( x − a ) − 2 pxy + 2 y + a = 0 . 7.

y2 + x2 (dy / dx)2 − 2 xy (dy / dx) = 4(dx / dy)2 .

8. cos y cos px + sin y sin px = p 9. 9 ( y + x p log p) = (2 + 3 log p) p3 . 10. x2 ( y − px) = yp2 .

(Gorakhpur 2006, 09)

[Hint. Put x2 = u and y2 = v] 11. e3 x ( p − 1) + p3 e2

y

[Hint. Put e x = u and e y = v .]

=0.

A nswers 4 1.

y = cx + a / x

2.

y = cx + ac (1 − c )

3.

y = cx + c − c 3

4. ( y − cx) (c − 1) = c 5. ( x − a) c 2 + ( x − y) c − y = 0 6. c 2 ( x2 − a2 ) − 2 cxy + y2 + a2 = 0

D-64

7. ( y − cx)2 = 4 / c 2 8. y = cx − cos −1 c 9. x =

1 2 1 p + cy −1, y = − xp log p + (2 + 3 log p) p3 3 9

10. y2 = cx2 + c 2 11. e y = ce x + c 3

2.7 Geometrical Meaning of a Differential Equation of the First Order (Avadh 2010)

Let

f ( x, y, p) = 0 ,

…(1)

be a differential equation of the first order, where p = dy / dx. Suppose the general solution of (1) is φ ( x, y, c ) = 0,

…(2)

where c is the arbitrary constant. For each real value of c the equation (2) represents a curve. Thus (2) is the equation of a family of curves; c being the parameter for the family. Hence every differential equation of the first order represents a family of curves. Let us take any point ( x1, y1) in the plane. If we substitute the coordinates of this point in (1) and (2) and solve the resulting equations in p and c, the values of c so obtained are the values of the parameter for the curves of the family (2) which pass through ( x1, y1) and the values of p are the slopes of the tangents to these curves at the point ( x1, y1). Naturally the degree of c in (2) must be equal to the degree of p in (1).

2.8 Singular Solutions (Gorakhpur 2009)

Sometimes a differential equation (for degree higher than one) possesses a solution which does not contain any arbitrary constant and which cannot be derived from the general solution of the differential equation by giving a particular value to the arbitrary constant. Such a solution is called a singular solution and it is generally not included in the general solution of the differential equation. The singular solution of a differential equation is given by the envelope of family of curves represented by that differential equation. Whenever the envelope of the family of curves φ ( x, y, c ) = 0 ,

…(1)

represented by the general solution of the differential equation f ( x, y, p) = 0

…(2)

exists, the equation of the envelope is the singular solution of the differential equation (2).

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Suppose that the family of curves (1) possesses an envelope. For any point P ( x, y) on the envelope, there exists a curve of the family (1), say φ ( x, y, c ) = 0, which touches the envelope at ( x, y). The values of x, y, dy / dx for the curve at P satisfy the differential equation (2). But the values of x, y, dy / dx at P for the envelope are the same as for the curve. Hence the values of x, y, dy / dx at each point of the envelope satisfy the differential equation (2). Consequently the envelope of (1) is also a solution of (2). This solution does not contain any arbitrary constant and in general, cannot be obtained from (1) by giving any particular value to the arbitrary constant c. Hence this envelope is the singular solution of (2).

2.9 Determination of Singular Solution with the Help of c-Discriminant and p-Discriminant Relations The discriminant:. Let F (c ) = 0 be an algebraic equation. The discriminant of this equation is the simplest function of the coefficients the vanishing of which represents the condition that the equation should have two equal roots. Thus the discriminant of the quadratic Ac 2 + Bc + C = 0 is B2 − 4 AC. The equation B2 − 4 AC = 0 is called the discriminant relation. Let the given differential equation be f ( x, y, p) = 0 .

…(1)

and let its general solution be φ ( x, y , c ) = 0 .

…(2)

The envelope of the family of curves (2) is contained in the locus obtained on eliminating c, between (2) and ∂φ ( x, y, c ) …(3) =0. ∂c Let the c-eliminant between (2) and (3) be ψ ( x, y) = 0 .

…(4)

Since the equation (4) may sometimes represent other loci besides the envelope, therefore only that part of the locus (4) is the singular solution which also satisfies the given differential equation (1). Therefore if the equation (4) fails to satisfy (1), we should resolve it into others that are simpler. Then we should try for each part whether it satisfies the differential equation (1) or not. Only those parts will constitute the singular solution which satisfy (1). From our knowledge of the theory of equations we know that the c-eliminant between (2) and (3) gives the condition for the equation (2) in c to have two equal roots. Therefore we shall call this c-eliminant the c-discriminant relation. The singular solution is contained in this c-discriminant relation. From the geometrical point of view the c-discriminant relation (4) is the locus of the points ( x, y) such that two of the

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curves of the family (2) through ( x, y) coincide. Now if two curves of the family coincide, their tangents must also coincide and so the corresponding values of p i. e., (dy / dx) given by (1) must also coincide. The condition for the equation (1) in p to have two equal roots is obtained on eliminating p between (1) and ∂f ( x, y, p) …(5) =0. ∂p The p-eliminant between (1) and (5) is called the p-discriminant relation and the envelope and hence the singular solution is also contained in it. Note 1: From the above discussion it is clear that if p occurs only in the first degree in the

differential equation, there will be no singular solution. Note 2: A differential equation which possesses a singular solution is not considered

completely solved until the singular solution also has been found. Therefore while solving a differential equation of the first order but higher degree we must find the general solution and also discuss the singular solution.

2.10 Working Rule for Finding the Singular Solution Let the given differential equation be f ( x, y, p) = 0 .

…(1)

Find the general solution of (1) and let it be φ ( x, y, c ) = 0 .

…(2)

Find the c-discriminant relation by eliminating c between ∂φ ( x, y, c ) φ ( x, y, c ) = 0 and =0. ∂c Also find the p-discriminant relation by eliminating p between ∂ f ( x, y, p) f ( x, y, p) = 0 and =0. ∂p The locus common to both the c-discriminant and the p-discriminant relations gives us the singular solution of (1) provided it satisfies (1). However, we can find only one of these two discriminant relations and only that part of the locus contained in it is the singular solution which also satisfies the given differential equation (1).

2.11 The Singular Solution of Clairaut’s Equation We know that the general solution of the Clairaut’s equation is

y = px + f ( p) y = cx + f (c )

Differentiating (2) partially w.r.t. c, we get 0 = x + f ′ (c ) .

…(1) …(2) …(3)

The singular solution, which is the envelope of (2), is obtained by eliminating c between (2) and (3).

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Now differentiating (1) partially w.r.t.p, we get 0 = x + f ′ ( p).

…(4)

The equation (1) and (4) differ from the equations (2) and (3) only in having p instead of c. Therefore the c-eliminant between (2) and (3) and the p-eliminant between (1) and (4) are identical and either of them gives us the singular solution. Hence the singular solution of the Clairaut’s equation (1) is obtained by eliminating p between (1) and (4). Note: The equation (4) is the same as the equation (3) of article 2.5 (a).

Example 19: Find the general and singular solutions of

y2 − 2 pxy + p2 ( x2 − 1) = m2 .

(Bundelkhand 2010)

Solution: The given differential equation can be written as

( y − px)2 = p2 + m2 or

y − px = ± √ ( p2 + m2 ) or y = px ± √ ( p2 + m2 ) ,

either of which is in Clairaut’s form. Hence the general solution is obtained on replacing p by the arbitrary constant c in the given differential equation. Thus the general solution is ( y − cx)2 = c 2 + m2

or c 2 ( x2 − 1) − 2 xyc + y2 − m2 = 0 .

…(1)

Now from the given differential equation and from the general solution (1) both the p and the c discriminant relations are 4 x2 y2 − 4 ( x2 − 1) ( y2 − m2 ) = 0 or

x2 m2 + y2 − m2 = 0

or

y2 + m2 x2 = m2 ,

which is the singular solution. Example 20: Find the general and singular solution of the differential equation.

( xp − y)2 = p2 − 1 .

(Bundelkhand 2008; Lucknow 06)

Solution: From the given differential equation, we have

xp − y = ± √ ( p2 − 1)

or

y = px ± √ ( p2 − 1)

either of which is in Clairaut’s form. Hence the general solution is obtained on replacing p by the arbitrary constant c in the given differential equation. Thus the general solution is ( xc − y)2 = c 2 − 1

or

c 2 ( x2 − 1) − 2 xyc + y2 + 1 = 0 . …(1)

The envelope of the family of curves (1) is the singular solution. From (1) the c-discriminant relation is 4 x2 y2 − 4 ( x2 − 1) ( y2 + 1) = 0 or

− x2 + y2 + 1 = 0

or

x2 − y2 = 1 ,

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which is obviously also the p-discriminant relation. Since the c-discriminant relation contains only one locus, therefore it gives us the envelope of the family of curves (1). Hence the singular solution is x2 − y2 = 1 . Example 21: Find the general and singular solution of

9 p2 (2 − y)2 = 4 (3 − y). Solution: Solving the given differential equation for p, we have

p= or

dy 2 √ (3 − y) =± dx 3 2− y

dx = ±

3 2− y dy , separating the variables. 2 √ (3 − y)

Integrating, we have 3 x+c=± 2 =± putting

3 2

∫ ∫

2− y dy √ (3 − y) 2 − (3 − t2 ) (−2 t) dt , t

2

3 − y = t so that − dy = 2 t dt 1 = ± 3 (t2 − 1) dt = ± 3 ( t3 − t) = ± t (t2 − 3) 3

∫

[∵ 3 − y = t2 ]

= ± √ (3 − y) (− y), ∴

( x + c )2 = y2 (3 − y),

…(1)

which is the general solution of the given differential equation. Differentiating (1) partially w.r.t. c, we get 2 ( x + c) = 0

i. e.,

x + c =0 .

…(2)

Eliminating c between (1) and (2), we get the c-discriminant relation as y2 (3 − y) = 0 . Now y = 0 gives dy / dx = p = 0 . Putting y = 0 and p = 0 in the given differential equation, we see that these values of y and p do not satisfy it. Therefore y = 0 is not a solution. Again 3 − y = 0 gives y = 3 and so dy / dx = p = 0 . Putting y = 3 and p = 0 in the given differential equation, we see that these values of y and p satisfy it. Therefore y = 3 is the singular solution. Example 22: Reduce the differential equation ( px − y) ( x − py) = 2 p to Clairaut’s form by

substituting x2 = u and y2 = v and find its complete primitive and the singular solution. Solution: We put u = x2 and v = y2 , so that

du / dx = 2 x

and

dv / dx = 2 y (dy / dx) = 2 py.

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∴

dv dv / dx 2 py py = = = du du / dx 2x x

or

P = py / x, where P = dv / du or p = xP / y.

Substituting this value of P in the given differential equation, we get   x2 P P  − y ( x − x P) = 2 x x y   or

( x2 P − y2 ) (1 − P) = 2 P

or

u P − v = 2 P /(1 − P)

or

x ( x2 P − y2 ) (1 − P) = 2 x P or

or

(u P − v) (1 − P) = 2 P

v = uP − 2 P /(1 − P),

which is in Clairaut’s form. Hence its general solution is v = uc − 2 c /(1 − c ) or

y2 = x2 c − 2 c /(1 − c )

or

c 2 x2 − c ( x2 + y2 − 2) + y2 = 0 .

or

y2 (1 − c ) = x2 c (1 − c ) − 2 c …(1)

Now (1) is a quadratic in c. So the c-discriminant relation is ( x2 + y2 − 2)2 − 4 x2 y2 = 0 or

( x2 − 2 xy + y2 − 2) ( x2 + 2 xy + y2 − 2) = 0

or

{( x − y)2 − 2} {( x + y)2 − 2} = 0

or

( x − y − √ 2)( x − y + √ 2)( x + y − √ 2)( x + y + √ 2) = 0 .

All the four equations

x − y − √ 2 = 0 , x − y + √ 2 = 0 , x + y − √ 2 = 0 and

x + y + √ 2 = 0 satisfy the given differential equation and are therefore the singular solutions.

Comprehensive Exercise 5 1.

Find the complete primitive and singular solution of y = px + √ (b2 + a2 p2 ). Interpret your results geometrically.

(Avadh 2010)

2.

Find the general and singular solutions of y = px + a / p .

3.

Solve and examine for singular solution the differential equation x3 p2 + x2 yp + a3 = 0 .

(Gorakhpur 2007) 3

4.

Find the general and singular solution of 27 y − 8 p = 0 .

5.

Solve and examine for singular solution the equation xp2 − ( x − a)2 = 0 .

6.

Find the general and singular solution of (a) ( y − px)2 + a2 p = 0 . (c)

7.

(b) 3 xy = 2 px2 − 2 p2 .

p2 + y2 = 1 .

Examine y2 (1 + p2 ) = r2 for singular solution.

(Bundelkhand 2009)

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8.

Reduce the equation x2 p2 + py (2 x + y) + y2 = 0 , where p = (dy / dx) to Clairaut’s form by putting u = x and v = xy and find its complete primitive and also its singular solution. (Gorakhpur 2010)

9.

Solve the differential equation ( px2 + y2 ) ( px + y) = ( p + 1)2 by reducing it to Clairaut’s form and find its singular solution.

(Purvanchal 2009)

2 3

10. Reduce the differential equation y = 2 px + y p to Clairaut’s form by putting y2 = v and hence find its general and singular solutions. 11. Solve and test for singular solutions p3 − 4 pxy + 8 y2 = 0 .

A nswers 5 1.

y = cx + √ (b2 + a2 c 2 ); x2 / a2 + y2 / b2 = 1

2.

y = cx + a / c ; y2 = 4 ax

3. c 2 + cxy + a3 x = 0 ; x = 0 , xy2 − 4 a3 = 0 4.

y2 = ( x + c )3 , y = 0

5. 9 ( y + c )2 = 4 x ( x − 3 a)2 , x = 0 6. (a) ( y − cx)2 + a2 c = 0 ; xy = a2 / 4 (c) 7.

(b) (2 y + 2 c )2 = 4 cx3 ; x3 − 6 y = 0

y = cos ( x + c ) ; y = ± 1

2

y + ( x + c )2 = r2 ; y = ± r

8. xy = cy + c 2 , y = 0 , y + 4 x = 0 9. c 2 ( x + y) − cxy − 1 = 0 ; x2 y2 + 4 ( x + y) = 0 10. y2 = cx + c 3 / 8 ; 27 y4 + 32 x3 = 0 11. y = c ( x − c )2 ; y = 0 , y =

4 3 x 27

O bjective T ype Q uestions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

Solution of the differential equation p2 − 8 p + 15 = 0 is (a) p = 5 , p = 3

(b) ( y − 5 x − c ) ( y − 3 x − c ) = 0

(c) ( y + 5 x) ( y + 3 x + c ) = 0

(d) None of these

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2.

3.

4.

Solution of the equation y2 log y = xyp + p2 is (a) log y = cx + x2

(b) log y = cx2 + e x

(c) log x = cy + y2

(d) None of these.

Solution of the equation y = px + log p is (a) y = e x + c

(b) y = cx + log c

(c) y = log cx

(d) x = e y + c .

Which of the following equations is Clairaut’s equation ? (a) x = py + f ( p)

(b) y = px + f (c )

(c) y = px + f ( p)

(d) y = x2 + c

Fill in The Blanks Fill in the blanks “……” so that the following statements are complete and correct. dy by the letter …… . dx

1.

We usually denote

2.

The differential equation y = px + f ( p), is known as …… . (Meerut 2003; Avadh 2005)

3.

n

n −1

p + A1 p

n−2

+ A2 p

+ ... + An − 1 p + An = 0 ,

where A1, A2 , … , An are some functions of x and y, is a differential equation of …… order and …… degree. 4.

Solution of the equation p2 − 7 p + 12 = 0 is …… .

5.

 dy  Solution of the equation   − ax3 = 0 is …… .  dx 

6.

Solution of the differential equation y = px + e p is …… .

7.

 dy   dy  Solution of the equation y = x   +   is …… .  dx   dx 

8.

Solution of the equation cos y cos px + sin y sin px = p is …… .

9.

Solution of the equation ( x − a) p2 + ( x + y) p − y = 0 is …… .

2

3

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

Every differential equation of the first order represents a family of curves.

2.

The singular solution of a differential equation is given by the envelope of the family of curves represented by that differential equation.

3.

Only those differential equations possess a singular solution in which p (or dy / dx) occurs only in the first degree.

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A nswers Multiple Choice Questions 1. 4.

(b) (c)

2.

(a)

3. (b)

Fill in the Blank(s) 1. 3. 5. 7. 9.

p first, nth 25 ( y + c )2 = 4 ax5 y = cx + c 3

2. Clairaut’s equation 4. ( y − 3 x − c ) ( y − 4 x − c ) = 0 6. y = cx + e c 8.

y = cx − cos −1 c

( x − a) c 2 + ( x − y) c − y = 0

True or False 1.

T

2. T

3. F

¨

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3 O rthogonal T rajectories

3.1 Trajectory (Lucknow 2005; Gorakhpur 06; Bundelkhand 08; Meerut 13, 13B)

efinition: A trajectory of a given system of curves is defined to be a curve which cuts all the members of the family according to a given law. Here we propose to find the equation of the trajectories (a family of curves) each member of which cuts each member of a given family of curves at a constant angle. If the angle is a right angle, the trajectories are called orthogonal trajectories, when it is other than a right angle they are called to be oblique trajectories.

D

3.2 Trajectories Cartesian Co-ordinates: To find the trajectories which cut every member of a given family of curves at a constant angle. Let the equation of the given family of curves be f ( x, y, c ) = 0 ,

…(1)

c being the arbitrary parameter. Let the required trajectories cut the given curves (1) at a constant angle α. Differentiating (1) with respect to x, we have

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∂f ∂f dy + ⋅ =0. ∂x ∂y dx

Y

…(2)

Eliminating c between (1) and (2), we get the differential equation of the given family of curves (1). Let it be dy   …(3) φ  x, y,  = 0 .  dx 

(x, y) Trajectory C

e urv

α

ψ2

ψ1

X O Let ( x, y) be the coordinates of a point of intersection of a member of the given curves (1) and a member of the required trajectories. If the tangent to the trajectory at this point makes an angle ψ1 with x-axis and if the tangent to the given curve at this point makes an angle ψ2 with x-aixs, we have tan ψ1 − tan α . dy / dx given by (3) = tan ψ2 = tan (ψ1 − α) = 1 + tan ψ1 tan α

Putting this value of (dy / dx) in (3), we get  tan ψ1 − tan α   =0. φ  x, y, 1 + tan ψ1 tan α  

…(4)

Now dy / dx for the trajectory is tan ψ1. So replacing tan ψ1 in the equation (4) by dy / dx, we get the differential equation of the required trajectories as  dy / dx − tan α   =0. φ  x, y, 1 + (dy / dx) tan α   Solving this differential equation, we shall obtain the equation of the required trajectories. Orthogonal trajectories: If α is a right angle, we have

∴

tan ψ1 × tan ψ2 = − 1 . dy 1 given by (3) = tan ψ2 = − . dx tan ψ1

Putting this value of dy / dx in (3) of the last article, we get  1   =0. φ  x, y, − tan ψ1  

…(5)

Now (dy / dx) for the trajectory is tan ψ1. So replacing tan ψ1 in the equation (5) by dy / dx, we get differential equation of the orthogonal trajectories of (1) as  dx  φ  x, y, −  = 0 . dy   Solving this differential equation, we shall get the equation of the orthogonal trajectories of the given family (1). Remember: To obtain the differential equation of the orthogonal trajectories, we have to write

− dx / dy for dy / dx in the differential equation of the original family of curves.

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3.3 Orthogonal Trajectories Polar Co-ordinates: Let the equation of the given family of curves be …(1)

f (r, θ, c ) = 0 ,

c being the parameter i. e., c can take any real value. Differentiating (1) w.r.t. θ and then eliminating c, let the differential equation of the given family (1) be dr   …(2) φ  r, θ,  = 0 .  dθ  Now if at a point of intersection P (r, θ), the tangents to the trajectory and the given curve make angles φ1 and φ2 with the common radius vector OP of that point, then tan φ1 − tan φ2 tan 90 ° = tan (φ1 − φ2) = 1 + tan φ1 tan φ2 and therefore tan φ1 tan φ2 = − 1 . dθ Now r given by (2) dr 1 = tan φ2 = − ⋅ tan φ1 ∴ ∴

dr / dθ given by (2) = − r tan φ1 .

Curve

Trajectory

P (r, θ) φ2

α φ1

O

X

putting this value of dr / dθ in (2), we have

φ (r, θ, − r tan φ1) = 0 dθ But tan φ1 = r for the trajectory. dr ∴

the differential equation of the required orthogonal trajectories is dθ   φ  r, θ, − r2  =0.  dr 

Solving this we shall get the required equation of the orthogonal trajectories. Remember: The differential equation of the orthogonal trajectories is obtained from the

differential equation (in polar co-ordinates) of a given family of curves by writing dθ dr dθ 1 dr for for or − r2 −r dr dθ dr r dθ 1 dr dθ or for r − ⋅ r dθ dr

Example 1: Find the orthogonal trajectories of the family of curves y = ax n. (Meerut 2011) Solution: The equation of the given family of curves is

y = ax n, a being the parameter Differentiating (1) with respect to x, we get

.…(1)

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dy / dx = a . nx n−1.

…(2)

Dividing (2) by (1) to eliminate a between (1) and (2), we have dy / dx anx n−1 n = = y x ax n or

x (dy / dx) = ny,

…(3)

which is the differential equation of the given family of curves. Now to obtain the differential equation of the orthogonal trajectories, write − dx / dy for dy / dx in (3). Hence the orthogonal trajectories are given by the differential equation x (− dx / dy) = ny.

…(4)

Separating the variables, we have − x dx = ny dy. 1 1 Integrating, we have − x2 = n y2 + c , where c is constant of integration. 2 2 Thus x2 + ny2 = 2 c or x2 + ny2 = c 2 is the required family of orthogonal trajectories. Example 2: Find the orthogonal trajectories of the family of parabolas y2 = 4 ax, where ‘a’ is the

variable parameter.

(Meerut 2013B)

Solution: Differentiating the given equation y2 = 4 ax,

with respect to x, we get 2 y (dy / dx) = 4 a . Eliminating a between (1) and (2), we get 2 x (dy / dx) = y,

…(1)

…(2) …(3)

which is the differential equation of the given family of parabolas. So to obtain the differential equation of the orthogonal trajectories, write − dx / dy for dy / dx in (3). Thus the differential equation of the required family of orthogonal trajectories is 2 x (− dx / dy) = y Integrating, we get 1 2 x2 = − y + c2 2

or

2 x dx = − y dy.

or

2 x2 + y2 = k 2 ,

which is the required family of orthogonal trajectories. Example 3: Find the orthogonal trajectories of the family of rectangular hyperbolas xy = c 2 . (Meerut 2003; Rohilkhand 06, 07; Kanpur 07, 12; Bundelkhand 09; Avadh 10, 11)

Solution: The equation of the given family of curves is

xy = c 2 , c being the parameter.

…(1)

Differentiating (1) w.r.t. x, we get x (dy / dx) + y = 0 ,

…(2)

which is independent of the parameter c. Hence the equation (2) is the differential equation of the given family of curves.

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So to obtain the differential equation of the orthogonal trajectories, write − dx / dy for dy / dx in (2). Thus the differential equation of the required family of orthogonal trajectories is x (− dx / dy) + y = 0 or ∴

x dx = y dy, in which the variables are separated. 1 1 2 Integrating, x2 = y + c , where c is constant of integration 2 2

or

x2 − y2 = 2 c = k 2 (say) is the required family of orthogonal trajectories.

Example 4: Find the orthogonal trajectories of the family of parabolas y2 = 4 a ( x + a), where ‘a’

is the parameter.

(Meerut 2006; Purvanchal 06; Lucknow 06; Avadh 07, 09)

Solution: The equation of the given family of parabolas is

y2 = 4 a ( x + a) ,

…(1)

a being the parameter. Differentiating (1) w.r.t. x, we get 2 y (dy / dx) = 4 a.

…(2)

Now eliminating a between (1) and (2), we get 1 y2 = 2 y (dy / dx)[ x + y (dy / dx)] 2 or

y = 2 x (dy / dx) + y(dy / dx)2 ,

…(3)

which is the differential equation of the given family of parabolas. So to obtain the differential equation of the orthogonal trajectories, write − dx / dy for dy / dx in (3). Thus the differential equation of the required family of orthogonal trajectories is y = 2 x (− dx / dy) + y (− dx / dy)2 or

y (dy / dx)2 + 2 x (dy / dx) = y.

…(4)

Now we observe that the differential equation (4) of the orthogonal trajectories is the same as the differential equation (3) of the given family of parabolas. Therefore the given family of parabolas (1) is self-orthogonal i. e., the orthogonal trajectories of the system belong to the system itself. Hence the equation of the orthogonal trajectories of (1) is y2 = 4 c ( x + c ), c being the parameter. Example 5: Find the orthogonal trajectory of the family of circles x2 + y2 = 2 ax, a being the

parameter.

(Meerut 2005, 10)

Solution: Differentiating x2 + y2 = 2 ax

…(1)

with respect ot x, we get or

2 x + 2 y (dy / dx) − 2 a = 0 x + y (dy / dx) − a = 0.

…(2)

Eliminating ‘a’ between (1) and (2), we get x2 + y2 = 2 x { x + y (dy / dx)} or x2 + 2 xy (dy / dx) − y2 = 0 ,

…(3)

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which is the differential equation of the given family of circles. So to obtain the differential equation of the orthogonal trajectories, write − dx / dy for dy / dx in (3). Thus the differential equation of the required family of orthogonal trajectories is x2 + 2 xy (− dx / dy) − y2 = 0 or

dx / dy = ( x2 − y2 ) / 2 xy.

…(4)

This is a homogeneous differential equation. To solve it putting x = vy, so that dx / dy = v + y (dv / dy), in (4), we get v + y (dv / dy) = y2 (v2 − 1) /(2 vy2 ) or

y

1 + v2 dv v2 − 1 , = −v=− dy 2v 2v

or

2 v dv 2

1+ v

=−

dy . y

Integrating, we have log (1 + v2 ) = − log y + log b or

or

1 + v2 = b / y

1 + ( x2 / y2 ) = b / y, since v = x / y.

Hence x2 + y2 = by is the equation of the required orthogonal trajectories. Example 6: Find the orthogonal trajectories of the family of curves

{ x2 /(a2 + λ )} + { y2 /(b2 + λ )} = 1 , where λ is the parameter.

(Meerut 2007B, 09B; Lucknow 10; Rohilkhand 10; Gorakhpur 11; Purvanchal 07, 09, 11)

Solution: Differentiating the given equation with respect to x, we get

2 x /(a2 + λ ) + {2 y /(b2 + λ )} ⋅ (dy / dx) = 0 or

x (b2 + λ ) + y (dy / dx)(a2 + λ ) = 0

or

λ { x + y (dy / dx)} = − { b2 x + a2 y (dy / dx)} .

∴

λ = − { b2 x + a2 y (dy / dx)} ÷ { x + y (dy / dx)}.

Thus

a2 + λ = (a2 − b2 ) x / { x + y (dy / dx)}

and

b2 + λ = − (a2 − b2 ) y (dy / dx) ÷ { x + y (dy / dx)} .

Substituting these values of (a2 + λ ) and (b2 + λ ) in the given equation, we get the differential equation of the given family of curves as x2 { x + y (dy / dx)} (a2 − b2 ) x or

−

y2 { x + y(dy / dx)} (a2 − b2 ) y (dy / dx)

 dy 1   = a2 − b2 . x2 − y2 + xy  −  dx dy / dx 

=1 …(1)

Hence putting − dx / dy for dy / dx in (1), the differential equation of the orthogonal trajectories is  dx dy  x2 − y2 + xy  − +  = a2 − b2 dy dx  

D-79

 dy 1   = a2 − b2 , x2 − y2 + xy  −  dx dy / dx 

or

which is the same as (1). Therefore solving it we shall get x2 / (a2 + µ ) + y2 (b2 + µ ) = 1 , µ being the parameter of the orthogonal trajectories. In other words the system of given confocal conics x2 / (a2 + λ ) + y2 / (b2 + λ ) = 1 is self-orthogonal. Example 7: Find the orthogonal trajectories of the cardioids r = a (1 − cos θ), a being the

parameter.

(Meerut 2006B; Gorakhpur 05; Lucknow 11; Kashi 11)

Solution: The given family of curves is r = a (1 − cos θ),

…(1)

the parameter being a. Differentiating (1) w.r.t. θ, we get dr / dθ = a sin θ .

…(2)

Now we shall eliminate a between (1) and (2). So dividing (1) by (2), we get r (dθ / dr) = (1 − cos θ) / sin θ,

…(3)

which is the differential equation of the given family of curves (1). To obtain the differential equation of the orthogonal trajectories, write − (1/ r) (dr / dθ) for r (dθ / dr) in (3). Thus we get 1 2 sin2 θ 1 dr 1 − cos θ 1 2 − = = = tan θ , 1 1 r dθ sin θ 2 2 sin θ cos θ 2 2 which is the differential equation of the orthogonal trajectories. To solve it, separating the variables, we get 1 1 1 − sin θ − sin θ dr 1 2 dθ = 2 ⋅ 2 2 dθ ⋅ = − tan θ dθ = 1 1 r 2 cos θ cos θ 2 2 Integrating, we have 1 θ + log c 2 1 1 1 or r / c = cos2 θ = (1 + cos θ) log (r / c ) = log cos2 θ 2 2 2 1 or r = k (1 + cos θ), r = c (1 + cos θ) 2 log r = 2 log cos or or

which is the required family of orthogonal trajectories. Example 8: Find the orthogonal trajectories of the system of curves r n = a n cos n θ, a being the

parameter. (Meerut 2007, 10B; Bundelkhand 05; Gorakhpur 06; Lucknow 08; Kanpur 08) Solution: The given family of curves is r n = a n cos nθ ,

the parameter being a. Taking logarithm of both sides of (1), we get n log r = n log a + log cos nθ .

…(1)

D-80

Now differentiating both sides w.r.t. θ, we get − n sin n θ n dr =0 + r dθ cos n θ

or

1 dr = − tan nθ r dθ

…(2)

which is the differential equation of the family of curves (1). Now putting − r

dθ 1 dr for in (2), the differential equation of the orthogonal dr r dθ

trajectories is −r

dθ = − tan nθ dr

or

dr = cot nθ dθ . r

Integrating, we get log r = (1 / n) log sin nθ + log c or

n log (r / c ) = log sin nθ

or

log (r / c )n = log sin nθ

or

r n = c n sin nθ,

which is the required family of orthogonal trajectories, the parameter being c.

Comprehensive Exercise 1 1.

Find the equation of the family of curves orthogonal to the family y = ax3 . (Avadh 2005; Rohilkhand 09; Kanpur 11)

2.

Find the orthogonal trajectories of the semi-cubical parabolas ay2 = x3 , where a is the variable parameter.

3.

Find the orthogonal trajectories of the family of circles x2 + y2 = a2 , where ‘a’ is the parameter.

4.

(Gorakhpur 2007; Kashi 13)

(Bundelkhand 2008; Kanpur 10)

Find the orthogonal trajectories of the family of curves x2 /3 + y2 /3 = a2 /3 , where a is parameter.

5.

Find the equation of the family of curves that is orthogonal to ax2 + y2 = 1 .

6.

Find the orthogonal trajectories of the system of curves (dy / dx)2 = a / x. (Bundelkhand 2007, 10; Agra 08; Purvanchal 14)

7.

Find the differential equation of the family of curves given by the equation x2 − y2 + 2 λxy = 1 , where λ is a parameter. Obtain the differential equation of its orthogonal trajectories and solve it.

8.

9.

Show that the orthogonal trajectories of the family of conics y2 − x2 + 4 xy − 2 cx = 0 consist of a family of cubics, with the common asymptote x + y = 0 . (Meerut 2009) Find the orthogonal trajectories of the family of circles x2 + y2 + 2 fy + 1 = 0 . where f is the parameter.

( Meerut 2004; Gorakhpur 08, 11)

j

D-82

3.

Orthogonal trajectories of the system of curves r n = a n cos nθ, a being the parameter, is ……

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

A family of curves is said to be self orthogonal if the differential equation of the orthogonal trajectories is the same as the differential equation of the given family of curves.

2.

Family of curves { x2 / (a2 + λ )} + { y2 / (b2 + λ )} = 1 , where λ is the parameter, is self orthogonal.

A nswers Fill in the Blank(s) 1.

−

dx dy

2. − r2

dθ dr

3. r n = c n sin nθ

True or False 1.

T

2. T

¨

D-83

4 L inear D ifferential E quations W ith C onstant C oefficients

4.1 Definitions

A

linear differential equation is an equation in which the dependent variable and its derivatives appear only in the first degree.

A linear differential equation of order n of the form dn y dx

n

+ a1

d n −1 y dx

n −1

+ a2

dn − 2 y dx

n−2

+ ... + an−1

dy + an y = Q , dx

…(1)

where a1, a2 , … , an−1, an are constants and Q is any function of x is called a linear differential equation with constant coefficients. For convenience, the operators

d d2 d3 dn , 2 , 3 , … , n are also denoted by D, dx dx dx dx

D2 , D3 , … , Dn respectively. Thus the equation (1) can also be written as Dn y + a1 Dn −1 y + ... + an −1 Dy + an y = Q or

[ Dn + a1 Dn −1 + … + an −1 D + an] y = Q .

If y = f ( x) is the general solution of

…(2)

D-84

[ Dn + a1 Dn − 1 + … + an−1 D + an] y = 0 ,

…(3)

and y = φ ( x) is any particular solution of the equation (2) not containing any arbitrary constant, then y = f ( x) + φ ( x) , is the general solution of (2). Thus the method of solving a linear equation is divided into two parts : First, we find the general solution of the equation (3). It is called the complementary function (C.F.). It must contain as many arbitrary constants as is the order of the given differential equation. Next, we find a solution of (2) which does not contain an arbitrary constant. This is called the particular integral (P.I.). If we add (C.F.) and (P.I.), we get the general solution of (2). Thus the general solution of (2) is y = C. F. + P. I.

(Gorakhpur 2005)

4.2 Determination of Complementary Function (C.F.) Consider a linear nth order differential equation with constant coefficients of the form f ( D) y = 0 i. e., [ Dn + a1 Dn −1 + a2 Dn − 2 + … + an] y = 0 .

…(1)

This is equivalent to [( D − m1) ( D − m2 ) … ( D − mn)] y = 0 .

…(2)

The solution of any one of the equations ( D − m1) y = 0 , ( D − m2 ) y = 0 , … , ( D − mn) y = 0

…(3)

is also a solution of (2) and we know that the general solution of ( D − m1) y = 0 is y = Ae m1 x . Hence we can assume that a solution of the equation (2) is of the form y = e mx . Then, substituting e mx for y in (1), so that Dy = me mx , D2 y = m2 e mx , … , Dn y = m ne mx , we get e mx (m n + a1m n − 1 + a2 m n − 2 + … + an) = 0 or

m n + a1m n − 1 + a2 m n − 2 + … + an = 0 , because e mx ≠ 0 .

Hence e mx will be a solution of (1) if m has the value obtained from the equation …(4) m n + a1m n − 1 + … + an = 0 . The equation (4) is called the auxiliary equation (A.E.) and is obtained by putting D = m in f ( D) = 0 . It will give in general n roots, say, m1, m2 , m3 , … , mn .

D-85

Case I: If all the roots of the Auxiliary equation (A.E.) are distinct: If the roots m1, m2 , m3 , …, mn are all distinct, then e m1 x , e m2 x , … , e mn x are all distinct and linearly independent. So the general solution of (1) in this case is y = c1e m1 x + c2 e m2 x + … + c n e mn x .

…(5)

Case II: Auxiliary equation having equal roots : If two roots are equal say m1 = m2 , then the solution (5) becomes y = c1e m1 x + c2 e m1 x + c3 e m3 x + … + c ne mn x y = (c1 + c2 ) e m1 x + c3 e m3 x + … + c ne mn x .

or

Now (c1 + c2 ) can be replaced by single constant say c. Therefore this solution has only (n − 1) arbitrary constants and so it is not the general solution. To obtain the general solution, consider the differential equation ( D − m1)2 y = 0 in which the two roots are equal. This can be written as ( D − m1) [( D − m1) y] = 0 . Now putting ( D − m1) y = v, we get ( D − m1) v = 0 , or dv / dx = m1v or dv / v = m1 dx, (variables being separated). integrating, ∴ or

log v = m1 x + log c1

log (v / c1) = m1 x

or

…(6)

v = c1 e m1 x .

Thus ( D − m1) y = v = c1 e m1 x and putting it in (6), we get ( D − m1) y = c1e m1 x Dy − m1 y = c1e m1 x

or

or

dy − m1 y = c1e m1 x . dx

d  ∵ D ≡ dx 

This is a linear equation of the first order. Hence the I. F. = e ∫ − m1 dx = e − m1 x . ∴

The solution of this equation is y ⋅ e − m1 x =

m1 x

− m1 x

∫ c1e ⋅ e dx + c2 = c1 dx + c2 = c1 x + c2 ∫

or

y = (c1 x + c2 ) e m1 x or y = (c2 + c1 x) e m1 x .

Hence the general solution of f ( D) y = 0 in this case is y = (c1 + c2 x) e m1 x + c3 e m3 x + … + c ne mn x . Similarly if three roots of the auxiliary equation are equal say, m1 = m2 = m3 , the general solution of f ( D) y = 0 will be y = (c1 + c2 x + c3 x2 ) e m1 x + c4 e m4 x + … + c ne mn x and so on. Case III: Auxiliary equation having complex roots:

(Meerut 2003)

Let the two roots of the auxiliary equation be complex, say m1 = α + iβ and m2 = α − iβ , (where i = √ − 1).

D-86

The solution corresponding to these two roots will be y = c1e(α + iβ) x + c2 e(α − iβ) x = c1e αx ⋅ e iβx + c2 e αx ⋅ e − iβx = c1e αx (cos βx + i sin βx) + c2 e αx (cos βx − i sin βx) = (c1 + c2 ) e αx cos βx + i (c1 − c2 ) e αx sin βx y = e αx [ A1 cos βx + A2 sin βx] ,

or

where A1 = c1 + c2 and A2 = i (c1 − c2 )

y = e αx (c1 cos βx + c2 sin βx), changing the constants.

or

If the imaginary roots are repeated, say α + iβ and α − iβ occur twice then the solution will be y = e αx [(c1 + c2 x)cos βx + (c3 + c4 x) sin βx] , and so on. Note 1: The expression e αx (c1 cos βx + c2 sin βx) can also be written as

c1e αx sin (βx + c2 )

c1e αx cos (βx + c2 ) .

or

Note 2: If a pair of the roots of the auxiliary equation are irrational i. e., they are α ± √ β ,

where β is positive, then the corresponding term in the C.F. will be e αx (c1 cosh √ βx + c2 sinh √ βx), or

c1e αx sinh (√ βx + c2 )

or

c1e αx cosh (√ βx + c2 ).

If these irrational roots are repeated twice, then the corresponding portion of the solution will be e αx {(c1 + c2 x) cosh √ βx + (c3 + c4 x) sinh √ βx} .

Example 1: Solve

d2 y dx2

−7

dy + 12 y = 0 . dx

Solution: The given differential equation is ( D2 − 7 D + 12) y = 0 .

∴ or

the auxiliary equation is m2 − 7 m + 12 = 0 (m − 3) (m − 4) = 0 .

∴

m = 3, 4 .

Hence the solution is y = c1e3 x + c2 e4 x . Example 2: Solve ( D3 + 6 D2 + 11D + 6) y = 0 . Solution: The auxiliary equation is

m3 + 6 m2 + 11m + 6 = 0 or

(m + 1) (m2 + 5 m + 6) = 0

or

(m + 1) (m + 2) (m + 3) = 0 .

(Meerut 2010)

D-87

∴

m = − 1, − 2, − 3 .

Hence the solution is y = c1e − x + c2 e −2 x + c3 e −3 x . Example 3: Solve

d2 x 2

dt

−3

dx + 2 x = 0 , given that when t = 0 , x = 0 and dx / dt = 0 . dt (Gorakhpur 2008) 2

Solution: The auxiliary equation is m − 3 m + 2 = 0

or

or

(m − 1) (m − 2) = 0 t

m = 1, 2 .

2t

Hence the solution is x = c1e + c2 e ,

…(1)

where c1 and c2 are arbitrary constants. Now x = 0 when t = 0 ; ∴

0 = c1 + c2 .

…(2)

Also dx / dt = c1 e t + 2 c2 e2 t , and dx / dt = 0 when t = 0 . ∴

…(3)

0 = c1 + 2 c2 .

Solving (2) and (3), we get c1 = 0 , c2 = 0 . Now putting the values of c1 and c2 in (1), we get the required solution as x = 0 . Example 4: Solve ( D3 − 3 D + 2) y = 0 . Solution: The auxiliary equation is m3 − 3 m + 2 = 0

or

(m − 1) (m2 + m − 2) = 0

∴

m = 1, 1, − 2.

or

(m − 1) {(m − 1) (m + 2)} = 0 .

Hence the solution is y = (c1 + c2 x) e x + c3 e −2 x . Example 5: Solve

d3 y dx3

−8y =0.

(Meerut 2010B)

Solution: The auxiliary equation is m3 − 8 = 0

or

(m − 2) (m2 + 2 m + 4) = 0 i. e.,

∴

m = 2 and m = − 1 ± i √ 3 .

m − 2 = 0 and m2 + 2 m + 4 = 0 .

Hence the solution is y = e − x (c1 cos √ 3 x + c2 sin √ 3 x) + c3 e2 x . Example 6: Solve

d4 y 4

dx

−2

d3 y dx3

−2

dy − y =0. dx

Solution: The auxiliary equation is

m4 − 2 m3 − 2 m − 1 = 0 or

(m4 − 1) − 2 m (m2 + 1) = 0

or

(m2 + 1) (m2 − 1) − 2 m (m2 + 1) = 0

D-88

or

(m2 + 1) (m2 − 2 m − 1) = 0

∴

m = 0 ± i ,1 ± √ 2 .

i. e., m2 = − 1

or

m2 − 2 m − 1 = 0 .

Hence the solution is y = e0 x (c1 cos x + c2 sin x) + e x (c3 cosh √ 2 x + c4 sinh √ 2 x) y = c1 cos x + c2 sin x + e x { c3 cosh √ 2 x + c4 sinh √ 2 x} .

or

Example 7: Solve ( D4 + k 4 ) y = 0 .

(Meerut 2003, 04B, 05B, 06) 4

4

Solution: The auxiliary equation is m + k = 0

or

(m2 + k 2 )2 − 2 k 2 m2 = 0

or

(m2 + k 2 )2 − (√ 2 km)2 = 0

or

(m2 + k 2 − √ 2 km) (m2 + k 2 + √ 2 km) = 0

or

m2 − √ 2 km + k 2 = 0 and m2 + √ 2 km + k 2 = 0

or

m=

√ 2 k ± √ (2 k 2 − 4 k 2 ) − √ 2 k ± √ (2 k 2 − 4 k 2 ) and m = 2 2 k k k k and − m= ±i ±i ⋅ √2 √2 √2 √2

or

Hence the solution is y = e kx / √2 { c1 cos (kx / √ 2) + c2 sin (kx / √ 2)} + e − kx / √2 { c3 cos (kx / √ 2) + c4 sin (kx / √ 2)} .

Comprehensive Exercise 1 Solve the following differential equations : 1.

d2 y 2

dx

+ (a + b)

3

3.

d y 3

dx

dy + aby = 0 . dx

2

+6

d y 2

dx

+3

dy − 10 y = 0 . dx

2. 4.

d2 y 2

dx

d2 y 2

dx

−3

dy −4y =0. dx

−4

dy +4y =0. dx

5. ( D3 − 4 D2 + 5 D − 2) y = 0 . 6.

d4 y 4

dx

+2

d3 y 3

dx

−3

d2 y 2

dx

7. (i) ( D4 + 1) y = 0 .

−4

dy +4y =0. dx

(Agra 2006)

8. ( D4 + 8 D2 + 16) y = 0 .

(ii)

d4 y dx4

− k4 y = 0 .

D-89

9.

( D2 ± µ 2 ) y = 0 .

10.

2

(Avadh 05, 14)

2

Solve (d y / dx ) + y = 0 , given y = 2 for x = 0 and y = − 2 for x =

1 π. 2

A nswers 1 1.

y = c1 e − ax + c2 e − bx

2.

y = c1e − x + c2 e4 x

3.

y = c1e x + c2 e −2 x + c3 e −5 x

4.

y = (c1 + c2 x) e2 x

5.

y = (c1 + c2 x) e x + c3 e2 x

6.

y = (c1 + c2 x) e x + (c3 + c4 x) e −2 x

7. (i) y = e x / √2 { c1 cos ( x / √ 2) + c2 sin ( x / √ 2)} + e − x / √2 { c3 cos ( x / √ 2) + c4 sin ( x / √ 2)} . (ii) y = c1e kx + c2 e − kx + c3 cos kx + c4 sin kx 8. 9.

y = (c1 + c2 x) cos 2 x + (c3 + c4 x) sin 2 x y = c1 cos µx + c2 sin µx and y = c3 e µx + c4 e − µx 1   10. y = 2 √ 2 cos  x + π  4 

4.3 The Particular Integral (P.I.) As already shown in 4.1 the complete solution of ( Dn + a1 Dn − 1 + a2 Dn − 2 + … + an − 1 D + an) y = Q or

…(1)

F ( D) y = Q is y = C. F. + P. I.,

where the C.F. consists of the general solution of the differential equation F ( D) y = 0 . In article 4.2 we have discussed different methods of finding the complementary function by taking the differential equation as F ( D) y = 0 . Methods of finding the particular integral will be discussed now. 1 The particular integral of the differential equation F ( D) y = Q is Q. It is F ( D) obviously a function of x which when operated by F ( D) gives Q. 1 1 Now as F ( D) can be regarded as the inverse operator of Q = Q, therefore F ( D) F ( D) F ( D). Similarly D and 1/ D are inverse operations. If D stands for differentiation then 1/ D will stand for integration. Hence the particular integral of the equation

D-90

F ( D) y = Q will be

1 Q F ( D)

because it satisfies the given equation.

4.4 Particular Integral in Some Special Cases Case I: To find P.I. when Q is of the form e ax , where a is any constant and F ( a) ≠ 0 . By simple differentiation we know that D (e ax ) = ae ax ; D2 (e ax ) = a2 e ax ; D3 (e ax ) = a3 e ax , … , Dn (e ax ) = a ne ax . It suggests that F ( D) e ax = F (a) e ax .

…(1)

Let F (a) ≠ 0 . Operating on both sides of (1) with 1 / F ( D), we have 1 1 F ( D) e ax = { F (a) e ax } F ( D) F ( D) 1 e ax , because F (a) is a constant F ( D)

or

e ax = F (a) ⋅

or

1 1 e ax = e ax , because F (a) ≠ 0 F (a) F ( D)

∴

P.I. =

1 1 ax e ax = e , provided F( a) ≠ 0. F ( D) F (a)

Working Rule : If P.I. = {1/ F ( D)} e ax , then put a for D in F ( D) and we get the P.I.,

provided F (a) ≠ 0 . Examples on case I: Example 9: Solve

d2 y 2

dx

−3

dy + 2 y = e5 x . dx

(Avadh 2008; Purvanchal 11; Kanpur 12)

Solution: The given equation can be written as

( D2 − 3 D + 2) y = e5 x , where d / dx ≡ D. Here

F ( D) = D2 − 3 D + 2 and Q = e5 x .

The auxiliary equation is m2 − 3 m + 2 = 0 or

(m − 1) (m − 2) = 0 ;

∴

m = 1, 2 .

∴ and

The C.F. = c1e x + c2 e2 x , the roots of the A.E. being distinct. P. I. =

1 1 e5 x e5 x e ax = 2 e5 x = 2 = ⋅ F ( D) D − 3D + 2 5 − 3 ⋅ 5 + 2 12 [We have put 5 for D in F ( D), because here a = 5]

D-91

Hence the complete solution is y = (C.F.) + (P.I.) 1 5x or y = c1e x + c2 e2 x + e . 12 Example 10: Solve

d2 y

+ 31

2

dx

dy + 240 y = 272 e − x . dx

(Meerut 2010B, 11)

Solution: The auxiliary equation is

m2 + 31m + 240 = 0

or

(m + 15) (m + 16) = 0 ;

m = − 15, − 16 .

∴

C. F. = c1e −15 x + c2 e −16 x .

∴ and

P. I. =

1 D2 + 31D + 240

= 272 ⋅

(272 e − x) = 272 ⋅

1 2

(− 1) + 31 (− 1) + 240

e− x =

1 D2 + 31D + 240

e− x

272 − x 136 − x e = e . 210 105

Hence the general solution is y = (C. F. ) + (P. I. ) 136 − x i. e., y = c1e −15 x + c2 e −16 x + e . 105 Example 11: Obtain the complete solution of the differential equation

d2 y dx2

−7

dy + 6 y = e2 x , dx

and determine the constant so that y = 0 when x = 0 . Solution: The given equation is [ D2 − 7 D + 6] y = e2 x .

∴

auxiliary equation is m2 − 7 m + 6 = 0

or

(m − 1) (m − 6) = 0

∴

C. F. = c1e x + c2 e6 x .

and

P. I. =

1 2

D − 7D + 6

or

e2 x =

m = 1, 6. 1 2

2 − 7 ⋅2 + 6

e2 x = −

e2 x ⋅ 4

the general solution is y = (C. F. ) + (P. I. ) 1 i. e., y = c1e x + c2 e6 x − e2 x . 4

∴

Now when y = 0 , x = 0 ; ∴ so that

0 = c1 + c2 −

1 , from (1) 4

1  c2 =  − c1 . 4 

Hence from (1), we have 1 1 1  y = c1e x +  − c1 e6 x − e2 x = c1 (e x − e6 x ) + (e6 x − e2 x ) 4  4 4 or

y = c1 (e x − e6 x ) +

1 2x 4x e (e − 1) is the required solution. 4

…(1)

D-92

Comprehensive Exercise 2 Solve the following differential equations : 1. 2.

d2 y 2

dx

d2 y 2

dx

− 2k

dy + k2 y = e x . dx

+2

dy + y = 2 e2 x . dx

2

3. 4. 5.

d y 2

dx

+

(Bundelkhand 2001)

dy + y = e x. dx

[ D2 + D + 1] y = e − x . d3 y 3

dx

+6

d2 y 2

dx

+ 11

dy + 6 y = e2 x . dx

A nswers 2 1.

y = (c1 + c2 x) e kx + e x /(1 − k )2

2.

y = (c1 + c2 x) e − x +

3.

2 2x e 9 1 1 1 y = e − x /2 { c1 cos ( x √ 3) + c2 sin ( x √ 3)} + e x 2 2 3

4.

y = c1e − x /2 cos { x (√ 3 / 2) + c2 } + e − x

5.

y = c1e − x + c2 e −2 x + c3 e −3 x +

1 2x e 60

Case II: To find P.I. when Q is of the form sin ax or cos ax and F ( − a2 ) ≠ 0 . By simple differentiation we know that D (sin ax) = a cos ax ; D2 (sin ax) = − a2 sin ax ; D3 (sin ax) = − a3 cos ax ; D4 (sin ax) = (− a2 )2 sin ax, ......,( D2 )n sin ax = (− a2 )n sin ax. It suggests that F ( D2 ) sin ax = F (− a2 ) sin ax. Let

F (− a2 ) ≠ 0 .

Now operating both sides of (1) with 1/ F ( D2 ), we get 1 F ( D2 ) or

F ( D2 ) sin ax =

sin ax = F (− a2 ) ⋅

1 F ( D2 )

1 F ( D2 )

{ F (− a2 ) sin ax}

sin ax.

…(1)

D-93

Thus Similarly

1 F( D2 ) 1 2

F( D )

1

sin ax =

F( − a2 )

sin ax, provided F (− a2 ) ≠ 0 .

1

cos ax =

F ( − a2 )

cos ax, provided F (− a2 ) ≠ 0 .

Working Rule: If P.I. = {1 / F ( D)} sin ax or cos ax, put − a2 for D2 , − a2 D for D3 ,

(− a2 )2 i. e., a4 for D4 , a4 D for D5 , − a6 for D6 etc. in F ( D) and calculate the P.I. Note: Linear factors in D of the form ( pD ± q) appearing in the denominator are

removed by first multiplying the Nr. and Deno. by the conjugate factors ( pD ∓ q) and then putting − a2 for D2 in the denominator. The operation left in the numerator can be easily worked out because the operator D stands for differentiation with respect to x.

Examples on Case II: Example 12: Solve

d2 y 2

dx

−

dy − 2 y = sin 2 x. dx

(Meerut 2008, 10)

2

Solution: The given equation is [ D − D − 2] y = sin 2 x.

∴

auxiliary equation is m2 − m − 2 = 0

or

(m + 1) (m − 2) = 0 ;

∴

C. F. = c1e − x + c2 e2 x .

And

P. I. =

1 2

D − D−2

∴

sin 2 x =

m = − 1, 2 . 1 sin 2 x, −4− D−2 putting − 22 i. e., − 4 for D2

∴ or

=−

( D − 6) ( D − 6) 1 sin 2 x = − sin 2 x = − 2 sin 2 x D+6 ( D + 6) ( D − 6) D − 36

=−

( D − 6) sin 2 x, putting −4 for D2 −4 − 36

=

1 1 { D (sin 2 x) − 6 sin 2 x} ( D − 6) sin 2 x = 40 40

=

1 1 3 (2 cos 2 x − 6 sin 2 x) = cos 2 x − sin 2 x . 40 20 20

the complete solution is y = (C. F. ) + (P. I. ) y = c1e − x + c2 e2 x +

1 3 cos 2 x − sin 2 x . 20 20

Example 13: Solve ( D2 − 5 D + 6) y = sin 3 x . Solution: Here the auxiliary equation is m2 − 5 m + 6 = 0 .

or

(m − 2) (m − 3) = 0 ;

∴

2x

C. F. = c1e

3x

+ c2 e .

∴

m = 2, 3.

D-94

And

P. I. = = = =

∴

1 D2 − 5 D + 6

sin 3 x =

1 − 32 − 5 D + 6

sin 3 x =

1 sin 3 x − 9 − 5D + 6

−1 − (5 D − 3) − (5 D − 3) sin 3 x = sin 3 x = sin 3 x 5D + 3 (5 D + 3)(5 D − 3) 25 D2 − 9 − (5 D − 3) 2

25 ⋅ (− 3 ) − 9

sin 3 x =

1 {5 D (sin 3 x) − 3 sin 3 x} 234

1 1 (5 ⋅ 3 cos 3 x − 3 sin 3 x) = (5 cos 3 x − sin 3 x) . 234 78

the complete solution is y = (C. F. ) + (P. I. ) y = c1e2 x + c2 e3 x + (1/ 78) (5 cos 3 x − sin 3 x) .

or

Example 14: Solve

d2 y dx2

+ 9 y = cos 2 x + sin 2 x .

Solution:. The auxiliary equation is

m2 + 9 = 0 ,

∴

m = ± 3i . or

C. F. = c1 cos 3 x + c2 sin 3 x

∴ And

P. I. = =

1 2

( D + 9) cos 2 x 2

−2 +9

(cos 2 x + sin 2 x) = +

sin 2 x 2

−2 +9

=

= c1 cos (3 x + c2 ) 1 2

D +9

cos 2 x +

1 2

D +9

sin 2 x

cos 2 x sin 2 x 1 + = (cos 2 x + sin 2 x) . 5 5 5

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 or y = c1 cos (3 x + c2 ) + (cos 2 x + sin 2 x) . 5

Comprehensive Exercise 3 Solve the following differential equations : 1. (d2 y / dx2 ) + 9 y = cos 4 x . 2. (i)

( D2 − 3 D + 2) y = sin 3 x.

3. (i)

( D2 − 3 D + 2) y = cos 3 x

(ii) ( D3 + D2 − D − 1) y = cos 2 x 2

4. 5.

d y 2

dx

d2 y 2

dx

−8

dy + 9 y = 40 sin 5 x. dx

−4

dy + y = a sin 2 x . dx

+2

dy + 10 y + 37 sin 2 x = 0 . dx

2

6.

d y 2

dx

(ii) ( D2 − 2 D + 5) y = sin 3 x.

(Avadh 2008; Kanpur 06)

(Meerut 2006B)

(Lucknow 2005; Gorakhpur 07)

D-95

A nswers 3 1.

y = c1 cos 3 x + c2 sin 3 x −

2. (i) (ii) 3. (i) (ii)

y = c1e x + c2 e2 x +

1 cos 4 x 7

1 (9 cos 3 x − 7 sin 3 x) 130

y = e2 x (c1 cos 2 x + c2 sin 2 x) + (1/ 26)(3 cos 3 x − 2 sin 3 x) y = c1e x + c2 e2 x −

1 (7 cos 3 x + 9 sin 3 x) 130

y = (c1 + c2 x) e − x + c3 e x − (1/ 25) (2 sin 2 x + cos 2 x) 5 (5 cos 5 x − 2 sin 5 x) 29

4.

y = c1e4 x cosh ( x √ 7 + c2 ) +

5.

y = c1e2 x cosh (√ 3 x + c2 ) + (1 / 73) a (8 cos 2 x − 3 sin 2 x)

6.

y = e − x (c1 cos 3 x + c2 sin 3 x) +

37 (2 cos 2 x − 3 sin 2 x) 26

Case III: To find P.I. when Q is of the form x m, where m is a positive integer. Consider first {1/( D − a)} x m . We have 1 1 1 xm = − xm = − xm ( D − a) (a − D) a {1 − ( D / a)} 1 1 − a 1 = − 1 + a =−

=−

D  a

−1

xm

 D D2 + 2 + … x m , expanding by the binomial theorem a a 

1 m 1 1  m −1 + 2 m (m − 1) x m − 2 + … .  x + mx a a a 

Here we observe that in the expansion by the binomial theorem, the terms of the expansion beyond the mth power of D need not be written since Dm + 1 x m = 0 , Dm + 2 x m = 0, etc. Working Rule: In order to evaluate {1 / F ( D)} x m , bring out common the lowest degree

term in D from F ( D) so that remaining factor in the denominator is of the form [1 + f ( D)] or [1 − f ( D)] which is taken in the numerator with a negative index. Next we expand [1 ± f ( D)]−1 in powers of D by trhe binomial theorem and operate upon x m with the expansion obtained. This expansion should be done upto the term Dm , since Dm + 1 x m = 0 and all higher differential coefficients of x m are zero. The whole process will be clear from the following examples. The following binomial expansions should be remembered well.

D-96

(i) (1 − x)−1 = 1 + x + x2 + x3 + x4 + … (ii) (1 + x)−1 = 1 − x + x2 − x3 + x4 − … (iii) (1 − x)−2 = 1 + 2 x + 3 x2 + 4 x3 + … (iv) (1 + x)−2 = 1 − 2 x + 3 x2 − 4 x3 + …

Examples on Case III: Example 15: Solve (d2 y / dx2 ) − 4 y = x2 .

(Lucknow 2010)

2

Solution: The auxiliary equation is m − 4 = 0 or m = ± 2 .

∴

C. F. = c1e2 x + c2 e −2 x .

And

P. I. =

1

x2 =

2

D −4

=−

1 4

1 − 4 [1 −

2

1 4

D ]

x2 = −

1  1 2 1− D  4  4 

−1

x2

 1 2  2 1 + 4 D + … x ,

expanding by binomial theorem upto the terms containing D2 1 1  = −  x2 + D2 ( x2 ) , because all the remaining terms vanish 4 4  1 2 1  1  2 1 = −  x + ⋅ 2 = −  x +  . 4 4  4 2 Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = c1e2 x + c2 e −2 x − ( x2 + ) . 4 2 Example 16: Solve

d3 y 3

dx

−

d2 y 2

dx

−6

dy = 1 + x2 . dx

(Gorakhpur 2006; Avadh 06)

Solution: The auxiliary equation is m3 − m2 − 6 m = 0

or

m (m + 2) (m − 3) = 0 . 0x

∴

C. F. = c1e

And

P. I. =

+ c2 e

−2 x

+ c3 e

1 4

2

3x

D − D − 6D

∴

m = 0 , − 2, 3.

= c1 + c2 e −2 x + c3 e3 x .

(1 + x2 ) =

1 − 6 D [1 +

1 6

D−

1 6

2

D ]

−1

(1 + x2 )

1  1  1 − (− D + D2 ) (1 + x2 ) 6 D  6  1  1 1  1 + (− D + D2 ) + (− D + D2 )2 + … (1 + x2 ) , =− 6 D  6 36 

=−

the terms in the expansion being needed only upto D2 1  1 1 1 2  =− 1 − D + D2 + D + … (1 + x2 ) 6 D  6 6 36  =−

1  1 7  1− D + D2 + … (1 + x2 ) 6 D  6 36 

D-97

=−

1  1 7  (1 + x2 ) − D (1 + x2 ) + D2 (1 + x2 ) , 6 D  6 36 

because all other terms vanish 1  1 7 1  25 1  2 − x + x2  =−  1 + x − x +  = −  6D  3 18  6 D  18 3 =−

1 6

 25

1

=−

1 25 1 1  x − x2 + x3  . 6 18 6 3 

2

∵ 1/ D ≡ 

∫ 18 − 3 x + x  dx ,

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 25 1 1  or y = c1 + c2 e −2 x + c3 e3 x −  x − x2 + x3  . 6 18 6 3  Example 17: Solve ( D3 + 8) y = x4 + 2 x + 1 . Solution: The auxiliary equation is m3 + 8 = 0

or

(m + 2) (m2 − 2 m + 4) = 0

∴

m = − 2, 1 ± i √ 3

Hence C.F. = c1e −2 x + e x { c2 cos ( x √ 3) + c3 sin ( x √ 3)} . And

P.I. =

1 D3 + 8

1 [1 + 8 1 = [1 − 8 =

( x4 + 2 x + 1) =

1 ( x4 + 2 x + 1)  1 3 8 1 + D   8 

1 8

D3 ]−1( x4 + 2 x + 1)

1 8

D3 + …] ( x4 + 2 x + 1),

the other terms in the expansion being of no need 1 1 = [( x4 + 2 x + 1)] − D3 ( x4 + 2 x + 1) 8 8 1 1 = [ x4 + 2 x + 1 − 3 x] = ( x4 − x + 1) . 8 8 Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1e −2 x + e x { c2 cos ( x √ 3) + c3 sin ( x √ 3)} +

1 4 ( x − x + 1) . 8

Example 18: Solve ( D2 + D − 2) y = x + sin x. Solution: The auxiliary equation is m2 + m − 2 = 0

or

(m − 1) (m + 2) = 0 .

∴

m = 1, − 2 .

And

P.I. = =

∴

1 D2 + D − 2

C.F. = c1e x + c2 e −2 x . ( x + sin x) =

1 ( D2 + D − 2)

x+

1 D2 + D − 2

1 1 x+ sin x 2 1 1   −1 + D − 2 − 2 1 − D − D2  2  2 

sin x

∫ dx

D-98 −1

( D + 3) sin x ( D − 3) ( D + 3)

=−

1 2

1 2  1 1 −  2 D + 2 D    

=−

1 2

( D + 3) D+3 1 1   1 1 + 2 D + … x + D2 − 9 sin x = − 2  x + 2  + − 1 − 9 sin x

x+

1 1 1 (x + ) − { D (sin x) + 3 sin x} . 2 2 10 1 1 1 =− x− − {cos x + 3 sin x} . 2 4 10 =−

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 1 3 or y = c1e x + c2 e −2 x − x − − cos x − sin x. 2 4 10 10 Example 19: Solve

d2 y 2

dx

dy + 4 y = x2 + e x + cos 2 x . dx

−4

(Meerut 2001)

2

Solution: The auxiliary equation is m − 4 m + 4 = 0 .

m = 2, 2 .

∴

Hence C. F. = (c1 + c2 x) e2 x , And

P. I. =

where

P1 = = P2 =

and

P3 =

1 2

( D − 4 D + 4) 1

D2 − 4 D + 4 1 4

x2 =

1 ( D − 2)2

1 ( D − 4 D + 4)

ex =

1 2

( D − 4 D + 4)] 1 4

 1  1 − 2 D

1 − 4 ⋅1 + 4

=

x2

ex = e x, 1

1 2

− 2 − 4D + 4

1 1

−2

3  2  x + 2 x + 2  ,

ex 2

cos 2 x =

1 4

x2 =

3 2  2 1  1 + D + 4 D + …  x = 4 2

=−

( x2 + e x + cos 2 x) = P1 + P2 + P3 , (say),

cos 2 x = −

1 cos 2 x 4D

1

∫ cos 2 x dx = − 4 ⋅ 2 sin 2 x = − 8 sin 2 x.

(Rohilkhand 2008)

Hence the complete solution is y = C. F. + P. I. or

y = C. F. + P1 + P2 + P3 1 3 1 y = (c1 + c2 x) e2 x + ( x2 + 2 x + ) + e x − sin 2 x. 4 2 8

or

Comprehensive Exercise 4 1. ( D2 + D − 6) y = x . 3

3

2

(Bundelkhand 2007) 2

2

2. (d y / dx ) + 3(d y / dx ) + 2(dy / dx) = x .

(Bundelkhand 2001)

D-99

3.

d3 y 3

dx

−4

d2 y 2

dx

+5

dy −2=0. dx

2

4. ( D + 2 D + 1) y = 2 x + x2 . 5. ( D2 + 3 D + 2) y = x2 . 4

2

(Bundelkhand 2001) 2

6. ( D + D + 16) y = 16 x + 256 .

(Lucknow 2005)

7. ( D2 − 5 D + 6) y = x + sin 3 x .

(Kanpur 2007)

8. ( D2 − 4 D + 3) y = e − x + 5 . 9. (d2 y / dx2 ) − 4 y = sin2 x.

(Gorakhpur 2007)

2

2

10. ( D − 2 D + 3) y = cos x + x . 11.

d3 y 3

dx

+2

d2 y dx2

+

(Meerut 2009B; Bundelkhand 11)

dy = e2 x + x2 + x . dx

(Meerut 2009; Gorakhpur 10)

A nswers 4 1 (6 x + 1) 36

1.

y = c1e2 x + c2 e −3 x −

2.

y = c1 + c2 e − x + c3 e −2 x +

3.

y = c1 + c2 e2 x cos ( x + c3 ) +

4.

y = (c1 + c2 x) e − x + ( x2 − 2 x + 2)

5.

y = c1 e − x + c2 e − 2 x +

6.

127 3  3  y = c1 e −(1 /2)√7. x sin  x + c2  + c3 e(1 /2)√7 x sin  x + c4  + x2 + 2  2  8

7.

y = c1e2 x + c2 e3 x +

1 6

8.

y = c1 e x + c2 e3 x +

1 −x 5 e + 8 3

9.

y = c1e2 x + c2 e −2 x −

1 x (2 x2 − 9 x + 21) 12 2 x 5

1 2 3 7 x − x+ ⋅ 2 2 4

5 1  (3 sin 3 x − 15 cos 3 x) x +  −  6  234

1 1 + cos 2 x 8 16

10. y = e x [c1 cos √ 2 x + c2 sin √ 2 x] + 11. y = c1 + (c2 + c3 x) e − x +

1 2 4 2 1 x + x+ + (cos x − sin x) 3 9 27 4

1 2x 1 3 3 2 e + x − x + 4x 18 3 2

D-100

4.5 P.I. when Q = eax V , where V is any Function of x By successive differentiation we notice that D (e axV ) = e ax D (V ) + Vae ax = e ax ( D + a) V , D2 (e ax V ) = D { e ax ( D + a) V } = e ax D ( D + a) V + ae ax ( D + a) V = e ax ( D + a)2 V . Similarly, D3 (e axV ) = e ax ( D + a)3 V , … , Dn (e ax V ) = e ax ( D + a)nV . ∴

F ( D) (e axV ) = e ax F ( D + a) V .

The result (1) is true for any function V of x. Taking

…(1) 1 V in place of V in (1), we F ( D + a)

have     1 1 F ( D)  e ax V  = e ax F ( D + a)  V F ( D + a ) F ( D + a )       1 F ( D)  e ax V  = e axV . F ( D + a)   1 on both sides of (2), we get Operating by F ( D)

i. e.,

e ax Thus,

…(2)

1 1 V = (e ax V ). F ( D + a) F ( D)

1 1 (e axV ) = e ax V. F ( D) F ( D + a)

Working Rule: Replace D by ( D + a) and take out e ax before the operator1 / F ( D). Then

determine {1 / F ( D + a)} V by the methods discussed in 4.4. This method also enables us to find {1 / F ( D)} e ax when F (a) is zero. We shall discuss it later on in article 4.6.

Example 20: Solve ( D2 − 2 D + 1) y = x2 e3 x . Solution: The auxiliary equation is

m2 − 2 m + 1 = 0 or

(m − 1)2 = 0 .

∴

m = 1, 1 .

∴

C.F. = (c1 + c2 x) e x .

(Agra 2005; Avadh 09)

D-101

And

P. I. =

1

x2 e3 x =

D2 − 2 D + 1

= e3 x

1 {( D + 3) − 1}2

1 ( D − 1)2

e3 x x2

x2 ,

putting D + 3 for D and bringing e3 x before the operator = e3 x

=

1 ( D + 2)2

x2 = e3 x

1 1 4 1 + D  2 

2

x2 =

1 3x  1 e 1 + D  4 2 

−2

x2

1 3x 1 1 e 1 − 2 ⋅ D + 3 ⋅ D2 + … x2 ,   4 2 4 

1 = e3 x 4 1 = e3 x 4

expanding by binomial theorem 3 1 − D + D2 + … x2   4 3 1 3 { x2 − 2 x + ⋅ 2} = e3 x { x2 − 2 x + } . 4 4 2

Hence the required solution is y = (C. F. ) + (P. I. ) 1 3 or y = (c1 + c2 x) e x + e3 x { x2 − 2 x + } . 4 2 Example 21: Solve ( D2 − 2 D + 5) y = e2 x sin x. Solution: The auxiliary equation is m2 − 2 m + 5 = 0 .

∴

m = {2 ± √ (4 − 20 )} / 2 = 1 ± 2 i.

Hence C.F. = e x (c1 cos 2 x + c2 sin 2 x) . And

P. I. =

1 2

D − 2D + 5

= e2 x = e2 x

e2 x sin x 1

2

sin x = e2 x

1 2

( D + 2) − 2 ( D + 2) + 5 D + 2D + 5 1 sin x, putting − 12 for D2 (− 12 + 2 D + 5)

=

D−2 1 2x 1 1 e sin x = e2 x sin x 2 D+2 2 ( D − 2) ( D + 2)

=

D−2 1 2x D − 2 1 e sin x = e2 x sin x 2 2 2 − 1− 4 ( D − 4)

sin x

1 2x 1 2x e ( D − 2) sin x = − e { D (sin x) − 2 sin x} 10 10 1 2x =− e (cos x − 2 sin x) . 10 =−

Hence the complete solution is y = (C.F.) + (P.I.) or

y = e x (c1 cos 2 x + c2 sin 2 x) −

1 2x e (cos x − 2 sin x) . 10

D-102

Comprehensive Exercise 5 Solve the following differential equations : 1. ( D + 1)3 y = x2 e − x .

2. ( D2 − 2 D + 1) y = x2 e x .

2

−x

2

2 2x

2

2x

3. ( D + 2 D + 1) y = e

/( x + 2).

2

4. ( D − 3 D + 2) y = xe . (Bundelkhand 2008)

5. ( D − 2 D + 1) y = x e . 6. ( D − 4 D + 4) y = e 4

(Avadh 2007)

x

(Purvanchal 2007)

sin 3 x .

(Purvanchal 2006)

x

7. ( D − 1) y = e cos x . 2

x

2

2x

(Lucknow 2010) 2

8. ( D − 2 D + 4) y = e cos x . 10. ( D − 5 D + 6) y = e

2x

9. ( D + 4 D − 12) y = ( x − 1) e .

sin 2 x.

A nswers 5 1 5 −x x e 60

1.

y = (c1 + c2 x + c3 x2 ) e − x +

3.

y = (c1 + c2 x) e − x + e − x { x log ( x + 2) − x + 2 log ( x + 2)}

4.

1 y = c1e x + c2 e2 x − e x  x2 + x 2 

5.

y = ( c1 + c2 x) e x + e2 x ( x2 − 4 x + 6)

6.

y = (c1 + c2 x) e2 x −

7.

y = c1 e x + c2 e − x + c3 cos x + c4 sin x −

8.

y = e x (c1 cos √ 3 x + c2 sin √ 3 x) +

9.

y = e2 x (c1 +

10. y = c1e2 x

2. y = (c1 + c2 x) e x +

1 4 x x e 12

1 2x e sin 3 x 9 1 x e cos x 5

1 x e cos x 2

1 2 9 x − x) + c2 e −6 x 16 64 1 2x + c2 e3 x + e (cos 2 x − 2 sin 2 x) 10

4.6 P.I. when Q = eax and F ( a ) = 0 In this case we use the method explained in article 4.5 to find the P.I. 1 1 We have P.I. = e ax = e ax . 1, if F (a) = 0 . F( D) F( D) Now this is of the form e ax V . Hence by the method of article 4.5, i. e., by putting ( D + a) for D in the operator and by making e ax free from the operator, we have

D-103

1 1. F ( D + a)

P. I. = e ax

Now this can be evaluated by using the method for finding P.I. in the case of x m .

Example 22: Solve

d2 y 2

dx

−3

dy + 2 y = e x. dx

Solution: The auxiliary equation is m2 − 3 m + 2 = 0

or

(m − 1) (m − 2) = 0 . x

∴

m = 1, 2 .

2x

Hence C. F. = c1e + c2 e . And

P. I. =

1 2

D − 3D + 2

ex =

1 1 ex = ex ( D − 2) ( D − 1) (1 − 2) ( D − 1)

[putting 1 for D in the factor D − 2 because it does not vanish by doing so] 1 =− e x ⋅1 D −1 [Note that D − 1 becomes zero by putting 1 for D ; so here we shall apply the method for e axV by taking 1 for V.] 1 1 = − ex 1= − ex 1= − ex x . ( D + 1) − 1 D Hence the complete solution is y = (C. F. ) + (P. I. ) or y = c1e x + c2 e2 x − x e x . Note: While finding P.I. in the case of e ax if F ( D) becomes zero by putting a for D, we

factorise F ( D) . Then we put D = a in the factors which do not vanish by doing so. The remaining operator is then dealt with by using the method for e axV on taking 1 for V. Example 23: Solve ( D3 − 7 D + 6) y = e2 x . Solution: The auxiliary equation is m3 − 7 m + 6 = 0

or

(m − 1) (m − 2) (m + 3) = 0 .

∴

x

And

2x

C. F. = c1e + c2 e P. I. = =

+ c3 e

−3 x

1 ( D3 − 7 D + 6)

∴

m = 1, 2, − 3 .

. e2 x =

1 e2 x ( D − 1) ( D − 2) ( D + 3)

1 e2 x , (2 − 1) ( D − 2) (2 + 3)

putting 2 for D in all the factors except D − 2 1 1 1 2x 1 1 1 1 2x = e ⋅1 = e 1 = e2 x 1 = xe2 x . 5 D−2 5 ( D + 2) − 2 5 D 5 Hence the complete solution is y = (C. F. ) + (P. I. )

or

y = c1e x + c2 e2 x + c3 e −3 x +

1 2x xe . 5

D-104 Example 24: Solve ( D2 + 4 D + 4) y = e2 x − e −2 x . Solution: The auxiliary equation is (m + 2)2 = 0 .

∴

m = − 2, − 2

Hence C. F. = (c1 + c2 x) e −2 x . And

Now Also

P. I. =

1 2

( D + 2)

1

e2 x

2

e2 x =

2

e −2 x =

( D + 2) 1

( D + 2) = e −2 x

(e2 x − e −2 x ) =

(2 + 2)2

=

1 ( D + 2)2

1 2

{( D − 2) + 2}

1 2

( D + 2)

e2 x −

1 ( D + 2)2

e −2 x .

1 2x e . 16

[∵ here, F (a) ≠ 0].

e −2 x ⋅ 1,

[∵ here F (a) = 0]

1 = e −2 x

1 2

D

1 = e −2 x

1 1 x = x2 e −2 x . D 2

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 2 x 1 2 −2 x or y = (c1 + c2 x) e −2 x + e − x e . 16 2 Example 25: Solve D2 y − 3 Dy + 2 y = cosh x .

(Rohilkhand 2010)

Solution: The auxiliary equation is

m2 − 3 m + 2 = 0 ∴

C. F. = c1e x + c2 e2 x .

And

P. I. = =

Now

or

1 2

D − 3D + 2

(m − 1) (m − 2) = 0

cosh x =

m = 1, 2 .

 e x + e− x    2  ( D − 3 D + 2)  1

2

1 1 1 1 ex + e− x . 2 D2 − 3 D + 2 2 D2 − 3 D + 2

1 1 1 1 ex = ex 2 2 D − 3D + 2 2 ( D − 1) ( D − 2) =

1 1 1 1 ex = − e x ⋅1 2 ( D − 1) (1 − 2) 2 D −1

=− Also

∴

1 x 1 1 1 1 e 1= − ex 1 = − xe x . 2 D + 1− 1 2 D 2

1 1 1 1 e− x = e − x , putting −1 for D 2 D2 − 3 D + 2 2 (− 1)2 − 3 (− 1) + 2 =

1 −x e . 12

∴

P. I. = −

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 −x or y = c1e x + c2 e2 x − xe x + e . 2 12

1 x 1 −x xe + e . 2 12

D-105 Example 26: Solve ( D2 − 4 D + 4) y = e2 x + sin 2 x . Solution: The auxiliary equation is

m2 − 4 m + 4 = 0

(m − 2)2 = 0 .

or

∴

m = 2, 2 .

2x

Hence C. F. = (c1 + c2 x) e . And

P.I. =

1 2

( D − 2)

1

Now

2

( D − 2)

e2 x =

and

2

D − 4D + 4 =

2

D − 4D + 4

1 2

( D − 2)

= e2 x 1

1

e2 x +

1 2

D

sin 2 x =

1 1 sin 2 x = − − 4D 4

sin 2 x .

e2 x ⋅ 1 = e2 x

(1) = e2 x

1 ( D + 2 − 2)2

x2 1 2 2 x 1 = x e , ( x) = e2 x ⋅ D 2 2

1 2

1

− 2 − 4D + 4

sin 2 x, putting − 22 for D2

1 1

1

∫ sin 2 x dx = − 4 ⋅ 2 (− cos 2 x) = 8 cos 2 x .

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = (c1 + c2 x) e2 x + x2 e2 x + cos 2 x . 2 8

Comprehensive Exercise 6 Solve the following differential equations : 1. ( D2 − 2 D + 1) y = e x .

2. ( D2 + D − 6) y = e2 x .

3. ( D2 − 4 D + 3) y = 2 e3 x .

4. ( D3 + 3 D2 + 3 D + 1) y = e − x .

5. ( D2 − a2 ) = cosh ax . 6. ( D3 − 5 D2 + 7 D − 3) y = e2 x cosh x . 2

2

2x

7. ( D − 4 D + 4) y = 8 ( x + e

(Rohilkhand 2005)

+ sin 2 x) .

A nswers 6 1 2 x x e 2

1.

y = (c1 + c2 x) e x +

3.

y = c1e x + c2 e3 x + xe3 x

5.

y = c1e ax + c2 e − ax + ( x / 2 a)sinh ax 1 1 y = (c1 + c2 x) e x + c3 e3 x + xe3 x − x2 e x 8 8 3 y = (c1 + c2 x) e2 x + 2 ( x2 + 2 x + ) + 4 x2 e2 x + cos 2 x 2

6. 7.

2. y = c1e2 x + c2 e −3 x +

1 2x xe 5

4. y = (c1 + c2 x + c3 x2 ) e − x +

1 3 −x x e 6

D-106

4.7 P.I. when Q = sin ax or cos ax and F ( − a 2 ) = 0 To find out these types of particular integrals, it is convenient to replace sin ax or cos ax by the exponential value and then apply 4.6. Thus the particular integral of cos ax 1 i. e., cos ax, when F (− a2 ) = 0 , can be written as F ( D2 ) =

1 F ( D2 )

(Real part of e iax ),

= Real Part of

1 F ( D2 )

[∵ e iax = cos ax + i sin ax]

e iax ⋅ 1, which is now of the form of article 4.5.

Similarly, the particular integral of sin ax 1 i. e., sin ax, when F (− a2 ) = 0 2 F (D ) =

1 F ( D2 )

(Imaginary part of e iax ),

= Imaginary part of

1 F( D2 )

[∵ e iax = cos ax + i sin ax]

e iax ⋅ 1,.

which is now of the form

1 e ax V of article 4.5. F ( D)

Now the method of article 4.5 is to be followed.

Example 27: Solve

d2 y dx2

+ a2 y = cos ax .

(Meerut 2003)

Solution: The auxiliary equation is

m2 + a2 = 0 , i. e., m = 0 ± ai . ∴

C. F. = e0 x (c1 cos ax + c2 sin ax) = c1 cos ax + c2 sin ax .

And

P. I. =

1 2

D + a2

cos ax

= the real part in = the real part in Now

1 D2 + a2 =

e iax

1 D2 + a2 1 2

2

D +a

(cos ax + i sin ax) e iax , by Euler’s theorem of trigonometry.

1 = e iax ( D + ia) ( D − ia)

1 e iax , putting ia for D in the factor D + ia (ia + ia) ( D − ia)

D-107

=

1 e iax ⋅ 1 2 ia ( D − ia)

=

1 iax 1 1 iax 1 e 1= e 1 2 ia ( D + ia) − ia 2 ia D

=

1 iax x e ⋅x= (cos ax + i sin ax), 2 ia 2 ai

=−

[∵ e iax = cos ax + i sin ax]  1 i  i = − i ∵ = 2 = − i 1 i  

ix (cos ax + i sin ax), 2a

x x [∵ i2 = − 1] cos ax + sin ax, 2a 2a x x 1 sin ax − i cos ax cos ax = the real part in   2a  2a D2 + a2

=−i ∴

=

x sin ax . 2a

(Agra 2006)

Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1 cos ax + c2 sin ax + ( x / 2 a)sin ax.

Remember : While solving the differential equation ( D2 + a2 ) y = cos ax , the P.I. 1 2

2

D +a

cos ax =

x x sin ax = 2a 2

Example 28: Solve ( D2 + a2 ) y = sin ax .

∫ cos ax dx . (Bundelkhand 2008; Rohilkhand 10)

Solution: Here as in Ex. 1, C.F. = c1 cos ax + c2 sin ax .

Also

P. I. =

1 D2 + a2

sin ax

= the coefficient of i in = the coefficient of i in Now ∴

1 D2 + a2 1 D2 + a2

e iax = − i

1 2

D + a2 1 D2 + a2

(cos ax + i sin ax) e iax .

x x cos ax + sin ax . 2a 2a

[Proceed as in Ex. 27]

sin ax

x x x = the coefficient of i in  − i cos ax + sin ax = − cos ax .  2a  2a 2a Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1 cos ax + c2 sin ax − ( x / 2 a)cos ax. 1 x x Remember : 2 sin ax = − cos ax = sin ax dx . 2 2 a 2 D +a

∫

D-108 Example 29: Solve ( D2 + 4) y = sin2 x . (Rohilkhand 2007; Gorakhpur 10; Avadh 11) 2

Solution: The auxiliary equation is m + 2 = 0 .

∴

m = 0 ± 2i . (∵ e0 x = 1)

Hence C. F. = c1 cos 2 x + c2 sin 2 x . And

P. I. = =

1 D2 + 4

sin2 x =

1 1 1 1 (2 sin2 x) = (1 − cos 2 x) 2 ( D2 + 4) D2 + 4 2

1 1 1 1 1− cos 2 x . 2 D2 + 4 2 D2 + 4 −1

Now

1 1 1 1 1 = 1 + D2  2   2 D +4 8 4

Again

1 1 1 1 cos 2 x = ⋅ x sin 2 x, (refer Ex. 27.) 2 D2 + 4 2 4

1=

1 1 2 1  1 − D + …  1 = ⋅   8 4 8

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = c1 cos 2 x + c2 sin 2 x + − x sin 2 x . 8 8 Example 30: Solve ( D2 + 1) y = sin x sin 2 x . Solution: The auxiliary equation is m2 + 1 = 0 or m = ± i .

∴ And

C. F. = c1 cos x + c2 sin x . 1 1 1 P. I. = 2 (sin x sin 2 x) = 2 (2 sin x sin 2 x) D +1 D +12 =

Now

1 1 1 1 1 1 (cos x − cos 3 x) = cos x − cos 3 x . 2 D2 + 1 2 D2 + 1 2 D2 + 1

1 2

D +1

cos x = the real part in

1 2

D +1

e ix

= the real part in

1 e ix ( D + i) ( D − i)

= the real part in

1 e ix (i + i) ( D − i)

= the real part in

1 1 1 e ix ⋅ 1 = the real part in e ix 1 2 i ( D − i) 2i D + i− i

1 ix 1 1 e 1 = the real part in e ix x 2i D 2i x = the real part in (cos x + i sin x) 2i 1 = the real part in − xi (cos x + i sin x) 2 1 1 1 = the real part in − i x cos x + x sin x = x sin x. 2 2 2 = the real part in

D-109

1

Again

D2 + 1

cos 3 x =

the P.I. =

∴

1 − 32 + 1

cos 3 x = −

1 cos 3 x . 8

1 1 1 1 1 1 ⋅ x sin x − ⋅ (− cos 3 x) = x sin x + cos 3 x . 2 2 2 8 4 16

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = c1 cos x + c2 sin x + x sin x + cos 3 x . 4 16

Comprehensive Exercise 7 Solve the following differential equations : 1.

( D2 + 4) y = sin 2 x .

2.

( D3 + a2 D) y = sin ax .

3.

2

(Agra 2005; Gorakhpur 09) (Gorakhpur 2005; Avadh 05)

2

( D + 9) ( D + 1) y = cos 3 x .

A nswers 7 2.

1 x cos 2 x 4 y = c1 + c2 cos ax + c3 sin ax − (1/ 2 a2 ) x sin ax

3.

y = c1 cos 3 x + c2 sin 3 x + c3 cos x + c4 sin x −

1.

y = c1 cos 2 x + c2 sin 2 x −

1 x sin 3 x 48

4.8 P.I. when Q = xV , where V is any Function of x Here P.I. =

1 ( xV ). F ( D)

By Leibnitz’s theorem, we have  d  Dn ( xV ) = x ⋅ DnV + nDn −1V = x DnV +  Dn V , d D  showing that F ( D) ( xV ) = xF ( D) V + F ′ ( D) V . …(1) 1 1 Taking , we get V in place of V in (1) and then operating on both sides by F ( D) F ( D) after transposition F ′ ( D) 1 1 ( xV ) = x V − V [Remember] F( D) F ( D) { F( D)}2 or

 d 1 1 1  V + ( xV ) = x  V. F ( D) F ( D)  d D F ( D)

D-110

Proceeding by repeated application of the above formula {1 / F ( D)} x mV can be determined. Note: The method of article 4.8 cannot be used when V is of the form cos ax or sin ax

and F (− a2 ) = 0 i. e., F ( D2 ) vanishes by putting − a2 for D2 . In that case P.I. is found by the method discussed in article 4.7.

d2 y

Example 31: Solve

2

dx

−2

dy + y = x sin x . dx

(Rohilkhand 2010)

2

Solution: The auxiliary equation is m − 2 m + 1 = 0 .

∴

m = 1, 1 .

And

P. I. = =x

1 2

( D − 1)

x sin x = x

1 2

D − 2D + 1

= x⋅ =−

Hence C.F. = (c1 + c2 x) e x . 1 2

( D − 1)

sin x −

2 ( D − 1)3

2

sin x −

3

D − 3 D2 + 3 D − 1

sin x, by article 4.8

sin x

1 2 sin x − sin x , putting − 12 i. e., −1 for D2 −2 D − D + 3 + 3D − 1

x 1 2 x 1 ⋅ sin x − sin x = cos x − sin x 2 D 2 ( D + 1) 2 D +1

=

( D − 1) x cos x D −1 x cos x − sin x = − 2 sin x 2 ( D + 1) ( D − 1) 2 D −1

=

1 1 1 1 x cos x + ( D − 1) sin x = x cos x + (cos x − sin x). 2 2 2 2

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = (c1 + c2 x) c x + x cos x + (cos x − sin x). 2 2 Example 32: Solve

d2 y dx2

+ 4 y = x sin x .

Solution: The auxiliary equation is m2 + 4 = 0 .

Therefore m = 0 ± 2 i . ∴ And

[∵ e0 x = 1]

C. F. = c1 cos 2 x + c2 sin 2 x . P. I. = = =

1 2

D +4 x sin x 2

−1 + 4

x sin x = x −

1 2

D +4

2D 2

2

(− 1 + 4)

sin x −

2D 2

( D + 4)2

sin x, by article 4.8

sin x, putting − 12 for D2

1 2 1 2 x sin x − D (sin x) = x sin x − cos x . 3 9 3 9

D-111

Hence the complete solution is y = (C.F.) + (P.I.) 1 2 or y = c1 cos 2 x + c2 sin 2 x + x sin x − cos x . 3 9 Example 33: Solve ( D4 − 1) y = x sin x .

(Lucknow 2006)

Solution: The auxiliary equation is m4 − 1 = 0 or (m2 − 1) (m2 + 1) = 0 .

∴

m = 1, − 1, 0 ± i .

∴

C. F. = c1e x + c2 e − x + e0 x (c3 cos x + c4 sin x) .

And

P. I. =

1 D4 − 1

x sin x .

Here we cannot use the method given in 4.8 because D4 − 1 vanishes by putting − 12 i. e., −1 for D2 . So here we shall proceed by the method given in article 4.7. ∴

P.I. = Imaginary part of = I. P. of e ix = I. P. of e ix = I. P. of e ix = I. P. of e ix

1 {( D + i)4 − 1}

1 4

D −1

[∵ e ix = cos x + i sin x]

x e ix ,

x, by using the method for e ax V 1

4

3

2

( D + 4 iD + 6 i D2 + 4 i3 D + i4 − 1) 1 4

3

2

D + 4 iD − 6 D − 4 iD 1

= I. P. of

1  3D e ix  ⋅ 1 +  − ....    −4i D 2i 

= I. P. of

e ix 1 3D ⋅ 1 − + ... x 2i −4i D 

= I. P. of

e ix 1  3 x −  , − 4i D  2 i

= I. P. of

e ix  x2 3 x  −  , − 4i  2 2i 

=

1 1 2 3   x cos x − x sin x  4 2 2

=

1 2 ( x cos x − 3 x sin x) . 8

x

−1

x

∵ D ≡ d   dx  [∵ (1/ D) stands for integration w.r.t. x]

1 ix  1 2 3  i e  x + i x ; 2 4 2  1 1 3 = I. P. of i (cos x + i sin x)  x2 + ix 2 4 2  = I. P. of

[∵i2 = − 1]

x,

− 4 iD [1 + (3 D / 2 i) − D2 − ( D3 / 4 i)]

x

[∵ i2 = − 1 ⇒ 1/ i = − i]

D-112

Hence the complete solution is y = (C. F. ) + (P. I. ) y = c1e x + c2 e − x + c3 cos x + c4 sin x +

or

Example 34: Solve ( D4 + 2 D2 + 1) y = x2 cos x .

1 2 ( x cos x − 3 x sin x). 8

(Gorakhpur 2010; Purvanchal 07, 08)

Solution: The auxiliary equation is m4 + 2 m2 + 1 = 0

(m2 + 1)2 = 0 giving m = ± i, ± i .

or ∴

C. F. = (c1 + c2 x) cos x + (c3 + c4 x) sin x .

And

P. I. =

1

x2 cos x

D4 + 2 D2 + 1

= Real part of

{( D + 1)2

1 2

( D + 2 iD)2

[∵ e ix = cos x + i sin x]

x2 e ix ,

{( D + i)2 + 1}2

= R. P. of e ix = R. P. of e ix

1 2

1

= R. P. of e ix

x2 , by article 4.5 [∵ i2 = − 1]

x2 ,

1 2

[∵ e0 x = 1]

2

2

4 i D [1 + ( D / 2 i)] −2

x2

= R. P. of −

D 1 ix 1  e 1+  2  4 2 i D 

= R. P. of −

1 ix 1 1 e [1 − iD]−2 x2 , 4 2 D2

= R. P. of −

1 ix 1  1 1 e 1 + 2 ⋅ iD + 3 ⋅ i2 D2 + … x2 , 2   4 2 4 D 

x2 ,

[∵ i2 = − 1] ∵ 1 = − i   i

expanding by binomial theorem = R. P. of −

1 ix 1  3 e 1 + iD − D2 + … x2 2   4 4 D 

= R. P. of −

1 ix  1 i 3  e + − + terms in D, D2 , and so on x2  D2  4 D 4

= R. P. of −

 1 ix  1 x4 1 3 e  + i x3 − x2 + terms in x1 , x 0 , 4 3 4 3 4  (∵ 1/ D stands for integration w.r.t. x)

= R. P. of − =−

1 1 4 1 3 3 2 (cos x + i sin x)  x + ix − x + terms in x1, x0  12  4 3 4

1  1 4 3 2 1 1 x − x  cos x +  x3  sin x  4 12 4  4 3  + terms already included in the C.F.

D-113

=−

1 4 1 3 ( x − 9 x2 ) cos x + x ⋅ sin x , 48 12

neglecting the terms already included in the C.F. Hence the complete solution is y = (C. F. ) + (P. I. ) 1 4 1 3 or y = (c1 + c2 x) cos x + (c3 + c4 x) sin x − ( x − 9 x2 ) cos x + x sin x . 48 12 Example 35: Solve ( D2 − 2 D + 1) y = xe x sin x .. (Purvanchal 2010; Kanpur 12; Avadh 13)

Solution: The auxiliary equation is

m2 − 2 m + 1 = 0 ∴

C. F. = (c1 + c2 x) e x .

And

P.I.= = ex = ex

1 2

D − 2D + 1 1

{( D + 1) − 1}2 1 2

D

(m − 1)2 = 0 .

or

xe x sin x =

∴

1 2

( D − 1)

m = 1, 1 .

e x x sin x

x sin x , by 4.5

( x sin x) = e x ⋅

1 D

∫ x sin x dx, [∵ (1 / D) stands for integration w.r.t. x]

= ex

1 (− x cos x + sin x) , integrating by parts D

= ex

x

∫ (− x cos x + sin x) dx = e [∫ − x cos x dx + ∫ sin x dx] = e x [− x sin x + sin x dx + sin x dx] = e x (− x sin x − 2 cos x) . ∫ ∫

Hence the complete solution is y = (C.F.) + (P.I.) or

y = (c1 + c2 x) e x − e x ( x sin x + 2 cos x) .

Comprehensive Exercise 8 Solve the following differential equations : 1. d2 y / dx2 + 9 y = x sin x . 2. ( D2 + 2 D + 1) y = x cos x . 3. ( D2 − 4 D + 4) y = 8 x2 e2 x sin 2 x .

(Gorakhpur 2011; Meerut 05B)

4. ( D2 + 4) y = x sin 2 x . 5. ( D2 + D) y = x cos x . 6. ( D2 − 1) y = x2 cos x . 7. ( D2 + 1) y = x2 sin 2 x .

(Meerut 2004, 07B; Lucknow 11)

D-114

8. ( D4 − 1) y = x sin 2 x . 2

(Meerut 2001, 06B)

2

9. ( D + m ) y = x cos mx . 10. (d2 y / dx2 ) − y = x sin x + (1 + x2 ) e x .

(Bundelkhand 2009)

A nswers 8 1 1 x sin x − cos x 8 32 1 1 1 + x sin x − sin x + cos x 2 2 2

1.

y = c1 cos 3 x + c2 sin 3 x +

2.

y = (c1 + c2 x) e − x

3.

y = (c1 + c2 x) e2 x + e2 x (− 2 x2 sin 2 x + 3 sin 2 x − 4 x cos x)

4.

y = c1 cos 2 x + c2 sin 2 x −

5.

y=

6.

y=

7.

y=

8.

y=

9.

y = c1 cos mx + c2 sin mx + ( x / 4 m2 ) cos mx + ( x2 / 4 m)sin mx

1 2 1 x cos 2 x + x sin 2 x 8 16 1 1 c1 + c2 e − x − x (cos x − sin x) + cos x + sin x 2 2 1 c1e x + c2 e − x − ( x2 − 1) cos x + x sin x 2 1 c1 cos x + c2 sin x − [24 x cos 2 x − (9 x2 − 26) sin 2 x] 27 1  32 c1 e x + c2 e − x + c3 cos x + c4 sin x + cos 2 x + x sin 2 x   15  15

10. y = c1e x + c2 e − x −

1 1 x (cos x + x sin x) + e (2 x3 − 3 x2 + 9 x) 2 12

4.9 The Operator

1 1 and , α being a Constant D −α D+α

If Q is any function of x, then 1 Q = eα x D−α

∫e

−α x

Q dx and

1 Q = e − α x e α x Q dx D+α

∫

(Meerut 20, 03; Rohilkhand 09)

1 1 Proof. We have Q= [e α x (e − α x Q)] D−α D−α = eα x = eα x

1 e − α x Q , by article 4.5 ( D + α) − α 1 −α x e Q = e α x e − α x Q dx . D

∫

Similarly proceed for second operator proof.

D-115

Example 36: Solve

d2 y dx2

+ a2 y = sec ax . (Agra 2006; Bundelkhand 07; Purvanchal 07, 08; Lucknow 09)

Solution: The auxiliary equation is

m2 + a2 = 0 ,

or

m = ± ia .

∴

C. F. = c1 cos ax + c2 sin ax .

And

P. I. =

1 D2 + a2

sec ax =

1 sec ax ( D + ia) ( D − ia)

=

1  1 1  −   sec ax [by resolving into partial fractions] 2 ia  D − ia D + ia 

=

 1  1 1 sec ax − sec ax  2 ia  D − ia D + ia 

=

1  iax e 2 ia 

=

cos ax + i sin ax  1  iax cos ax − i sin ax e dx − e − iax dx ,  2 ia  cos ax cos ax 

∫e

− iax

sec ax dx − e − iax e iaxsec ax dx , by article 4.9 

∫

∫

∫

(∵ e − iax = cos ax − i sin ax and e iax = cos ax + i sin ax) =

1  iax  i e  x + log cos ax − e − iax 2 ia   a 

=

 e iax + e − iax  x  e iax − e − i ax  1    + 2 (log cos ax)  a 2i 2   a 

=

x 1 sin ax + 2 (log cos ax) cos ax . a a

 x − i log cos ax    a   

Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1 cos ax + c2 sin ax +

x 1 sin ax + 2 cos ax ⋅ log (cos ax) . a a

Comprehensive Exercise 9 Solve the following differential equations : 1.

(d2 y / dx2 ) + 9 y = sec 3 x.

(Meerut 2007)

D-116

2.

( D2 + a2 ) y = cosec ax .

3.

( D2 + a2 ) y = tan ax .

(Lucknow 2008)

A nswers 9 1 1 x sin 3 x + cos 3 x log (cos 3 x) 3 9

1.

y = c1 cos 3 x + c2 sin 3 x +

2.

y = c1 cos ax + c2 sin ax + (1/ a2 )sin ax log(sin ax) − ( x / a)cos ax

3.

1 1 y = c1 cos ax + c2 sin ax − (1/ a2 )cos ax log tan  π + ax 4 2 

O bjective T ype Q uestions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

The solution of differential equation

d2 y dx2

+ y = 0 is

(a) c1 e − x + c2 e x

(b) c1 cos x + c2 sin x

(c) (c1 + c2 x) cos x + (c3 + c4 x) sin x

(d) None of these. (Bundelkhand 2001)

2.

2

x

P.I. of the differential equation ( D + D + 1) y = e is (a)

1 x e 3

(b) 3e x

(c) e x 3.

(d) None of these. ax

For the differential equation F ( D) y = e , if F (a) = 0, then P.I. = given by 1 (a) e ax F (a) (c) e ax

4.

(Avadh 2005)

(b)

1 1 F ( D + a)

For the particular integral following is correct

1 e ax is F ( D)

1 e ax F (− a)

(d) None of these (Rohilkhand 2005)

1 f ( D2 )

sin ax when f (− a2 ) = 0 , which one of the

D-117

1

(a)

1

(c)

2

1

sin ax =

f ( D2 )

2

D +a

f (− a2 )

sin ax =

(b)

sin ax

x cos ax 2a

1 D2 + a2

(d) None of these (Bundelkhand 2001; Agra 08) 2

5.

6.

x cos ax 2a

sin ax = −

The solution of the differential equation

d y dx2

−3

dy + 2 y = e x is dx

(a) y = c1e x + c2 e3 x + x

(b) y = (c1 + c2 ) e x − xe x

(c) y = c1 e x + c2 e2 x − xe x

(d) y = c1 x + c2 e x − e2 x .

The P.I. of the differential equation ( D2 + 1) y = cos x is 1 sin x 2 x (c) sin x 2 (a)

(b) − (d)

x sin x 2

x cos x 2

(Rohilkhand 2007)

Fill in the Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

The general solution of the equation f ( D) y = 0 is called the …… 7 of the equation f ( D) y = Q . d2 y

+ a2 y = 0 is ……

2.

The solution of the differential equation

3.

The complementary function (C.F.) of the differential equation d2 y dx2

4.

dx

dy + 4 y = 0 is …… dx

The particular integral (P.I.) of the differential equation d3 y 3

dx 5.

−5

2

−4

d2 y 2

dx

+5

dy − 2 = 0 is …… dx

The particular integral of the differential equation f ( D) y = Q will be ……

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

A linear differential equation is an equation in which the dependent variable and its derivatives appear only in the first degree.

2.

d4 y 4

dx

−5

d3 y 3

dx

−2

dy + 4 xy = 0 is a linear differential equation with constant dx

coefficients. 3.

e

−x

(Meerut 2003) 2x

(c1 cos √ 3 x + c2 sin √ 3 x) + c3 e

d3 y dx3

− 8y = 0 .

is the solution of the differential equation

D-118

A nswers Multiple Choice Questions 1. 6.

(b) (c)

2.

(a)

3. (c)

4.

(b)

5.

1 Q F ( D)

5.

(c)

Fill in the Blank(s) 1. C.F. 3.

x

4x

c1 e + c2 e

2. y = c1 cos ax + c2 sin ax . 2 4. x 5

True or False 1.

T

2.

F

3.

T

¨

D-119

5 H omogeneous L inear D ifferential E quations

5.1 Homogeneous Linear Differential Equations (Meerut 2009B)

D

efinition: A differential equation of the form xn

dn y dx

n

+ a1 x n−1

dn −1 dx

n−1

+ … + an−1 x

dy + an y = Q, dx

…(1)

where a1, a2 , … , an are constants and Q is either a constant or a function of x is called a homogeneous linear differential equation.

5.2 Method of Solution To solve such equations, we introduce a new independent variable z such that x = e z or log x = z , so that 1/ x = dz / dx. We have

dy dy dz 1 dy = ⋅ = ⋅ dx dz dx x dz

∴

x

dy dy d d so that x = ≡ ≡ D , say. dx dz dx dz

D-120

d dx

Now

x

or

xn

n−1  n−1 d n−1 y  dn y y n−1 d  = xn x + ( − 1 ) n x n−1 n−1  n  dx dx  dx 

dn y dx n

d n−1 y  d  = x − n + 1 x n−1 n−1  dx  dx = ( D − n + 1) x n−1

d n−1 y dx n−1

…(2)

⋅

Putting n = 2, 3, 4,… etc. in (2), we have x2

d2 y 2

dx

= ( D − 1) x

dy = ( D − 1) Dy dx

d   ∵ x dx ≡ D

= D ( D − 1) y, because the operators can be interchanged x3

d3 y dx3

= ( D − 2) x2

d2 y dx2

= ( D − 2) ( D − 1) Dy = D ( D − 1) ( D − 2) y .

Whence, generalising, we have xn

dn y dx n

= D ( D − 1) ( D − 2) … ( D − n + 1 y .

n dy 2 d2 y n d y in (1) and thus changing the ,x , … , x dx dx2 dx n independent variable from x to z, we have

Substituting these values of x

[{ D ( D − 1) … ( D − n + 1)} + a1 { D ( D − 1) … ( D − n + 2)}] + … + an−1 D + an] y = Q , or

f ( D) y = Q,

…(3)

where Q has now become a function of z. In the differential equation (3) the independent variable is z and the operator D stands for d / dz . Obviously (3) is a linear differential equation with constant coefficients and so it can be solved by the methods given in the previous chapter i. e., chapter 3. Thus as in the case of the linear equations with constant coefficients, the general solution of (3) is the sum of any particular integral of (3) and the complementary function i. e., the general solution of f ( D) y = 0 .

…(4)

To find the Complementary function (C.F.): (i)

If m1, m2 , …, mn are the roots of the auxiliary equation of (4), and no two of them are equal, the general solution of (4) i. e., the C.F. of the solution of (3) is easily seen to be y = c1e m1z + c2 e m2 z + ... + c ne mnz , or

(ii)

y = c1 x m1 + c2 x m2 + … + c n x mn .

[∵ e z = x]

In case there are r roots a like, each equal to m, and the rest all different, then the

D-121

C.F. = (c1 + c2 z + … + c r z r −1) e mz + c r +1e mr +1z + ... + c ne mn z = [c1 + c2 log x + … + c r (log x)r −1] x m + c r +1 x mr +1 + ... + c n x mn . (iii) In case the roots are imaginary, say of the form α ± iβ, then the C. F. = e α x (c1 cos βz + c2 sin βz ) = x α [c1 cos (β log x) + c2 sin (β log x)] . In this case we can also write the C. F. = c1e α z cos (βz + c2 ) = c1 x α cos (β log x + c2 ) . (iv) In case the roots α ± iβ occur r times, the C.F. corresponding ot these roots will be e α z [(c1 + c2 z + … + c r z r −1)cos βz + (c1 ′ + c2 ′ z + … + c r ′ z r −1)sin βz ] = x α [{ c1 + c2 log x + … + c r (log x)r −1} cos (β log x) + { c1 ′ + c2 ′ log x + … + c r ′ (log x)r −1} sin (β log x)]. To find the Particular integral. (P.I.): The particular integral (P.I.) of (3) is given by 1 Q. f ( D) If α is a constant, we have 1 1 Q= { e α z (e − αz Q)} D−α D−α = eα z

1 e−α z Q ( D + α) − α

= eα z

1 − αz e Q = e α z e − α z Q dz . D

∫

Methods to find the P.I: General Methods: (i) We resolve the operator f ( D) into linear factors. Thus we write f ( D) = ( D − m1) ( D − m2 ) … ( D − mn) . Then the 1 1 P. I. = Q= Q f ( D) ( D − m1) ( D − m2 ) .. ( D − mn−1)( D − mn) = by operating

1 e mn z ( D − m1) ( D − m2 ) … ( D − mn−1)

∫e

− mn z

Q dz ,

1 upon Q as explained above. D − mn

Similarly we operate with other remaining factors in succession and thus we find the P.I. (ii)

First we resolve f ( D) into linear factors as in (i) and then we break up { f ( D)} −1 into partial fractions. Then the

D-122

 A1 1 A2 An  Q= + + ... + Q f ( D) D − m D − m D − mn  1 2 

P. I. =

= A1e m1z

∫e

− m1z

Q dz + … + An e mnz

∫e

− mn z

Q dz .

Special short Methods. (i)

When Q is of the form e a z , then 1 1 az ea z = e , provided f (a) ≠ 0 . f ( D) f (a)

P. I. =

(ii) When Q is of the form cos az or sin az, then the P.I. is given by 1 2

cos az =

2

sin az =

f (D ) and

1 f (D )

1 f (− a2 ) 1 f (− a2 )

cos az , sin az ,

(iii) If Q is of the form z m , we have P.I. =

[provided f (− a2 ) ≠ 0]

1 z m. f ( D)

We expand { f ( D)} −1 in ascending powers of D retaining terms as far as Dm and operate each term on z m . (iv) If Q is of the form e a zV , where V is any function of z, we have the 1 1 P.I. = e a zV = e a z V. f ( D) f ( D + a) (v) If Q is of the form zV, where V is any function of z, we have the  d 1 1 1  V + (zV ) = z  V. f ( D) f ( D) dD f ( D) 

P.I. =

Example 1: Solve x2 D2 y + 5 x Dy + 4 y = 0 , where D ≡ d / dx. Solution: The given differential equation is

x2

d2 y 2

dx

+ 5x

dy +4y =0. dx

…(1)

Now (1) is a homogeneous linear differential equation of order 2. So putting x = e z and denoting d / dz by D′, we have x ∴

dy d2 y = D′ y, x2 = D′ ( D′ − 1) y . dx dx2

the differential equation (1) transforms to { D′ ( D′ − 1) + 5 D′ + 4} y = 0

or

( D′2 + 4 D′ + 4) y = 0

D-123

or

( D′ + 2)2 y = 0 ,

which is a linear differential equation with constant coefficients, the independent variable being z. ∴

auxiliary equation is (m + 2)2 = 0 , giving m = − 2, − 2 .

∴

the complete solution is y = (c1 + c2 z ) e −2 z

or

y = (c1 + c2 z ) (e z )−2

or

y = (c1 + c2 log x) x −2 .

Example 2: Solve ( x3 D3 + 3 x2 D2 − 2 xD + 2) y = 0 , where D ≡ d / dx. Solution: Putting x = e z and denoting d / dz by D ′, we have

x Dy = x and ∴

dy d2 y = D ′ y, x2 D2 y = x2 = D ′ ( D ′ − 1) y dx dx2

x3 D3 y = x3

d3 y dx3

= D ′ ( D ′ − 1) ( D ′ − 2) y.

the given differential equation transforms to [ D ′ ( D ′ − 1) ( D ′ − 2) + 3 D ′ ( D ′ − 1) − 2 D ′ + 2] y = 0

or

[ D ′ ( D ′ − 1) ( D ′ − 2) + 3 D ′ ( D ′ − 1) − 2 ( D ′ − 1)] y = 0

or

( D ′ − 1) [ D ′ ( D ′ − 2) + 3 D ′ − 2] y = 0

or

( D ′ − 1) ( D ′2 + D ′ − 2) y = 0

or

( D ′ − 1) ( D ′ + 2) ( D ′ − 1) y = 0

or

( D ′ − 1)2 ( D ′ + 2) y = 0 .

∴

the auxiliary equation is (m − 1)2 (m + 2) = 0 , giving m = 1, 1, − 2 .

∴

the general solution is y = (c1 + c2 z ) e z + c3 e −2 z = (c1 + c2 z ) e z + c3 (e z )−2

or

y = (c1 + c2 log x) x + c3 x −2 ,

or

y = (c1 + c2 log x) x + c3 ⋅ (1/ x2 ).

[∵ x = e z and z = log x]

Example 3: Solve

x4

d3 y 3

dx

+ 2 x3

d2 y 2

dx

− x2

dy + xy = 1 . dx

Solution: Dividing out by x, the given equation may be written as

x3

d3 y 3

dx

+ 2 x2

d2 y 2

dx

−x

dy 1 + y= , dx x

…(1)

D-124

which is a linear homogeneous equation of order 3. Putting x = e z in (1) and denoting d / dz by D, we have [ D ( D − 1) ( D − 2) + 2 D ( D − 1) − D + 1] y = e − z ( D − 1)2 ( D + 1) y = e − z .

or ∴

auxiliary equation is (m − 1)2 (m + 1) = 0 , giving m = 1, 1, − 1 .

∴

C. F. = (c1 + c2 z ) e z + c3 e − z = (c1 + c2 log x) x + c3 x −1.

And

[∵ e z = x and z = log x]

 1  1 e−z   2 ( D + 1) ( D − 1) ( D − 1) ( D + 1)  1 1 1 1 1 1 = ⋅ e−z = e−z ⋅ 1 = e−z 1 ( D + 1) 4 4 D +1 4 D − 1+ 1 1

P. I. =

=

e−z =

2

1 −z 1 1 1 e 1 = ze − z = log x. 4 D 4 4x

Hence the complete solution is c3 1 + log x . x 4x

y = (c1 + c2 log x) x +

d3 y d2 y dy Example 4: Solve x3 + 2 x2 + 3x − 3 y = x2 + x. 3 dx dx dx2 (Meerut 2001, 07; Gorakhpur 05, 08) z

Solution: Putting x = e and denoting d / dz by D, the given differential equation

becomes [ D ( D − 1) ( D − 2) + 2 D ( D − 1) + 3 D − 3] y = e2 z + e z or

[ D ( D − 1) ( D − 2) + 2 D ( D − 1) + 3 ( D − 1)] y = e2 z + e z

or

( D − 1) { D ( D − 2) + 2 D + 3} y = e2 z + e z

or

( D − 1) ( D2 + 3) y = e2 z + e z .

Hence the auxiliary equation is (m − 1) (m2 + 3) = 0 , giving m = 1, 0 ± √ 3 i. ∴

C. F. = c1e z + e0 z (c2 cos √ 3 z + c3 sin √ 3 z ) = c1e z + c2 cos √ 3 z + c3 sin √ 3 z .

Also

P. I. = =

1 ( D − 1) ( D2 + 3) e2 z 2

(2 − 1) ( D + 3)

e2 z +

+

1 ( D − 1) ( D2 + 3) 1

2

(1 + 3) ( D − 1)

ez ⋅ 1

ez

D-125

∴

=

1 2z 1 z 1 1 1 1 e + e 1 = e2 z + e z 1 7 4 D + 1− 1 7 4 D

=

1 2z 1 z e + e .z . 7 4

the complete solution is 1 2z 1 z e + ze . 7 4 1 1 y = c1 x + c2 cos (√ 3 log x) + c3 sin (√ 3 log x) + x2 + x log x . 7 4 y = c1 e z + c2 cos √ 3 z + c3 sin √ 3 z +

or

[∵ e z = x and z = log x] d2 y dy Example 5: Solve x2 −x − 3 y = x2 log x . dx dx2 (Meerut 2004B, 10; Gorakhpur 07; Kanpur 07, 14; Lucknow 09, 11; Avadh 08; Bundelkhand 04)

Solution: Putting x = e z or z = log x and denoting d / dz by D the equation becomes

[ D ( D − 1) − D − 3] y = e2 z . z ( D2 − 2 D − 3) y = e2 z . z .

or ∴

auxiliary equation is (m2 − 2 m − 3) = 0

or

(m − 3) (m + 1) = 0 i. e., m = − 1, 3 .

∴

3z

And

C.F. is c1e

P. I. =

+ c2 e − z . 1 2

( D − 2 D − 3)

1

= e2 z . = e2 z = e2 z

2

[( D + 2) − 2( D + 2) − 3] 1 z 2 ( D + 2 D − 3) 1 − 3 [1 −

= e2 z − =− ∴

or

e2 z . z

2 3

D−

1 3

D2 ]

z

z

1  2 1  1 −  D + D2    3   3 3

−1

z

1 2z  2 1 1 2  e 1 + D + D2 + … z = − e2 z (z + ).   3 3 3 3 3

complete solution is y = c1 e3 z + c2 e − z −

1 2z 2 e (z + ) 3 3

y = c1 x3 + c2 x −1 −

1 2 2 x (log x + ). 3 3

D-126

5.3 Equations Reducible to Homogeneous Form A differential equation of the form (a + bx)n

dn y dx

n

+ P1 (a + bx)n − 1

dn −1 y dx

n −1

+ … + Pn − 1 (a + bx)

dy + Pn y = Q, dx

where P1, P2 , …, Pn are constants and Q is a function of x, can be reduced to the homogeneous linear form by putting a + bx = t and then can be solved by the method explained in article 5.2. We can also solve this differential equation directly by making the substitution. e z = a + bx or

z = log (a + bx).

If we denote the operator d / dz by D, we can easily see that (a + bx)

dy d2 y = b Dy, (a + bx)2 2 = b2 D (d − 1) y, and so on. dx dx

Example 6: Solve ( x + a)2

d2 y dx2

− 4 ( x + a)

dy +6y = x. dx (Meerut 2001, 06, 11; Kanpur 11)

Solution: Putting ( x + a) = e z and denoting d / dz by D, the given equation becomes

[ D ( D − 1) − 4 D + 6] y = e z − a or

( D − 3) ( D − 2) y = e z − a .

∴

auxiliary equation is (m − 3) (m − 2) = 0 , giving m = 3, 2 .

∴

C. F. = c1e3 z + c2 e2 z = c1 (e z )3 + c2 (e z )2 = c1 ( x + a)3 + c2 ( x + a)2

and

P. I. =

1 1 ez − ae0 z , ( D − 3) ( D − 2) ( D − 3) ( D − 2)

=

1 1 ez − a e0 z , ( D − 3) ( D − 2) ( D − 3) ( D − 2)

=

1 1 ez − a e0 z (1 − 3) (1 − 2) (0 − 3) (0 − 2)

=

1 z 1 1 1 e − a = ( x + a) − a . 2 6 2 6

Hence the complete solution is y = C. F. + P. I. or

y = c1 ( x + a)3 + c2 ( x + a)2 +

1 1 ( x + a) − a . 2 6

[∵ e0 z = 1]

D-127

Comprehensive Exercise 1 Solve the following differential equations : 1. x2 (d2 y / dx2 ) − 4 x (dy / dx) + 6 y = x . 2.

2

2

2

2

2

2

(Bundelkhand 2010)

4

x (d y / dx ) − 4 x (dy / dx) + 6 y = x .

(Gorakhpur 2009)

3. x (d y / dx ) − 2 x (dy / dx) + 2 y = 1/ x . 4. x2 (d2 y / dx2 ) + x (dy / dx) − 4 y = x2 . 2

2

(Rohilkhand 2009)

2

2

5. x (d y / dx ) + 2 x (dy / dx) − 20 y = ( x + 1) . 6. x2 (d2 y / dx2 ) + 4 x (dy / dx) + 2 y = e x . (Meerut 2005; Avadh 08, 11; Rohilkhand 07, 11) 2

2

2

7. x (d y / dx ) + 7 x (dy / dx) + 13 y = log x. 2

2

(Bundelkhand 2006)

2

8. (i) x (d y / dx ) − x (dy / dx) + 2 y = x log x. (Rohilkhand 2010; Gorakhpur 06, 10, 11; Purvanchal 09; Avadh 06; Bundelkhand 13) 3

3

3

2

2

2

(ii) x (d y / dx ) + 3 x (d y / dx ) + x (dy / dx) + y = x log x. (Kanpur 2009; Rohilkhand 08; Purvanchal 08)

9. x3 (d3 y / dx3 ) + 3 x2 (d2 y / dx2 ) + x (dy / dx) + y = x + log x. 2

2

(Avadh 2006, 14)

2

10. x (d y / dx ) − x (dy / dx) + 4 y = cos (log x) + x sin (log x). 11. x3 (d3 y / dx3 ) + 2 x2 (d2 y / dx2 ) + 2 y = 10 [ x + (1/ x)]. (Meerut 2007B; Bundelkhand 09; Agra 2006; Lucknow 06) 2

dy log x sin (log x) + 1 + y= ⋅ dx x dx 1 13. ( x2 D2 + 3 xD + 1) y = ⋅ (1 − x)2 12. x2

d y 2

− 3x

(Rohilkhand 2008; Purvanchal 08; Bundelkhand 01)

14. Solve (3 x + 2)2

2

d y 2

dx

+ 3 (3 x + 2)

dy − 36 y = 3 x2 + 4 x + 1. dx (Bundelkhand 2009; Agra 06; Lucknow 06)

15. Solve (1 + x)2

2

d y 2

dx

+ (1 + x)

dy + y = 4 cos log (1 + x). dx

D-128

A nswers 1 1.

y = c1 x2 + c2 x3 +

1 x 2

2.

y = c1 x2 + c2 x3 +

1 4 x 2

3.

y = c1 x + c2 x2 +

1 ⋅ (1/ x) 6

4.

y = c1 x2 + (c2 / x2 ) +

5.

y = c1 x4 + c2 x −5 −

6.

y = (c1 / x) + (c2 / x2 ) + (e x / x2 )

7.

y = c1 {cos (log x2 + c2 )} / x3 +

8. (i) (ii) 9.

1 2 x log x 4

1 2 1 1 x − x− 14 9 20

1 6 log x − 13 169

y = c1 x cos (log x + c2 ) + x log x y = c1 x −1 + c2 √ x cos {(√ 3 / 2) log x + c3 } +

y = c1 x − 1 + c2 √ x cos {(√ 3 / 2) log x + c3 } +

1 3 x log x − x 2 4

1 x + log x 2

10. y = x [c1 cos (√ 3 log x) + c2 sin (√ 3 log x)] 1 1 + [3 cos (log x) − 2 sin (log x)] + x sin (log x) 13 2 11. y = cx −1 + c2 x cos (log x + c3 ) + 5 x + 2 x −1 log x 12. y = x2 (c1 x √3 + c2 x −√3 ) + +

1 6x

(5 sin log x + 6 cos log x) 382 cos log x + 54 sin log x  1 log x ⋅ +   x 61 3721

13. y = (c1 + c2 log x)

1 1 x + log x x 1− x

14. y = c1 (3 x + 2)2 + c2 (3 x + 2)−2 +

1 [(3 x + 2)2 log (3 x + 2) + 1] 108

15. y = c1 cos log (1 + x) + c2 sin log (1 + x) + 2 log (1 + x) sin log (1 + x)

D-129

O bjective T ype Q uestions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

The complementary function (C.F.) of the differential equation x2

2.

d2 y 2

dx

−x

dy − 3 y = x2 log x is dx

(a) c1 cos x + c2 sin x

(b) c1 e3 z + c2 e − z

(c) c1 e z + c2 e −3 z

(d) None of these d2 y

C.F. of the differential equation x2

2

dx

(c) c1 e − x + c2 e x

(d) None of these 3

d y

C.F. of the differential equation x4 x

(a) (c1 + c2 x) e + c3 e (c) (c1 + c2 log x) x + 4.

3

dx

+ 2 x3

d2 y 2

dx

dy + xy = 1 is dx

− x2

−x

(b) c1e x + c2 e − x + c3 e2 x

c3 1 log x + x 4x

(d) None of these.

C.F. of the differential equation ( x + a)2 2

dy + 2 y = e x is dx

(b) c1 x −1 + c2 x −2

(a) c1 + c2 x

3.

+ 4x

3

d2 y 2

dx

dy + 6 y = x is dx

− 4 ( x + a)

(a) c1 x + c2 x

(b) c1 ( x + a)3 + c2 ( x + a)2

(c) c1 e3 x + c2 e2 x

(d) None of these.

Fill in the Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

Homogeneous linear differential equation x2

d2 y 2

dx

+ 5x

dy +4y =0 dx

will reduce to a linear differential equation with constant coefficients by putting x = …… 2.

To reduce the homogeneous linear differential equation ( x + a)3

d3 y 3

dx

− 2 ( x + a)2

d2 y 2

dx

− 5 ( x + a)

dy + 3 y = 2x dx

to a linear differential equation with constant coefficients, we have to put ……

D-130

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

2.

3.

d2 y

dy −x − 3 y = x2 log x is a linear homogeneous differential equation of dx dx2 order two. (Meerut 2003) x2

x4

d3 y 3

+ 2 x3

d2 y 2

− x2

dy + xy = 1 is not a linear homogeneous differential dx

+ 2x

dy + xy = x2 is a linear homogeneous differential dx

dx dx equation of order 3. x3

d3 y 3

+ 4 x2

d2 y 2

dx dx equation of order 3.

A nswers Multiple Choice Questions 1. 4.

(b) (b)

2. (b)

3. (c)

Fill in the Blank(s) 1.

ez

2 ( x + a) = e z

True or False 1.

T

2. F

3. F

¨

D-131

6 O rdinary S imultaneous D ifferential E quations

6.1 Introduction n the present chapter, we shall discuss differential equations containing one idependent variable and two or more than two dependent variables. To completely solve such equations we need as many simultaneous equations as there are dependent variables.

I

6.2 Methods of Solving Simultaneous Linear Differential Equations with Constant Coefficients Let x and y be the two dependent variables and t be the independent variable. Thus the equations will contain differential coefficients of x, y with respect to t. Let D ≡ d / dt. Then such equations can be put in the form

and

f1( D) x + f2 ( D) y = T1

…(1)

φ1( D) x + φ2 ( D) y = T2 ,

…(2)

D-132

where T1 and T2 are some functions of the independent variable t. Here f1( D), f2 ( D), φ1( D) and φ 2( D) are all rational integral functions of D with constant coefficients. There are two methods to solve such equations. First method: Method of Elimination (Symbolic method): To eliminate y between (1) and (2), operating both sides of (1) by φ 2( D) and of (2) by by f2 ( D), we get φ 2 ( D) f1 ( D) x + φ 2 ( D) f2 ( D) y = φ 2 ( D) T1 and

f2 ( D) φ 1 ( D) x + f2 ( D) φ2 ( D) y = f2 ( D) T2 .

…(3) …(4)

Subtracting (4) from (3), we get [φ2 ( D) f1 ( D) − f2 ( D) φ1 ( D)] x = φ2 ( D) T1 − f2 ( D) T2 , which is of the form F ( D) x = T .

…(5)

The equation (5) is a linear differential equation with constant coefficients in x and t. Solving it we can find the value of x in terms of t. Putting this value of x in either (1) or (2), we get the value of y. Note 1: The equations can also be solved by first finding the value of y by eliminating x

between (1) and (2) and then solving the resulting linear differential equation in y and t. The value of x can now be obtained from (1) or (2) after putting the value of y. Note 2: Since f2 ( D) and φ2 ( D) are functions of D with constant coefficients, so

f2 ( D) φ2 ( D) = φ2 ( D) f2 ( D) . Second method: Method of differentiation: Sometimes x or y can be conveniently eliminated if we differentiate the equations (1) and (2). For example suppose the two given equations (1) and (2) connect four quantities x, y, dx / dt and dy / dt. Differentiating (1) and (2) w.r.t. t, we get in all four equations containing x, y,

dy dx dy d2 x d2 y and ⋅ Eliminating three quantities y, , , , dt dt dt2 dt2 dt

d2 y

from these four equations, we get an equation of the second order with x as the dt2 dependent and t as the independent variable. Solving this equation we find the value of x in terms of t. Then the value of the other variable can be found.

6.3 Number of Arbitrary Constants In the general solution of (1) and (2) the number of arbitrary constants is equal to the degree of D in the determinant f1( D) f2 ( D) ∆= , if ∆ ≠ 0 . φ 1( D) φ 2( D) If ∆ vanishes then the system is dependent. Here we shall not consider such cases.

D-133

Example 1: Solve the simultaneous equations

dy dx − 7x + y = 0 , − 2x − 5 y = 0 . dt dt (Purvanchal 2008; Rohilkhand 09; Lucknow 10)

Solution: Writing D for d / dt, the given equations are

and

( D − 7) x + y = 0

…(1)

− 2 x + ( D − 5) y = 0 .

…(2)

Let us eliminate y between (1) and (2). Operating on both sides of (1) by D − 5 and then subtracting (2) from it, we get {( D − 5) ( D − 7) + 2} x = 0 or

( D2 − 12 D + 37) x = 0 ,

…(3)

which is a linear differential equation of second order in x and t with constant coefficients. A.E. is m2 − 12 m + 37 = 0 ⇒ m = 6 ± i . ∴ ∴

the solution of (3) is x = e6 t (c1 cos t + c2 sin t)

…(4)

dx = 6 e6 t (c1 cos t + c2 sin t) + e6 t (− c1 sin t + c2 cos t) . dt

Putting the values of x and dx / dt in (1), we get y = 7 x − Dx = 7 x − (dx / dt) = 7 e6 t (c1 cos t + c2 sin t) − 6 e6 t (c1 cos t + c2 sin t) − e6 t (− c1 sin t + c2 cos t) or

y = e6 t [(c1 − c2 ) cos t + (c1 + c2 ) sin t] .

…(5)

The required general solution of the given equations consists of the equations (4) and (5). Example 2: Solve the simultaneous equations

d2 y dt2

− 3x − 4 y = 0 ,

d2 y dt2

+ x + y =0.

(Kanpur 2003; Meerut 09)

Solution: Writing D for (d / dt), the given equations are

( D2 − 3) x − 4 y = 0 and

x + ( D2 + 1) y = 0 .

…(1) …(2)

Let us eliminate y from (1) and (2). Multiplying both sides of (2) by 4 and operating on both sides of (1) by D2 + 1 and adding, we get {( D2 + 1) ( D2 − 3) + 4} x = 0 or

( D4 − 2 D2 + 1) x = 0 .

…(3)

D-134

A.E. is m4 − 2 m2 + 1 = 0 , m = ± 1, ± 1.

∴ ∴

(m2 − 1)2 = 0 ,

or

the solution of (3) is x = (c1 + c2 t) e t + (c3 + c4 t) e − t

…(4)

dx = (c1 + c2 t) e t − (c3 + c4 t) e − t + c2 e t + c4 e − t , dt

∴

d2 x

and

2

dt

= (c1 + c2 t) e t + (c3 + c4 t) e − t + 2 c2 e t − 2 c4 e − t

Putting these values in (1), we get 4 y = D2 x − 3 x = (d2 x / dt2 ) − 3 x = (c1 + c2 t) e t + (c3 + c4 t) e − t + 2 c2 e t − 2 c4 e − t − 3 (c1 + c2 t) e t − 3 (c3 + c4 t) e − t = − 2 (c1 + c2 t) e t − 2 (c3 + c4 t) e − t + 2 c2 e t − 2 c4 e − t 1 t 1 …(5) e (c2 − c1 − c2 t) − e −1 (c4 + c3 + c4 t). 2 2 The required general solution of the given equations consists of the equations (4) and (5).

or

y=

Example 3: Solve the simultaneous equations

d2 x dt2

+ 4 x + y = te3 t ,

d2 y

+ y − 2 x = cos2 t .

dt2

Solution: Writing D for (d / dt), the given equations are

( D2 + 4) x + y = te3 t

…(1)

2

and

2

…(2)

− 2 x + ( D + 1) y = cos t

Eliminating y from (1) and (2), we get {( D2 + 1) ( D2 + 4) + 2} x = ( D2 + 1)(te3 t ) − cos2 t ( D4 + 5 D2 + 6) x = 10 te3 t + 6 e3 t − cos2 t.

or 4

2

A.E. is m + 5 m + 6 = 0 , ∴ ∴

or

2

…(3)

2

(m + 3) (m + 2) = 0 .

m = ± i √ 3, ± i √ 2 . C.F. = (c1 cos √ 3 t + c2 sin √ 3 t) + (c3 cos √ 2 t + c4 sin √ 2 t) . 10 6 1 P. I. = 4 te3 t + 4 e3 t − 4 cos2 t D + 5 D2 + 6 D + 5 D2 + 6 D + 5 D2 + 6 = 10 e3 t

1 4

2

( D + 3) + 5 ( D + 3) + 6 −

= 10 e3 t

2

3 + 5 .3 + 6

⋅ e3 t

1 (1 + cos 2 t) D + 5D + 6 2 4

132 + 138 D + 52 D + … −

1 4

1

1 2

t+6

2

t+

1 3t e 22

1 1 1  1 cos 2 t −    6 + 5 D2 + D4 2 D4 + 5 D2 + 6  2

D-135

= 10 e3 t ⋅

= ∴

5 e3 t 66

−1

1  23 1 3t D + … t + e 1 +  132  22 22 cos 2 t 1 1 1 − ⋅ e0 t − 2 6 + 5 D2 + D4 2 (− 4)2 + 5 (− 4) + 6

3t  t − 23  + e − 1 − 1 cos 2 t .    22  22 12 4

x = (c1 cos √ 3 t + c2 sin √ 3 t) + (c3 cos √ 2 t + c4 sin √ 2 t) 5 3t 49 3 t 1 1 + te − e − cos 2 t − ⋅ 66 1452 4 12

Now from (4), we find

…(4)

dx d2 x and 2 . dt dt

Putting the values of x and d2 x / dt2 in (1), we get y=− or

d2 x dt2

− 4 x + te3 t

y = − (c1 cos √ 3 t + c2 sin √ 3 t) − 2 (c3 cos √ 2 t + c4 sin √ 2 t) 1 33 3 t 1 + t . e3 t − e + ⋅ 66 2452 3

…(5)

The required solution consists of the equations (4) and (5). Example 4: Solve the simultaneous differential equations

x

dy + z =0 dx

…(1)

x

dz + y =0. dx

…(2) (Meerut 2001, 09B)

Solution: Differentiating (1) with respect to x, we get

dy dz d2 y dy dz or x2 + =0 +x +x =0. dx dx dx dx dx dx2 dz Putting the value of x from (2), we get dx x

d2 y 2

x2

+

d2 y dx2

+x

dy − y =0, dx

…(3)

which is a linear homogeneous equation. To solve (3), putting x = e t and denoting d / dt by D, we have x ∴

dy d2 y = Dy, x2 = D ( D − 1) y . dx dx2

the equation (3) becomes { D ( D − 1) + D − 1} y = 0

or

∴

y = c1e t + c2 e − t = c1 x + c2 x −1.

∴

dy dy c c = c1 − 22 ⇒ x = c1 x − 2 ⋅ dx dx x x

( D2 − 1) y = 0 . …(4)

D-136

dy = − c1 x + c2 x −1. dx The required solution consists of the equations (4) and (5).

∴

from (1), we get z = − x

…(5)

Example 5: Solve the simultaneous differential equations

d2 x 2

dt

+ m2 y = 0 ,

d2 y dt2

− m2 x = 0 .

Solution: Writing D for (d / dt), the given equations are

D2 x + m2 y = 0 and

2

…(1)

2

D y − m x =0 .

…(2)

Eliminating y from (1) and (2), we get ( D4 + m4 ) x = 0 . A.E. is M 4 + m4 = 0 ⇒ ( M 2 + m2 )2 − 2 M 2 m2 = 0 . or

( M 2 − √ 2 Mm + m2 ) ( M 2 + √ 2 Mm + m2 ) = 0 .

∴

M 2 − √ 2 Mm + m2 = 0 ,

Solving these, we get m m M= ±i √2 √2

M 2 + √ 2 Mm + m2 = 0 .

or

and M = −

m m ±i ⋅ √2 √2

∴

m m x = c1e(m / √2)t cos  t + c2  + c3 e −(m / √2)t cos  t + c4  √2  √2 

∴

dx m   m t + c  − sin  m t + c   = c1e(m / √2)t .  cos  2  2   √2 dt √2  √2  − c3 e −(m / √2)t .

and

d2 x 2

dt

= c1e(m / √2)t .

m2 2

…(3)

m   m t + c  + sin  m t + c    cos  4  4   √2 √2  √2 

  m t + c  − 2 sin  m t + c   cos  2  2  √2 √2 

m m2  t + c2   + c3 e −(m / √2)t . − cos  √2  2

  m t+c  cos  4  √2 

m m  + 2 sin  t + c4  − cos  t + c4   √2  √2  m m   = m2  c3 e −(m / √2)t sin  t + c4  − c1e(m / √2)t sin  t + c2   .     √2 √2  From the equation (1), we get y = −

1 d2 x m2 dt2

⋅

Putting the value of d2 x / dt2 , we have m m y = c1e(m / √2)t sin  t + c2  − c3 e −(m / √2)t .sin  t + c4  . √2  √2 

…(4)

The equations (3) and (4) constitute the required general solution of the given equations.

D-137 Example 6: Solve t dx = (t − 2 x) dt, t dy = (tx + ty + 2 x − t) dt .

(Rohilkhand 2001)

Solution: The given equations are

t dx = (t − 2 x) dt and t dy = (tx + ty + 2 x − t) dt . dx 2 From (1), we have + x = 1, which is linear. dt t

…(1) …(2)

I.F. = e ∫ (2 / t)dt = e2 log t = t2 . xt2 =

∴

2

∫t

⋅ 1 dt + c1 =

1 3 t + c1 3

1 t + c1 t −2 . 3 Now adding (1) and (2), we get or

…(3)

x=

t (dx + dy) = t ( x + y) dt

or

dx + dy = dt . x+ y

Integrating, log ( x + y) = t + log c2 . ∴ x + y = c2 e t or y = c2 e t − x . Putting in (4) the value of x found in (3), we get 1 y = c2 e t − t − c1t −2 . 3

…(4)

…(5)

The equations (3) and (5) give the required general solution of the given equations. Example 7: Solve the simultaneous equations

t2

d2 x dt2

+t

d2 y dy dx + 2 y = 0 , t2 2 + t − 2x = 0 . dt dt dt

Solution: Here both the given equations are linear homogeneous equations. First we

shall change them into linear differential equations with constant coefficients. For this, we put t = e z . Then denoting d / dz by D, we have t ∴

dy d2 y dx d2 x = Dx, t2 2 = D ( D − 1) x, t = Dy, t2 2 = D ( D − 1) y . dt dt dt dt

the given equations transform to { D ( D − 1) + D} x + 2 y = 0

i. e.,

D2 x + 2 y = 0

and

{ D ( D − 1) + D} y − 2 x = 0

i. e.,

D2 y − 2 x = 0 .

…(1)

Eliminating y from (1) and (2), we get ( D4 + 4) x = 0 . A.E. is or

m4 + 4 = 0 ,

or

(m2 + 2)2 − 4 m2 = 0

(m2 − 2 m + 2) (m2 + 2 m + 2) = 0 .

…(2)

D-138

∴

m2 − 2 m + 2 = 0

∴

m=

i. e.,

m = 1 ± i, − 1 ± i .

∴

x = e z (c1 cos z + c2 sin z ) + e − z (c3 cos z + c4 sin z )

∴

dx = e z (c1 cos z + c2 sin z ) + e z (− c1 sin z + c2 cos z ) dz

2 ± √ (4 − 8) 2

m2 + 2 m + 2 = 0 .

or

and m =

− 2 ± √ (4 − 8) 2 …(3)

− e − z (c3 cos z + c4 sin z ) + e − z (− c3 sin z + c4 cos z ) and

d2 x dz 2

= e z (c1 cos z + c2 sin z ) + 2 e z (− c1 sin z + c2 cos z ) − e z (c1 cos z + c2 sin z ) + e − z (c3 cos z + c4 sin z ) − 2 e − z (− c3 sin z + c4 cos z ) − e − z (c3 cos z + c4 sin z ) = 2 e z (− c1 sin z + c2 cos z ) − 2 e − z (− c3 sin z + c4 cos z ) .

From (1), we have y=− ∴

1 2 1 d2 x D x=− ⋅ 2 2 dz 2

y = e z (c1 sin z − c2 cos z ) + e − z (− c3 sin z + c4 cos z ) .

…(4)

Now changing the variable z into t in the equations (3) and (4) by the relation t = e z or z = log t, we get x = t (c1 cos log t + c2 sin log t) + t −1 (c3 cos log t + c4 sin log t) and

y = t (c1 sin log t − c2 cos log t) + t −1 (− c3 sin log t + c4 cos log t) .

These equations give the required general solution of the given equations. dy dx dz Example 8: Solve = ny − mz , = lz − nx , = mx − ly. dt dt dt (Kanpur 2012) Solution: Multiplying the given equations by x, y, z respectively and then adding, we

get dy dx dz + y +z =0. dt dt dt Integrating, we get 1 2 1 2 1 2 x + y + z =k 2 2 2 x

or

x2 + y2 + z 2 = c1.

…(1)

Now multiplying the given equations by l, m, n respectively and then adding, we get dy dx dz l +m +n =0 . dt dt dt Integrating, we get lx + my + nz = c2 . …(2) The equations (1) and (2) constitute the complete solution of the given differential equations.

D-139 Example 9: Solve the simultaneous equations

3 given that x =

dy dx dx dy +2 − 4 x + 3 y = 8 e −3 t , 4 + + 3 x + 4 y = 8 e −3 t dt dt dt dt

1 , y = 0 , when t = 0 . 5

Solution: Writing D for (d / dt), the given equations can be written as

(3 D − 4) x + (2 D + 3) y = 8 e −3 t and

(4 D + 3) x + ( D + 4) y = 8 e

…(1)

−3 t

…(2)

Eliminating y from (1) and (2), we get [( D + 4)(3 D − 4) − (2 D + 3) (4 D + 3)] x = 8 [( D + 4) e −3 t − (2 D + 3) e −3 t ] or

− 5 ( D2 + 2 D + 5) x = 8 (− 3 e −3 t + 4 e −3 t + 6 e −3 t − 3 e −3 t )

or

( D2 + 2 D + 5) x = −

32 −3 t e . 5

A.E. is m2 + 2 m + 5 = 0 , giving m = − 1 ± 2 i . ∴

C.F. = e −1 (c1 cos 2 t + c2 sin 2 t) .

Also

P.I. =

∴

1  − 32 e −3 t  = − 4 e −3 t .    5 D2 + 2 D + 5  5

x = e − t (c1 cos 2 t + c2 sin 2 t) −

4 −3 t e . 5

Given that x = 1/ 5, when t = 0, we have 1 4 or c1 = 1 . = c1 − , 5 5 4 ∴ x = e − t (cos 2 t + c2 sin 2 t) − e −3 t 5 dx 12 −3 t −t ⇒ = − e (cos 2 t + c2 sin 2 t) + e − t (− 2 sin 2 t + 2 c2 cos 2 t) + e . dt 5 Now multiplying (2) by 2 and then subtracting (1) from it, we get dx 5y = −5 − 10 x + 8 e −3 t . dt Putting the values of x and dx / dt, we get 4 y = e − t (− cos 2 t − 2 c2 cos 2 t + 2 sin 2 t − c2 sin 2 t) + e − 3 t . 5 Given that y = 0, when t = 0, we have 4 1 or c2 = − 0 = − 1 − 2 c2 + , ⋅ 5 10 1 −t 4 …(3) ∴ x= e (10 cos 2 t − sin 2 t) − e −3 t 10 5 1 −t 4 and …(4) y= e (21 sin 2 t − 8 sin 2 t) + e −3 t . 10 5 The required solution consists of equations (3) and (4) .

D-140

Comprehensive Exercise 1

1. 2. 3. 4. 5. 6. 7. 8. 9.

Solve the following simultaneous differential equations : dy dx + 5 x + y = et , − x + 3 y = e2 t . dt dt (Meerut 2004B) dy dx dx dy t 2t +2 − 2 x + 2 y = 3e , 3 + + 2 x + y = 4e . dt dt dt dt (Meerut 2007) dy dx 2t + 2 x − 3 y = t, − 3x + 2 y = e . dt dt (Lucknow 2011)  d + 2 x + 3 y = 0 , 3 x +  d + 2 y = 2 e3 t .      dt   dt  dy dx = 3x + 2 y , + 5x + 3 y = 0 . dt dt (Gorakhpur 2009) dy dx = ax + by , = a ′ x + b ′ y. dt dt dy dx + ωy = 0 , − ωx = 0 . dt dt (Kanpur 2003, 05) dy dx t + 4 x + 3 y = t, + 2x + 5 y = e . dt dt dx dy dx dy + − 2 y = 2 cos t − 7 sin t, − + 2 x = 4 cos t − 3 sin t . dt dt dt dt (Lucknow 2005, 09; Avadh 10)

10. ( D − 17) y + (2 D − 8) z = 0 , (13 D − 53) y − 2 z = 0 . dy dy dx dx 11. 4 +9 + 11x + 31 y = e t , 3 +7 + 8 x + 24 y = e2 t . dt dt dt dt dy dy dx 12. +2 + x + 7 y = et − 3 , − 2 x + 3 y = 12 − 3 e t . dt dt dt 13.

d2 x dt2

+ 4 x + y = te t ,

d2 y dt2

+ y − 2 x = sin2 t.

d2 x

dy dy dx − = 2 x + 2t , +4 =3y. dt dt dt dt2 dy 1 dx 2 15. + ( x − y) = 1, + ( x + 5 y) = t . dt t dt t 14.

A nswers 1 1. x = (c1 + c2 t) e −4 t + 2. x =

4 t 1 2t 7 2t 1 t e − e , y = − (c1 + c2 + c2 t) e −4 t + e + e 25 36 36 25

1 2t 3 t 15 t e − e + c1 e −6 t /5 , y = e − 8 c1 e −6 t /5 + c2 e − t 2 11 22

3. x = c1 e −5 t + c2 e t +

3 2t 2 13 4 3 12 e − t− , y = − c1e −5 t + c2 e t + e2 t − t − 7 5 25 7 5 25

D-141

4. x = c1 e t + c2 e −5 t −

3 3t 5 e , y = − c1e t + c2 e −5 t + e3 t 8 8

5. x = c1 cos t + c2 sin t, y = 6. x = c1e m1t + c2 e m2 t , y = m1 =

1 1 (c2 − 3 c1) cos t − (c1 + 3 c2 ) sin t 2 2

1 (m1 − a) c1 e m1t + (m2 − a) c2 e m2 t , where b

{

}

(a + b ′ ) + √ [(a − b ′ )2 + 4 a ′ b] (a + b ′ ) + √ [(a − b ′ )2 + 4 a ′ b] and m2 = 2 2

7. x = c1 cos ωt + c2 sin ωt, y = c1 sin ωt − c2 cos ωt 5 1 31 8. x = c1e −2 t + c2 e −7t + t − et − , 14 8 196 y=

1 3

 − 3 t − 2 c e −2 t + 3 c e −7t + 5 e t + 27    1 2 8 98   7

9. x = c1e √2 t + c2 e −√2 t + 3 cos t, y = (√ 2 + 1) c1e √2 t + (1 − √ 2) c2 e −√2 t + 2 sin t 10. y = c1e3 x + c2 e5 x , z = − 7 c1e3 x + 6 c2 e5 x 31 t 49 2 t 19 2 t 11 t e − e , y = − (c2 + c1 + c2 t) e −4 t + e − e . 25 36 36 25 31 t 93 12. x = e −4 t (c1 cos t + c2 sin t) + e − ⋅ 26 17 2 t 6 y = − (c1 + c2 ) e −4 t cos t + (c1 − c2 ) e −4 t sin t − e + ⋅ 13 17 1 1 1 1 13. x = c1 cos (√ 3 t + c2 ) + c3 cos (√ 2 t + c4 ) + e t  t −  − + cos 2 t, 6  6  12 4 1 7 1 y = − c1 cos (√ 3 t + c2 ) − 2 c3 cos (√ 2 t + c4 ) + e t  t −  + ⋅ 6  6 3 1 1 14. x = (c1 + c2 t) e t + c3 e −3 /2 t − t, y = (3 c2 − c1 − c2 t) e t − c3 e −(3 /2)t − ⋅ 6 3 11. y = (c1 + c2 t) e −4 t +

15. x =

c1 t3

+

c2 t4

+

t2 3t c c 2 t2 1 + , y = − 13 − 42 + − t⋅ 15 10 15 20 2t t

6.4 Simultaneous Equations of the Form P1 dx + Q1 dy + R1 dz = 0 P2 dx + Q2 dy + R2 dz = 0 ,

…(1) …(2)

where the coefficients P1, P2 , Q1, Q2, R1 and R2 are functions of x, y, z . Above is the general form of a system of simultaneous equations of the first order in three variables.

D-142

Taking z as the independent variable, the equations (1) and (2) can be written in the form dy dx P1 + Q1 + R1 = 0 dz dz dy dx P2 + Q2 + R2 = 0 . dz dz Solving these equations for dx / dz and dy / dz by the method of cross- multiplication, we get dx / dz dy / dz 1 = = Q1 R2 − Q2 R1 R1 P2 − R2 P1 P1Q2 − P2 Q1 or

dy dx dz , = = Q1 R2 − Q2 P1 R1 P2 − R2 P1 P1Q2 − P2 Q1

which is of the form dy dz dx = = , P Q R

…(3)

where P, Q, R are functions of x, y and z. Thus the equations (1) and (2) can always be expressed in the form (3). To solve the simultaneous equations in the form (3) we can use the following methods. First Method: Suppose one variable is absent from two members of (3), say, z is absent from P and Q. Then the solution of dx dy = P Q gives a relation between x and y and we thus get one equation constituting the complete solution of the given equations. If necessary, we can use this relation to eliminate x or y from one of the remaining equations in (3) and we can find the other integral relation which gives us the second equation of the complete solution. Second Method: By a property of ratio and proportion in algebra, we know that each dx dy dz of the equal fractions in (3), is also equal to , , P Q R l dx + m dy + n dz L dx + M dy + N dz , etc. , lP + mQ + nR LP + MQ + NR ∴

dx dy dz l dx + m dy + n dz L dx + M dy + N dz = = = = ⋅ P Q R lP + mQ + nR LP + MQ + NR

…(4)

The relations between x, y, z that satisfy (3) also satisfy (4) . By a proper choice of multipliers l, m, n ; L, M , N etc., it is possible to find equations which can be easily solved. Particularly l, m, n can be so chosen that lP + mQ + nR = 0. which makes l dx + m dy + n dz = 0 . Let l dx + m dy + n dz be an exact differential, say du.

…(5)

D-143

Then (5) ⇒ du = 0 ⇒ u = a , which gives one equation constituting the complete solution. Again L, M , N may be found such that LP + MQ + NR = 0

⇒

L dx + M dy + N dz = 0 .

If L dx + M dy + N dz be an exact differential, say dv , then dv = 0 ⇒ v = b, which gives the second equation constituting the complete solution. Hence the complete solution consists of the equations u = a and v = b which must of course be independent of each other. Note 1: Sometimes only one set of multipliers may serve the purpose. Note 2: We can find one relation, say u = a, by the first method and the second relation,

say v = b, by the second method.

6.5 Geometrical Interpretation of the Differential Equations dx = dy = dz ⋅ P

Q

R

(Gorakhpur 2005)

From solid geometry, we know that the direction cosines of the tangent to a curve in three dimensional Euclidean space at any point ( x, y, z ) are dx dy dz , , ds ds ds i. e., the direction cosines are in the ratio of dx : dy : dz . Hence the given differential equations represent a system of curves in space, the direction cosines of the tangent to any member of this system at any point ( x, y, z ) are proportional to P, Q and R. Let u = a and v = b be two component solutions of the given equations. Then the curves are obtained by the interesection of the surfaces u = a and v = b. Since the arbitrary constants a and b can take infinite values, the curves are doubly infinite in number.

x dx

dy

dz

Example 10: Solve 2 = = ⋅ y+z y−z z − 2 yz − y2 (Meerut 2001, 04; Avadh 08)

Solution: Taking the last two members of the given equations, we get

( y − z ) dy = ( y + z ) dz or

or

y dy − z dz − ( y dz + z dy) = 0

2 y dy − 2 z dz − 2 ( y dz + z dy) = 0 .

Integrating, we get y2 − z 2 − 2 yz = c1 .

…(1)

D-144

Again choosing 1, y, z as multipliers, we get x dx dy dz = = y+z y−z z 2 − 2 yz − y2 = ∴

x dx + y dy + z dz 2

2

1 . (z − 2 yz − y ) + y ( y + z ) + z ( y − z )

=

x dx + y dy + z dz ⋅ 0

x dx + y dy + z dz = 0 .

Integrating, we get x2 + y2 + z 2 = c2 .

…(2)

The required solution consists of the equations (1) and (2). Example 11: Solve

dx 2

2

2

y +z −x

=

dy dz = ⋅ − 2 xy − 2 xz

Solution: Taking the last two members of the given equations, we get

dy dz = ⋅ y z Integrating, we get log y = log z + log c1 or

log y = log (c1 z ) or y = c1z .

…(1)

Again choosing x, y, z as multipliers, we get dy x dx + y dy + z dz dx dz = = = ⋅ y2 + z 2 − x2 − 2 xy − 2 xz − x ( x2 + y2 + z 2 ) Taking the last two fractions, we get dz 2 x dx + 2 y dy + 2 z dz = ⋅ z x2 + y2 + z 2 Integrating, log z + log c2 = log ( x2 + y2 + z 2 ) or

x2 + y2 + z 2 = zc2 .

…(2)

The required solution consists of the equations (1) and (2) . dy dx dz Example 12: Solve = = ⋅ 1 − 2 3 x2 sin ( y + 2 x) (Meerut 2001, 10B; Gorakhpur 07) Solution: From the first two members, we get

2 x + y = c1.

…(1)

Taking the first and the last members, we get dx dz dz = = , using (1). 1 3 x2 sin ( y + 2 x) dx2 sin c1 ∴

dz = 3 x2 sin c1 dx .

Integrating, z = x3 sin c1 + c2 or

z − x3 sin ( y + 2 x) = c2 .

The required solution consists of the equations (1) and (2).

…(2)

D-145 Example 13: Solve

dx 2

x(y − z )

Solution: Choosing

∴

2

=

dy 2

2

− y (z + x )

=

dz 2

z ( x + y2 )

⋅ (Meerut 2005, 11)

1 1 1 , − , − as multipliers, we have each fraction x y z

dx dy dz − − x y z = ⋅ 0 dx dy dz − − =0 x y z

⇒

dx dy dz = + ⋅ x y z

Integrating, log x + log c1 = log y + log z or

yz = c1 x.

…(1)

Again choosing x, y, z as multipliers, we get x dx + y dy + z dz each fraction = ⋅ 0 ∴

x dx + y dy + z dz = 0 .

Integrating,

x2 + y2 + z 2 = c2 .

…(2)

The equations (1) and (2) give the required solution of the given equations. Example 14: Solve

dy dx dz = = ⋅ cos ( x + y) sin ( x + y) z (Meerut 2006, 07B; Gorakhpur 08, 11)

Solution: From the given equations, we have

dx + dy dx − dy dz = = ⋅ cos ( x + y) + sin ( x + y) cos ( x + y) − sin ( x + y) z Taking the first two members, we get cos ( x + y) − sin ( x + y) (dx + dy) = dx − dy . cos ( x + y) + sin ( x + y) Integrating, log [cos ( x + y) + sin ( x + y)] = x − y + log c1 or

[cos ( x + y) + sin ( x + y)] e y− x = c1.

Again taking the first and the third members, we get dx + dy dz = cos ( x + y) + sin ( x + y) z (1/ √ 2)(dx + dy) dz = 1 z sin ( x + y + π) 4 dz 1 or √2 = cosec ( x + y + π)(dx + dy) . z 4 1 1 Integrating, √ 2 log z = log tan ( x + y + π) + log c2 2 4 or

…(1)

D-146

or

y π x z √2 cot  + +  = c2 . 2 2 8

…(2)

The required solution consists of the equations (1) and (2). Example 15: Solve

dx dy dz = = ⋅ x y z − a √ ( x2 + y2 + z 2 )

(Meerut 2005B, 10B)

Solution: From the first two members of the given differential equations, we have

…(1)

x = c1 y.

Again choosing x, y, z as multipliers, we have x dx + y dy + z dz dx dy dz = = = 2 2 2 2 2 x y z − a √ ( x + y + z ) x + y + z 2 − az √ ( x2 + y2 + z 2 ) or

dy u du dz , putting x2 + y2 + z 2 = u2 = = y z − au u2 − azu

or

dy du + dz dz du = = = ⋅ y z − au u − az (u + z ) (1 − a)

Taking the first and the last members and integrating, we get (1 − a) log y = log (u + z ) + log c2 or

y(1− a) = (u + z ) c2

or

y(1− a) = c2 { √ ( x2 + y2 + z 2 ) + z } .

…(2)

The required solution consists of the equations (1) and (2). dy dx dz Example 16. Solve 3 = = ⋅ y x − 2 x4 2 y4 − x3 y 9 z ( x3 − y3 ) Solution: From the first two members of the given differential equations, we get

(2 y4 − x3 y) dx = ( y3 x − 2 x4 ) dy or

2 y  1 1 2x   3 − 2  dx =  2 − 3  dy , dividing by x3 y3 y  y  x x

or

2y   1 2x   1  2 dy − 3 dx +  2 dx − 3 dy = 0 . x  x y  y 

Integrating, Again taking

y 2

x

+

x y2

= c1 .

1 1 1 as multipliers, we get , , x y 3z 1 1 1 dx + dy + dz x y 3z

1 1 1 dx + dy + dz x y 3z each fraction = 3 = ⋅ 0 ( y − 2 x3 ) + (2 y3 − x3 ) + 3 ( x3 − y3 ) ∴

1 1 1 dx + dy + dz = 0 . x y 3z

…(1)

D-147

Integrating, log x + log y +

1 log z = log c2 3

xyz1 /3 = c2 .

or

…(2)

The required solution consists of the equations (1) and (2). Example 17: Solve

dy dx dz = 2 = ⋅ xy y zxy − 2 x2

(Purvanchal 2006)

Solution: From the first two members of the given differential equations, we get

dx dy = ⋅ x y Integrating,

log x = log y + log c1

or

x = c1 y.

…(1)

Again taking the last two members, we get dy dy dz dz or = = y2 xyz − 2 x2 y2 c1 y2 z − 2 c12 y2 or

dy =

dz

or

c1z − 2 c12

[∵ x = c1 y]

dz ⋅ z − 2 c1

c1 dy =

Integrating, c1 y = log (z − 2 c1) + c2 or

 x x = log  z − 2  + c2 y 

or

x = log ( yz − 2 x) − log y + c2 .

[∵ x = c1 y] …(2)

The required solution consists of the equations (1) and (2). dx

dy

dz

Example 18: Solve 2 = = ⋅ x + y2 2 xy ( x + y) z Solution: From the given equations, we have

dx + dy x2 + y2 + 2 xy

=

dx − dy x2 + y2 − 2 xy

=

dz ⋅ ( x + y) z

Taking the first two members, we have dx + dy dx − dy = ⋅ 2 ( x + y) ( x − y)2 Integrating, − ( x + y)−1 = − ( x − y)−1 + c1 or

1 1 − = c1, x− y x+ y

or

2 y = c1 ( x2 − y2 )

or

2y x2 − y2

= c1

Taking the first and the last members, we have dx + dy dx + dy dz dz or = = ⋅ 2 ( x + y ) z x+ y z ( x + y)

…(1)

D-148

Integrating, log ( x + y) = log z + log c2 or

…(2)

( x + y) = c2 z .

The required solution consists of the equations (1) and (2). dy dx dz Example 19: Solve 2 = 2 = ⋅ nxy x y dx

dy

Solution: We have 2 = 2 . x y

Integrating, or

1 1 = + c1 x y y − x = c1 xy.

…(1)

Also using 1/ x, − 1/ y and c1 / n as multipliers, each fraction (1/ x) dx − (1/ y) dy + (c1 / n) dz = x − y + c1 xy =

(1/ x) dx − (1/ y) dy + (c1 / n) dz , using (1). 0

(1/ x) dx − (1/ y) dy + (c1 / n) dz = 0 .

∴

Integrating, log x − log y + (c1 / n) z = c2 nxy c1  y  or z = log   + c2 or z = log   x  n y−x

y  + c2 . x

…(2)

The required solution consists of the equations (1) and (2).

Comprehensive Exercise 2 Solve the following simultaneous differential equations : dx dy dz 1. = = ⋅ yz zx xy 2.

dy dx dz . = = mz − ny nx − lz ly − mx

Also give geometrical interpretation of the solution of these equations. dy dx dz 3. = = ⋅ z − z z 2 + ( x + y)2 4. 5. 6.

a dx b dy c dz = = ⋅ (b − c ) yz (c − a) zx (a − b) xy dx xz (z 2 + xy)

=

dy − yz (z 2 + xy)

=

dz x4

dy dx dz = = 2 ⋅ z ( x + y) z ( x − y) x + y2

⋅

(Lucknow 2006)

D-149

7.

− dx dy dz = = ⋅ x ( x + y) y ( x + y) ( x − y) (2 x + 2 y + z )

8.

dx dy dz = = ⋅ 1 3 5 z + tan ( y − 3 x)

9. 10.

11. 12.

(Meerut 2006B; Gorakhpur 08, 11)

dy dx dz = = ⋅ x ( y − z ) y (z − x) z ( x − y) dx x ( y2 − z 2 )

=

dy y (z 2 − x2 )

=

dz z ( x2 − y2 )

⋅ (Gorakhpur 2005, 09)

dy dx dz = = ⋅ 1+ y 1+ x z dx y2 + yz + z 2

=

dy z 2 + zx + x2

=

dz x2 + xy + y2

⋅ (Avadh 2007)

  dy − dx dz − dy dz − dx = = Hint.  ( x − y) ( x + y + z ) ( y − z ) ( x + y + z ) ( x − z ) ( x + y + z )  dy dx dz 13. 2 = 2 = 2 ⋅ x − yz y − zx z − xy (Gorakhpur 2006)   dx − dy dy − dz dz − dx = = Hint. ⋅ ( x − y ) ( x + y + z ) ( y − z ) ( x + y + z ) ( z − x ) ( x + y + z )   14.

dx x2 + y2 + yz

=

dy x2 + y2 − xz

=

dz ⋅ z ( x + y)

(Kanpur 2009)

 dx − dy x dx + y dy  dz = =  Hint. z ( x + y) z ( x + y) ( x + y) ( x2 + y2 )  dy dx dz 15. = = ⋅ y+z z+x x+ y  dx − dy dy − dz dx + dy + dz  = = Hint. . y−x z−x 2 ( x + y + z ) 

A nswers 2 1. x2 − y2 = c1, x2 − z 2 = c2 2. lx + my + nz = c1, x2 + y2 + z 2 = c2 These equations constituting the general solution of the given differential equations represent a family of circles. 3. x + y = c1, log { z 2 + ( x + y)2 } − 2 x = c2

D-150

4. a2 x2 + b2 y2 + c 2 z 2 = c1, ax2 + by2 + cz 2 = c2 5. xy = c1, (z 2 + xy)2 − x4 = c2 6. x2 − y2 − z 2 = c1, 2 xy − z 2 = c2 7. xy = c1, x2 + y2 + ( x + y) z = c2 8.

y − 3 x = c1, 5 z + tan ( y − 3 x) = c2 e5 x

9. x + y + z = c1, xyz = c2 10. x2 + y2 + z 2 = c1, xyz = c2 11. z ( x − y) = c1, z = c2 ( x + y + 2) 12. ( y − x) = c1 (z − x), ( y − x) = c2 (z − y) 13. ( x − y) = c1 ( y − z ), ( y − z ) = c2 (z − x) 14. x − y − z = c1, x2 + y2 = z 2 c2 15. ( y − x) = c1 (z − y), ( x − y)2 ( x + y + z ) = c2

O bjective T ype Q uestions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

dy dx + 5 x + y = et , − x + 3 y = e2 t , for x, we get x = dt dt 4 t 1 2t (a) (c1 + c2 t) e −4 t + e − e 25 36 2 (b) c1 e −4 t + c2 e t − 5

Solving the equations

(c) c1e t + c2 e −4 t + c3 e −2 t (d) None of these. 2.

3.

Solving the equations

dy dx dy + + 2x + y = 0 , + 5 x + 3 y = 0 , we get x = dt dt dt

(a) c1 e t + c2 e − t

(b) c1 cos t + c2 sin t

(c) (c1 + c2 t) cos t

(d) c1 e t + c2 e − t .

Solving the equations

dy dx + ωy = 0 , − ωx = 0 for x, we get x = dt dt

(a) c1e ωt + c2 e − ωt

(b) (c1 + c2 t) e ωt

(c) c1 cos ωt + c2 sin ωt

(d) None of these.

D-151

4.

dy dx dz is = = y+z z+x x+ y

Solution set of the equations (a) ( y − x) = c1 (z − y)

(b) x + y = c1 ( y + z )

2

( x − y)2 ( x + y + z ) = c2 .

( x − y) ( x + y + z ) = c2 (c) ( y − z ) = c1 ( x − y)

(d) None of these.

2

( x + y) ( x − y + z ) = c2 5.

Solution set of the equations

dx dy dz is = = x y z

(a) xy = c1, yz = c2

(b) x = c1 z , y = c2 z

(c) x = c1 y, y = c2 z

6.

7.

(d) None of these. dx dy dz Solution set of the equations is = = yz zx xy (a) x2 + y2 = c1, y2 − z 2 = c2

(b) x2 − z 2 = c1, y2 + z 2 = c2

(c) x2 − y2 = c1, x2 − z 2 = c2

(d) None of these.

Solution set of the simultaneous equations

dx dy dz is = = y x z

(a) x2 − y2 = c1, x + y = c2 z

(b) x2 + y2 = c1, x + y = c2 z

(c) x2 − y2 = c1, x − y = c2 z

(d) x2 − y2 = c1, y2 − z 2 = c2 .

Fill in the Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

Eliminating y between the equations dy dx − 7x + y = 0 , − 2x − 5 y = 0 dt dt we obtain the differential equation……

2.

Eliminating y between the simultaneous equations d2 x dt2

+ m2 y = 0 ,

d2 y dt2

− m2 x = 0

we obtain the differential equation …… 3.

The complete solution of the simultaneous equations dy dx dz = ny − mz , = lz − nx, = mx − ly is …… dt dt dt

(Meerut 2003)

D-152

A nswers Multiple Choice Questions 1. (a)

2.

(b)

3. (c)

4.

(a)

5. (b)

6.

(c)

7. (a)

Fill in the Blank(s) 1. ( D2 − 12 D + 37) x = 0 2. ( D4 + m4 ) x = 0 3. x2 + y2 + z 2 = c1, lx + my + nz = c2

¨

D-153

7 L inear E quations of S econd O rder W ith V ariable C oefficients

7.1 Linear Equation of Second Order

A

n equation of the form d2 y 2

+P

dy + Qy = R , dx

dx where P, Q, R are functions of x only, is called the ‘linear equation of second order’. There is no general method for solving such equations. Here we shall discuss certain methods by which the solutions of such equations can be found.

7.2 The Complete Solution in Terms of a Known Integral The following theorem relating to the linear differential equation of the second order is of great importance : If an integral included in the complementary function of such an equation be known, then the complete primitive (or the general solution) can be found in terms of the known integral. Let y = u be a known integral in the complementary function of

D-154

d2 y 2

+P

dy + Qy = R dx

+P

dy + Qy = 0 . dx

+P

du + Qu = 0 . dx

dx i. e., it is a solution of d2 y 2

dx ∴

d2 u 2

…(1)

…(2)

dx Let y = uv be the solution of (1). dy du dv Putting y = uv, we get =v +u dx dx dx d2 y

du dv d2 v +u 2 ⋅ dx dx dx dx dx Substituting these values the equation (1) becomes  d2 u du dv d2 v  dv  du  v 2 + 2 + u  + Q vu = R + u 2  + P  v  dx dx dx dx  dx   dx

and

2

=v

d2 v

d2 u 2

+2

+

  d2 u dv  du du  + Pu + v  2 + P + Qu = R 2  dx  dx dx   dx

+

dv  du  + Pu + v. 0 = R , using (2) 2  dx  dx

or

u

or

u

or

2 du dv R  = ⋅ + P +   u dx  dx u dx

Putting

2

dx

d2 v 2

dx d2 v

…(3)

2

dv d2 v dp , (3) becomes = p, 2 = dx dx dx dp  2 du R + P +  p= ,   dx u dx u

…(4)

which is linear with p as dependent variable. I. F. =

2 du   dx ∫  P + u dx  e 2 log u + ∫ P dx

=

2   ∫  P dx + du u e

= u2 e ∫ P dx

=e

Hence solution of (4) is pu2 e ∫ P dx =

R

2

∫  u u

 e ∫ P dx  dx + c1. 

…(5)

dv c1e − ∫ P dx e − ∫ P dx = + u Re ∫ P dx dx. dx u2 u2 Integrating this, we get  e − ∫ P dx  e − ∫ P dx y = c2 + c1 dx + u Re ∫ P dx dx dx.  2 2 u  u  Hence the solution of (1) is  e − ∫ P dx  e − ∫ P dx y = uv = c2 u + c1 u dx + u  ⋅ u Re ∫ P dx dx dx. …(6) 2 2 u  u 

∴

∫

p=

∫

∫

∫

∫

∫

∫

D-155

It contains the given solution y = u and since it contains two arbitrary constants so it is the complete primitive or the general solution of (1). It is evident from (6) that the second part of the complementary function is u u

∫ ∫

e − ∫ P dx u2

dx and the particular integral is

 e − ∫ P dx ⋅  2  u

∫ u Re ∫

P dx

 dx dx. 

7.3 To Find one Integral In C.F. by Inspection d2 y

Consider 1.

2

dx

+P

dy + Q y = 0. dx

y = e mx is a solution of (1) if m2 + Pm + Q = 0 .

We have y = e mx ⇒ If

dy d2 y = me mx , 2 = m2 e mx . dx dx

y = e mx is a solution of (1), then (m2 + Pm + Q) e mx = 0 ⇒ m2 + Pm + Q = 0 .

Deductions. (i) y = e x is a solution of (1) if 1 + P + Q = 0 . (ii)

y = e − x is a solution of (1) if 1 − P + Q = 0 .

2.

y = x m is a solution of (1) if m (m − 1) + Pmx + Qx2 = 0 .

We have y = x m ⇒ If

dy d2 y = mx m −1, 2 = m (m − 1) x m −2 . dx dx

y = x m is a solution of (1), then m (m − 1) x m −2 + Pmx m −1 + Qx m = 0

or m (m − 1) + Pmx + Qx2 = 0 . Deductions. (i) y = x is a solution of (1) if P + Qx = 0. (ii) y = x2 is a solution of (1) if 2 + 2 Px + Qx2 = 0 .

All the above results are summarised below : (i)

y = e x is a part of C.F. if 1 + P + Q = 0 .

(ii)

y = e − x is a part of C.F. if 1 − P + Q = 0 .

(iii) y = e mx is a part of C.F. if m2 + m P + Q = 0 . (iv) y = x is a part of C.F. if P + Qx = 0 . (v)

y = x2 is a part of C.F. if 2 + 2 P x + Qx2 = 0 .

(vi) y = x m is a part of C.F. if m (m − 1) + Pmx + Qx2 = 0 .

…(1)

D-156

Example 1: Solve ( x + 2)

d2 y dx2

− (2 x + 5)

dy + 2 y = ( x + 1) e x . dx

(Gorakhpur 2009; Lucknow 07; Rohilkhand 09; Kanpur 07, 09)

Solution: The given equation can be written as

d2 y 2

dx

(2 x + 5) dy ( x + 1) x 2 + y= e . x + 2 dx x + 2 x +2

−

…(1)

Comparing (1) with the standard form d2 y dx2

+P

P=−

dy + Qy = R, we have dx

2x + 5 x +1 x 2 ,Q = ,R= e . x +2 x +2 x +2

Here 22 + 2 P + Q = 0 , ∴ Putting and

y = e2 x is a part of the C.F. of the solution of the equation (1). y = ve2 x , d2 y 2

dx

dy dv = e2 x + 2 v e2 x , dx dx = e2 x

d2 v 2

+ 4 e2 x

dv + 4 v e2 x in (1), we get dx

dx 2 x + 3 dv x + 1 − x + = e x + 2 dx x + 2 dx2 dp 2 x + 3 x + 1 −x + p= e , dx x +2 x +2 d2 v

or

dv   where p =   dx 

which is a linear equation with p as dependent variable. I. F. = e ∴

p⋅ = =

∫

2 x +3 x +2

e2 x = x +2

∫

x +1

dx

=e

 1   dx ∫ 2 − x + 2 

= e2 x − log ( x +2) =

x + 1 − x e2 x e ⋅ dx + c1 x +2 x +2  1

x



1

∫ ( x + 2)2 e dx + c1 = ∫  x + 2 − ( x + 2)2  e 1 ex + x +2

e2 x x +2

∫e

x

⋅

1 ( x + 2)2

dx −

∫e

x

1 ( x + 2)2

x

dx + c1

dx + c1 ,

integrating the first integral by parts. dv −x −2 x ∴ p= = e + c1 e ( x + 2) . dx Integrating this, we get 1 1 v = − e − x − c1e −2 x ( x + 2) − c1 e −2 x + c2 2 4

D-157

1 c1 (2 x + 5) e −2 x + c2 . 4 The general solution of (1) is 1 y = v e2 x = − e x − c1 (2 x + 5) + c2 e2 x . 4 = − e− x −

∴

Example 2: Solve x

d2 y

dy + ( x − 1) y = 0 . dx

− (2 x − 1)

dx2

(Meerut 2010; Gorakhpur 08)

Solution: The given equation can be written in the standard form as

d2 y

1 dy  1  − 2 −  + 1 −  y = 0 .  x  dx  x

2

dx ∴

Here

1 + P + Q = 0, y = e x is a part of the C.F. of the solution of (1).

Putting

y = v e x, d2 y

and

=

2

dx

…(1)

dy dv x = e + v e x, dx dx

d2 v 2

dx

ex + 2

dv x e + v ex dx

in (1), we get d2 v

+

2

dx dp dx =− ⇒ log p = − log x + log c1 p x

⇒

p=

⇒ ∴

dp p 1 dv dv =0 ⇒ + = 0 , where p = x dx dx x dx

dv c1 = ⇒ v = c1 log x + c2 . dx x

The complete solution of (1) is y = v e x = (c1 log x + c2 ) e x .

d2 y dy Example 3: Solve x2 − ( x2 + 2 x) + ( x + 2) y = x3 e x . 2 dx dx (Meerut 2010B; Bundelkhand 01; Gorakhpur 06; Purvanchal 07)

Solution: The standard form of the given equation is

d2 y

2  dy  1 2   x − 1 +  + +  y = xe .  x  dx  x x2  2  1 2   Here P + Qx = − 1 +  +  + 2  x = 0 ,  x  x x  dx2

∴

…(1)

y = x is a part of the C.F. of the solution of (1).

Putting y = vx,

dy d2 y dv d2 v dv =x + v, = x 2 +2 2 dx dx dx dx dx

in (1), we get d2 v 2

dx

−

which is linear in p.

dv = ex dx

or

dp − p = e x, dx

dv   where p =   dx 

D-158

I. F. = e ∫ ∴

pe − x =

∴

p=

− dx

∫e

x

= e− x .

⋅ e − x dx + c1 = x + c1.

dv = x e x + c1 e x . dx

Integrating this, we get v = x e x − e x + c1e x + c2 . ∴

The complete solution of (1) is y = v x = x2 e x − x e x + c1 x e x + c2 x .

Example 4: Solve

d2 y 2

dx

− cot x

dy − (1 − cot x) y = e x sin x . dx

(Meerut 2007B)

Solution: The given differential equation is

d2 y

− cot x

dx2

dy − (1 − cot x) y = e x sin x . dx

Here 1 + P + Q = 0 , ∴ y = ex

is a part of the C.F. of (1).

Putting y = ve x , and

d2 y 2

dx

=

dy dv x = e + ve x , dx dx d2 v 2

dx

ex + 2

dv x e + ve x dx

in (1), we get d2 v 2

or

+ (2 − cot x)

dv = sin x dx

dx dp + (2 − cot x) p = sin x, dx

dv   where p =   dx 

which is linear in p. I. F. = e ∫ (2 − cot e2 x = sin x

∴

p

∴

p=

∫

x) dx

= e2 x − log sin x =

e2 x ⋅ sin x

e2 x 1 sin x dx + c1 = e2 x + c1 . sin x 2

dv 1 = sin x + c1 e −2 x sin x . dx 2

Integrating this, we get v=− ∴

1 c cos x + 1 e −2 x (− 2 sin x − cos x) + c2 . 2 5

The complete solution of (1) is y = v ex = −

1 x c e cos x − 1 e − x (2 sin x + cos x) + c2 e x . 2 5

D-159

d2 y dy Example 5: Solve x2 − 2 x (1 + x) + 2 (1 + x) y = x3 . dx dx2 (Rohilkhand 01; Gorakhpur 10; Lucknow 05, 06)

Solution: The given equation can be written in the standard form as

d2 y

 1 1  1  dy − 2  + 1 +2 2 +  y = x.  x  dx x x

2

dx

…(1)

Here P + Q x = 0 , y = x is a part of the C.F. of (1). dy dv y = v x, =x +v dx dx

∴ Putting and

d2 y 2

dx

= x

d2 v 2

dx

+2

dv dx

in (1), we have d2 v 2

dx

−2

dv =1 dx

or

dp − 2 p = 1, dx

dv   where p =   dx 

which is linear in p. I. F. = e ∫ ∴

− 2 dx

= e −2 x .

pe −2 x = 1 ⋅ e −2 x dx + c1 = −

∫

1 −2 x e + c1. 2

dv 1 = − + c1 e2 x . dx 2 1 c Integrating, we get v = − x + 1 e2 x + c2 . 2 2

∴

∴

p=

The complete solution of (1) is 1 c y = v x = − x2 + 1 x e2 x + c2 x. 2 2

Example 6: Solve sin2 x

d2 y dx2

= 2 y, given y = cot x is a solution. (Purvanchal 2010; Lucknow 08; Kanpur 07, 08)

dy dv Solution: Putting y = v cot x, = cot x − v cosec2 x, dx dx d2 y 2

dx

=

d2 v 2

dx

cot x − 2

dv cosec2 x + 2 v cosec2 x cot x, dx

the given differential equation becomes d2 v dv sin2 x cot x 2 − 2 =0 dx dx d2 v 2 dv or − =0 dx2 sin x cos x dx dp dv 2 or − p = 0 , where p = dx sin x cos x dx

D-160

or

dp 2 = dx = 4 cosec 2 x dx . p sin x cos x

Integrating, we get log p = 4 ⋅

dv = c1 tan2 x = c1 (sec2 x − 1) . dx

or

p=

∴

v = c1

∴

1 log tan x + log c1 2

∫ (sec

2

x − 1) dx + c2 = c1 (tan x − x) + c2 .

the complete solution of the given equation is y = v cot x = c1 (1 − x cot x) + c2 cot x .

Example 7: Solve ( x sin x + cos x)

d2 y 2

dx

− x cos x

dy + y cos x = sin x ( x sin x + cos x)2 . dx

Solution: The given equation in the standard form is

d2 y 2

dx

−

x cos x dy cos x + y = sin x ( x sin x + cos x) . x sin x + cos x dx x sin x + cos x …(1)

Here P + Qx = 0, ∴

y = x is a part of the C.F. of the solution of (1).

Putting y = vx and the corresponding values of

dy d2 y in (1), we get , dx dx2

d2 v

2  dv sin x ( x sin x + cos x) x cos x  = + − x dx  x x sin x + cos x  dx 2

or

 dp  2 x cos x sin x ( x sin x + cos x)  p= ⋅ + − dx  x x sin x + cos x  x

which is linear in p where p =

I. F. = e

dv ⋅ dx

2  x cos x  dx ∫ −  x x sin x + cos x 

= e2 log

x − log ( x sin x + cos x)

x2 = x sin x + cos x

=

x2 x sin x + cos x

∫ x sin x dx + c1 = − x cos x + sin x + c1.

∴

p⋅

∴

p=

or

dv 1 1 1 1  = − sin x cos x − cos 2 x + 2 sin x cos x + c1  sin x + 2 cos x x  dx x x x

dv 1 c = (− x cos x + sin x) ( x sin x + cos x) + 12 ( x sin x + cos x) dx x2 x

D-161

or

1 2

v=−

1

2

1

 sin x cos x  + 2  dx + c2 x x 

∫ 

+ c1

∴



∫ sin 2 x dx − 2 ∫  x cos 2 x − x2 sin 2 x dx d  sin 2 x   dx − c1 x 

=

1 1 cos 2 x − 4 2

=

cos x 1 1 sin 2 x cos 2 x − − c1 + c2 . 4 2 x x

∫ dx 

d  cos x   dx + c2 x 

∫ dx 

The complete solution of (1) is 1 1 y = vx = x cos 2 x − sin 2 x − c1 cos x + c2 x . 4 2   d2 y dy − y = ( x − 1)  2 − x + 1 . dx   dx

Example 8: Solve x

Solution: The given equation in the standard form is

d2 y

y x dy + = x − 1. x − 1 dx x − 1

−

2

dx

…(1)

Here P + Qx = 0 , ∴

y = x is a part of the C.F. of the solution of (1).

Putting y = vx and the corresponding values of d2 v

or

dx2 dp + dx

dy d2 y and 2 in (1), we get dx dx

x  dv x − 1 2 = + −   x x − 1 dx x x  x −1 2 ,  p=  −  x x − 1 x

which is linear in p. I. F. = e

2 x   dx ∫ −  x x − 1

= e2 log ∴

p⋅

∴ or

x − x − log ( x −1)

x2 e− x = ( x − 1)

=

∫

=e

∫

1  2  dx ∫  −1 − x x −1

=

x2 − x e . x −1

x − 1 x2 − x ⋅ e dx + c1 x x −1

xe − x dx + c1 = − xe − x − e − x + c1 .

x − 1 ( x − 1) c1 ( x − 1) x dv =− − + e dx x x2 x2 1 dv 1 1  = − 1 + 2 + c1  − 2  e x . x x  dx x

p=

dv   where p =   dx 

D-162

Integrating, we get v=− x− ∴

1 1 + c1 e x + c2 . x x

The complete solution of (1) is y = vx = − x2 − 1 + c1 e x + c2 x y = c1e x + c2 x − (1 + x2 ) .

or

Comprehensive Exercise 1 Solve the following differential equations : 1. ( x + 1)

d2 y

− 2 ( x + 3)

2

dx

2

2. x 3. x 4.

d y dx2 d2 y dx2

d2 y

− 2 ( x + 1)

dy + ( x + 2) y = ( x − 2) e x . dx

− (2 x + 1)

dy + ( x + 1) y = ( x2 + x − 1) e2 x . dx

+ (1 − cot x)

dx2

5. ( x − x2 )

d2 y 2

dx

dy − y cot x = sin2 x . dx

− (1 − 2 x)

2

6. (3 − x)

d y

− (9 − 4 x)

2

dx

2

7. x

d y 2

dx

8. (2 x − 1)

d y 2

dx

−2

2

d y 2

dx

+ (1 − x)

dy + (1 − 3 x + x2 ) y = (1 − x)3 . dx

dy + (6 − 3 x) y = 0 . dx

dy − 2 y = x3 . dx

+ ( x − 2) 2

9. x

dy + ( x + 5) y = e x . dx

(Kanpur 2011)

dy + (3 − 2 x) y = 2 e x . dx

dy − y = e x. dx

10. Solve x2 y2 + xy1 − 9 y = 0 , given that y = x3 is a solution. 11. Solve

d2 y 2

dx

2 2  + 1 + cot x − 2  y = x cos x ,  x x 

(Gorakhpur 2008, 11)

sin x given that is an integral included in the C.F. x 12.

Solve (1 − x2 )

d2 y 2

dx

−x

−1 dy − a2 y = 0 , given that y = ce a sin x is an integral. dx

D-163

A nswers 1 1. 2.

1 1 x e x + c1e x ( x + 1)5 + c2 e x 4 5 1 2 x 1 y = − x e + xe x + c1 x3 e x + c2 e x 2 3 y=−

3.

y = x e2 x + c1 x2 e x + c2 e x

4.

y=−

5.

1 (sin 2 x − 2 cos 2 x) + c1 (sin x − cos x) + c2 e − x 10 1 y = − x + c1 x2 e − x + c2 e x 2

6.

y = c1e3 x (4 x3 − 42 x2 + 150 x − 183) + c2 e x

7.

y = x3 + (c1 − 3) x2 − 2 (c1 − 3) ( x − 1) + c2 e − x

8.

y = − c1 xe − x − xe x + e x

9.

y = e x log x + c1 e x

10. y = c1 x 11. y =

−3

∫

∫ (2 x − 1) log (2 x − 1) dx − 2 e x [e −2 x (2 x − 1) e2 x log (2 x − 1) dx] dx + c2 e x ∫ ∫

e− x dx + c2 e x x

3

+ c2 x

x2 sin x 2 sin x  + c1 − x cos x + 2 sin x log sin x − 6 x 



∫ log sin x dx + c2

sin x x

−1 −1  c  12. y = − 1 e −2 a sin x + c2  ⋅ e a sin x  2a 

7.4 Removal of the First Derivative (Reduction to Normal Form) (Agra 2007; Gorakhpur 05)

If we are unable to obtain a part of the C.F. of the solution of the differential equation d2 y 2

dx

+P

dy + Qy = R, dx

…(1)

then we cannot solve (1) by the method given in article 7.2. In such cases the equation (1) can sometimes be solved by reducing it into the form in which the term containing the first derivative is absent. For this we first change the dependent variable from y to v in the equation (1) by putting y = uv, where u is some function of x.

D-164

Then

dy du dv d2 y =v +u , dx dx dx dx2 =v

d2 u dx2

Putting the values of y ,

+2

dv du d2 v ⋅ +u 2 ⋅ dx dx dx

dy d2 y in (1), we get , dx dx2

 d2 u dv du d2 v  dv  du + u  + Q vu = R ⋅ + u 2  + P v v 2 + 2 dx dx dx   dx dx   dx d2 v

  d2 u 2 du dv du  + + Qu = R. + u P + v  2 +P 2   u dx  dx dx  dx   dx dv To remove the term of the first derivative we choose u such that dx 2 du du 1 or P+ =0 = − P dx u dx u 2

or

u

or

log u = −

1 2

∫ P dx

or

u=e

− 1 ∫ P dx 2

…(2)

…(3)

Now the equation (2) becomes d2 v dx2 But from (3),

+

 v  d2 u du R +P + Qu = ⋅  u  dx2 dx u 

…(4)

du 1 d2 u 1  du dP  = − Pu, 2 = −  P +u dx 2 2  dx dx  dx dP    1   P  − 2 Pu + u dx    1 1 dP = P2 u − u ⋅ 4 2 dx

=−

1 2

Putting these values in (4), we get d2 v

1 dP 1 1  + v  P2 − − P ⋅ P + Q = 4 2 dx 2   dx

R

2

or or where

e

− 1 ∫ P dx 2

d2 v

1 P dx  P2 1 dP  ∫ + v Q − −  = R e2 dx 4 2 dx  

2

d2 v dx2

+ Xv = Y ,

X =Q−

1 P dx 1 dP 1 2 ∫ . − P and Y = Re2 2 dx 4

The reduced equation (5) may easily be integrated. The equation (5) is called the normal form of the equation (1). Note: The students should remember the values of u, X and Y . They can write the

reduced equation (5) directly.

D-165

Example 9: Solve

d2 y 2

dx

− 4x

2 dy + (4 x2 − 1) y = − 3 e x sin 2 x . dx

(Rohilkhand 2001; Meerut 05; Agra 07; Gorakhpur 09) 2

Solution: Here P = − 4 x, Q = 4 x2 − 1, R = − 3 e x sin 2 x .

We choose u = e

− 1 ∫ P dx 2

=e

− 1 ∫ −4 x dx 2

2

= ex .

Putting y = uv in the given equation, it reduces to its normal form d2 v dx2

…(1)

+ Xv = Y ,

where

X =Q−

and

Y = Re2

1 dP 1 2 1 1 − P = 4 x2 − 1 − (− 4) − ⋅ 16 x2 = 1 , 2 dx 4 2 4

1 P dx ∫

2

2

= −3 e x sin 2 x ⋅ e − x = −3 sin 2 x.

Hence the normal form (1) of the given equation is d2 v dx2

+ v = − 3 sin 2 x

( D2 + 1) v = − 3 sin 2 x.

or

…(2)

Now (2) is a linear differential equation with constant coefficients. A.E. is m2 + 1 = 0 ⇒ m2 = − 1 ⇒ m = ± i . ∴ C.F. of the solution of (2) = c1 cos x + c2 sin x . −3 1 P. I. = 2 (− 3 sin 2 x) = sin 2 x = sin 2 x. D +1 − 22 + 1 ∴

the solution of (2) is v = c1 cos x + c2 sin x + sin 2 x.

Hence the general solution of the given equation is 2

y = uv = e x (c1 cos x + c2 sin x + sin 2 x) . Example 10: Solve

d2 y dx2

− 2 tan x

dy + 5 y = sec x . e x . dx (Agra 2006; Gorakhpur 06; Kanpur 09) x

Solution: Here P = − 2 tan x, Q = 5, R = e sec x .

We choose u = e

− 1 ∫ P dx 2

= e ∫ tan x dx = e log

sec x

= sec x.

Substituting y = uv in the given equation, it reduces to its normal form d2 v

+ Xv = Y , dx2 1 dP 1 2 1 where X = Q − − P = 5 − (− 2 sec2 x) − tan2 x = 6 2 dx 4 2 and

1 P dx ∫

Y = Re2

= e x sec x (sec x)−1 = e x .

…(1)

D-166

Hence the equation (1) is d2 v dx2

+ 6v = e x

( D2 + 6) v = e x .

or

…(2)

A.E. is m2 + 6 = 0 ⇒ m = ± √ 6 i . ∴

C.F. of the solution of (2) = c1 cos √ 6 x + c2 sin √ 6 x. 1 1 P. I. = 2 e x = e x. 7 D +6

∴

the solution of (2) is v = c1 cos √ 6 x + c2 sin √ 6 x +

1 x e . 7

Hence the complete solution of the given equation is y = uv = sec x (c1 cos √ 6 x + c2 sin √ 6 x + Example 11: Solve

d2 y

+

dx2

1 dy  1 1 6 + − − 2 y = 0. x1 /3 dx  4 x2 /3 6 x4 /3 x  1

Solution: Here P = x −1 /3 , Q =

We choose u = e

− 1 ∫ P dx 2

1 x e ). 7

=e

2 /3

4x

− 1 ∫ x −1 /3 dx 2

−

1 4 /3

6x

=e

− 3 x2 /3 4

6

−

x2

,R=0.

.

Substituting y = uv in the given equation, it reduces to its normal form d2 v dx2

…(1)

+ Xv = Y ,

1 dP 1 2 − P 2 dx 4 1 1 6 1 1 6  1 = 2 / 3 − 4 / 3 − 2 −  − x −4 / 3  − x −2 / 3 = − 2 ,   2 3 4 4x 6x x x

where

X =Q−

and

Y = Re2

1

∫ P dx

=0.

Hence the normal form (1) of the given equation is d2 v 2

dx

−

6 2

x

v = 0 or x2

d2 v dx2

− 6v = 0 ,

which is a homogeneous linear equation. In order to solve it, putting x = e z , the differential equation (2) becomes [ D ( D − 1) − 6] v = 0 , where D ≡ d / dz or

( D2 − D − 6) v = 0 .

A.E. is m2 − m − 6 = 0 , giving m = − 2, 3 . ∴

the solution of (2) is v = c1e −2 z + c2 e3 z = c1 x −2 + c2 x3 .

Hence the complete solution of the given equation is y = uv = e(−3 /4) x

2 /3

(c1 x −2 + c2 x3 ).

…(2)

D-167 Example 12: Solve

d2 y 2

dx

2 dy + (4 x2 − 3) y = e x . dx (Gorakhpur 2005, 07; Lucknow 09)

− 4x

2 Solution: Here P = − 4 x, Q = 4 x2 − 3, R = e x .

− 1 ∫ P dx 2

We choose u1 = e

2

= e ∫ 2 x dx = e x .

Substituting y = uv in the given equation, it reduces to its normal form d2 v dx2

+ Xv = Y ,

where

X =Q−

and

Y = R e2

…(1)

1 dP 1 2 − P = −1 2 dx 4

1 P dx ∫

= 1.

Hence the normal form (1) of the given equation is d2 v dx2

− v = 1 or ( D2 − 1) v = 1 .

…(2)

A.E. is m2 − 1 = 0 , giving m = ± 1. ∴

C.F. of the solution of (2) = c1 e x + c2 e − x . 1

P. I. = ∴

D2 − 1

(1) =

1 D2 − 1

e0 x =

1 02 − 1

e0 x = − 1 .

the solution of (2) is v = c1 e x + c2 e − x − 1.

Hence the complete solution of the given equation is 2

y = uv = e x (c1e x + c2 e − x − 1) . Example 13: Solve

d2 y 2

dx

+ 2x

dy + ( x2 + 1) y = x3 + 3 x . dx

Solution: Here P = 2 x, Q = x2 + 1, R = x ( x2 + 3) .

We choose u = e

− 1 ∫ P dx 2

= e− x

2

/2

.

Substituting y = uv in the given equation, it reduces to its normal form d2 v dx2

…(1)

+ Xv = Y .

where

X =Q−

and

Y = R e2

1 dP 1 2 − P =0 2 dx 4

1 P dx ∫

= x( x2 + 3) e x

2

/2

.

Hence the normal form (1) of the given equation is d2 v 2

dx

= x ( x2 + 3) e x

2

/2

.

…(2)

D-168

Integrating (2), we get 2 dv = x3 e x /2 dx + 3 dx =

∫

∫xe

∫

2   x2  x e x /2  dx + 3  

= x2 e x

2

/2

−2

∫

x ex

2

/2

x2 / 2

∫

dx + c1

x ex

dx + 3

2

/2

∫

dx + c1

x ex

2

/2

dx + c1,

integrating the first integral by parts = x2 e x

2

/2

+ ex

2

/2

+ c1 .

Integrating again, we get v=

2 x2 / 2

∫x e

dx +

∫e

x2 / 2

dx + c1 x + c2 = x e x

2

/2

+ c1 x + c2 .

Hence the complete solution of the given equation is y = uv = x + (c1 x + c2 ) e − x

2

/2

.

2

d y

1 dy 1 + ( x + x1 /2 − 8) y = 0 . x1 /2 dx 4 x2 1 1 Solution: Here P = − 1 /2 , Q = ( x + x1 /2 − 8), R = 0 . x 4 x2 Example 14: Solve

We choose u = e

−

dx2

− 1 ∫ P dx 2

1

= e2

−1 /2 dx ∫x

= ex

1 /2

= e√ x .

Substituting y = uv in the given equation, it reduces to its normal form d2 v dx2

…(1)

+ Xv = Y ,

where

X =Q−

and

Y = R e2

1 dP 1 2 2 − P =− 2 2 dx 4 x

1 P dx ∫

= 0.

Hence the normal form (1) of the given equation is d2 v dx2 x2

or

 2 + v − 2  = 0  x 

d2 v dx2

…(2)

− 2v = 0 .

This is a homogeneous linear equation. Putting x = e z , the differential equation (2) becomes { D ( D − 1) − 2} v = 0

( D2 − D − 2) v = 0 ,

or

where D ≡ d / dz . A.E. is ∴

m = 2, − 1 .

m2 − m − 2 = 0

or

(m − 2) (m + 1) = 0 .

D-169

Thus the solution of (2) is v = c1 e2 z + c2 e − z = c1 x2 + c2 x −1. Hence the complete solution of the given equation is y = uv = e √ x (c1 x2 + c2 x −1).

Comprehensive Exercise 2 Solve the following equations by removing the first derivative : d2 y

1. x2

2

dx

− 2 ( x2 + x)

2

2. 3. 4.

d y dx2 d2 y

 2 dy  2 cos x  + y cos x = 0 .  dx 

(Rohilkhand 2007)

d y dx2

6. x

7.

dy + (4 x2 − 1) y = − 3 e x (sin 2 x + 5 e −2 x + 6) . dx (Kanpur 2009)

dx

−

2

5.

(Kanpur 2008; Bundelkhand 01)

2

2 dy  2 + 1 + 2  y = x e x .  x dx x 

2

d dx

− 4x

dy + ( x2 + 2 x + 2) y = 0 . dx

+

2 dy ± n2 y = 0 . x dx

(Agra 2008)

dy d  dy  − y − 2 x + 2 y + x2 y = 0 . x   dx dx dx

d2 y dx2

−

(Rohilkhand 2008)

2 dy  2 2 + a + 2  y = 0 . x dx  x 

  d2 y  dy  8.  2 + y cot x + 2  + y tan x = sec x .   dx   dx 9. x2 (log x)2

d2 y dx

2

10.

d y 2

dx

2

− 2 tan x

− 2 x log x

dy + [2 + log x − 2 (log x)2 ] y = x2 (log x)3 . dx

dy +5y =0. dx

(Agra 2008)

A nswers 2 1.

y = x e x (c1 x + c2 )

2.

y = e x (c1 cos x + c2 sin x + sin 2 x − 3 e −2 x − 18)

3.

y = x (c1 cos x + c2 sin x +

4.

y = (c1 cos √ 2 x + c2 sin √ 2 x) sec x

2

1 x e ) 2

D-170

1 c1 cos (nx + c2 ) x

5.

y=

or

6.

y = x (c1 cos x + c2 sin x)

7.

y = x c1 cos (ax + c2 )

8.

y=

9.

y = { c1 x2 + c2 x −1 +

y=

1 (c1 e nx + c2 e − nx ) x

1 sin x + (c1 x + c2 ) cos x 2 1 3

x2 log x} . log x

10. y = (sec x) . { c1 cos √ 6 x + c2 sin √ 6 x}

7.5 Transformation of the Equation by Changing the Independent Variable Sometimes the equation may become easily integrable by changing the independent variable. Let the linear equation of second order be d2 y dx2

+P

dy + Q y = R. dx

…(1)

Let the independent variable be changed from x to z with the help of a relation of the form z = f ( x). Then

dy dy dz d2 y d2 y  dz  2 dy d2 z = ⋅ , = 2   + ⋅ ⋅ dx dz dx dz dx2 dx2 dz  dx 

Substituting these values in (1), we get d2 y  dz  2 dy d2 z  dy dz  + ⋅ 2 + P  ⋅ + Qy = R 2  dx  dz dx dz  dz dx  or

d2 y  dz  2 dy   + dz dz 2  dx  2

or

or where

d y dz 2 d2 y dz 2

 d2 z dz   2 + P  + Q y = R dx   dx

d2 z

dz +P 2 dx ⋅ dy + Q y = R dx + 2 2 dz  dz  2  dz   dz         dx   dx   dx  + P1

dy + Q1 y = R1, dz

 d2 z dz  P1 =  2 + P  dx   dx

 dz     dx 

…(2) 2

 dz  , Q1 = Q    dx 

2

2

 dz  and R1 = R   .  dx 

Here P1, Q 1 and R1 are functions of x and may be expressed as functions of z with the help of the relation z = f ( x) .

D-171

How to choose z ? We would like to choose z in such a way that the equation (2) can be easily integrated. Case I: Let us choose z in such a way that P1 vanishes i. e., d2 z 2

dx

+P

dz =0 dx

⇒

d  dz  dz =0   +P dx  dx  dx

⇒

dz = e − ∫ P dx ⇒ z = dx

∫ [e

Then the equation (2) is reduced to

− ∫ P dx

d2 y dz 2

] dx .

+ Q 1 y = R1 .

Now this equation can be easily solved, if (i)

Q 1 is a constant because then it is a linear equation with constant coefficients,

(ii)

Q1 is of the form (constant)/z 2 because then it is a linear homogeneous equation

with variable coefficients. Case II: Suppose we choose z such that Q Q1 = = ± a2 (i. e., a constant) 2  dz     dx  or

a

dz = √ (± Q) dx

or

az =

∫ √ (± Q) dx ,

(+ ive or – ive sign is taken to make the expression under the radical sign + ive). With this choice of z, the equation (2) is reduced to d2 y dz

2

+ P1

dy + a2 y = R1. dz

This equation can be easily solved if P1 comes out to be a constant because then it is a linear equation with constant coefficients. Note 1: Students should remember the values of P1, Q1 and R1 in equation (2). Note 2: There are only two choices for z, either P1 = 0 or Q 1 = a2 . Sometimes it is

possible to make both the choices to get the solution of the given equation.

Example 15: Solve x

d2 y 2

dx

−

dy − 4 x3 y = 8 x3 sin x2 . dx (Meerut 2004, 13B; Garhwal 09; Agra 05; Gorakhpur 09)

Solution: The given equation written in the standard form is

D-172

d2 y dx2

−

1 dy − 4 x2 y = 8 x2 sin x2 . x dx

…(1)

Here P = − 1/ x, Q = − 4 x2 , R = 8 x2 sin x2 . Changing the independent variable from x to z by a relation of the form z = f ( x), the given equation is transformed into d2 y dz

2

+ P1

dy + Q 1 y = R1, dz

…(2)

d2 z

dz +P 2 dx , Q = Q , R = R ⋅ dx P1 = 1 1 2 2 2 dz  dz   dz           dx   dx   dx 

where

Choosing z such that Q 1 =

− 4 x2 (dz / dx)2

(dz / dx)2 = 4 x2 .

= constant = −1 (say), we have dz / dx = 2 x ⇒ z = x2 .

∴

1 ⋅2x x = 0 , R1 = 2 sin x2 = 2 sin z . 4 x2

2− Now ∴

P1 =

The transformed equation (2) is d2 y dz 2

or

− y = 2 sin z

( D2 − 1) y = 2 sin z .

…(3)

A.E. is m2 − 1 = 0 ⇒ m2 = 1 ⇒ m = ± 1 . ∴

C.F. of the solution of (3) = c1 e z + c2 e − z . P. I. =

∴

1 D2 − 1

2 sin z =

2 sin z = − sin z . − 1− 1

Solution of the equation (3) is y = c1e z + c2 e − z − sin z .

Hence the complete solution of the equation (1) is 2

2

y = c1e x + c2 e − x − sin x2 . Example 16: Solve

d2 y 2

dx

+ cot x

dy + 4 y cosec2 x = 0 . dx (Meerut 2001; Agra 06; Gorakhpur 05)

Solution: Here P = cot x, Q = 4 cosec2 x, R = 0 .

Changing the independent variable from x to z by a relation of the form z = f ( x), the given equation is transformed into d2 y dz 2

+ P1

dy + Q 1 y = R1, dz

…(1)

D-173

d2 z where

P1 = dx

dz dx , Q = 1 2

+P

2

 dz     dx 

Choosing z such that Q1 =

4 cosec2 x (dz / dx)2

Q  dz     dx 

2

, R1 =

R  dz     dx 

2

⋅

= constant =1 (say), we have

(dz / dx)2 = 4 cosec2 x dz = 2 cosec x, dx

or Now P1 = ∴

or

z = 2 log tan

− 2 cosec x cot x + 2 cosec x cot x 4 cosec2 x

x ⋅ 2

= 0 , R1 = 0

the transformed equation (1) is d2 y dz 2

+ y = 0 or ( D2 + 1) y = 0 .

…(2)

A.E. is m2 + 1 = 0 ⇒ m2 = − 1 ⇒ m = ± i . ∴

Solution of the equation (2) is y = c1 cos z + c2 sin z .

Hence the complete solution of the given equation is x x   y = c1 cos 2 log tan  + c2 sin 2 log tan  .    2 2 Example 17: Solve cos x

d2 y 2

dx

+ sin x

dy − 2 y cos3 x = 2 cos5 x . dx

(Meerut 2006; Rohilkhand 07; Agra 07; Avadh 11; Purvanchal 06, 10)

Solution: The given equation can be written in the standard form as

d2 y 2

dx

+ tan x

dy − 2 y cos2 x = 2 cos4 x. dx

…(1)

Here P = tan x, Q = − 2 cos2 x, R = 2 cos4 x. Changing the independent variable from x to z by a relation of the form z = f ( x), the given equation is transformed into d2 y dz

2

+ P1 d2 z

where

dy + Q1 y = R1, dz

dz +P 2 dx , Q = Q , R = R ⋅ dx P1 = 1 1 2 2 2 dz  dz   dz           dx   dx   dx 

…(2)

D-174

Let us choose z such that Q1 = 2

− 2 cos2 x (dz / dx)2

= constant = − 2 (say).

Then

 dz  2   = cos x,  dx 

Now

P1 = 0 , R1 = 2 cos2 x = 2 (1 − sin2 x) = 2 (1 − z 2 ) .

∴

or

dz = cos x dx

or

z = sin x .

The transformed equation (2) is d2 y dz

2

− 2 y = 2 (1 − z 2 ) .

…(3)

A.E. is m2 − 2 = 0 , giving m = ± √ 2. ∴

C.F. of the solution of (3) = c1 e √2 z + c2 e −√2 z .  D2   2 (1 − z ) = − 1 − P. I. = 2 2  D −2  1

2

= − (1 + ∴

−1

(1 − z 2 )

1 2 1 D + … ) (1 − z 2 ) = − (1 − z 2 ) + ⋅ 2 = z 2 . 2 2

The solution of the equation (3) is y = c1 e √2 z + c2 e −√2 z + z 2.

Hence the complete solution of the given equation is y = c1e √2 sin x + c2 e −√2 sin x + sin2 x. d2 y dy 1 Example 18: Solve x6 + 3 x5 + a2 y = 2 ⋅ 2 dx dx x Solution: The given equation can be written in the standard form as

d2 y dx2 Here

P=

+

3 dy a2 1 + y= 8 ⋅ x dx x6 x

3 a2 1 ,Q = 6 , R= 8 ⋅ x x x

To solve this differential equation we change the independent variable from x to z by choosing z such that Q 2

(dz / dx)

=

a2 / x6 2

(dz / dx)

2

Then

1  dz    = 6,  dx  x

or

= constant = a2 (say).

dz 1 = 3 dx x

or

z = − 1/(2 x2 ).

Now by the substitution z = − 1/(2 x2 ), the given differential equation is transformed into d2 y dz 2

+ P1

dy + Q1 y = R1, where dz

D-175

d2 z

dz +P 4 3 2 dx = (−3 / x ) + (3 / x)(1/ x ) = 0 , dx P1 = 2 2 (dz / dx) (dz / dx) Q

Q1 = ∴

2

(dz / dx)

R

= a2 , R1 =

2

(dz / dx)

=

1/ x8 6

1/ x

=

1 x2

= −2 z .

the transformed equation is d2 y dz 2

+ a2 y = − 2 z

( D2 + a2 ) y = − 2 z .

or

…(1)

The C.F. of the solution of (1) = c1 cos az + c2 sin az . 1

P. I. =

D2 + a2

D2  1  1 + 2  2  a  a 

−1

(−2 z )

1  1 2 1  1 − 2 D + … (−2 z ) = 2 (−2 z ).  a2  a a

= ∴

(− 2 z ) =

the solution of the equation (1) is y = c1 cos az + c2 sin az −

1 a2

⋅ 2z .

Hence the complete solution of the given equation is a a 1 y = c1 cos 2 − c2 sin 2 + 2 2 2x 2x a x or

y = c1 cos

Example 19: Solve

d2 y 2

dx

a

+ c2 sin

2 x2

+ (tan x − 1)2

a 2 x2

+

1 a2 x2

⋅

dy − n (n − 1) y sec 4 x = 0 . dx

Solution: Comparing the given differential equation with the standard form

d2 y 2

dx

dy + Qy = R, we have dx

+P

P = (tan x − 1)2 , Q = − n (n − 1) sec4 x, R = 0 . To solve the given differential equation we change the independent variable from x to z by choosing z such that Q (dz / dx)2

=

− n (n − 1) sec4 x (dz / dx)2

2

Then

 dz  4   = sec x,  dx 

or

= constant = − n (n − 1), say.

dz = sec2 x, dx

or

z = tan x.

Now by the substitution z = tan x,the given differential equation is transformed into d2 y dz 2

+ P1

dy + Q 1 y = R1, where dz

D-176

d2 z

dz +P 2 2 2 2 dx = 2 sec x tan x + (tan x − 1) sec x = 1, dx P1 = 4 2 (dz / dx) sec x Q1 = ∴

Q 2

(dz / dx)

= − n (n − 1), R1 =

R (dz / dx)2

=0.

the transformed equation is d2 y dz

dy − n (n − 1) y = 0 dz

+

2

{ D2 + D − n (n − 1)} y = 0 .

or

…(1)

A.E. is m2 + m − n (n − 1) = 0 or

(m + n) { m − (n − 1)} = 0 .

∴

m = − n, n − 1 .

∴

the solution of the equation (1) is y = c1e − nz + c2 e(n − 1)z .

Hence the complete solution of the given equation is y = c1e − n tan x + c2 e(n − 1) tan x .

Comprehensive Exercise 3 Solve the following equations : 1. 2.

d2 y

+ (3 sin x − cot x)

dx2 d2 y

− (1 + 4 e x )

2

dx 2

3. 4.

d y

(a2 − x2 ) d y

−

2

dx

d2 y

6.

x

7.

(1 + x)2

2

dx

2

d y 2

dx

d y 2

dx

−

a2 dy x2 + y =0. x dx a

(Gorakhpur 2008, 11)

(Rohilkhand 2006)

(Bundelkhand 2001)

1 dy + 4 x2 y = x4 . x dx + (4 x2 − 1) d2 y 2

dx

2

8.

x dy + 3 e2 x y = e2( x + e ). dx

1 dy  + 1 −  + 4 x2 ye − 2 x = 4( x2 + x3 ) e −3 x .  x  dx

dx2

2

5.

dy + 2 y sin2 x = e − cos x sin2 x. dx

− cot x

dy + 4 x3 y = 2 x3 . dx

+ (1 + x)

dy + y = 4 cos log (1 + x) . dx

dy − y sin2 x = 0 . dx

(Gorakhpur 2010)

D-177

9. x4

d2 y dx2

10. (1 + x2 )2

+ 2 x3 d2 y

dy + n2 y = 0 . dx + 2 x (1 + x2 )

2

dx

dy +4y =0. dx

A nswers 3 1 − cos x e 6

1.

y = c1e cos x + c2 e2 cos x +

2.

y = c1e3 e + c2 e e − e2 e

3.

y = c1 cos {2 e − x (1 + x)} − c2 sin {2 e − x (1 + x)} + e − x (1 + x)

4. 5. 6.

x

x

x

 √ (a2 − x2 )  √ (a2 − x2 ) y = c1 cos    + c2 sin  √a √a     1 y = c1 cos ( x2 + c2 ) + x2 4 2 1 y = (c1 + c2 x2 ) e − x + 2

7.

y = c1 cos {log (1 + x) + c2 } + 2 log (1 + x) ⋅ sin log (1 + x)

8.

y = c1e − cos

9.

n  y = c1 cos  + c2  x 

x

+ c2 e cos x

10. y = c1 cos (2 tan−1 x) + c2 sin (2 tan−1 x), or y (1 + x2 ) = c1 (1 − x2 ) + 2 c2 x.

7.6 Method of Variation of Parameters Now we shall discuss an elegant but somewhat artificial method for finding the complete primitive of a linear equation whose complementary function is known. Let the linear equation of second order be d2 y 2

dx

+P

dy + Qy = R . dx

…(1)

Suppose y = Au + Bv is the complementary function of (1), where A and B are arbitrary constants and u and v are functions of x. Then y = Au + Bv is the solution of d2 y 2

dx ∴

+P

dy + Qy = 0 . dx

( Au2 + Bv2 ) + P ( Au1 + Bv1) + Q ( Au + Bv) = 0

or

A (u2 + Pu1 + Qu) + B (v2 + Pv1 + Qv) = 0

⇒

u2 + Pu1 + Qu = 0 and v2 + Pv1 + Qv = 0 .

…(2)

D-178

Now, suppose that …(3)

y = Au + Bv

is the complete primitive of (1) where A and B are not constants but functions of x, so chosen that (1) will be satisfied. Differentiating (3), dy = A1u + B1v + Au1 + Bv1. dx Let us choose A and B such that …(4)

A1u + B1v = 0 . dy = Au1 + Bv1 dx

Then

d2 y

and

dx2

= Au2 + Bv2 + A1u1 + B1v1.

Putting these values in (1), we get ( Au2 + Bv2 + A1u1 + B1v1) + P ( Au1 + Bv1) + Q ( Au + Bv) = R or

A (u2 + Pu1 + Qu) + B (v2 + Pv1 + Qv) + A1u1 + B1v1 = R

or

A1u1 + B1v1 = R1,

[using 2] …(5)

Solving (4) and (5), we get vR uR and B1 = , which on integration give A and B. A1 = u1v − uv1 v1u − vu1 A=

v R dx

∫ u1v − uv1 + c1 and

B=

u R dx

∫ v1u − vu1 + c2 ,

where c1 and c2 are parameters. Substituting these values of A and B in (3), we get the complete primitive of (1). Note : Since the general solution is obtained by varying the arbitrary constants of the

complementary function, therefore the method is known as method of variation of parameters. Alternate method to find C.S: Consider the linear equation of second order d2 y 2

dx

+P

dy + Qy = R . dx

The P.I. of (1) is given by vR P.I. = − u dx + v W

∫

uR

…(1)

vR

uR

∫ W dx = − u ∫ uv1 − vu1 dx + v ∫ uv1 − vu1 dx , …(2)

where u and v are defined by the C.F. of d2 y 2

dx i. e.,

+P

dy + Qy = 0 dx

by y = c1 u + c2 v, where c1 and c2 are arbitrary constants

…(3)

D-179

u

v

u1

v1

is called the Wronskian of u and v

and

W=

∴

C. S. = C. F. + P. I.

Note : The method of variation of parameters must be used, if instructed to do so in the

question.

Example 20: Solve by the method of variation of parameters

d2 y dx2

+ 4 y = 4 tan 2 x. (Rohilkhand 2001, 11; Lucknow 09; Avadh 05; Meerut 04B)

Solution: The C.F. of the given equation i. e., the solution of the equation

d2 y dx2

+ 4 y = 0 is

y = c1 cos 2 x + c2 sin 2 x, where c1 and c2 are constants. Let

y = A cos 2 x + B sin 2 x

…(1)

be the complete primitive of the given equation where A and B are functions of x, so chosen that the given equation will be satisfied. dy dA dB Then = − 2 A sin 2 x + 2 B cos 2 x + cos 2 x + sin 2 x . dx dx dx Let us choose A and B such that dA dB cos 2 x + sin 2 x = 0 . dx dx dy Then = − 2 A sin 2 x + 2 B cos 2 x dx and

d2 y 2

dx

= −2

dA dB sin 2 x + 2 cos 2 x − 4 A cos 2 x − 4 B sin 2 x. dx dx

Putting these values in the given equation, we get dA dB − 2 sin 2 x + 2 cos 2 x = 4 tan 2 x dx dx dA dB or − sin 2 x + cos 2 x = 2 tan 2 x . dx dx Solving (2) and (3), we get 2 sin2 2 x dB dA =− , = 2 sin 2 x. dx cos 2 x dx Integrating these, we get A= −2

∫

…(2)

(1 − cos2 2 x) dx + c1 cos 2 x

…(3)

D-180

∫

= − 2 (sec 2 x − cos 2 x) dx + c1 = − log (sec 2 x + tan 2 x) + sin 2 x + c1 and

B = − cos 2 x + c2 .

Putting the values of A and B in (1), the complete primitive of the given equation is y = c1 cos 2 x + c2 sin 2 x − [ log (sec 2 x + tan 2 x)] ⋅ cos 2 x. Alternate method to find P.I.: Here u = cos 2 x, v = sin 2 x, R = 4 tan 2 x u v cos 2 x and W= = u1 v1 − 2 sin 2 x Thus,

vR

sin 2 x 2 cos 2 x

=2.

uR

∫ W dx + v ∫ W dx = − cos 2 x 2 sin 2 x tan 2 x dx + sin 2 x 2 cos 2 x tan 2 x dx ∫ ∫ = − 2 cos 2 x (sec 2 x − cos 2 x) dx + 2 sin 2 x sin 2 x dx ∫ ∫

P.I. = − u

= − cos 2 x [log (sec 2 x + tan 2 x) − sin 2 x] − sin 2 x cos 2 x . = − cos 2 x log (sec 2 x + tan 2 x) . Hence, the C.S. is y = c1 cos 2 x + c2 sin 2 x − (cos 2 x) log (sec 2 x + tan 2 x) . Example 21: (i) Solve by the method of variation of parameters

d2 y dx2

−2

dy + y = e x log x . dx

Solution: First we shall find the C.F. of the given equation i. e., the solution of the

equation d2 y 2

dx

−2

dy + y =0 dx

A.E. is m2 − 2 m + 1 = 0

or

( D2 − 2 D + 1) y = 0 .

or

(m − 1)2 = 0 , giving m = 1, 1 .

C.F. is y = (c1 + c2 x) e x .

∴

Now assume A and B as the functions of x in such a way that the given equation is satisfied by y = ( A + Bx) e x = Ae x + Bxe x . ∴

∴

dy dA dB = Ae x + B (e x + xe x ) + e x + xe x dx dx dx dA dB = Ae x + B (e x + xe x ), assuming e x + xe x =0. dx dx d2 y dx2

= Ae x + B (2 e x + xe x ) + e x

dA dB + (e x + xe x ) ⋅ dx dx

…(1)

…(2)

D-181

Now putting these values in the given equation, we get dA dB ex + e x (1 + x) = e x log x . dx dx

…(3)

Solving (2) and (3), we get dA dB = − x log x and = log x . dx dx Integrating these, we get A= −

 x2 x log x dx + c1 = −  log x − 2

∫

∫

1 x2  ⋅ dx + c1 x 2 

 x2 x2   + c1, = −  log x − 4  2 B=

1

∫ log x dx + c2 = x log x − ∫ x ⋅ x dx + c2 = x log x − x + c2 .

Putting these values of A and B in (1), the general solution of the given equation is  x2  x2 y = − e x  + c1 + x e x ( x log x − x + c2 ) log x − 4 2  = (c1 + c2 x) e x +

1 2 x x e (2 log x − 3) . 4

Alternate method to find P.I. Here u = e x , v = x e x , R = e x log x and Thus

W= P.I. = − u = − ex

u

v

u1

v1

=

ex

xex

ex

(1 + x) e x

uR

= e2 x .

uR

∫ W dx + v ∫ W dx xe x ⋅ e x log x

dx + xe x

e x ⋅ e x log x

∫ e2 x ∫ = − e x x log x dx + xe x log x dx ∫ ∫

 x2 = − e x  log x − 2

∫

e2 x

dx

1 x2   ⋅ dx + x ⋅ e x  x log x −  x 2 

1



∫ x ⋅ x dx

 x2 x2 = − e x  log x − 4 2

 1  + x e x ( x log x − x) = x2 e x (2 log x − 3) . 4  1 Hence, the C.S. is y = (c1 + c2 x) e x + x2 e x (2 log x − 3). 4 Example 22: (ii) Solve by the method of variation of parameters

d2 y dx2

+ n2 y = sec nx .

(Meerut 2003, 07; Gorakhpur 11; Lucknow 10; Avadh 09, 10, 13)

Solution: The C.F. of the given equation i. e. the solution of the equation

D-182

d2 y

+ n2 y = 0 is

dx2

y = c1 cos nx + c2 sin nx, where c1 and c2 are constants. Let

…(1)

y = A cos nx + B sin nx

be the general solution of the given equation where A and B are functions of x, so chosen that the given equation will be satisfied. dy dA dB Then = − An sin nx + Bn cos nx + cos nx + sin nx . dx dx dx Let us choose A and B such that dA dB cos nx + sin nx = 0 dx dx dy Then = − An sin nx + Bn cos nx dx and

d2 y dx2

= − An2 cos nx − Bn2 sin nx − n

…(2)

dA dB sin nx + n cos nx . dx dx

Putting these values in the given equation, we get dA dB −n sin nx + n cos nx = sec nx . dx dx Solving (2) and (3), we get dA 1 = − tan nx dx n

and

…(3)

dB 1 = ⋅ dx n

Integrating these, we get 1 x A = 2 log cos nx + c1, B = + c2 . n n Putting these values of A and B in (1), the general solution of the given equation is 1 x y = c1 cos nx + c2 sin nx + 2 cos nx log cos nx + sin nx . n n Example 23: Apply the method of variation of parameters to solve

d2 y dx2

+ y = tan x.

Solution: The given differential equation is

d2 y dx2

+ y = tan x

A.E. is m2 = − 1 ∴

or

or

( D2 + 1) y = tan x

m=±i

C.F. is y = A cos x + B sin x

Here u = cos x, v = sin x, R = tan x and

W = uv1 − u1 v = cos2 x + sin2 x = 1

∴

P.I. = −

∫

vR dx + v W

uR

∫ W dx .

…(1)

D-183

We have

∫

− vR dx = − W

sin x tan x dx = − 1

∫

∫

sin2 x dx = − cos x

∫

1 − cos2 x dx cos x

∫ (cos x − sec x) dx = sin x − log (sec x + tan x) cos x ⋅ tan x uR ∫ W dx = ∫ 1 = ∫ sin x dx = − cos x =

and

P. I. = [sin x − log (sec x + tan x)] cos x − cos x sin x

∴

= − cos x log (sec x + tan x) Hence the complete solution is y = A cos x + B sin x − cos x log (sec x + tan x) Example 24: Solve by the method of variation of parameters

x2

d2 y 2

dx

+x

dy − y = x2 e x . dx

(Rohilkhand 2006, 11; Meerut 07;

Gorakhpur 06, 10; Purvanchal 06; Lucknow 08, 11; Avadh 14)

Solution: First we shall find the C.F. of the given equation i.e., the solution of the

equation x2

d2 y 2

dx

+x

dy − y =0. dx

…(1)

This is a homogeneous linear equation. Putting x = e z , the differential equation (1) becomes { D ( D − 1) + D − 1} y = 0 , where D ≡

d dz

( D2 − 1) y = 0

or

whose solution is y = c1 e z + c2 e − z . ∴

solution of (1) is y = c1 x + c2

Let

y = Ax +

1 ⋅ x

B x

…(2)

be the complete primitive of the given equation where A and B are functions of x, so chosen that the given equation will be satisfied. dy B dA 1 dB Then = A− 2 + x+ ⋅ ⋅ dx dx x dx x Let us choose A and B such that dA 1 dB x + ⋅ =0. dx x dx Then

dy B = A− 2 dx x

and

…(3) d2 y 2

dx

=

dA 1 dB 2 − 2 + 3 B. dx x dx x

Putting these values in the given equation, we get dA 1 dB − 2 = ex . dx x dx

…(4)

D-184

Solving (3) and (4), we get dA 1 x dB x2 x = e and =− e . dx 2 dx 2 Integrating these, we get 1 A = e x + c1 2

and

B=−

1 2 x x e + xe x − e x + c2 . 2

Putting the values of A and B in (2), the complete primitive of the given equation is c 1 y = c1 x + 2 + e x − e x . x x Example 25: Solve by the method of variation of parameters

(1 − x)

d2 y 2

dx

+x

dy − y = (1 − x)2 . dx

Solution: First we shall find the C.F. of the given equation i. e., the solution of the

equation (1 − x) 2

or

d y dx2

+

d2 y 2

dx

+x

dy − y =0 dx

y x dy − =0. 1 − x dx 1 − x

…(1)

Here P + Qx = 0. ∴

y = x is a part of the C.F.

Putting y = vx in (1), the reduced equation is d2 v  x 2  dv + +  =0 2 dx 1 − x x  dx or

dp  x 2 dv + +  p = 0, where p = dx 1 − x x  dx

or

dp  1 + − 1 − + dx  x −1

2  p=0 x

or

dp  1 = 1 + − p  x −1

2  dx . x

Integrating, log p = x + log ( x − 1) − 2 log x + log c1 dv c1 ( x − 1) e x 1 1  = = c1  − 2  e x . x x  dx x2 1 1  Integrating again, v = c1  − 2  e x dx + c2 x x  or

p=

∫

or ∴ Let

v=

c1 x e + c2 . x

solution of (1) is y = vx = c1 e x + c2 x . y = Ae x + Bx

…(2)

be the complete primitive of the given equation where A and B are functions of x, so chosen that the given equation will be satisfied.

D-185

dy dA dB = Ae x + B + e x +x ⋅ dx dx dx Let us choose A and B such that dA dB ex +x =0. dx dx dy Then = Ae x + B dx

Then

d2 y

and

2

…(3)

dA dB + + ex A . dx dx

= ex

dx Putting these values in the given equation, we get dA dB ex + = 1 − x. dx dx Solving (3) and (4), we get dA dB = − xe − x and = 1. dx dx Integrating these, we get

…(4)

A = (1 + x) e − x + c1 and B = x + c2 . Putting the values of A and B in (2) the general solution of the given equation is y = c1 e x + c2 x + 1 + x + x2 . Example 26: Solve by the method of variation of parameters

d2 y 2

dx

+ (1 − cot x)

dy − y cot x = sin2 x. dx

Solution: First we shall find the C.F. of the given equation i.e., the solution of the

equation d2 y 2

dx

+ (1 − cot x)

Here 1 − P + Q = 0 ,

∴

dy − y cot x = 0 . dx

y = e − x is a part of the C.F.

Putting y = ve − x , the equation (1) reduces to d2 v

or or

dv − (1 + cot x) =0 dx dx2 dp dv − (1 + cot x) p = 0 , where p = dx dx dp = (1 + cot x) dx . p

Integrating, log p = x + log sin x + log c1. dv ∴ p= = c1e x sin x dx or ∴

1 x e (sin x − cos x) + c2 . 2 the solution of (1) i.e., the C.F. of the given equation is 1 y = ve − x = c1 ⋅ (sin x − cos x) + c2 e − x . 2 v = c1

∫e

x

sin x dx + c2 = c1 ⋅

…(1)

D-186

y = A (sin A − cos x) + Be − x

Let

…(2)

be the complete solution of the given equation where A and B are functions of x, so chosen that the given equation will be satisfied. dy dA dB − x ∴ = A (cos x + sin x) − Be − x + (sin x − cos x) + e . dx dx dx Let us choose A and B such that dA dB − x (sin x − cos x) + e =0. dx dx dy = A (cos x + sin x) − Be − x ∴ dx d2 y

and

dx2

=

…(3)

dA dB − x (cos x + sin x) − e + A (− sin x + cos x) + Be − x . dx dx

Putting these values in the given equation, we get dA dB − x (cos x + sin x) − e = sin2 x. dx dx

…(4)

Solving (3) and (4), we get dA 1 = sin x dx 2 dB 1 x and = e (sin x cos x − sin2 x) . dx 2 Integrating these, we get 1 A = − cos x + c1 2 1 and B= e x (sin 2 x − 1 + cos 2 x) dx + c2 4

∫

=

1 ex ex 1 ex ⋅ (sin 2 x − 2 cos 2 x) − + ⋅ (cos 2 x + 2 sin 2 x) + c2 4 5 4 4 5

=

ex ex (3 sin 2 x − cos 2 x) − + c2 . 20 4

Putting the values of A and B in (2), the general solution of the given equation is 1 y = c1 (sin x − cos x) + c2 e − x − (sin 2 x − 2 cos 2 x) . 10

Comprehensive Exercise 4 Solve the following differential equations by the method of variation of parameters: 1.

d2 y dx2

+ y = x.

(Agra 2007)

2.

d2 y dx2

+ a2 y = cosec ax .

D-187

3.

x2

d2 y

− 2 x (1 + x)

dx2

4.

(1 − x2 )

5.

(1 − x2 )

6.

( x + 2)

2

d y

− 4x

2

dx

d2 y

+x

2

dx

d2 y

dy + 2 ( x + 1) y = x3 . dx

dy − (1 + x2 ) y = x . dx

dy − y = x (1 − x2 )3 /2 . dx

− (2 x + 5)

2

dx

(Gorakhpur 2007)

dy + 2 y = ( x + 1) e x . dx

2

7.

d y dx2

+ y = cosec x .

2

8. 9.

d y dx2

− y=

2 1+ ex

(Purvanchal 2007, 10)

⋅ (Agra 2008; Gorakhpur 05; Purvanchal 09; Kanpur 07, 14)

xy1 − y = ( x − 1) ( y2 − x + 1) .

A nswers 4 1.

y = c1 cos x + c2 sin x + x

2.

y = c1 cos ax + c2 sin ax −

3.

y = c1 x + c2 x e2 x

4.

y=

5.

y = c1 [√ (1 − x2 ) + x sin−1 x] + c2 x − ( x / 9) (1 − x2 )3 /2

6.

y = c1 (2 x + 5) + c2 e2 x − e x

7.

y = c1 cos x + c2 sin x − x cos x + sin x log sin x 1 + e x  y = c1 e x + c2 e − x + e x log  x  − 1 − e − x log (1 + e x )  e 

8. 9.

1 (1 − x2 )

x 1 cos ax + 2 sin ax log sin ax a a 1 2 1 − x − x 2 4

⋅ (c1 cos x + c2 sin x + x)

y = c1 e x + c2 x − (1 + x2 )

7.7 Method of Operational Factors In some cases the method of operational factors easily solves the equation. Let the linear equation of second order d2 y 2

dx

+P

dy + Qy = R , dx

be expressed in the form f ( D) y = R

D-188

or

f1 ( D) f2 ( D) y = R

i. e., f ( D) can be resolved into a product of two factors f1( D) and f2 ( D), such that, if f2 ( D) operates upon y, and then f1( D) operates upon the result of the operation, the same result is obtained as if f ( D) operates upon y. Note: With the exception of the classes of equations, the factors of f ( D) are generally

not commutative. Hence a great care is to be taken in writ ing them in the right order.

d2 y

Example 27: Solve x

2

dx

+ (1 − x)

dy − y = e x. dx

Solution: In the symbolic form, the given equation is

[ xD2 + (1 − x) D − 1] y = e x , where D ≡ d / dx ( xD + 1) ( D − 1) y = e x .

or

…(1)

[Note that here the factors are not commutative because on expansion {( D − 1)( xD + 1)} y gives { xD2 + (2 − x) D − 1} y] . Let

…(2)

( D − 1) y = v . x

Then (1) gives ( xD + 1) v = e . or

x

dv + v = ex dx

dv 1 ex + v= , dx x x

or

which is a linear differential equation of the first order. Now

I.F. = e ∫ (1 / x) dx = e log

x

= x.

x

∴

vx =

∫x

e dx + c1 = e x + c1 x

or

v = e x x −1 + c1 x −1.

Putting the value of v in (2), we get ( D − 1) y = e x x −1 + c1 x −1 or Here ∴

dy − y = e x x −1 + c1 x −1, which is again linear. dx I.F. = e ∫ − dx = e − x . ye − x = =

∴

∫

∫ (e

x −1

x

1 dx + c1 x

∫

y = e x log x + c1e x

∫

+ c1 x −1) e − x dx + c2 e− x dx + c2 = log x + c1 x e

−x

x

∫

e− x dx + c2 . x

dx + c2 e x ,

which is the required general solution of the given differential equation.

D-189

d2 y

Example 28: Solve 3 x2

2

dx

+ (2 + 6 x − 6 x2 )

dy −4y =0. dx

Solution: In the symbolic form, the given equation is

[3 x2 D2 + (2 + 6 x − 6 x2 ) D − 4] y = 0 or

[(3 x2 D2 + 6 x D + 2 D) − (6 x2 D + 4)] y = 0

or

[ D (3 x2 D + 2) − 2 (3 x2 D + 2)] y = 0

or

( D − 2) (3 x2 D + 2) y = 0 .

…(1)

[Here factors are not commutative because on expansion (3 x2 D + 2) ( D − 2) y gives {3 x2 D2 + (2 − 6 x2 ) D − 4} y]. Let

(3 x2 D + 2) y = v .

Then (1) gives ( D − 2) v = 0 ⇒ dv = 2v dx

…(2)

dv − 2v = 0 dx v = c1e2 x .

⇒

Putting the value of v in (2), we get (3 x2 D + 2) y = c1 e2 x or

3 x2

dy dy 2 c + 2 y = c1e2 x ⇒ + 2 y = 12 e2 x . dx dx 3 x 3x

I. F. = e −2

∫

2 dx 3 x2

c1 3

∫

∴

ye

or

y = c2 e3 x +

3x

=

2

( ) = e−32x .

2 −1 x

= e3

x −2 e2 x ⋅ e c1 32x e 3

∫

−2

dx + c2

3x

2 x− 2

x −2 e

3x

dx,

which is the required solution. Example 29: Solve x

d2 y dx2

− ( x + 2)

dy + 2 y = x3 . dx

Solution: In the symbolic form, the given equation is

[ xD2 − ( x + 2) D + 2] y = x3 or

( xD − 2) ( D − 1) y = x3

…(1)

[Here factors are not commutative because on expansion {( D − 1) ( xD − 2)} y gives { xD2 − ( x + 1) D + 2} y]. Let

…(2)

( D − 1) y = v . 3

Then (1) gives ( xD − 2) v = x or

x

dv − 2 v = x3 dx

D-190

or

dv 2 − v = x2 , which is linear. dx x 1 I. F. = e ∫ (−2 / x) dx = e −2 log x = 2 x ⋅ 1



2

∫  x

∴

v⋅

∴

v = x3 + c1 x2 .

x2

=

⋅

1  dx + c1 = x + c1. x2 

Putting the value of v in (2), we get ( D − 1) y = x3 + c1 x2 or

dy − y = x3 + c1 x2 , which is again linear. dx I. F. = e ∫ − dx = e − x .

∴

ye − x =

∫e

−x

( x3 + c1 x2 ) dx + c2

= − ( x3 + c1 x2 ) e − x − (3 x2 + 2 c1 x) e − x − (6 x + 2 c1) e − x − 6 e − x + c2 y = − ( x3 + c1 x2 ) − (3 x2 + 2 c1 x) − (6 x + 2 c1) − 6 + c2 e x

or

= − x3 − (c1 + 3) x2 − 2 (c1 + 3) x − 2 (c1 + 3) + c2 e x y = − x3 − (c1 + 3) { x2 + 2 x + 2} + c2 e x .

or

which is the required general solution of the given differential equation. Example 30: Solve [( x + 3) D2 − (2 x + 7) D + 2] y = ( x + 3)2 e x . Solution: The given equation can be written as

{( x + 3) D − 1} ( D − 2) y = ( x + 3)2 e x .

…(1)

[Here factors are not commutative because on expansion ( D − 2) {( x + 3) D − 1} y gives {( x + 3) D2 − (2 x + 5) D + 2} y] Let

…(2)

( D − 2) y = v . 2 x

Then (1) gives {( x + 3) D − 1} v = ( x + 3) e or or

dv − v = ( x + 3)2 e x , dx dv 1 − v = ( x + 3) e x , which is linear. dx x + 3

( x + 3)

I. F. = e

−∫

dx x +3

1 = x +3

= e − log ( x +3) = 1

∫ x + 3 ⋅ ( x + 3) e

∴

v⋅

∴

v = e x ( x + 3) + c1 ( x + 3) .

1 ⋅ x +3 x

dx + c1 = e x + c1.

D-191

Putting the value of v in (2), we get ( D − 2) y = e x ( x + 3) + c1 ( x + 3) dy − 2 y = e x ( x + 3) + c1 ( x + 3), which is again linear. dx

or

I. F. = e ∫ −2 dx = e −2 x . ye −2 x =

∴

∫e

−2 x

( x + 3) dx + c1

∫e

−2 x

( x + 3) dx + c2

= − ( x + 3) e − x − e − x −

1 1 c1 ( x + 3) e −2 x − c1e −2 x + c2 2 4

or

y = − ( x + 3) e x − e x −

1 1 c1 ( x + 3) − c1 + c2 e2 x 2 4

or

y = − xe x − 4 e x −

or

y = − xe x − 4 e x + A (2 x + 7) + Be2 x ,

1 c1 (2 x + 7) + c2 e2 x 4

which is the required general solution of the given differential equation.

Comprehensive Exercise 5 Solve the following equations by the method of factorization of the operator: d2 y

+ ( x − 2)

dy − 2 y = x3 . dx

+ ( x − 1)

dy − y = x2 . dx

1.

x

2.

x

3.

3 x2 D2 y + (2 − 6 x2 ) Dy − 4 y = 0 .

4.

x2

5.

[ xD2 + (1 − x) D − 2(1 + x)] y = e − x (1 − 6 x) .

2

dx

d2 y dx2

d2 y dx2

+

dy − (1 + x2 ) y = e − x . dx

A nswers 5 1.

y = x3 + a ( x2 − 2 x + 2) + be − x

2.

y = c1 ( x − 1) + c2 e − x + x2

3.

y = c1e2 x

(−2 x) + 2 /(3 x)

∫e

dx + c2 e2 x

D-192

4.

y = c1e x

5.

y = xe − x + c1e2 x

∫e

− 2 x + (1 / x)

dx + c2 e x −

1

∫xe

−3 x

1 −x e 2

dx + c2 e2 x

Objective Type Questions

Fill in The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

y = x m is a part of C.F. of equation d2 y 2

dx 2.

y = e x is a solution of d2 y dx2

3.

dy + Qy = R if …… dx

+P

y = e − x is a part of C.F. of d2 y 2

dx 4.

dy + Qy = R if …… dx

+P

y=e

mx

dy + Qy = R if …… dx

+P

is a part of C.F. of  d2 y   dy     dx2  + P  dx  + Qy = R if ……  

5.

y = x is a solution of  d2 y   dy     dx2  + P  dx  + Qy = R if ……  

6.

y = x2 is a part of C.F. of equation  d2 y   dy     dx2  + P  dx  + Qy = R if ……  

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

y = x2 is a part of C.F. of the solution of equation (1 − x)

d2 y 2

dx

+x

dy − y = (1 − x)2 . dx

(Meerut 2003)

D-193

2.

y = e x is a part of C.F. of the solution of equation x

3.

2

dx

− (2 x − 1)

dy + ( x − 1) y = 0 . dx

y = x is a part of C.F. of the solution of equation x

4.

d2 y

d2 y 2

dx

 d2 y  dy − y = ( x − 1)  2 − x + 1 . dx  dx 

2 2  + 1 + cot x − 2  y = x cos x is a linear differential equation of second  x x 

order.

A nswers Fill in the Blank(s) 1.

m (m − 1) + Pmx + Qx2 = 0

2. 1 + P + Q = 0

3.

1− P + Q = 0

4. m2 + Pm + Q = 0

5.

P + Qx = 0

6. 2 + 2 Px + Qx2 = 0

True or False 1.

F

2. T

3.

T

4. T

7.8 Guidelines of the Procedure for the Solution of Linear Differential Equations of Second Order Before closing this chapter we give the guidelines of the procedure to be applied for the solution of linear differential equation of the second order. 1.

First of all put the differential equation in the standard form d2 y 2

dx

+P

dy + Qy = R . dx

2.

Then try to find by inspection an integral belonging to the complementary function of the given differential equation. If it is found, then proceed as in article 7.2.

3.

If a part of the C.F. is not found by inspection, then find the value of 1 dP 1 2 Q− − P . If it is a constant or a constant divided by x2 , then the normal 2 dx 4 form of the equation is easily integrable and so proceed as in article 7.4.

D-194

4.

In case the methods given in (2) and (3) do not succeed, the method of change of independent variable given in article 7.5 may be tried. Q In this method we often choose z such that = some suitable constant. (dz / dx)2

5.

In some cases the method of operational factors solves the equation easily.

6.

If the students are instructed in the question to solve the equation by the method of variation of parameters, then only the method of variation of parameters must be applied.

¨

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8 P artial D ifferential E quations of T he F irst O rder

8.1 Partial Differential Equations quations which contain one or more partial derivatives are called Partial Differential Equations. Such equations arise in geometry and physics when the number of independent variables in the problem under consideration is two or more. Whenever we consider the case of two independent variables, x and y will usually be taken as the independent variables and z as the dependent variable. The partial differential ∂z ∂z coefficients will be denoted by p and q respectively. The second order partial , ∂x ∂y

E

derivatives are denoted by r, s, t, so that ∂2 z ∂x2

= r,

∂2 z ∂2 z = s, = t. ∂x ∂y ∂ y2

Some examples of partial differential equations are : pz − qz = z 2 + ( x + y)2 ,

…(1)

p tan x + q tan y = tan z ,

…(2)

p2 + q2 − 2 px − 2 q y + 2 xy = 0 ,

…(3)

D-196

p2 − q y2 = y2 − x2 ,

…(4) …(5)

r − 2 s + t = sin (2 x + 3 y), y2

∂2 z 2

∂x

2

− 2 xy

y ∂z x2 ∂z ∂2 z ∂2 z + x2 = + ⋅ 2 x ∂x y ∂y ∂x ∂y ∂y

…(6)

Order and Degree of a Partial Differential equation. As in the case of ordinary differential equations, we define the order of a partial differential equation to be the order of the highest order derivative occurring in the equation. The degree of a partial differential equation is the degree of the highest order derivative which occurs in it after the differential equation has been rationalised (i. e., made free from radicals and fractions so far as derivatives are concerned). In the above mentioned examples equations (1) and (2) are of first order and first degree. Equations (3) and (4) are of first order and second degree. Equations (5) and (6) are of second order and first degree. In the present chapter we shall discuss the partial differential equations of the first order.

8.2 Derivation of a Partial Differential Equation Partial differential equations can be derived in two ways : (a)

By the elimination of arbitrary constants from a relation between x, y and z

(b)

By the elimination of arbitrary functions of these variables.

Now we illustrate these methods. (a) By the elimination of arbitrary constants: Let z be a function of x and y such that f ( x, y, z , a, b) = 0 ,

…(1)

where a and b denote arbitrary constants. Differentiating (1) partially with respect to x and y, we get the relations ∂f ∂f ∂z ∂f ∂f + ⋅ = 0 i. e., + p=0 ∂x ∂z ∂x ∂x ∂z and

∂f ∂f ∂z ∂f ∂f + ⋅ = 0 i. e., + q = 0. ∂y ∂z ∂y ∂y ∂z

…(2) …(3)

By means of the three equations (1), (2) and (3) two constants a and b can be eliminated and we obtain a relation of the form …(4) F ( x, y, z , p, q) = 0 . This shows that the system of surfaces (1) gives rise to a partial differential equation of the first order given by (4).

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We observe that if the number of constants to be eliminated is equal to the number of independent variables then the derived partial differential equation is of the first order. But if the number of constants to be eliminated is greater than the number of independent variables then, in general, the derived partial differential equations will be of the second or higher order. Note:

(b) By the elimination of arbitrary functions: Suppose we have a relation between x, y and z of the type …(1)

f (u, v) = 0,

where u and v are known functions of x, y and z and f is an arbitrary function of u and v. This relation can also be expressed in the form u = φ (v), where φ is arbitrary. Differentiating (1) partially with respect to each of the independent variables x and y regarding z as dependent variable, we get

and Eliminating

∂f  ∂u ∂u  ∂f  ∂v ∂v  + p + p =0   + ∂u  ∂x ∂z  ∂v  ∂x ∂z 

…(2)

∂f  ∂u ∂u  ∂f  ∂v ∂v  + + q + q  = 0.   ∂u  ∂y ∂z  ∂v  ∂y ∂z 

…(3)

∂f ∂f and between (2) and (3), we get ∂u ∂v ∂u   ∂u +p    ∂x ∂z 

 ∂v ∂v  ∂v   ∂u ∂u   ∂v +q  = +q   +p    x ∂ z ∂ ∂ ∂ ∂ ∂ y z y z    

or

 ∂u ∂v ∂v ∂u  −   p+  ∂y ∂z ∂y ∂z 

or

Pp + Qq = R

where

P=

∂ (u, v) ∂u ∂v ∂v ∂u − = , ∂ y ∂ z ∂ y ∂ z ∂ ( y, z )

Q=

∂v ∂u ∂u ∂v ∂ (u, v) − = , ∂x ∂z ∂x ∂z ∂ (z , x)

R=

∂u ∂v ∂u ∂v ∂ (u, v) − = ⋅ ∂x ∂y ∂y ∂x ∂ ( x, y)

and

∂u ∂v ∂u ∂v  ∂v ∂u ∂u ∂v  ⋅ − − ⋅   q=  ∂x ∂z ∂x ∂z  ∂x ∂y ∂y ∂x …(4)

The equation (4) is a partial differential equation of the first order. If the given relation between x, y, z contains two arbitrary functions then the derived partial differential equation will contain partial derivatives of an order higher than two, except in particular cases. Note 1:

It should be observed that the partial differential equation (4) derived in (b) is a linear equation i. e., powers of p and q are both unity while the partial differential equation (4) derived in (a) need not be linear. Note 2:

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Example 1:

Form a partial differential equation by the elimination of the constants h and k

from ( x − h)2 + ( y − k )2 + z 2 = c 2 . Solution:

…(1)

Differentiating (1) partially w.r.t. x and y, we get and

x − h + zp = 0

y − k + zq = 0.

Putting the values of x − h and y − k from the last two equations in the given equation (1), we get z 2 ( p2 + q2 + 1) = c 2 , which is the required partial differential equation. Example 2:

Form a partial differential equation by eliminating a, b, c from x2 a2

Solution:

and

y2

+

+

b2

z2 c2

= 1.

Differentiating (1) partially w.r.t. x and y, we get x z + p=0 2 a c2 y z + q = 0. b2 c 2

…(1)

…(2) …(3)

Since the relations (1), (2) and (3) are not sufficient to eliminate the constants a, b and c so we require one more relation. Differentiating (2) partially w.r.t. x, we get 1 2

a

+

p2 c

2

+

z

r = 0.

c2

…(4)

Multiplying (4) by x and then subtracting (2) from it, we get 1 {xzr + xp2 − pz } = 0 c2 or

pz = xp2 + xzr.

Thus after the elimination of a, b and c we obtain a partial differential equation of order 2. Note:

In this case one more partial differential equation can also be obtained.

Differentiating (3) partially w.r.t. y, we get 1 2

b

+

q2 c

2

+

z c2

t = 0.

Multiplying it by y and then subtracting (3) from it, we get qz = yq2 + yzt, which is also a partial differential equation of order 2.

D-199 Example 3:

Form a partial differential equation by eliminating the arbitrary function φ from z = e ny φ ( x − y).

Solution:

…(1)

Differentiating (1) partially w.r.t. x and y, we get p = e ny φ′ ( x − y),

and

q=ne

ny

…(2)

φ ( x − y) − e

ny

φ′ ( x − y).

…(3)

From (1), (2) and (3), we get q = nz − p

p + q = nz ,

or

which is the required partial differential equation of order one. Example 4:

Form a partial differential equation by eliminating the functions f and F from …(1)

z = f ( x + iy) + F ( x − iy). Solution:

and

Differentiating (1) partially w.r.t. x and y, we get ∂z = f ′ ( x + iy) + F ′ ( x − iy) ∂x ∂z = if ′ ( x + iy) − iF ′ ( x − iy) ∂y

…(2) …(3)

Differentiating (2) and (3) partially w.r.t. x and y respectively, we get ∂2 z ∂x2 ∂2 z

and

∂ y2 ∂2 z

Hence

∂y

2

= f ′′ ( x + iy) + F ′′ ( x − iy), = − f ′′ ( x + iy) − F ′′ ( x − iy).

=−

∂2 z 2

∂x

or

∂2 z 2

∂x

+

∂2 z ∂ y2

= 0,

which is the required partial differential equation of the second order.

Comprehensive Exercise 1 Form partial differential equations by eliminating arbitrary constants from the following relations : 1.

z = ( x + a) ( y + b).

2.

z = ax + a2 y2 + b.

3. 4.

(Lucknow 2011; Bundelkhand 13) (Lucknow 2008)

1 z = ax e + a2 e2 y + b. 2 z = A e pt sin px. y

Form partial differential equations by eliminating the arbitrary functions from the following equations : 5.

z = f ( y / x).

(Kanpur 2010)

D-200

6. 7. 8.

z = f ( x + ay) + F ( x − ay). 1 z = y2 + 2 f  + log y ⋅ x 

(Lucknow 2009)

f ( x + y + z , x2 + y2 − z 2 ) = 0 .

(Agra 2001)

A nswers 1 1. 4.

z = pq ∂2 z ∂x2

7. x2

+

∂2 z ∂t2

=0

∂z ∂z + y = 2 y2 ∂x ∂y

2. q = 2 yp2

3. q = px + p2

5.

6.

px + qy = 0

∂2 z

∂2 z = a2 ∂ y2 ∂x2

8. ( y + z ) p − (z + x) q = x − y

8.3 Linear Partial Differential Equation Definition: A partial differential equation is said to be linear if the dependent variable z and all its partial differential coefficients occur in it in first degree. A partial differential equation is said to be non-linear if some or all the partial differential coefficients appearing in it do not occur in first degree. Every linear partial differential equation is necessarily of first degree but a partial differential equation of first degree may or may not be linear. For example, the partial differential equation 2

∂2 z

∂z  ∂z  +4 + 6z = 9  +5 2   ∂ ∂y x ∂x

is of first degree but it is not linear. Below we give some examples of linear partial differential equations. (i)

The partial differential equation ∂z ∂z ( x2 + y2 ) + 3x y + ( x + y) z = e x + y ∂x ∂y is a linear partial differential equation of first order.

(ii) The partial differential equation x

∂2 z ∂x2

+ (3 x + 4 y)

∂2 z ∂2 z ∂z ∂z + ex + 5x +6y + 9 z = x2 + y2 2 ∂x ∂y ∂ x ∂y ∂y

is a linear partial differential equation of second order. (iii) The partial differential equation ∂2 u 2

∂x

+

∂2 u ∂ y2

=0

is a linear partial differential equation of second order.

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(iv) The partial differential equation ∂2 u ∂x2

+

∂2 u ∂ y2

+

∂2 u ∂z 2

=0

is a linear partial differential equation of second order. (v) The partial differential equation ∂3 z

x4

∂x3

+ y2

∂2 z ∂z ∂z +x + y + 6 z = 9 ( x2 + y2 ) ∂x ∂y ∂x ∂y

is a linear partial differential equation of third order. (vi) The partial differential equation ∂z ∂z x + y = 3x + 4 y ∂x ∂y is a linear partial differential equation of first order. (vii) The partial differential equation ∂4 z 4

∂x

−

∂4 z ∂ y4

=0

is a linear partial differential equation of order 4. (viii) The partial differential equation ∂u ∂u ∂u x + y +z =0 ∂x ∂y ∂z is a first-order linear partial differential equation in three variables. Here are some examples of partial differential equations which are not linear. (i)

The partial differential equation p2 x + q2 y = z is not linear.

(ii)

The partial differential equation ( x2 − y2 ) pq − x y ( p + q) − 1 = 0 is not linear.

(iii) The partial differential equation

∂2 z

+4

∂x2 However the partial differential equation ∂2 z ∂x2

+4

∂z ∂z ∂2 z ⋅ +5 = 0 is not linear. ∂x ∂y ∂ y2

∂2 z ∂2 z +5 = 0 is linear. ∂x ∂y ∂ y2

8.4 Classification of Partial Differential Equations of First Order into Linear, Semi-Linear, Quasi-Linear and Non-Linear A first order partial differential equation in two variables in its most general form is given by F ( x, y, z , p, q) = 0 ,

…(1)

D-202

where and

∂z ∂x ∂z q= ⋅ ∂y p=

In this differential equation z is dependent variable and x, y are independent variables. When the function F is not a linear expression in p and q, the equation (1) is said to be non-linear. (a) Non-linear partial differential equation of first order: A first order partial differential equation in which dependent variable is z and is a function of two independent variables x and y is called a non-linear equation if the partial ∂z ∂z derivatives i. e., p and i. e., q do not occur in it in first degree. ∂x ∂y For example, the partial differential equations of the first order such as ( x + y) ( p + q)2 + ( x − y) ( p − q)2 = 1, x2 y3 p2 q = z 3 , ( x2 + y2 ) ( p2 + q2 ) = 1, z ( p2 − q2 ) = x − y, z 2 ( p2 + q2 ) = x2 + y2 etc., are all non-linear equations. (b) Quasi-linear partial differential equation of first order: A first order partial differential equation F ( x, y, z , p, q) = 0 in which dependent variable is z and is a function of two independent variables x and y is called a quasi-linear equation if the function F is a linear expression in p and q but not necessarily linear in z. A quasi-linear partial differential equation of first order is of the form ∂z ∂z f ( x, y, z ) + g ( x, y, z ) = h ( x, y, z ) ∂x ∂y where the functions f and g depend on z also. Thus the first order partial differential equations such as ∂z ∂z (x + y + z) + xy + xz = 3 x2 + 5 y2 + 6 z 2 , ∂x ∂y ( x2 + y2 ) 2z

∂z ∂z + 4 xyz = 3z + e x + y , ∂x ∂y

∂z ∂z +5y = 6 z 2 + log x + e y etc., ∂x ∂y

are all quasi-linear equations. (c) Semi-linear partial differential equation of first order: A first order partial differential equation F ( x, y, z , p, q) = 0 in which dependent variable is z and is a function of two independent variables x and y is called a semi-linear equation if it is of the form ∂z ∂z f ( x, y) + g ( x, y) = h ( x, y, z ). ∂x ∂y

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In a semi-linear partial differential equation of first order the coefficients of

∂z ∂z and ∂x ∂y

are functions of x and y only and they do not depend on z. The terms that do not involve ∂z ∂z and contain some terms that are not of first degree in z. ∂x ∂y Thus the first order partial differential equations such as ∂z ∂z ( x + y2 ) + x log y = 2 z 2 x + xy + e x , ∂x ∂y p cos ( x + y) + q sin ( x + y) = z 3 + sin x + e y , ∂z ∂z ( y3 x − 2 x4 ) + (2 y4 − x3 y) = 9 ( x2 − y2 ) log z , etc., ∂x ∂y are all semi-linear equations. (d) Linear partial differential equation of first order: A first order partial differential equation F ( x, y, z , p, q) = 0 in which dependent variable is z and is a function of two independent variables x and y is called a linear equation if in this differential equation the dependent variable z and its partial ∂z ∂z differential coefficients and all occur in first degree. ∂x ∂y Thus a linear partial differential equation of first order is of the form f ( x, y)

∂z ∂z + g ( x, y) + h ( x, y) z = c ( x, y) ∂x ∂y

where f ( x, y), g ( x, y), h ( x, y) and c ( x, y) are functions of x and y only and they do not contain any term of z. For example, the first order partial differential equations such as ∂z ∂z ( x2 + y3 ) + (3 x + 5 y2 ) + ( x − y) z = sin ( x + y), ∂x ∂y p cos ( x + y) + q sin ( x + y) = z + e y sin x, log x

∂z ∂z +e y = x ( y − z ), ∂x ∂y

p + 3 q = 5 z + tan ( y − 3 x), xy 2 ∂u ∂u ∂u ∂u ∂u ∂u x + y +z = 5u + ,x + y2 + z3 = u + xyz, ∂x ∂y ∂z z ∂x ∂y ∂z ( y3 x − 2 x4 ) p + (2 y4 − x3 y) q = 9 z ( x3 − y3 ), etc., are all linear equations. Remark:

In many standard text books on partial differential equations the concepts of

quasi-linear and semi-linear differential equations of first order are not introduced. There we find only two categories of first order partial differential equations— one

D-204

linear and the other non-linear. According to this classification a partial differential equation is said to be linear if all the partial derivatives occurring in it appear only in first degree and there is no restriction on the dependent variable, it may or may not occur in first degree. Also, a differential equation is said to be non-linear if it is not linear. Thus a first order partial differential equation F ( x, y, z , p, q) = 0 is linear if in this differential equation the partial derivatives p and q occur only in first degree and there is no restriction on the dependent variable z, it may occur in any form. Accordingly every first order partial differential equation of the form f ( x, y, z )

∂z ∂z + g ( x, y, z ) = h ( x, y, z ) ∂x ∂y

is linear where f , g, h are any functions of x, y, z . Thus the partial differential equations such as ( x2 − yz ) p + ( y2 − zx) q = z 2 − xy, (mz − ny) p + (nx − lz ) q = ly − mx, ( y2 + z 2 − x2 ) p − 2 xyq + 2 zx = 0 , x ( y − z ) p + y (z − x) q = z ( x − y), etc., are all linear. On the other hand the partial differential equations such as x2 p2 + y2 q2 = z 2 , p2 + q2 = 3 pq, pq = 5, ( y − x) (qy − px) = ( p − q)2 , p2 = z 2 (1 − pq), p3 + q3 = 27 z etc., are all non-linear.

8.5 Lagrange’s Linear Partial Differential Equation The partial differential equation Pp + Qq = R , where P, Q and R are any functions of x, y, and z is called Lagrange’s linear partial differential equation of first order. Thus Lagrange’s linear partial differential equation is of the form ∂z ∂z f ( x, y, z ) + g ( x, y, z ) = h ( x, y, z ), ∂x ∂y where f , g, h are any functions of x, y, z . The partial differential equations such as ( y2 + z 2 − x2 ) p − 2 xyq + 2 zx = 0 , p cos ( x + y) + q sin ( x + y) = z , ∂z ∂z x ( y − z) + y (z − x) = z ( x − y), etc., ∂x ∂y are all Lagrange’s linear partial differential equations of first order.

D-205

Comprehensive Exercise 2 Tell whether the following partial differential equations are linear or non-linear: 1.

2.

3.

4.

5.

∂2 u

x2

∂x2

∂3 z 3

∂x 4

+3

∂2 z ∂x2

∂2 z 2

∂x

∂2 z ∂x2

+ y2 ∂2 z 2

∂x

+x

−4

∂2 u

∂2 u + z2 = 3 u. ∂ y2 ∂z 2 +4

∂2 z ∂ y2

∂z ∂z ⋅ + 5z = 9e x . ∂x ∂y

+9

∂2 z ∂z ∂z +5 +6 = log x. ∂x ∂y ∂x ∂y

y ∂2 z 4x = − ⋅ 2 ∂x ∂y y x2 2

−

 ∂z  ∂2 z − 2   = ( y − 1) e x . ∂x ∂y  ∂y 

Which of the following partial differential equations of first order are non-linear, quasi-linear, semi-linear or linear ? ∂z ∂z + ( x − y) + xz = 3 x + 4 y. ∂x ∂y

6.

( x2 + y2 )

7.

zy2 p − xyq = x (z − 2 y).

8.

∂z ∂z ( y2 − x2 ) − xy = x2 + y2 − xz 2 . ∂x ∂y

9.

∂z ∂z ( y3 + x3 ) + 4 x2 y = x3 z − 4 y2 + yz . ∂x ∂y ∂z ∂z ∂z ∂z ⋅ +5 +6 + 3 z = 9 ( x2 + y2 ). ∂x ∂y ∂x ∂y

10.

4

11.

( x2 − yz ) p + ( y2 − zx) q = z 2 − xy.

12.

(mx − ny) p + (nx + my) q = z + ly − mx2 .

13.

x2

14.

x

15.

(z 2 − 2 yz − y2 ) p + x ( y + z ) q = x ( y − z ).

∂z ∂z + ( y3 + x 3 ) = x 2 + y2 + z 2 . ∂x ∂y 2

 ∂u  ∂u ∂u + y   +3 = x2 + y2 + z 2 . ∂x ∂z  ∂y 

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A nswers 2 1. 4. 7. 10. 13.

Linear Linear Quasi-linear Non-linear Semi-linear

2. 5. 8. 11. 14.

Non-linear Non-linear Semi-linear Quasi-linear Non-linear

3. 6. 9. 12. 15.

Linear Linear Linear Linear Quasi-linear

8.6 Equation Equivalent to the Linear Equation A partial differential equation which is linear in p and q is of the type Pp + Qq = R,

…(1)

where P, Q, R are functions of x, y, z . Let any relation u = a be an integral of (1). Differentiating it partially with respect to x and y, we get ∂u ∂u ∂u ∂u + p = 0 and + q = 0. ∂x ∂z ∂y ∂z These give

p= −

∂u ∂u ∂u ∂u / ,q=− / ⋅ ∂x ∂z ∂y ∂z

Putting these values of p and q in (1) it changes to ∂u ∂u ∂u P +Q +R = 0. ∂x ∂y ∂z

…(2)

Hence, if u = a satisfies (1), u = a also satisfies (2). ∂u Conversely, dividing (2) by and substituting p and q for their values given above we ∂z see that if u = a is an integral of (2), it is also an integral of (1). Thus equation (2) can be taken as equivalent to equation (1).

8.7 Lagrange’s Solution of the Linear Partial Differential Equation of First Order The first systematic theory of the linear partial differential equations was given by Lagrange. For that reason the equation Pp + Qq = R is referred to as Lagrange’s equation. The method of solving a linear equation of this form is contained in the theorem given below : Theorem: is

The general solution of the linear partial differential equation Pp + Qq = R

…(1)

f (u, v) = 0,

…(2)

D-207

where f is an arbitrary function and …(3)

u ( x, y, z ) = c1 and v ( x, y, z ) = c2 form a solution of the equations dx dy dz = = ⋅ P Q R

…(4) (Lucknow 2010)

Proof: We shall give a purely analytic proof of this theorem. If the relation u ( x, y, z ) = c1 satisfies the equations (4) then the equations dx dy dz ∂u ∂u ∂u dx + dy + dz = 0 and = = ∂x ∂y ∂z P Q R must be compatible i. e., we should have ∂u ∂u ∂u P +Q +R = 0. ∂x ∂y ∂z

…(5)

Similarly we should have ∂v ∂v ∂v P +Q +R = 0. ∂x ∂y ∂z

…(6)

Solving the equations (5) and (6), we get P Q R ⋅ = = ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v − − − ∂y ∂z ∂z ∂y ∂z ∂x ∂x ∂z ∂x ∂y ∂y ∂x

…(7)

Earlier we have seen that the relation f (u, v) = 0 gives the partial differential equation  ∂u ∂v ∂u ∂v  −   p+  ∂y ∂z ∂z ∂y 

∂u ∂v ∂u ∂v  ∂u ∂v ∂u ∂v  − ⋅ −   q=  ∂z ∂x ∂x ∂z  ∂x ∂y ∂y ∂x

…(8)

Substituting from the equations (7) into the equation (8), we find that (2) is a solution of the equation (1) if u and v are given by the equations (3). In place of f (u, v) = 0 the functional relation can be written as u = φ (v), where φ denotes an arbitrary function. Note:

It is called Lagrange’s solution of the linear equation (1) and the equations (4) are called Lagrange’s auxiliary equations or Lagrange’s subsidiary equations.

8.8 The Linear Equation Containing more than two Independent Variables The generalisation of Lagrange’s method is as follows : Let the linear equation with n independent variables x1, x2 , ... , xn be P1 p1 + P2 p2 + P3 p3 + ... + Pn pn = R.

…(1) ∂z where P1, P2, ..., Pn and R are functions of x1, x2, ... , x n and z. Here pi denotes for ∂x i i = 1, 2, ..., n.

D-208

Then the general solution of (1) is given by f (u1, u2 , ... , un) = 0 where ui ( x1, x2 , ... , x n, z ) = c i, i = 1, 2, ..., n are any independent integrals of the auxiliary equations dx1 dx2 dx dz = = ... = n = ⋅ P1 P2 Pn R

8.9 Geometrical Interpretation of Lagrange’s Linear Equation (Lucknow 2011)

Lagrange’s linear equation is Pp + Qq = R or Pp + Qq + (− 1) R = 0 .

…(1)

Since d.r.’s of the normal at the point ( x, y, z ) on the surface f ( x, y, z ) = c are given by ∂f ∂f ∂f ∂f ∂f ∂f ∂f , , or − / ,− / ,− 1 ∂x ∂y ∂z ∂x ∂z ∂y ∂z i. e.,

∂z ∂z , , − 1 or p, q, − 1, ∂x ∂y

hence the equation (1) shows that the normal to a certain surface is perpendicular to the line whose d.r.’s are P, Q, R. But we know that the simultaneous equations dx dy dz = = P Q R

…(2)

represent a family of curves in space such that the direction ratios of the tangent to any member of this family at any point ( x, y, z ) are P, Q, R. If u = a and v = b are two independent integrals of (2) then f (u, v) = 0 represents a surface through such curves. Through every point of such a surface passes a curve of the family, lying wholly on the surface. Thus the general solution of (1) is the family of surfaces such that the normal to a surface at any point must be perpendicular to the tangent to a curve of the family represented by (2).

8.10 The Equation Pp + Qq = R Represents a Family of Surfaces Orthogonal to the Family of Surfaces Represented by P dx + Q dy + R dz = 0 (If Integrable) We know that d.r.’s of the normal at the point ( x, y, z ) to a surface of the family represented by Pp + Qq = R are p, q, − 1.

D-209

Also d.r.’s of the normal at the point ( x, y, z ) to a surface of the family represented by P dx + Q dy + R dz = 0 are P, Q, R. Since the equation Pp + Qq + (− 1) R = 0 shows that the two lines whose d.r.’s are p, q, − 1 and P, Q, R are prependicular, hence the surfaces represented by Pp + Qq = R are orthogonal to the surfaces represented by P dx + Q dy + R dz = 0.

8.11 Integral Surfaces Passing Through a Given Curve Now we shall indicate how the general solution of a linear partial differential equation may be used to find the integral surface passing through a given curve. Suppose that we have obtained two solutions u = a and v = b

…(1)

of the auxiliary equations (4) of 8.7. Then we know that the general solution of the corresponding linear equation is of the form f (u, v) = 0

…(2)

arising from a relation f (a, b) = 0

…(3)

between the constants a and b. We have to consider the problem of determining the function f in special cases. If we want to find the integral surface passing through the curve whose parametric equations are x = x (t), y = y (t), z = z (t), t being a parameter, then the solutions in (1) must be such that u ( x (t), y (t), z (t)) = a, v ( x (t), y (t), z (t)) = b. Eliminating the single variable t from these two equations we find a relation of the type (3). The required solution is then given by the equation (2).

8.12 Working Rule to Find the Solution of Lagrange’s Equation Pp + Qq = R Form the auxiliary equations dx dy dz = = ⋅ P Q R Find two independent integrals of these auxiliary equations, say u = a and v = b. Then the general solution of the partial differential equation Pp + Qq = R is given by f (u, v) = 0 , where f is an arbitrary function or it can also be written as u = φ (v), where φ is an arbitrary function.

D-210

Solve (z 2 − 2 yz − y2 ) p + ( xy + zx) q = xy − zx.

Example 5:

…(1) (Lucknow 2007)

Solution:

The Lagrange’s auxiliary equations of the given differential equation (1) are dy dx dz = = ⋅ 2 2 xy + zx xy − zx z − 2 yz − y

Taking the last two members, we get ( y − z ) dy = ( y + z ) dz 2

or

y dy − (z dy + y dz ) − z dz = 0 .

2

Integrating, y − 2 yz − z = c1. Again choosing x, y, z as multipliers, we get x dx + y dy + z dz each fraction = 0 Integrating,

⇒

x dx + y dy + z dz = 0.

x2 + y2 + z 2 = c2 .

Hence the general solution of (1) is given by f ( y2 − 2 yz − z 2 , x2 + y2 + z 2 ) = 0 or Example 6:

y2 − 2 yz − z 2 = φ ( x2 + y2 + z 2 ). Solve (mz − ny) p + (nx − lz ) q = ly − mx.

…(1)

(Rohilkhand 2008; Lucknow 09; Purvanchal 10; Avadh 10)

Solution:

The Lagrange’s auxiliary equations of the given differential equation (1) are dy dx dz = = ⋅ mz − ny nx − lz ly − mx

Choosing x, y, z as multipliers, we get x dx + y dy + z dz each fraction = 0

⇒ x dx + y dy + z dz = 0.

Integrating, x2 + y2 + z 2 = c1. Again choosing l, m, n as multipliers, we get l dx + m dy + n dz each fraction = 0

⇒ l dx + m dy + n dz = 0.

Integrating, l x + my + nz = c2 . Hence the general solution of (1) is given by f ( x2 + y2 + z 2 , l x + my + nz ) = 0 , where f is an arbitrary function. Example 7: Solve ( y2 + z 2 − x2 ) p − 2 xyq + 2 zx = 0 .

…(1) (Lucknow 2009; Avadh 10, 11)

Solution:

The Lagrange’s auxiliary equations of (1) are

D-211

dx 2

2

2

y +z −x

=

dy dz = ⋅ − 2 xy − 2 zx

Taking the last two members, we get dy dz = ⋅ y z Integrating, log

y = log c1 z

or

y = c1. z

Again choosing x, y, z as multipliers, we get x dx + y dy + z dz each fraction = ⋅ − x ( x2 + y2 + z 2 ) ∴

x dx + y dy + z dz 2

2

2

− x (x + y + z ) 2

2

=

dz − 2 zx

⇒ 2

2

Integrating, log ( x + y + z ) = log z + log c2

or

x dx + y dy + z dz 2

2

x + y +z

2

=

dz ⋅ z

x2 + y2 + z 2 = zc2 .

Hence the general solution of (1) is  y x2 + y2 + z 2   = 0 , or f  ,  z  z 

 y x2 + y2 + z 2 = zφ  ,  z 

where f and φ are arbitrary functions. Example 8:

Solve ( y + z ) p + (z + x) q = x + y.

…(1)

(Rohilkhand 2008; Kanpur 08; Purvanchal 07; Bundelkhand 13)

Solution:

∴

The Lagrange’s auxiliary equations of (1) are dy dx dz = = ⋅ y+z z+x x+ y dx − dy dy − dz dx + dy + dz = = ⋅ y−x z− y 2 (x + y + z)

Taking the first two members, we get log ( y − x) = log (z − y) + log c1 or y − x = c1 (z − y). Again taking the first and the last members, we get − 2 log ( x − y) = log ( x + y + z ) − log c2 or

( x − y)2 ( x + y + z ) = c2 .

Hence the general solution of (1) is given by  y−x  f  , ( x − y)2 ( x + y + z )  = 0 . z − y   ( y − z) (z − x) x− y Example 9: Solve p+ q= ⋅ yz zx xy Solution:

The auxiliary equations of (1) are dy dx dz = = ⋅ x ( y − z ) y (z − x) z ( x − y)

…(1) (Rohilkhand 2009)

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1 1 1 dx + dy + dz dx + dy + dz x y z Each fraction = and also = ⋅ 0 0 1 1 1 ∴ dx + dy + dz = 0 and dx + dy + dz = 0 . x y z Integrating, x + y + z = c1 and xyz = c2 . Hence the general solution of (1) is given by f ( x + y + z , xyz ) = 0 . Example 10:

Solve p + 3 q = 5 z + tan ( y − 3 x).

…(1) (Rohilkhand 2007, 10; Avadh 14)

Solution:

The auxiliary equations of (1) are dx dy dz = = ⋅ 1 3 5 z + tan ( y − 3 x)

Taking the first two members, we get dy − 3 dx = 0. y − 3 x = c1.

∴

Again taking the first and the last members, we get dz dx = ⋅ 5 z + tan c1 log (5 z + tan c1) = 5 x + log c2 e −5 x {5 z + tan ( y − 3 x)} = c2 .

∴ or

Hence the general solution of (1) is given by e −5 x {5 z + tan ( y − 3 x)} = f ( y − 3 x). Example 11: Solution:

Solve x2 ( y − z ) p + y2 (z − x) q = z 2 ( x − y).

The auxiliary equations of (1) are dx dy dz = = ⋅ x2 ( y − x) y2 (z − x) z 2 ( x − y) dx

dy

dz

dx dy dz + + x y z Each fraction = and also = ⋅ 0 dx dz dx dy dz + + = 0. ∴ + + = 0 and 2 2 2 x y z x y z x2

Integrating,

+

y2 0 dy

+

z2

1 1 1 + + = c1 and x y z = c2 . x y z

Hence the general solution of (1) is given by  1 1 1  f  + + , x y z  = 0. y z  x 

…(1)

(Agra 2001)

D-213 Example 12:

Solve x ( y2 + z ) p − y ( x2 + z ) q = z ( x2 − y2 ).

…(1)

(Lucknow 2011; Purvanchal 07, 11, 14)

Solution:

The auxiliary equations of (1) are dy dx dz = = ⋅ 2 2 2 x ( y + z ) − y ( x + z ) z ( x − y2 )

Choosing 1/ x, 1/ y, 1/ z as multipliers, we get dx dy dz + + = 0. x y z Integrating, log x + log y + log z = log c1 or

xyz = c1. Again choosing x, y, − 1 as multipliers, we get x dx + y dy − dz = 0. 2

Integrating, x + y2 − 2 z = c2 . Hence the general solution of (1) is given by Example 13:

f ( x y z , x2 + y2 − 2 z ) = 0 .

Find the equation of the integral surface of the differential equation 2 y (z − 3) p + (2 x − z ) q = y (2 x − 3), 2

…(1)

2

which passes through the circle z = 0 , x + y = 2 x. Solution:

The auxiliary equations of (1) are dy dx dz = = ⋅ 2 y (z − 3) 2 x − z y (2 x − 3)

Taking the first and the third members, we get (2 x − 3) dx = 2 (z − 3) dz . 2

Integrating, x − 3 x − z 2 + 6 z = a. 1 Again, using , y, − 1 as multipliers, we get 2 1 dx + y dy − dz = 0 . 2 1 1 2 1 Integrating, x+ y − z = b or x + y2 − 2 z = b. 2 2 2 The parametric equations of the circle are x = t, y = √ (2 t − t2 ), z = 0 . Putting these values in the equations (2) and (3), we get t2 − 3 t = a and t + (2 t − t2 ) = b. Eliminating t from these, we find the relation a + b = 0, showing that the required integral surface is ( x2 − 3 x − z 2 + 6 z ) + ( x + y2 − 2 z ) = 0 or

x2 + y2 − z 2 − 2 x + 4 z = 0 .

…(2)

…(3)

D-214

Comprehensive Exercise 3 Solve the following equations : 1.

xzp + yzq = xy.

2.

p tan x + q tan y = tan z . 2

2

(Avadh 2006; Lucknow 10) (Kanpur 2007; Avadh 08, 10)

2

3.

x p+ y q = z .

4.

yzp + zxq = xy.

(Avadh 2007) (Lucknow 2011)

2

5. 6.

y z p + zxq = y2 . x 2

(Lucknow 2007; Avadh 10, 11)

2

2

( x − yz ) p + ( y − z x) q = z − xy. 2

2

(Lucknow 2010)

2

7.

z − px − qy = a √ ( x + y + z ).

8.

z ( xp − yq) = y2 − x2

9.

(3 x + y − z ) p + ( x + y − z ) q = 2 (z − y).

(Rohilkhand 2011)

(Avadh 2014)

 dy dx dz = = ⋅ Hint. 3 x + y − z x + y − z 2 ( z − y)  (i) Choose −1, 3, 1 as multipliers. dx − dy + dz dx + dy − dz  (ii) =  … …  10. p2 + p3 = 1 + p1. ∂u ∂u ∂u 11. x + y +z = xyz . ∂x ∂y ∂z

(Lucknow 2006; Kanpur 12)

12. Find the integral surface of the linear partial differential equation x ( y2 + z ) p − y ( x2 + z ) q = ( x2 − y2 ) z which contains the line x + y = 0 , z = 1. 13. Find the general integral of the partial differential equation (2 xy − 1) p + (z − 2 x2 ) q = 2 ( x − yz ) and also the particular integral which passes through the line x = 1, y = 0 . 14. Find the family of surfaces orthogonal to the family of surfaces given by the differential equation ( y + z ) p + (z + x) q = x + y. 15. Find the inegral surface of the partial differential equation x2 p + y2 q + z 2 = 0 xy = x + y, z = 1.

which passes through the hyperbola (Avadh 2012)

D-215

A nswers 3  sin x  sin z = f   sin y  sin y 

1.

y  f  z 2 − xy,  = 0  x

2.

3.

1 1 1 1 − = f  −   x z x y

4. f ( x2 − y2 , x2 − z 2 ) = 0

5.

f ( x3 − y3 , x2 − y2 ) = 0

x − y z − x 6. f  ,  =0  y − z y − z

7.

 x  z1− a = {z + √ ( x2 + y2 + z 2 )} f    y

8. f ( x y, x2 + y2 + z 2 ) = 0

9.

( x − y + z )2 = ( x + y − z ) f ( x − 3 y − z )

10. f ( x1 + z , x1 + x2 , x1 + x3 ) = 0

 x y  11. f  , , xyz − 3 u = 0  y z 

12. x2 + y2 + 2 xyz − 2 z + 2 = 0

13. x2 + y2 − xz − y − z − 1 = 0

14. xy + yz + zx = c

15.

1 1 2 + + =3 x y z

8.13 The Integrals of the Non-Linear Equation The complete and particular integrals: We have seen that the relation of the type f ( x, y, z , a, b) = 0

…(1)

gives rise to a partial differential equation of the first order of the form F ( x, y, z , p, q) = 0

…(2)

on the elimination of arbitrary constants a and b. Here x, y are independent variables and z is dependent variable. The relation (1) is a solution of (2). Any such relation which contains as many arbitrary constants as there are independent variables and is a solution of a partial differential equation of the first order is called a complete solution or a complete integral of that equation. A particular integral of (2) can be obtained by giving particular values to a and b in (1). The singular integral: If the envelope of the doubly infinite system of surfaces represented by (1) exists, it is also a solution of the equation (2). The reason is that the envelope of all the surfaces represented by (1) is touched at each of its points by some one of these surfaces. Hence the coordinates of any point on the envelope of surfaces (1) with the corresponding values of p and q, being indentical with the x, y, z , p, q of some point on one of these surfaces, must satisfy (2). The equation of the envelope of the

D-216

surfaces represented by (1) can be obtained by eliminating a and b between the three equations f = 0,

∂f ∂f = 0 and = 0. ∂a ∂b

This equation of the envelope is called the singular integral of the differential equation (2). It differs from a particular integral in the sense that it is not contained in the complete integral, i. e.,it cannot be obtained from the complete integral by giving particular values to the constants. The general integral: If in the equation (1), one of the constants is a function of the other, say b = φ (a), then this equation becomes …(3)

f ( x, y, z , a, φ (a)) = 0 .

It is a one-parameter subfamily of the family (1). The equation of the envelope of the family of surfaces represented by (3) is also a solution of the equation (2). It is called the general integral of (2) corresponding to the complete integral (1). The equation of the envelope of the surfaces represented by (3) is obtained by eliminating a between ∂f f ( x, y, z , a, φ (a)) = 0 and = 0. ∂a

8.14 General Method of Solution of a Non-Linear Partial Differential Equation of Order One with two Independent Variables. (Charpit’s Method) (Lucknow 2006, 11; Avadh 06)

Let the given partial differential equation be f ( x, y, z , p, q) = 0 .

…(1)

Since z depends upon x and y, it gives that dz = p dx + q dy.

…(2)

The fundamental idea in Charpit’s method is to introduce a second partial differential equaion of the first order F ( x, y, z , p, q, a) = 0 , …(3) containing an arbitrary constant a and which is such that : (i) The equations (1) and (3) can be solved to find p = p ( x, y, z , a), q = q ( x, y, z , a). (ii) Substituting these values of p and q in (2), the equation dz = p ( x, y, z , a) dx + q ( x, y, z , a) dy is integrable. If such a relation (3) has been found, the solution of equation (4), φ ( x, y, z , a, b ) = 0

…(4)

…(5)

D-217

containing two arbitrary constants a and b will be a solution of the equation (1). Also it is a complete integral of the equation (1). Now the main problem is to devise a method of finding the relation (3). Let us assume that (3) is the relation which when taken along with (1) gives those values of p and q which make (2) integrable. Differentiating (1) and (3) with respect to x, we get ∂f ∂f ∂f ∂p ∂f ∂q  + p+ + = 0 ∂x ∂z ∂p ∂x ∂q ∂x  …(6)  ∂F ∂F ∂F ∂p ∂F ∂q  and + p+ + =0  ∂x ∂z ∂p ∂x ∂q ∂x Again, differentiating (1) and (3) w.r.t. y, we get ∂f ∂f ∂f ∂p ∂f ∂q  + q+ + = 0 ∂y ∂z ∂p ∂y ∂q ∂y  …(7)  ∂F ∂F ∂F ∂p ∂F ∂q  and + q+ + =0  ∂y ∂z ∂p ∂y ∂q ∂y ∂p ∂q Eliminating from the equations in (6) and from the equations in (7), ∂x ∂y we get  ∂f ∂F ∂f ∂F   ∂f ∂F ∂f ∂F  ∂q  ∂f ∂F ∂f ∂F  − − −   + p  +   = 0,  ∂x ∂p ∂p ∂x   ∂z ∂p ∂p ∂z  ∂x  ∂q ∂p ∂p ∂q 

…(8)

 ∂f ∂F ∂f ∂F   ∂f ∂F ∂f ∂F  ∂p  ∂f ∂F ∂f ∂F  and  − − −  +q  +   = 0. …(9) ∂y  ∂p ∂q ∂q ∂p   ∂y ∂q ∂q ∂y   ∂z ∂q ∂q ∂z  Since

∂q ∂x

=

∂2 z ∂x ∂y

=

∂p ∂y

, hence adding (8) and (9) and re-arranging, we get

∂f  ∂F  ∂f ∂f  ∂F  ∂f ∂f  ∂F  ∂f +p + +q + −p −q      ∂x ∂z  ∂p  ∂y ∂z  ∂q  ∂p ∂q  ∂z  ∂f  ∂F  ∂f  ∂F + − + − = 0. …(10)    ∂p  ∂x  ∂q  ∂y This is a linear partial differential equation of the first order with x, y, z , p, q as independent variables and F as dependent variable. Lagrange’s auxiliary equations of (10) are dp dq dz dx dy dF …(11) = = = = = ⋅ ∂f ∂f ∂f ∂f ∂f ∂f ∂f ∂f 0 +p +q −p −q − − ∂x ∂z ∂y ∂z ∂p ∂q ∂p ∂q These equations are known as Charpit’s auxiliary equations. Any of the integrals of (11) will satisfy (10). If such an integral contains p or q, it can be taken as the required second relation. It should be noted that not all of Charpit’s equations (11) need be used, but that p or q must occur in the solution obtained. Of course, the simpler the integral containing p or q, or both p and q that is derived from (11), the easier will be the subsequent labour in finding the solution of (1).

D-218

Example 14:

Find a complete integral of the equation 2 zx − px2 − 2 qxy + pq = 0 .

Solution:

The given differential equation is f ( x, y, z , p, q) ≡ 2 zx − px2 − 2 qxy + pq = 0 .

…(1)

Charpit’s auxiliary equations are dp dq dz dx dy = = = = 2 2 2 z − 2 qy 0 px − pq + 2 qxy − pq x − q 2 xy − p from which it follows that

dq = 0 or q = a (a constant).

…(2)

Solving the equations (1) and (2) for p and q, we get 2 x (z − ay) p= , q = a. x2 − a Putting these values of p and q in dz = p dx + q dy, we get 2 x (z − ay) dz = dx + a dy x2 − a dz − a dy 2x or = dx. 2 z − ay x −a Integrating,

log (z − ay) = log ( x2 − a) + log b

or

z − ay = b ( x2 − a)

or

z = ay + b ( x2 − a),

which is a complete integral of (1). Example 15:

Find a complete integral of px + qy = pq. (Rohilkhand 08, 10, 14; Avadh 10, 11; Kanpur 12; Purvanchal 09)

Solution:

Here f ( x, y, z , p, q) ≡ px + qy − pq = 0 .

…(1)

Charpit’s auxiliary equations are dp dq dy dz dx = = = = p q − p ( x − q) − q ( y − p) − ( x − q) − ( y − p) from which it follows that dp dq = , or p = aq. p q Solving the equations (1) and (2) for p and q, we get y + ax q= , p = y + ax. a Putting these values of p and q in dz = p dx + q dy, we get ( y + ax) dz = ( y + ax) dx + dy a or a dz = ( y + ax) (dy + a dx). 1 Integrating, az = ( y + ax)2 + b, which is a complete integral of (1). 2

…(2)

D-219 Example 16:

Find a complete integral of ( p2 + q2 ) y = qz . (Lucknow 2007, 08; Purvanchal 10, 14; Avadh 12; Kanpur 14)

Solution:

The given differential equation is f ≡ ( p2 + q2 ) y − qz = 0 .

…(1)

Charpit’s auxiliary equations are dy dq dp dz dx = = = = − pq ( p2 + q2 ) − q2 −2 p2 y − 2 q2 y + qz −2 py −2 qy + z from which it follows that dp dq = − pq p2

or

p dp + q dq = 0

p2 + q2 = a2 (say)

or

…(2)

Solving the equations (1) and (2) for p and q, we get q=

a2 y z

, p=

q z

√ (z 2 − a2 y2 ).

Putting these values of p and q in dz = p dx + q dy, we get dz =

a2 y a √ (z 2 − a2 y2 ) dx + dy z z

z dz − a2 y dy

or

√ (z 2 − a2 y2 )

= a dx.

Integrating, √ (z 2 − a2 y2 ) = ax + b z 2 = a2 y2 + (ax + b )2 , which is a complete integral of (1).

or Example 17:

Find a complete integral of p = (qy + z )2 . (Meerut 2008; Lucknow 11; Purvanchal 08, 11; Bundelkhand 13)

Solution:

The given differential equation is

f ≡ p − (qy + z )2 = 0 .

…(1)

Charpit’s auxiliary equations are dp dq dy = = = ...... 2 p (qy + z ) 4 q (qy + z ) − 2 y (qy + z ) from which it follows that

dp dy =− ⋅ p y

Integrating, py = a.

…(2)

Putting p = a / y in (1), we get a (qy + z )2 = ⋅ y ∴

q=

1 y

  a    −z.     y  

D-220

Putting these values of p and q in dz = p dx + q dy, we get  a 1   a  dz = dx +  − z  dy   y y  y    a  y dz + z dy = a dx +   dy.  y

or

Integrating, yz = ax + 2 √ (ay) + b, which is a complete integral of (1). Find a complete integral of

Example 18:

px + qy = z (1 + pq)1 /2 . (Avadh 2013; Kanpur 14)

Solution:

Here f ≡ px + qy − z (1 + pq)1 /2 = 0 .

Charpit’s auxiliary equations are dp p − p (1 + pq)1 /2

=

dq q − q (1 + pq)1 /2

…(1) = ......

from which it follows that dp dq = ⇒ p = aq. p q

…(2)

Putting p = aq in (1), we get q (ax + y) = z (1 + aq2 )1 /2 or ∴

q2 [(ax + y)2 − az 2 ] = z 2 . z az and p = aq = q= ⋅ 2 2 1 /2 2 [(ax + y) − az ] [(ax + y) − az 2 ]1 /2

Putting these values of p and q in dz = p dx + q dy, we get z (a dx + dy) dz = √ {(ax + y)2 − az 2 } a dx + dy dz or = z √ {(ax + y)2 − az 2} Let ax + y = √ au so that a dx + dy = √ a du. √ a du dz ∴ = z √ (au2 − az 2 ) or

du √ (u2 − z 2 ) = dz z

To solve this, putting u = vz , we get v + z or or or

dv = √ (v2 − 1) − v dz dz dv = z √ (v2 − 1) − v

z

dz = − {√ (v2 − 1) + v} dv. z

dv 1 = √ (v2 z 2 − z 2 ) dz z

D-221

Integrating, 2 1  v  v log z = −  √ (v2 − 1) − log {v + √ (v2 − 1)} − +b 2 2  2

v2 v 1 + √ (v2 − 1) − log {v + √ (v2 − 1)} = b, 2 2 2 u ax + y which is a complete integral of (1), where v = = ⋅ z z√a or

log z +

Example 19: Solution:

Find a complete integral of p2 x + q2 y = z .

(Lucknow 2009)

The given differential equation is f ≡ p2 x + q2 y − z = 0.

…(1)

Charpit’s auxiliary equations are dp 2

− p+ p

=

dq −q+q

2

=

dy dx = = ... − 2 px − 2 qy

from which it follows that p2 dx + 2 px dp

=

p2 x

q2 dy + 2 qy dq q2 y

⋅

Integrating , log p2 x = log q2 y + log a. ∴

p2 x = aq2 y.

…(2)

Solving (1) and (2) simultaneously for p and q, we get 1 /2

  z q=  a y ( 1 + )  

 az and p =   (1 + a)

1 /2

  x

⋅

Putting these values of p and q in dz = p dx + q dy, we get 1 /2

or

1 /2

 az    z dz =  dx +     (1 + a) x   (1 + a) y  dy dz dx √ (1 + a) = √a + ⋅ √z √x √y

dy

Integrating , √ {(1 + a) z} = √ (ax) + √ y + b, which is a complete integral of (1).

Comprehensive Exercise 4 Apply Charpit’s method to find the complete integrals of the following equations: 1.

z 2 ( p2 z 2 + q2 ) = 1.

2. 3.

( p + q) ( px + qy) = 1. p xy + pq + qy = yz .

(Lucknow 2006, 07, 08; Meerut 07; Purvanchal 09, 11) (Kanpur 2010; Purvanchal 07) (Avadh 2009)

D-222

4.

z = px + qy + p2 + q2 .

5.

z = pq.

6.

(Rohilkhand 2009)

p + q = 2 x.

(Rohilkhand 2004)

2

7.

q = − px + p .

8.

q = 3 p2 .

9.

2

10.

(Kanpur 2009; Purvanchal 07)

(Rohilkhand 2014)

yz p − q = 0 . zpq = p + q.

11. 2 (z + px + qy) = yp2 . 12.

(Lucknow 2011)

2

z = pqxy.

(Meerut 2006; Lucknow 10)

A nswers 4 1.

(a2 z 2 + 1)3 = 9 a4 (ax + y + b)2 a

y

3.

(z − ax) ( y + a) = be

5.

2 √ z = √ ax + (1 / √ a) y + b

7.

z=

8.

z = ax + 3 x2 y + b

2. √ (1 + a) z = 2 √ ( x + ay) + b 4. z = ax + by + a2 + b2 6. z =

1 (2 x − a)3 + a2 y + b 6

x2 1  x  ± √ ( x2 + 4 a) + 2 a log {x + √ ( x2 + 4 a)} + ay + b 4 2 2 

10.

z 2 = 2 (a + 1) {x + ( y / a)} + b

12.

z = bx a y1/ a

9. z 2 = 2 ax + a2 y2 + b 11. z =

ax y

2

−

a2 4 y3

+

b y

8.15 Special Methods of Solution Applicable to Certain Standard Forms. There are a few standard forms to which many first order differential equations are reducible and which can be integrated by methods which are sometimes shorter than the general method. Standard I:

Equations involving only p and q and no x, y, z

Let the equation be written as f ( p, q) = 0 .

…(1)

The complete integral is given by z = ax + by + c ,

…(2)

D-223

where a and b are connected by f (a, b) = 0 .

…(3)

Obviously, from (2) we have ∂z p= =a ∂x ∂z and q= = b. ∂y Substituting p = a and q = b in (3), we get (1). From (3) we may find b in terms of a b = φ (a) say.

i. e.,

The complete integral of (1) is z = ax + φ (a). y + c .

…(4)

General Integral: Putting c = ψ(a) in (4), where ψ denotes an arbitrary function, we get

z = ax + φ (a) y + ψ(a).

…(5)

Differentiating (5) partially w.r.t. a, we get 0 = x + φ′ (a) y + ψ ′ (a).

…(6)

The general integral is obtained by eliminating a between (5) and (6). Singular Integral: The singular integral, if it exists, is obtained by eliminating a and c between the complete integral (4) and the equations formed by differentiating (4) partially w.r.t. a and c i. e., between the equations z = ax + φ (a) y + c , 0 = x + φ′ (a) . y and

0 = 1.

Since 1 = 0 is inconsistent, therefore, in this case there is no singular integral. Note.

In many cases, using some transformations, equations can be reduced to the

form of the standard I.

Example 20: Solution:

Solve p2 + q2 = m2 .

…(1)

The given equation is of the form f ( p, q) = 0 .

Therefore a complete integral is given by z = ax + by + c where a2 + b2 = m2 i. e.,

z = ax + √ (m2 − a2 ) y + c .

…(2)

To find the general integral put c = φ (a) in (2). Then

z = ax + √ (m2 − a2 ) y + φ (a).

Differentiating (3) partially w.r.t. a, we get

…(3)

D-224

0=x−

a y + φ′ (a). √ (m2 − a2 )

…(4)

Eliminating a from (3) and (4) the general integral is obtained. Example 21:

Find a complete integral of ( x + y) ( p + q)2 + ( x − y) ( p − q)2 = 1.

Solution:

…(1)

Put x + y = X 2 , x − y = Y 2 ∂z ∂z ∂X ∂z ∂Y 1 = ⋅ + ⋅ = ∂x ∂X ∂x ∂Y ∂x 2 X ∂z ∂z ∂X ∂z ∂Y 1 q= = ⋅ + ⋅ = ∂y ∂X ∂y ∂Y ∂y 2 X

so that

p=

and

These give p + q =

∂z 1 ∂z + ⋅ ∂X 2Y ∂Y ∂z 1 ∂z − ⋅ ∂X 2Y ∂Y

⋅

1 ∂z 1 ∂z , p− q = ⋅ X ∂X Y ∂Y

Substituting these values in the equation (1), we get 2

2

 ∂z   ∂z    +  = 1,  ∂X   ∂Y 

…(2)

which is of the form of standard I. Hence, a complete integral of (2) is given by z = aX + bY + c , where a2 + b2 = 1. a complete integral of (1) is given by

∴

z = a √ ( x + y) + √ (1 − a2 ) √ ( x − y) + c , where a and c are arbitrary constants. Example 22: Solution:

Solve x2 p2 + y2 q2 = z 2 .

(Purvanchal 2009)

The given equation can be written as 2

2  y ∂z   x ∂z   = 1.   +  z ∂x   z ∂y 

Put

…(1)

dy dz dx = dX , = dY , = dZ x y z

i. e., so that ∴ Similarly

X = log x, Y = log ∂z ∂z ∂X p= = ⋅ + ∂x ∂X ∂x ∂Z ∂Z ∂z 1 = ⋅ = ⋅ ∂X ∂z ∂X z y ∂z ∂Z = ⋅ ∂Y z ∂y

y, Z = log z ∂z ∂Y ∂z 1 ⋅ = ⋅ ⋅ ∂Y ∂x ∂X x x ∂z p. x = ⋅ ⋅ z ∂x

Substituting these values in (1), we get 2

2

 ∂Z   ∂Z    +  = 1,  ∂X   ∂Y  which is of the form of standard I i. e., of the form

…(2)

D-225

 ∂Z ∂Z  f  ,  = 0.  ∂X ∂Y  Hence, a complete integral of (2) is given by Z = aX + bY + c1, where a2 + b2 = 1. ∴

a complete integral of (1) is given by log z = a log x + √ (1 − a2 ) log y + c1.

If we take a = cos α, c1 = log c then complete integral can be written as log z = cos α log x + sin α log y + log c z = c x cos α . ysin α , where α and c are arbitrary constants.

or

General integral is obtained by eliminating α from z = φ (α) x cos α . ysin α , where c = φ (α) 0 = φ′ (α) x cos α . ysin α + φ (α) {x cos α . log x. (− sin α). ysin α

and

+ x cos α . y sin α . log y. cos α} Singular integral is obtained by eliminating α and c between the equations z = c x cos α . y sin α , 0 = c {x cos α . log x. (− sin α). ysin α + x cos α . ysin α . log y. cos α} 0 = x cos α . y sin α .

and

Hence the singular integral is z = 0. Example 23:

Find a complete integral of ( y − x) (qy − px) = ( p − q)2 . (Agra 2002; Kanpur 07, 08; Purvanchal 11)

Solution:

and

Put x + y = X , xy = Y , so that ∂z ∂z ∂X ∂z ∂Y ∂z p= = ⋅ + ⋅ = ⋅1 + ∂x ∂X ∂x ∂Y ∂x ∂X ∂z ∂z ∂X ∂z ∂Y ∂z q= = ⋅ + ⋅ = ⋅1 + ∂y ∂X ∂y ∂Y ∂y ∂X

∂z ⋅ y ∂Y ∂z ⋅ x. ∂Y

Substituting these values of p and q in the given equation, we get   ∂z ∂z  ∂z    ∂z 2  ∂z  ( y − x)  +x + y   y−  x = ( y − x)   ∂Y   ∂X ∂Y  ∂Y     ∂X ∂z  ∂z  = ( y − x)2    ∂Y  ∂X

2

or

( y − x)2

or

∂z  ∂z  =  , which is of the form of standard I. ∂X  ∂Y 

2

Hence, a complete integral is given by z = aX + bY + c , where a = b2 . a complete integral of the given equation is z = a ( x + y) + √ a ( xy) + c , where a and c are arbitrary constants.

∴

2

D-226 Example 24: Solution:

Find a complete integral of pq = x m y nz 2 l .

(Avadh 2014)

The given equation can be written as pz − l qz − l ⋅ = 1. xm yn yn + 1 xm + 1 z1 − l , Y = , Z = , m +1 n +1 1− l

Put

X=

so that

∂Z ∂Z dx 1 ∂Z ∂Z dy 1 = ⋅ = z− l . p m , = ⋅ = z− l q n ⋅ ∂X ∂x dX ∂ Y ∂ y dY x y

Putting these values in (1), we get ∂Z ∂Z ⋅ = 1, which is of the form of standard I. ∂X ∂Y Hence a complete integral is given by ∴

Z = aX + bY + c , where ab = 1. a complete integral of the given equation is yn + 1 z1 − l xm + 1 =a + + c, m + 1 a (n + 1) 1− l

where a and c are arbitrary constants. Standard II: Equations involving only p, q and z i. e., quations of the form …(1) f ( z, p, q ) = 0. Let us assume z = f ( x + ay) as a trial solution of given equation (1), where a is an arbitrary constant. ∴ ∴ and ∴

z = f ( X ) where X = x + ay. ∂z dz ∂X dz p= = = ∂x dX ∂x dX ∂z dz ∂X dz q= = =a ⋅ ∂y dX ∂y dX dz   dz Equation (1) reduces to the form f  z , ,a  =0  dX dX 

which is an ordinary differential equation of order one. Integrating it we may get the complete integral. The general and the singular integrals are to be found in the usual way. Rule: The method of solving the equations of the Standard II can be formulated in the following rule : dz dz Put for q, where X = x + ay. for p, a dX dX Now solve the resulting ordinary differential equation in the variables z and X. Then substitute x + ay for X. This gives a complete solution. Note: Sometimes using transformations equations reduce to the form of standard II.

D-227

Example 25:

Find a complete integral of 9 ( p2 z + q2 ) = 4.

The given equation is of the form f (z , p, q) = 0 . dz dz Putting for q, the given equation becomes for p, a dX dX 2   dz  2 2  dz  9z   +a    = 4, where X = x + ay  dX     dx   Solution:

or

2  dz  9 (z + a2 )   = 4,  dX 

or

3 √ (z + a2 ) dz = dX . 2

Integrating,

(z + a2 )3 /2 = X + b

or

(z + a2 )3 /2 = x + ay + b

or

(z + a2 )3 = ( x + ay + b)2 ,

or

dz 2 = dX 3 √ (z + a2 )

which is a complete integral of the given equation. Example 26:

Find a complete integral of z 2 ( p2 z 2 + q2 ) = 1.

(Lucknow 2006; Meerut 08)

dz dz Solution: Putting for q, the given equation becomes for p, a dX dX 2  dz  2 2 2  dz  z 2   .z + a    = 1, where X = x + ay  dX    dX   2

or or Integrating, or

 dz  z 2 (z 2 + a2 )   =1  dX  z √ (z 2 + a2 ) dz = dX . 1 2 (z + a2 )3 /2 = X + b 3 9 ( x + ay + b)2 = (z 2 + a2 )3 ,

which is a complete integral of the given equation. Example 27:

Find a complete integral of pz = 1 + q2 .

The given equation is of the form f ( p, q, z ) = 0 . dz dz Putting for p, a for q, the given equation becomes dX dX Solution:

z or

2 dz  dz  = 1 + a2   , where X = x + ay  dX  dX

2 dz  dz  a2  + 1 = 0.  −z  dX  dX

D-228

∴ or or

z ± √ (z 2 − 4 a2 ) dz = dX 2 a2 dz z ± √ (z 2 − 4 a2 )

=

dX 2 a2

or

z + √ (z 2 − 4 a2 ) 4 a2

dz =

dX 2 a2

{z + √ (z 2 − 4 a2 )} dz = 2 dX .

Integrating, we get z2  z 1  +  √ (z 2 − 4 a2 ) − ⋅ 4 a2 log {z + √ (z 2 − 4 a2 )} = 2 X + b 2 2 2  or

z 2 + [z √ (z 2 − 4 a2 ) − 4 a2 log {z + √ (z 2 − 4 a2 )}] = 4 ( x + ay) + b,

which is a complete integral of the given equation. Example 28:

Find a complete integral of p2 = z 2 (1 − pq).

The given equation is of the form f ( p, q, z ) = 0 . dz dz Putting for q, the given equation becomes for p, a dX dX Solution:

2 dz dz   dz  2  a   = z 1 −  , where X = x + ay  dX  dX dX  

or

2  dz  2 2   (1 + az ) = z  dX 

or

√ (1 + az 2 ) dz = dX z

or

or Integrating, or

1 + az 2 z √ (1 + az 2 )

dz = dX

1 az   + dz = dX .  2 2   z √ (1 + az ) √ (1 + az )  1 log [z √ a + √ (1 + az 2 )] + √ (1 + az 2 ) = X + b √a 1 log [z √ a + √ (1 + az 2 )] + √ (1 + az 2 ) = x + ay + b, √a

which is a complete integral of the given equation. Standard III: Equation of the form f ( x, p) = F ( y, q). As a trial solution let us put each side equal to an arbitrary constant i. e.,

f ( x, p) = F ( y, q) = a

from which we obtain p = f1 ( x, a)

and

q = f2 ( y, a).

Now from dz = p dx + q dy, we have dz = f1 ( x, a) dx + f2 ( y, a) dy. Integrating , we get z = ∫ f1 ( x, a) dx + ∫ f2 ( y, a) dy + b which is the complete integral.

…(1)

D-229

The general integral can be obtained in the usual way. As in the case of standard I, there is no singular integral. Write the differential equation in the form (1). Put both the sides of the equation equal to an arbitrary constant. Solving them find the values of p and q. Substitute the values of p and q in dz = p dx + q dy and integrate to find a complete integral. Rule:

Note:

Sometimes using transformations equations reduce to the form of standard III.

Example 29: Solution:

Find a complete integral of p2 + q2 = x + y.(Meerut 2007; Rohilkhand 03)

Separating q and y from p and x, the given equation can be written as p2 − x = y − q2 = a, (say).

∴

p = √ ( x + a) and q = √ ( y − a).

Putting the values of p and q in dz = p dx + q dy, we get dz = √ ( x + a) dx + √ ( y − a) dy. 2 2 Integrating, z = ( x + a)3 /2 + ( y − a)3 /2 + b, 3 3 which is a complete integral of the given equation. Example 30: Solution:

Find a complete integral of yp = 2 y x + log q.

(Lucknow 2010)

The given equation can be written as 1 p = 2x + log q y 1 log q = a, (say). y

or

p − 2x =

∴

p = 2 x + a, log q = ay i. e., q = e ay .

Putting the values of p and q in dz = p dx + q dy, we get dz = (2 x + a) dx + e ay dy. Integrating,

1 z = x2 + ax + e ay + b, a

which is a complete integral of the given equation. Example 31:

Find a complete integral of z 2 ( p2 + q2 ) = x2 + y2 .

1 Solution: Put z dz = dZ ; i. e., Z = z 2 , so that 2 ∂z ∂Z ∂z ∂Z z = = P (say), z = = Q (say). ∂x ∂x ∂y ∂y Putting these values the given equation becomes

(Avadh 2006)

D-230

P2 + Q2 = x2 + y2 or P2 − x2 = y2 − Q2 = a2 (say). ∴

P = √ (a2 + x2 ), Q = √ ( y2 − a2 ).

Now

dZ = P dx + Q dy = √ (a2 + x2 ) dx + √ ( y2 − a2 ) dy.

Integrating , Z =

1 a2 x √ (a2 + x2 ) + log {x + √ (a2 + x2 )} 2 2 +

1 a2 y √ ( y2 − a2 ) − log { y + √ ( y2 − a2 )} + b 2 2

z 2 = x √ (a2 + x2 ) + a2 log {x + √ (a2 + x2 )} + y √ ( y2 − a2 )

or

− a2 log { y + √ ( y2 − a2 )} + b, which is a complete integral of the given equation. Example 32:

Find a complete integral of x2 y3 p2 q = z 3 .

The given equation can be written as

Solution:

2  1 ∂z   1 ∂z   = 1. x2 y3     z ∂x   z ∂y 

Put

…(1)

1 dz = dZ i. e., Z = log z . Then z 1 ∂z ∂Z 1 ∂z ∂Z = = P (say), = = Q (say). z ∂x ∂x z ∂y ∂y

The equation (1) reduces to x2 y3 P2 Q = 1 or

x2 P2 =

∴

P=

1 Q y3

= a2 , (say).

a 1 , Q= ⋅ 2 3 x a y

Now dZ = P dx + Q dy =

a 1 dx + dy. x a2 y3 1

Integrating,

Z = a log x −

or

log z = a log x −

2 2

2a y

+b

1 2 2

2a y

+ b,

which is a complete integral of the given equation. Example 33: Solution:

Find a complete integral of z ( p2 − q2 ) = x − y.

(Lucknow 2006)

The given equation can be written as 2 2      √ z ∂z  −  √ z ∂z   = x − y.  ∂x  ∂y     

…(1)

D-231

Putting √ z dz = dZ i. e., Z = so that

2 3 /2 z , 3

∂Z ∂z ∂Z ∂z = √z = P (say), = √z = Q (say), ∂x ∂x ∂y ∂y

the equation (1) reduces to P2 − Q2 = x − y or P2 − x = Q2 − y = a (say). ∴

P = √ (a + x), Q = √ (a + y).

Now

dZ = P dx + Q dy = √ (a + x) dx + √ (a + y) dy. 2 2 Z = (a + x)3 /2 + (a + y)3 /2 + b 3 3

Integrating, or

z 3 /2 = (a + x)3 /2 + (a + y)3 /2 + b,

which is a complete integral of the given equation. Standard IV:

Equation of the form …(1)

z = px + qy + f ( p, q ) (analogous to Clairaut’s form) We know that the solution of Clairaut’s equation dy is y = cx + f (c ). y = px + f ( p) where p = dx Similarly the complete integral of Clairaut’s equation (1) is z = ax + by + f (a, b).

Rule: To get the complete integral of the equation of this type replace p and q by a, b (two arbitrary constants) respectively.

General Integral is obtained as in other cases. Singular Integral. The complete integral is F = z − ax − by − f (a, b) = 0 ∂f ∂F = 0, = 0, gives x + ∂a ∂a ∂f ∂F and = 0. = 0, gives y + ∂b ∂b

…(1) …(2) …(3)

Singular integral is obtained by eliminating a, b from (1), (2) and (3).

Example 34:

Find a complete integral of z = px + q y + p2 + q2 .

(Kanpur 2009)

The given equation is of the form of standard IV i. e., of the form z = px + q y + f ( p, q). Hence a complete integral is given by Solution:

z = ax + by + a2 + b2 .

D-232 Example 35:

Find the singular integral of z = px + q y + log pq.

Solution:

(Lucknow 2009)

The complete integral of the given equation is z = ax + by + log ab.

…(1)

Differentiating (1) partially w.r.t. a and b, we get 1 0=x+ a 1 and 0= y+ ⋅ b ∴

a = − 1/ x

and

b = − 1/ y.

…(2)

Eliminating a and b between (1) and the equations (2), we get  z = x −  or

 1  1 1  + y  −  + log    x  y  xy 

z = − 2 − log xy, which is the required singular integral.

Example 36:

Find a complete integral and the singular integral of 4 xyz = pq + 2 px2 y + 2 qxy2 .

Solution:

Put x2 = X , y2 = Y , so that p=

∂z ∂z dX ∂z ∂z = = 2 X1 /2 , q = 2Y 1 /2 ⋅ ∂x ∂X dx ∂X ∂Y

The given equation then reduces to ∂z ∂z ∂z ∂z z=X +Y + ⋅ ⋅ ∂X ∂Y ∂X ∂Y ∴

a complete integral is z = aX + bY + ab

or

z = ax2 + by2 + ab.

…(1)

Differentiating (1) partially w.r.t. a and b, we get 0 = x2 + b and 0 = y2 + a.

…(2)

Eliminating a and b between (1) and the equations (2), the singular integral is z = − x2 y2 − x2 y2 + x2 y2 i. e., z + x2 y2 = 0 . Example 37:

Find the singular integral of z = px + q y + c √ (1 + p2 + q2 ). (Rohilkhand 2011; Avadh 14)

Solution:

The complete integral of the given equation is z = ax + by + c √ (1 + a2 + b2 ). Differentiating (1) partially w.r.t. a and b, we get ac bc 0=x+ , 0= y+ , 2 2 √ (1 + a + b ) √ (1 + a2 + b 2)

…(1)

Singular integral.

…(2)

D-233

a2 c 2 + b2 c 2

so that

x2 + y2 =

i. e.,

c 2 − x2 − y 2 =

i. e.,

1 + a2 + b2 =

1 + a2 + b2 c2 1 + a2 + b2 c2

⋅

c 2 − x2 − y2

…(3)

Also, from (2) x √ (1 + a2 + b2 ) −x = 2 c √ (c − x2 − y2 )

a=−

and

y √ (1 + a2 + b2 )

b=−

=

c

…(4)

− y …(5)

2

√ (c − x2 − y2 )

Putting the values from (3), (4) and (5) in (1), we get the singular solution as x2

z=−

or

z=

2

2

2

√ (c − x − y ) c 2 − x2 − y2

√ (c 2 − x2 − y2 )

−

y2 2

2

2

√ (c − x − y )

+

c2 2

√ (c − x2 − y2 )

i. e., z 2 = c 2 − x2 − y2 i. e., x2 + y2 + z 2 = c 2 .

Comprehensive Exercise 5 Find complete integrals of the following equations : 1.

q = e − p /α .

2.

p 2 − q 2 = λ.

3.

√ p + √ q = 1.

4.

p 2 = zq.

5.

p 3 + q 3 − 3 pqz = 0.

6.

p (1 + q ) = qz.

7.

√ p + √ q = 2x.

8.

pq = xy.

9.

3

10.

2

p − 3x = q − y.

11.

z = px + qy − 2 √ ( pq ).

12.

z = px + qy − p 2 q. 2

2

2

13.

p + q = z ( x + y ).

15.

z = px + qy + 3 p1/ 3 q1/ 3 .

z = px + qy + pq. (Lucknow 2010) (Lucknow 2008)

14.

q z = px + qy + − p. p

A nswers 5 1.

z = ax + ye− a /α + c

2.

3.

z = ax + (1 − √ a )2 y + c

4. z = ax + ( a 2 − λ ) 1/ 2 + c

z = ax + ( a 2 − λ )1/ 2 + c

D-234

5. 7. 9. 11. 13. 14.

3a ( x + ay ) + b = (1 + a 3 ) log z 1 z = ( a + 2x )3 + a 2 y + b 6 2 z = x3 + ax ± ( y + a )3 / 2 + b 3

6. az − 1 = bex+ ay 8. 2az = a 2 x2 + y2 + 2ab 10.

12. z = ax + by − 2 √ ( ab ) 3 log z = ( a + x )3 / 2 + ( y − a )3 / 2 + b 2 b 15. z = ax + by + − a a

z = ax + by + ab. z = ax + by − a 2 b

z = ax + by + 3 ( ab )1/ 3

8.16 Compatible Systems of First Order Partial Differential Equations Definition:

If every solution of the first order partial differential equation …(1)

f ( x, y, z , p, q) = 0 is also a solution of the first order partial differential equation

…(2)

g ( x, y, z , p, q) = 0 , then the differential equations (1) and (2) are said to be compatible.

To find the condition for the pair of equations (1) and (2) to be compatible. ∂ ( f , g) If J = ≠ 0 , then the equations (1) and (2) can be solved to obtain the explicit ∂ ( p, q) expressions

…(3)

p = φ ( x, y, z ), q = ψ ( x, y, z )

for p and q. The condition that the pair of equations (1) and (2) should be compatible reduces then to the condition that the system of equations (3) should be completely integrable. If z is a function of x and y, then we know that ∂z ∂z dz = dx + dy = p dx + q dy. ∂x ∂y Therefore, the system of equations (3) will be completely integrable if and only if the single differential equation …(4) dz = φ dx + ψ dy i. e., φ dx + ψ dy − dz = 0 is integrable. We know that the necessary and sufficient condition for the single differential equation P dx + Q dy + R dz = 0 to be integrable is that  ∂Q ∂R  ∂P ∂Q   ∂R ∂P   +Q  = 0. P − − −  + R  ∂x ∂y  ∂z  ∂x   ∂z  ∂y Therefore, the single differential equation (4) is integrable if and only if φ (ψ z − 0 ) + ψ (0 − φz) − 1 (φ y − ψ x ) = 0

[ ∵ P = φ, Q = ψ, R = − 1]

D-235

…(5) i. e., if and only if ψ x + φψ z = φ y + ψ φz . Substituting from equations (3) into equation (1) and differentiating with respect to x and z, respectively, we obtain the equations

and

f x + f p φx + f q ψ x = 0

…(6)

f z + f p φz + f q ψ z = 0.

…(7)

Multiplying (7) by φ and adding to (6), we get f x + φ f z + f p (φx + φ φ z) + f q (ψ x + φ ψ z ) = 0 .

…(8)

In a similar manner, equation (2) will give …(9)

g x + φ gz + g p (φx + φ φz) + gq (ψ x + φ φz ) = 0 . Now solving the equations (8) and (9), we obtain ψ x + φ ψz = where

J=

∂ ( f , g) 1  ∂ ( f , g) +φ   J  ∂ ( x, p) ∂ (z , p) 

…(10)

∂ ( f , g) ⋅ ∂ ( p, q)

If we had differentiated the given pair of equations (1) and (2) with respect to y and z after substituting in them from equations (3), then we would have obtained φ y + ψ φz = −

1 J

 ∂ ( f , g) ∂ ( f , g) +ψ  ⋅ ∂ (z , q)   ∂ ( y , q)

…(11)

Substituting from equations (10) and (11) in (5), we get the desired condition that the two equations (1) and (2) should be compatible as …(12)

[ f , g] = 0 , where

[ f , g] =

∂ ( f , g) ∂ ( f , g) ∂ ( f , g) ∂ ( f , g) +p + +q ⋅ ∂ ( x, p) ∂ (z , p) ∂ ( y , q) ∂ (z , q)

…(13)

[ ∵ p = φ and q = ψ] Particular Case: Theorem:

The first order partial differential equations

p = P ( x, y), q = Q ( x, y) ∂P ∂Q are compatible if and only if = ⋅ ∂y ∂x Proof:

We know that if z is a function of two independent variables x and y, then ∂z ∂z dz = dx + dy = p dx + q dy. ∂x ∂y

So, the partial differential equations p = P ( x, y), q = Q ( x, y) are compatible if and only if the single differential equation dz = P dx + Q dy is integrable.

D-236

Now we know that if P and Q are functions of two variables x and y, then P dx + Q dy is an exact differential d φ ( x, y) if and only if ∂P ∂Q = ⋅ ∂y ∂x ∴

The single differential equation dz = P dx + Q dy is integrable if and only if ∂P ∂Q = ⋅ ∂y ∂x

Hence, the differential equations p = P ( x, y), q = Q ( x, y) are compatible if and only if ∂P ∂Q = ⋅ ∂y ∂x

Example 38:

Show that the equations xp − yq = x, x2 p + q = xz

are compatible and find their solution. Solution:

We know that the first order partial differential equations f ( x, y, z , p, q) = 0

and

g ( x, y, z , p, q) = 0

are compatible if and only if [ f , g] = 0 , ∂ ( f , g) ∂ ( f , g) ∂ ( f , g) ∂ ( f , g) where [ f , g] = +p + +q ⋅ ∂ ( x, p) ∂ (z , p) ∂ ( y , q) ∂ (z , q) Here, the given equations are f ( x, y, z , p, q) ≡ xp − yq − x = 0 and

g ( x, y, z , p, q) ≡ x2 p + q − x z = 0 .

∴

∂f ∂ ( f , g) ∂x = ∂ ( x, p) ∂g  ∂x ∂f ∂ ( f , g) ∂z = ∂ (z , p) ∂g  ∂z

∂f  x  p−1 ∂p   = x2 ( p − 1) − x (2 xp − z ), =  ∂g   2 xp − z x2  ∂p ∂f  x 0 ∂p   = 0 − (− x2 ) = x2 , =  ∂g  2  − x x  ∂p

∂f ∂ ( f , g) ∂y = ∂ ( y, q) ∂g  ∂y

∂f  − q ∂q  = ∂g   0 ∂q 

− y  = − q,  1

D-237

and

∂f ∂ ( f , g) ∂z = ∂ (z , q) ∂g  ∂z

∂f   0 − y ∂q  = − xy. =    ∂g   − x 1 ∂q

Therefore, [ f , g] = x2 ( p − 1) − 2 x2 p + xz + px2 − q − q x y = x2 p − x2 − 2 x2 p + x z + px2 − q − q x y = − x2 + xz − q − q x y = − x2 + x2 p + q − q − q x y [∵ from the given differential equations, xz = x2 p + q] = − x2 + x ( xp − yq) = − x2 + x . x [∵ from the given differential equations, xp − yq = x] 2

2

= − x + x = 0. Hence, the given equations are compatible. Now, let us find p and q by solving the equations x p − yq = x and

2

x p + q = xz

Multiplying (2) by y and adding to (1), we get yx2 p + xp = xyz + x or or ∴

xp (1 + xy) = x (1 + yz ) 1 + yz p= ⋅ 1 + xy x2 + x2 yz from (2), q = xz − x2 p = xz − 1+ x y =

xz (1 + xy) − x2 − x2 yz 1 + xy

=

x (z − x) ⋅ 1 + xy 1 + yz x (z − x) ,q = ⋅ 1 + xy 1 + xy

Thus,

p=

Now

dz =

∂z ∂z dx + dy = p dx + q dy ∂x ∂y

 x (z − x) 1 + yz   dx +  =  dy. 1 xy +    1 + xy 

…(1) …(2)

D-238

So we have to solve the single differential equation (1 + xy) dz − (1 + yz ) dx − x (z − x) dy = 0 ⇒

(1 + xy) dz − (1 + xy) dx + (1 + xy) dx − (1 + yz ) dx − x (z − x) dy = 0

⇒

(1 + xy) (dz − dx) − y (z − x) dx − x (z − x) dy = 0 (1 + xy) (dz − dx) − (z − x) ( y dx + x dy) =0 (1 + xy)2

⇒

(1 + xy) (dz − dx) − (z − x) d (1 + xy)

⇒

(1 + xy)2

=0

⇒

 z − x  =0 d 1 + xy 

⇒

z−x = c , where c is an arbitrary constant. 1 + xy

Hence, the required solution of the given equations is z − x = c (1 + xy) or

z = x + c (1 + xy), where c is a constant.

Example 39: Show that the differential equations

p = 5 x − 4 y + 3, q = 4 x + 5 y + 2 do not

possess any common solution. Solution:

We know that the first order partial differential equations

p = P ( x, y), q = Q ( x, y) ∂P ∂Q are compatible if and only if = ⋅ ∂y ∂x Here, P = 5 x − 4 y + 3, Q = 4 x + 5 y + 2. ∂P ∂Q We have = − 4 and = 4. ∂y ∂x Since

∂P ∂Q ≠ , therefore the given differential equations are not compatible. ∂y ∂x

Hence, the given differential equations do not possess any common solution. Example 40:

Show that the differential equations ∂z ∂z = 6 x + 3 y, = 3x − 4 y ∂x ∂y

are compatible and find their solution. Solution:

We know that the first order partial differential equations p = P ( x, y), q = Q ( x, y)

are compatible if and only if ∂P ∂Q = ⋅ ∂y ∂x Here, P ( x, y) = 6 x + 3 y, Q ( x, y) = 3 x − 4 y.

D-239

We have

∂P =3 ∂y

and

∂Q = 3. ∂x

Since

∂P ∂Q = , therefore the given differential equations are compatible. ∂y ∂x

Now dz =

∂z ∂z dx + dy = p dx + q dy ∂x ∂y = (6 x + 3 y) dx + (3 x − 4 y) dy = 6 x dx + 3 ( y dx + x dy) − 4 y dy = 6 x dx + 3 d ( xy) − 4 y dy.

Integrating both sides, we get z =6⋅

y2 x2 + 3 xy − 4 ⋅ +c 2 2

z = 3 x2 + 3 xy − 2 y2 + c

or

as the required solution of the given differential equations. Example 41:

Show that the differential equations p = x2 − ay, q = y2 − ax

are compatible and find their common solution. Solution:

We know that the first order partial differential equations

p = P ( x, y), q = Q ( x, y) ∂P ∂Q are compatible if and only if = ⋅ ∂y ∂x Here, P ( x, y) = x2 − ay, Q ( x, y) = y2 − ax. We have Since Now

∂P =−a ∂y

and

∂Q = − a. ∂x

∂P ∂Q = , therefore the given differential equations are compatible. ∂y ∂x dz =

∂z ∂z dx + dy = p dx + q dy ∂x ∂y

= ( x2 − ay) dx + ( y2 − ax) dy = x2 dx + y2 dy − a ( x dy + y dx) = x2 dx + y2 dy − ad ( xy). Integrating both sides, we get z=

y3 x3 + − axy + c 3 3

as the required common solution of the given differential equations.

D-240 Example 42:

Solution:

Solve the simultaneous differential equations ∂z ∂z = x4 − 2 xy2 + y4 , = − (2 x2 y − 4 xy3 + sin y). ∂x ∂y

We know that the first order partial differential equations p = P ( x, y), q = Q ( x, y)

are compatible if and only if ∂P ∂Q = ⋅ ∂y ∂x Here,

P = x4 − 2 xy2 + y4 , Q = − 2 x2 y + 4 xy3 − sin y.

We have

∂P ∂Q = − 4 xy + 4 y3 , = − 4 xy + 4 y3 . ∂y ∂x

Since Now

∂P ∂Q = , therefore the given differential equations are compatible. ∂y ∂x ∂z ∂z dx + dy = p dx + q dy ∂x ∂y

dz =

= P dx + Q dy = ( x4 − 2 xy2 + y4 ) dx + (− 2 x2 y + 4 xy3 − sin y) dy. Integrating both sides, we get z= (treating

=

4

∫ (x

∫

P dx

∫

y as a constant)

− 2 xy2 + y4 ) dx +

(treating

+

Q dy

+c

(taking in Q only those terms which do not contain x )

∫

(− sin y) dy + c

y as a constant)

1 = x 5 − x2 y2 + x y 4 + cos y + c . 5 Hence, the required common solution of the given simultaneous equations is 1 z = x5 − x2 y2 + xy4 + cos y + c , 5 where c is an arbitrary constant.

Comprehensive Exercise 6 1.

Show that the differential equations ∂z ∂z = 7 x + 8 y − 5, = 9 x + 11 y + 3 ∂x ∂y are not compatible.

2.

Show that the differential equations p = 4 x + 3 y + 1, q = 3 x + 2 y + 1 are compatible and find their solution.

D-241

3.

Show that the differential equations p = 1 + 4 xy + 2 y2 , q = 1 + 4 xy + 2 x2 are compatible and find their common solution.

4.

Show that the differential equations in each of the following systems are compatible and find their solution : (i)

p = ax + hy + g, q = hx + by + f

(ii) p = 2 x − y, q = 5 − x − 2 y (iii) p = (2 ax + by) y, q = (ax + 2 by) x y x (iv) p = x − , q= y+ 2 2 2 x + y x + y2  x (v) p = 1 + e x / y , q = e x / y 1 −  y  2 2 (vi) p = y2 e xy + 4 x3 , q = 2 xy e xy − 3 y2

(vii) p = (e y + 1) cos x, q = e y sin x (viii) p = sin x cos y + e2 x , q = cos x sin y + tan y. 5.

Show that the equations f ( x, y, p, q) = 0 , g ( x, y, p, q) = 0 ∂ ( f , g) ∂ ( f , g) are compatible if + = 0. ∂ ( x, p) ∂ ( y , q)

A nswers 6 1.

z = 2 x2 + 3 xy + x + y2 + y + c

2.

z = x + 2 x2 y + 2 xy2 + y + c

4.

(i) z =

1 2 1 ax + hxy + gx + by2 + fy + c 2 2

(ii) z = x2 − xy − y2 + 5 y + c (iii) z = ayx2 + bxy2 + c (iv) z =

 x  1 2 1 2  + x − tan−1  y +c 2  y 2

(v) z = x + ye x / y + c (vi)

2 z = e xy + x4 − y3 + c

(vii) z = (e y + 1) sin x + c . (viii) z = − cos x cos y +

1 2x e + log sec y + c . 2

D-242

Objective Type Questions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

Equation p tan y + q tan x = sec2 z is of order (a) one

(b) two

(c) zero 2.

(d) None of these. 2

2

Equation r + 2 s − t = 0 is of order (a) one

(b) two

(c) three

(d) None of these. (Rohilkhand 2003) 2

3.

Equation

∂ z 2

∂x

3

2

 ∂z  ∂ z +   = 0 is of degree ∂x ∂y  ∂y 

−2

(a) one

(b) two

(c) three 4.

5.

(d) None of these. 3

2

4

Equation p + qx + z = 0 is of degree (a) two

(b) three

(c) four

(d) one.

The equation Pp + Qq = R is known as (a) Charpit’s equation

(b) Lagrange’s equation

(c) Bernoulli’s equation

(d) Clairaut’s equation. (Rohilkhand 2003)

6.

Out of the following four partial differential equations, the differential equation which is linear is (a)

(b)

∂3 z 3

∂x

−3

∂2 z ∂z ∂2 z ⋅ +8 = sin x 2 ∂y ∂x ∂ y2 2

∂2 z

∂z  ∂z  +  + + 9z = 0 2  ∂x  ∂y ∂x

(c) 4

∂2 z ∂x2

+5

∂2 z ∂2 z ∂z ∂z +6 +7 +8 + 3 z = ( x3 + y3 ) sin x 2 ∂x ∂y ∂x ∂y ∂y 2

2  ∂z   ∂z  (d)   +   = x2 + y2 .  ∂x   ∂y 

7.

The differential equation (2 x + 3 y) p + 4 xq − 8 pq = x + y is (a) linear

(b) non-linear

(c) quasi-linear

(d) semi-linear.

D-243

8.

The differential equation ( x2 + y2 )

∂z ∂z + ( x + y) + (3 x − 4 y) z = x2 + y2 is ∂x ∂y

(a) linear

(b) quasi-linear

(c) semi-linear 9.

(d) non-linear. ∂z ∂z The differential equation ( x + y − 3 z ) + (3 x + 4 y) + 2 z = x + y is ∂x ∂y (a) linear

(b) quasi-linear

(c) semi-linear 10.

11.

12.

(d) non-linear. ∂ z ∂ z The differential equation x2 + y2 = ( x + y) z 2 + 4 x is ∂x ∂y (a) linear

(b) quasi-linear

(c) semi-linear

(d) non-linear.

Out of the following four pairs of first order partial differential equations, mention the pair in which the differential equations are compatible : (a) p = 4 y − 7 x + 3, q = 7 x + 4 y + 2

(b) p = x + y, q = y − x

(c) p = 3 x + 8 y, q = 8 x − 3 y

(d) p = x + y + 1, q = 2 x − y + 1.

Lagrange’s auxiliary equations of Pp + Qq = R are given by dx dy dz dx dy dz (a) (b) = = = = R Q P P Q R dx dy dz (c) (d) None of these. = = Q R P (Rohilkhand 2004)

Fill In The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

The order of the differential equation 4

∂3 z ∂x2 ∂y

+9

∂2 z ∂x2

+8

∂2 z ∂ y2

+6

∂z ∂z +3 +8=0 ∂x ∂y

is … . 2.

In a linear partial differential equation all the partial derivatives occurring in it are in … degree.

3.

Lagrange’s linear equation is of the form … .

4.

Lagrange’s auxiliary equations or Lagrange’s subsidiary equations are … .

5.

Lagrange’s auxiliary equations of ( y2 + z 2 − x2 ) p − 2 xyq + 2 zx = 0 are … .

6.

Lagrange’s subsidiary equations of z ( xp − yq) = y2 − x2 are … .

7.

Lagrange’s auxiliary equations of x

8.

xy ∂z ∂z ∂z are … . + y +t = az + ∂x ∂y ∂t t dp dq dy dz dx dF Equations = = = = = ∂f ∂f ∂f ∂f ∂f ∂f ∂f ∂f 0 +p +q −p −q − − ∂x ∂z ∂y ∂z ∂p ∂q ∂p ∂q

are known as … .

D-244

9.

For the differential equation ( p2 + q2 ) y = qz , Charpit’s auxiliary equations are… .

10.

For the differential equation p 2 x + q2 y = z ,Charpit’s auxiliary equations are … .

11.

The first order partial differential equations ∂z ∂z ∂P = P ( x, y), = Q ( x, y) are compatible if =… ∂x ∂y ∂y

12.

If every solution of the differential equation f ( x, y, z , p, q) = 0 is also a solution of the differential equation g ( x, y, z , p, q) = 0, then the two differential equations are said to be …

True or False 1.

Write ‘ T ’ for true and ‘F’ for false statement. ∂z ∂z The differential equation ⋅ = 3 xy is non-linear. ∂x ∂y ∂2 u

∂2 u ∂u ∂u +5 +6 + 7 u = x2 is linear. ∂x ∂y ∂x ∂y

2.

The differential equation

3.

The differential equation

4.

The differential equation z = px + qy is linear.

5.

The differential equation p2 + q2 = ( x2 + y2 )2 is quasi-linear.

6.

A complete integral of the equation z = px + qy + p2 + q2 is given by

∂x2 ∂2 v 2

∂x

+

+2

∂2 v ∂ y2

= 0 is non-linear.

z = ax + by + a2 + b2 . 7.

A

complete integral of q = 3 p2 is z = ax + 3 a2 y + c , a and c are arbitrary

constants. 8.

p (1 + q) = qz is a linear equation.

9.

pz − qy = z 2 + ( x + y)2 is a linear differential equation.

10.

The differential equations

∂z ∂z = 5 x − 7 y, = 6 x + 8 y possess a common ∂x ∂y

solution. 11. 12.

The differential equations p = 12 x + 7 y + 1, q = 7 x + 4 y + 1 are compatible. y x The differential equations p = are compatible. − 1, q = − 2 2 2 x + y x + y2

13.

The differential equations

∂z ∂z = y sin 2 x, = − (1 + y2 + cos2 x) ∂x ∂y

are not compatible. 14.

The differential equations p = cos x (cos x − sin α sin y), q = cos y (cos y − sin α sin x) are compatible.

D-245

∂P ∂Q = ⋅ ∂x ∂y

15.

The differential equations p = P ( x, y), q = Q ( x, y) are compatible if

16.

Singular integral of a differential equation is obtained by giving particular values to arbitrary constants in its general solution.

17.

The equation of envelope to a surface is the singular integral of its differential equation.

18.

Any relation which contains as many arbitrary constants as there are independent variables and is a solution of a partial differential equation of the first order is called a complete integral of that equation.

A nswers Multiple Choice Questions 1. 4. 7. 10.

(a) (b) (b) (c)

2. 5. 8. 11.

(b) (b) (a) (c)

3. 6. 9. 12.

Fill in the Blank(s) 1.

3

2.

first

3.

Pp + Qq = R

4.

dx dy dz = = P Q R

5.

dx 2

2

2

y +z −x

=

dy dz = − 2 xy − 2 zx

6.

dy dx dz = = 2 zx − yz y − x2

7.

dx dy dt dz = = = x y t az + ( xy / t)

8. 9.

10.

Charpit’s auxiliary equations. dy dp dq dz dx = = = = − pq p2 − 2 p2 y − 2 q2 y + qz − 2 py −2 qy + z dp 2

−p+ p

=

dq −q + q

11.

∂P ∂x

12.

compatible.

2

=

dz 2

2

− 2 p x − 2q y

=

dy dx = − 2 px − 2 qy

(a) (c) (b) (b)

D-246

True or False 1.

T

2.

T

3.

F

4.

T

5.

F

6.

T

7.

T

8.

F

9.

T

10.

F

11.

T

12.

T

13.

F

14.

T

15.

F

16.

F

17.

T

18.

T

¨

D-247

9 L inear P artial D ifferential E quations of S econd and H igher O rder with C onstant C oefficients

9.1 The General Linear Partial Differential Equation of an Order Higher than the First partial differential equation in which the dependent variable and its derivatives appear only in the first degree and are not multiplied together, their coefficients all being constants or functions of x and y, is called a linear partial differential equation. The general form of such an equation is

A

∂ nz ∂x n

+ A1

∂ nz ∂x n−1 ∂y

+ ... + An

∂ nz ∂y n

+ B0

+M

∂ n−1z ∂x n−1

+ ...

∂z ∂z +N + P z = f ( x, y), …(1) ∂x ∂y

where the coefficients A1, ..., An , B 0 ,..., M , N , P are constants or functions of x and y. If the coefficients of various terms are constants then it is called a linear partial differential equation with constant coefficients.

D-248

9.2 The Homogeneous Linear Partial Differential Equation with Constant Coefficients In this equation all the partial derivatives appearing in the equation are of the same order. A linear homogeneous partial differential equation of order n with constant coefficients is of the form ∂ nz ∂x n

+ A1

∂ nz ∂x n−1 ∂y

+ ... + An

∂ nz ∂y n

= f ( x, y),

…(1)

where A1, ..., An are constants. ∂ ∂ Denoting the operators and by D, D ′ respectively, the equation (1) can also be ∂x ∂y written as or where

( Dn + A1 Dn−1 D ′ + ... + An D ′ n ) z = f ( x, y) …(2)

F ( D, D ′ ) z = f ( x, y), n

F ( D, D ′ ) ≡ D + A1 D

n−1

n

D ′ + ... + An D ′ .

Note that F ( D, D ′ ) is a homogeneous function in D, D ′ of degree n.

9.3 Solution of a Linear Homogeneous Partial Differential Equation with Constant Coefficients. As in the case of ordinary linear differential equations the basic theorem is : Theorem 1: If u is the complementary function and z 1 a particular integral of a linear partial differential equation F ( D, D ′ ) z = f ( x, y) then u + z1 is a general solution of the equation. Proof:

The complementary function of the equation F ( D, D ′ ) z = f ( x, y)

is the most general solution of the equation F ( D, D ′ ) z = 0.

…(1) …(2)

It must contain as many arbitrary constants as is the order of the differential equation (2). Any solution of (1) is called a particular integral of (1). It does not contain any arbitrary constant. Since the equations (1) and (2) are of the same order, the solution u + z1 will contain as many arbitrary constants as the general solution of (1) requires. Also F ( D, D ′ ) u = 0 , F ( D, D ′ ) z1 = f ( x, y) so that

F ( D, D ′ ) (u + z1) = f ( x, y).

This shows that, in fact, u + z1 is a solution of (1). This completes the proof. The next result is of extensive use in the solution of the partial differential equations.

D-249

Theorem 2:

If u1 , u2 , ... , un , are solutions of the homogeneous linear partial differential

equation F ( D, D ′ ) z = 0 then

n

is also a solution, where c r ’s are arbitrary

Σ c r ur

r =1

constants. Proof:

The proof of this theorem is obvious.

We have F ( D, D ′ ) (c r ur ) = c r F ( D, D ′ ) ur . n

n

r =1

r =1

Also

F ( D, D ′ ) Σ vr = Σ F ( D, D ′ ) vr , for any set of functions vr .

Hence

F ( D, D ′ ) Σ c r ur =

n

r =1

n

Σ F ( D, D ′ ) (c r ur )

r =1

n

=

Σ c r F ( D, D ′ ) ur =

r =1

n

Σ

r =1

c r . 0 = 0.

9.4 Determination of the Complementary Function (C.F.) of the Linear Homogeneous Partial Differential Equation with Constant Coefficients Consider a linear homogeneous nth order partial differential equation with constant coefficients of the form F ( D, D ′ ) z = f ( x, y). …(1) The complementary function of (1) is the general solution of F ( D, D ′ ) z = 0 i. e.,

( D n + A1 Dn − 1 D ′ + A2 Dn − 2 D ′2 + ... + An D ′ n ) z = 0

…(2)

This is equivalent to [( D − m1 D ′ ) ( D − m2 D ′ )...( D − mn D ′ )] z = 0 ,

…(3)

where m1, m2 , ... , mn are some constants. The solution of any one of the equations ( D − m1 D ′ ) z = 0 , ( D − m2 D ′ ) z = 0 , ... , ( D − mn D ′ ) z = 0

…(4)

is also a solution of (3) and we know that the general solution of ( D − mD ′ ) z = 0 is z = φ ( y + mx), where φ is an arbitrary function. Hence we can assume that a solution of the equation (3) is of the form z = φ ( y + mx). Differentiation will give Dz = mφ′ ( y + mx), D n z = m nφ(n) ( y + mx), D ′ n z = φ(n) ( y + mx) and, in general, D r D ′ s z = m r φ(r + s) ( y + mx). Therefore, the substitution of φ ( y + mx) for z in (2) gives (m n + A1m n − 1 + ... + An) φ(n) ( y + mx) = 0 .

D-250

This is true if m is a root of the equation m n + A1m n − 1 + ... + An = 0 .

…(5)

The equation (5) is called the auxiliary equation (A.E.) and is obtained by putting D = m, D ′ = 1 in F ( D, D ′ ) = 0 . It will give in general n roots, say, m1, m2 , ... mn. Each value of m will give a solution of (2). Hence if all the roots of the Auxiliary equation are distinct, the general solution of (2) i. e., the complementary function of (1) is z = φ 1 ( y + m1 x) + φ 2 ( y + m2 x) + ... + φn ( y + mn x)

…(6)

where φ1, φ 2 , ... , φ n are arbitrary functions.

Solution when the auxiliary equation has equal roots i. e., the roots of the A.E. are repeated. The equation corresponding to two repeated roots each equal to m is ( D − mD ′ )( D − mD ′ ) z = 0 . Putting ( D − mD ′ ) z = u, this becomes ( D − mD ′ ) u = 0, the solution of which is u = φ ( y + mx). Hence ( D − mD ′ ) z = φ ( y + mx) or

p − mq = φ ( y + mx).

Lagrange’s auxiliary equations of this linear equation are dx dy dz = = ⋅ 1 − m φ ( y + mx) Taking the first two members, we get dy + m dx = 0. y + mx = a.

∴

Taking the first and the third members, we get dz = φ ( y + mx) dx = φ(a) dx. ∴

z = φ (a). x + b.

Hence

z = xφ ( y + mx) + ψ ( y + mx).

Proceeding in the same way it can be shown that when a root m is repeated r times, the corresponding part of the complementary function is φ1 ( y + mx) + x φ 2 ( y + mx) + x2 φ 3 ( y + mx) + ... + x r − 1 φ r ( y + mx).

Example 1: Solution:

Solve 2 r + 5 s + 2 t = 0 .

…(1)

We know that r=

∂2 z 2

∂x

= D2 z , s =

∂2 z ∂2 z = DD ′ z , t = 2 = D ′2 z . ∂x ∂y ∂y

D-251

Hence the given equation can be written as (2 D2 + 5 DD ′ + 2 D ′2 ) z = 0 . A.E. is 2 m2 + 5 m + 2 = 0 ∴

m=−

or

(2 m + 1)(m + 2) = 0 .

1 , − 2. 2

Therefore the general solution of (1) is 1 z = f ( y − x) + ψ ( y − 2 x) or z = φ (2 y − x) + ψ ( y − 2 x). 2 Example 2: Solution:

Solve ( D3 − 6 D2 D ′ + 11 DD ′2 − 6 D ′3 ) z = 0 . The auxiliary equation is m3 − 6 m2 + 11m − 6 = 0

∴

…(1)

or (m − 1)(m − 2)(m − 3) = 0 .

m = 1, 2, 3.

Therefore the general solution of (1) is z = φ1 ( y + x) + φ 2 ( y + 2 x) + φ 3 ( y + 3 x). Example 3: Solution:

Solve ( D3 − 3 D2 D ′ + 2 DD ′2 ) z = 0 .

(Lucknow 2010)

The auxiliary equation is m3 − 3 m2 + 2 m = 0 or m (m − 1)(m − 2) = 0 .

∴

m = 0 , 1, 2 .

Therefore the general solution of (1) is z = φ 1 ( y) + φ 2 ( y + x) + φ 3 ( y + 2 x). Example 4: Solution:

Solve

∂4 z ∂x 4

−

∂4 z ∂y 4

= 0. (Avadh 2012) …(1) 4

4

The equation (1) can be written as ( D − D′ ) z = 0 .

A.E. is m4 − 1 = 0 or (m − 1)(m + 1)(m2 + 1) = 0 . ∴

m = 1, − 1, ± i.

Therefore the general solution of (1) is z = φ1 ( y + x) + φ 2 ( y − x) + φ 3 ( y + ix) + φ 4 ( y − ix). Example 5: Solve ( D4 − 2 D3 D′ + 2 DD′3 − D′4 ) z = 0 .

(Lucknow 2006, 09)

…(1)

Solution: The auxiliary equation is

m4 − 2 m3 + 2 m − 1 = 0 ∴

or (m + 1) (m − 1)3 = 0 .

m = − 1, 1, 1, 1.

Therefore the general solution of (1) is z = φ1 ( y − x) + φ 2 ( y + x) + xφ 3 ( y + x) + x2 φ 4 ( y + x). Example 6: Solution:

Solve 25 r − 40 s + 16 t = 0 . The equation (1) can be written as

…(1)

D-252

(25 D2 − 40 DD ′ + 16 D ′2 ) z = 0 . A.E. is

25 m2 − 40 m + 16 = 0

∴

m = 4 / 5, 4 / 5 .

or (5 m − 4)2 = 0 .

Therefore the general solution of (1) is 4  4    z = φ1  y + x  + x φ 2  y + x    5  5  or

z = f1 (5 y + 4 x) + x f2 (5 y + 4 x).

Comprehensive Exercise 1 1.

Solve r = a2 t.

(Rohilkhand 2010, Avadh 10)

3

2

2

2.

Solve ( D − 4 D D ′ + 4 DD ′ ) z = 0 .

3.

Solve

4.

Solve ( D2 − 3 aDD ′ + 2 a2 D ′2 ) z = 0 .

5.

Solve 2

6.

Solve

7.

Solve r + t + 2 s = 0.

∂3 z

−7

∂x3

∂2 z 2

∂x

∂2 z ∂x2

−

∂3 z ∂ x ∂ y2

−3 ∂2 z ∂ y2

+6

∂3 z ∂ y3

= 0.

∂2 z ∂2 z − 2 2 = 0. ∂x ∂y ∂y = 0.

2

(Kanpur 2009) 2

8.

Solve (4 D + 12 DD ′ + 9 D ′ ) z = 0 .

9.

Solve ( D 4 + D ′4 − 2 D 2 D ′2 ) z = 0 .

10.

Solve ( D 4 + D ′4 ) z = 0 .

A nswers 1 1.

z = φ1 ( y + ax) + φ 2 ( y − ax)

2.

z = φ1 ( y) + φ 2 ( y + 2 x) + x φ 3 ( y + 2 x)

3.

z = φ1 ( y + x) + φ 2 ( y + 2 x) + φ 3 ( y − 3 x)

4.

z = φ1 ( y + ax) + φ 2 ( y + 2 ax)

5.

z = φ1 (2 y − x) + φ 2 ( y + 2 x)

6.

z = φ1 ( y + x) + φ 2 ( y − x)

7.

z = φ1 ( y − x) + x φ 2 ( y − x)

8.

z = φ1 (2 y − 3 x) + x φ 2 (2 y − 3 x)

D-253

9. 10.

z = φ1 ( y + x) + x φ 2 ( y + x) + φ 3 ( y − x) + x φ4 ( y − x) z = φ1 ( y + α x) + φ2 ( y + α x) + φ 3 ( y + β x) + φ4 ( y + β x) where

α=

1 1 1 1 +i and β = − +i ⋅ √2 √2 √2 √2

9.5 Determination of the Particular Integral (P.I.) The particular integral of the equation (1) of 9.4 will be denoted by

1 f ( x, y). F ( D, D ′ )

1 V is defined as the function which gives V when it is operated upon by F ( D, D ′ ) F ( D, D ′ ). The symbolic function F ( D, D ′ ) can be treated as an algebraic function of D and D ′. It can be factorised, resolved into partial fractions or can be expanded in ascending powers of D or D ′ . 1 1 Note: means integration w.r.t. x, means integration w.r.t y, and so on and D D′ particular integral would be different according as F ( D, D ′ ) is expanded in ascending powers of D or D ′.

Example 7: Solution:

Solve ( D2 + 3 DD ′2 + 2 D ′2 ) z = x + y.

(Avadh 2012; Kanpur 14)

The auxiliary equation is m2 + 3 m + 2 = 0

or (m + 2)(m + 1) = 0 .

∴

m = − 1, − 2.

∴

C. F. = φ1 ( y − x) + φ 2 ( y − 2 x).

Now P.I. = =

1 D2 + 3 DD ′ + 2 D ′2

( x + y) =

3 D ′ 2 D ′2 1  1 + + D D2  D2

   

−1

3D ′  1  1 3 1− + ...  ( x + y) = 2 ( x + y) − 3 D ′ ( x + y) 2   D D D D =

3 1 1 x3 x2 x3 1 2 x3 + y⋅ − 3 1= + x y −3⋅ = − x3 + x2 y. 6 2 6 2 6 3 2 D

Hence the general solution of the given equation is z = C. F. + P.I. = φ 1 ( y − x) + φ 2 ( y − 2 x) − Example 8: Solution:

( x + y)

Solve ( D2 − 6 DD ′ + 9 D ′2 ) z = 12 x2 + 36 xy. A.E. is m2 − 6 m + 9 = 0

or (m − 3)2 = 0 .

1 3 1 2 x + x y. 3 2

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m = 3, 3.

∴

C. F. = φ1 ( y + 3 x) + x φ 2 ( y + 3 x). 1 P.I. = 2 (12 x2 + 36 xy) D − 6 DD ′ + 9 D ′2

∴ Now

=

1 ( D − 3 D ′ )2

=

(12 x2 + 36 xy)

3D ′  1  1−  2  D  D

−2

(12 x2 + 36 xy)

 6D ′ D ′2 1  1 27 + + + ...  (12 x2 + 36 xy)  2 2 D D  D  1 6 = 2 (12 x2 + 36 xy) + 3 D ′ (12 x2 + 36 xy) D D 6 4 3 = x + 6 x y + 3 (36 x) D =

= x 4 + 6 x3 y + 9 x 4 = 10 x 4 + 6 x3 y. Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y + 3 x) + x φ 2 ( y + 3 x) + 10 x 4 + 6 x3 y. Solve r + (a + b) s + abt = xy.

Example 9: Solution:

(Purvanchal 2014)

The given equation can be written as {D2 + (a + b) DD ′ + ab D ′2 } z = xy.

A.E. is m2 + (a + b) m + ab = 0 m = − a, − b.

∴ ∴ Now

or (m + a)(m + b) = 0 .

C. F. = φ1 ( y − ax) + φ 2 ( y − bx). 1 P.I. = 2 xy D + (a + b) DD ′ + abD ′2 D′ D ′2 1  = 2  1 + (a + b) + ab D D2 D  =

  

−1

xy

D′  1  1 1 1 − (a + b) ...  ( xy) = 2 ( xy) − (a + b) 3 {D ′ ( xy)} 2  D D  D D  x3 y

x4 ⋅ 6 24 Hence the general solution of the given equation is =

− (a + b)

z = C. F. + P.I. = φ1 ( y − ax) + φ 2 ( y − bx) + Example 10:

Solve (2 D2 − 5 DD ′ + 2 D ′2 ) z = 24 ( y − x).

Solution:

A.E. is 2 m2 − 5 m + 2 = 0

x3 y 6

− (a + b)

x4 ⋅ 24

D-255

or

(2 m − 1) (m − 2) = 0 . 1 m = , 2. 2

∴

C. F. = φ1 (2 y + x) + φ 2 ( y + 2 x). 1 Now P.I. = 24 ( y − x) 2 D2 − 5 DD ′ + 2 D ′2

∴

−1

=

 5 D ′ D ′2  1− + 2  2D 2D  D2

=

5D ′   1 5 ...  24 ( y − x) = 24 ( y − x) + 24  1+ 2   2 D 2D 2D 4 D3

1

   

24 ( y − x)

1

2

 x2 x3  5 x3  + ⋅ 24 ⋅ = 12  y − = 6 x2 y + 3 x3 . 2 6 4 6   Hence the general solution of the given equation is z = C. F. + P.I. = φ1 (2 y + x) + φ 2 ( y + 2 x) + 6 x2 y + 3 x3 . Example 11: Solution:

Solve

∂2 z 2

∂x

∂2 z

−

∂ y2

= x − y. (Lucknow 2008)

The given equation can be written as ( D2 − D ′2 ) z = x − y.

A.E. is m2 − 1 = 0 . ∴

m = 1, − 1.

∴

C. F. = φ1 ( y + x) + φ 2 ( y − x).

Now

P.I. =

1 D2 − D ′2

( x − y) =

D′2  1    1 − D2  D2 

−1

( x − y)

 D ′2 1  1 +  ( x − y) + ...  D2  D2 

=

1

=

D2

( x − y) =

1 3 1 x − y ⋅ x2 . 6 2

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y + x) + φ 2 ( y − x) + Example 12:

1 3 1 2 x − yx . 6 2

Solve ∂3 u 3

∂x

+

∂3 u ∂y

3

+

∂3 u ∂z

3

−3

∂3 u = x3 + y3 + z 3 − 3 xyz . ∂x ∂y ∂z

(Avadh 2012)

…(1) Solution:

The given equation can be written as ( D13 + D23 + D33 − 3 D1 D2 D3 ) u = x3 + y3 + z 3 − 3 xyz

D-256

( D1 + D2 + D3 ) ( D12 + D22 + D32 − D1 D2 − D2 D3 − D3 D1) u

or

= x3 + y3 + z 3 − 3 xyz ( D1 + D2 + D3 ) ( D1 + ωD2 + ω2 D3 ) ( D1 + ω2 D2 + ωD3 ) u

or

= x3 + y3 + z 3 − 3 xyz where ω is an imaginary cube root of unity. Now consider ( D1 + ωD2 + ω2 D3 ) u = 0 .

…(2)

Auxiliary equations are dx dy dz = = 2 ⋅ ω 1 ω These give y − ωx = a, z − ω2 x = b. Hence solution of (2) is φ1 ( y − ωx, z − ω2 x) = 0 . Similarly ( D1 + D2 + D3 ) u = 0 and ( D1 + ω2 D2 + ωD3 ) u = 0 give φ 2 ( y − x, z − x) = 0 and φ 3 ( y − ω2 x, z − ωx) = 0 respectively. ∴

C. F. of (1) = φ1 ( y − ωx, z − ω2 x) + φ 2 ( y − x, z − x) + φ 3 ( y − ω2 x, z − ωx).

Now P.I. corresponding to x3 is =

1 x3 D13 + D23 + D33 − 3 D1 D2 D3

=

 1  D23  1 + + ...  3 3 D1  D1 

−1

x3 =

1 3 x6 x6 = = ⋅ x 4 ⋅ 5 ⋅ 6 120 D13

Similarly particular integrals corresponding to y3 and z 3 are

y6 z6 and respectively 120 120

and P.I. corresponding to (− 3 xyz ) 1 = (− 3 xyz ) 3 3 D1 + D2 + D33 − 3 D1 D2 D3 =

  1 D12 1 − ...  − 3 D1 D2 D3  D2 D3 

=−

−1

(− 3 xyz )

x2 y2 z 2 1 (− 3 xyz ) = ⋅ 3 D1 D2 D3 8

Hence the general solution of the equation (1) is z = C. F. + P.I.

Comprehensive Exercise 2 1.

Solve ( D2 − 2 DD ′ + D ′2 ) z = 12 xy.(Lucknow 2008; Avadh 10, 11; Kanpur 12)

2.

Solve

∂3 z ∂x3

−

∂3 z ∂ y3

= x3 y3 .

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3.

Solve ( D2 − a2 D ′2 ) z = x2 .

4.

Solve

5.

Solve

6.

Solve ( D2 − DD ′ − 6 D ′2 ) z = xy.

∂2 z 2

∂x

∂2 z ∂x2

+3

∂2 z ∂2 z + 2 2 = 2 x + 3 y. ∂x ∂y ∂y

+3

∂2 z ∂2 z + 2 2 = 6 ( x + y). ∂x ∂y ∂y

A nswers 2 1.

z = φ1 ( y + x) + x φ 2 ( y + x) + 2 x 3 y + x 4

2.

z = φ1 ( y + x) + φ 2 ( y + ωx) + φ 3 ( y + ω2 x) +

3. 4.

x9 1 6 3 x y + 120 10080

1 4 x 12 7 3 z = φ1 ( y − x) + φ 2 ( y − 2 x) − x 3 + x 2 y 6 2 z = φ1 ( y + ax) + φ 2 ( y − ax) +

5.

z = φ1 ( y − x) + φ 2 ( y − 2 x) + 3 x 2 y − 2 x 3

6.

z = φ1 ( y − 2 x) + φ 2 ( y + 3 x) +

1 3 1 4 x y+ x 6 24

9.6 Short Method for Finding the Particular Integral When f ( x, y) is a function of ax + by, we have a shorter method for determining the particular integral. Let

f ( x, y) = φ (ax + by).

Then

Dr φ (ax + by) = a r φ(r) (ax + by), D ′ r φ (ax + by) = b r φ(r) (ax + by),

and

Dr D ′ s φ (ax + by) = a r b s φ(r + s) (ax + by),

where φ(r) is the rth derivative of φ w.r.t. ax + by as a whole. Since F ( D, D ′ ) is homogeneous in D and D ′ of degree n, hence or

F ( D, D ′ ) φ (ax + by) = F (a, b) φ(n) (ax + by) 1 1 φ(n) (ax + by) = φ (ax + by), F ( D, D ′ ) F (a, b)

provided that F (a, b) ≠ 0 . 1 1 Putting ax + by = t, we get φ(n) (t) = φ (t). F ( D, D ′ ) F (a, b) Now integrating both the sides n times w.r.t. ‘t’, we get

D-258

1 1 φ (t) = F ( D, D′ ) F (a, b)

∫ ∫… ∫

φ (t) dt … dt, where t = ax + by .

To find the particular integral of an equation F ( D, D ′ ) z = φ (ax + by), where F ( D, D ′ ) is a homogeneous function of D, D ′ of degree n, proceed as follows : Working Rule:

(i)

Put ax + by = t and integrate φ (t), n times with respect to t.

(ii)

Find F (a, b), replacing D, D ′ by a, b respectively in F ( D, D ′ ). 1 (iii) Now P.I. = × nth integral of φ (t) w.r.t. ‘t’, where t = ax + by. F (a, b) Exceptional case when F ( a, b) = 0. If F (a, b) = 0 then the above method fails. Now F (a, b) = 0 if and only if (bD − aD ′ ) is a factor of F ( D, D ′ ). Let us consider the equation (bD − aD ′ ) z = x

r

φ (ax + by)

…(1)

r

or

bp − aq = x φ (ax + by).

Lagrange’s auxiliary equations are dy dx dz = = r ⋅ b − a x φ (ax + by) Taking the first two members, we get a dx + b dy = 0. ax + by = c (constant).

∴

Again taking the first and the last members, we get dz = z=

∴

xr xr φ (ax + by) dx = φ (c ) dx. b b

x r +1 x r +1 φ (c ) = φ (ax + by). b (r + 1) b (r + 1)

It gives the solution of (1). 1 From (1), z = x r φ (ax + by). bD − aD ′ Thus

1 x r +1 x r φ ( ax + by) = φ ( ax + by) . ( bD − aD ′ ) b ( r + 1)

Hence if z =

then

1 (bD − aD ′ )n z= =

φ (ax + by),

1 n −1

(bD − aD ′ ) 1

n −1

(bD − aD ′ )

  1 φ (ax + by)    (bD − aD ′ )  x φ (ax + by), using (4) b

…(2)

…(3) …(4)

D-259

=

=

1 n−2

(bD − aD ′ ) 1

n−2

(bD − aD ′ )

  1 x φ (ax + by)    (bD − aD ′ ) b  1 x2 ⋅ φ (ax + by), using (4) b 2b

… … … … … … … … … … … … … … … … … … … … … … … … = Thus

x

b n!

1 ( bD − aD ′ )

Example 13: Solution:

n

n

n

φ (ax + by), by repeated application of (4).

φ ( ax + by) =

xn n

b n!

φ ( ax + by) .

Solve 4 r − 4 s + t = 16 log ( x + 2 y).

(Lucknow 2007; Purvanchal 09; Rohilkhand 14)

The given equation can be written as (4 D2 − 4 DD ′ + D ′2 ) z = 16 log ( x + 2 y).

A.E. is 4 m2 − 4 m + 1 = 0 or (2 m − 1)2 = 0 . ∴ Now

C. F. = φ1 (2 y + x) + x φ 2 (2 y + x). 1 P.I. = 16 log ( x + 2 y) (2 D − D ′ )2 =

x2 2

2 ⋅2 !

16 log ( x + 2 y) = 2 x 2 log ( x + 2 y).

[ ∵ F (a, b) = 0 ]

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 (2 y + x) + x φ 2 (2 y + x) + 2 x2 log ( x + 2 y). Example 14: Solution:

∴ ∴ Now

Solve ( D3 − 4 D2 D ′ + 4 DD ′2 ) z = 4 sin (2 x + y). 3

2

A.E. is m − 4 m + 4 m = 0

2

or m (m − 2) = 0 .

m = 0 , 2, 2. C. F. = φ1 ( y) + φ2 ( y + 2 x) + x φ 3 ( y + 2 x). 1 P.I. = 3 4 sin (2 x + y) 2 D − 4 D D′ + 4 DD′2 1 1 = ⋅ 4 sin (2 x + y) 2 D (D − 2 D ′ ) 1

{− 2 cos (2 x + y)} ( D − 2 D ′ )2 1 = −2 cos (2 x + y) ( D − 2 D ′ )2 =

(Rohilkhand 2011)

D-260

= −2⋅

x2 cos (2 x + y) = − x2 cos (2 x + y). 2!

[ ∵ F (a, b) = 0 ]

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y) + φ 2 ( y + 2 x) + x φ 3 ( y + 2 x) − x 2 cos (2 x + y). Example 15:

Solve ( D2 − 2 DD ′ + D ′2 ) z = e

x +2 y

+ x3 . (Lucknow 2010; Rohilkhand 14)

2

A.E. is m − 2 m + 1 = 0

Solution:

2

or (m − 1) = 0 .

m = 1, 1.

∴

C. F. = φ1 ( y + x) + x φ 2 ( y + x). 1 1 Now P.I. = e x +2 y + x3 2 (D − D ′ ) ( D − D ′ )2

∴

=

1 2

(1 − 2)

=e

x +2 y

⋅e

x +2 y

+

D′  1  1−  2  D  D

−2

x3

2D ′  1  + ...  x3 = e  1+  D D2 

+

x +2 y

+

1 D2

x3 = e

x +2 y

+

1 5 x . 20

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y + x) + x φ 2 ( y + x) + e Example 16: Solution:

Solve

∂2 z ∂x 2

+

∂2 z ∂y 2

x +2 y

+

1 5 x . 20

= cos mx cos ny. (Lucknow 2006; Kanpur 07; Avadh 14)

The given equation can be written as 1 ( D2 + D ′2 ) z = [cos (mx + ny) + cos (mx − ny)]. 2

A.E. is m2 + 1 = 0 . ∴ ∴ Now

m = ± i. C. F. = φ1 ( y + ix) + φ 2 ( y − ix). 1 1 1 1 P.I. = ⋅ 2 cos (mx + ny) + cos (mx − ny) 2 D + D ′2 2 D2 + D ′2 =

1 1 1 1 {− cos (mx + ny)} + ⋅ 2 {− cos (mx − ny)} 2 m2 + n2 2 m + n2

=−

1 2

2 (m + n2 )

{cos (mx + ny) + cos (mx − ny)}.

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y + ix) + φ 2 ( y − ix) 1 − [cos (mx + ny) + cos (mx − ny)]. 2 2 (m + n2 )

D-261 Example 17: Solution:

Solve 2 r − s − 3 t = 5 e x / e y.

(Rohilkhand 2009)

The given equation can be written as (2 D2 − DD ′ − 3 D ′2 ) z = 5 e

x− y

.

A.E. is 2 m2 − m − 3 = 0 or (2 m − 3) (m + 1) = 0 . ∴ Now

C. F. = φ1 (2 y + 3 x) + φ 2 ( y − x). 1 1 P.I. = 5e x − y = 5e ( D + D ′ ) (2 D − 3 D ′ ) 2 D2 − DD ′ − 3 D ′2 =

1 1 ⋅ ⋅ 5e D+ D′ 2+3

=

x e 1!

x− y

= xe

x− y

x− y

=

1 e D+ D′

x− y

x− y

.

[ ∵ F (a, b) = 0 ]

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 (2 y + 3 x) + φ 2 ( y − x) + x e Example 18:

x− y

.

Solve ( D2 + 3 DD ′ + 2 D ′2 ) z = x + y. (Rohilkhand 2010; Agra 02; Avadh 14; Purvanchal 14)

Solution:

2

A.E. is m + 3 m + 2 = 0

or (m + 2) (m + 1) = 0 .

m = − 1, − 2.

∴

C. F. = φ1 ( y − x) + φ 2 ( y − 2 x). 1 P.I. = 2 ( x + y) D + 3 DD ′ + 2 D ′2

∴ Now

=

( x + y)3 ( x + y)3 = ⋅ (1 + 3 ⋅ 1 ⋅ 1 + 2 ⋅ 1) 6 36 1

⋅

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y − x) + φ 2 ( y − 2 x) + Example 19: Solution:

or ∴ Now

Solve ( D2 − 5 DD′ + 4 D′2 ) z = sin (4 x + y).

A.E. is m2 − 5 m + 4 = 0 (m − 1) (m − 4) = 0 ∴ m = 1, 4. C. F. = φ1 ( y + x) + φ 2 ( y + 4 x). 1 P.I. = 2 sin (4 x + y) D − 5 DD ′ + 4 D ′2 1 1 = ⋅ sin (4 x + y) (D − 4 D ′ ) (D − D ′ ) 1 1 = ⋅ {− cos (4 x + y)} ( D − 4 D ′ ) (4 − 1) 1 1 = {− cos (4 x + y)} D − 4D ′ 3

( x + y)3 ⋅ 36 (Lucknow 2006)

D-262

=−

1 ⋅ x cos (4 x + y). 3

[ ∵ F (a, b) = 0 ]

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y + x) + φ 2 ( y + 4 x) −

1 x cos (4 x + y). 3

Comprehensive Exercise 3 1.

Solve ( D2 − 6 DD ′ + 9 D ′2 ) z = 6 x + 2 y.

2.

Solve

3.

Solve ( D3 − 4 D2 D ′ + 4 DD ′2 ) z = cos ( y + 2 x).

4.

Solve r − 2 s + t = sin (2 x + 3 y).

∂2 z ∂x 2

+

∂2 z ∂y 2

= 12 ( x + y).

2

(Kanpur 2010; Meerut 13)

2

5.

Solve (2 D − 5 DD ′ + 2 D ′ ) z = 5 sin (2 x + y).

6.

Solve ( D2 + 2 DD ′ + D ′2 ) z = e 2 2

(Purvanchal 2011)

x +3 y

.

(Purvanchal 2007)

2 2

7.

Solve ( D + D ′ ) z = 30 (2 x + y).

8.

Solve ( D3 − 2 D2 D ′ − DD ′2 + 2 D ′3 ) z = e

9.

Solve ( D3 − 7 DD ′2 − 6 D ′3 ) z = x2 + x y2 + y3 + cos ( x − y).

10. 11.

x+ y

.

Solve r + s − 2 t = √ (2 x + y). 3

2

(Lucknow 2010) 2

3

Solve ( D − 4 D D ′ + 5 DD ′ − 2 D ′ ) z = e 2

2

2x − y

y +2 x

x+ y

+ ( y + x)

12.

Solve ( D − 3 DD ′ + 2 D ′ ) z = e

13.

Solve log s = x + y.

14.

Solve ( Dx3 − 7 Dx D y2 − 6 D y3 ) z = sin ( x + 2 y) + e 3 x + y.

+e

1 2 x (3 x + y) 4

1.

z = φ1 ( y + 3 x) + x φ 2 ( y + 3 x) +

2.

z = φ1 ( y + ix) + φ 2 ( y − ix) + ( x + y)3

3.

z = φ1 ( y) + φ 2 ( y + 2 x) + x φ 3 ( y + 2 x) +

4.

z = φ 1 ( y + x) + x φ2 ( y + x) − sin (2 x + 3 y) 1 z = φ1 (2 y + x) + φ 2 ( y + 2 x) − ⋅ 5 x cos (2 x + y) 3 1 2 x +3 y z = φ1 ( y − x) + x φ 2 ( y − x) + e 25

6.

.

+ cos ( x + 2 y).

A nswers 3

5.

1 /2

1 2 x sin ( y + 2 x) 4

(Meerut 2013)

D-263

7.

z = φ1 ( y + ix) + φ 2 ( y − ix) + (2 x + y)3

8.

z = φ1 ( y − x) + φ 2 ( y + x) + φ 3 ( y + 2 x) −

9.

z = φ1 ( y − x) + φ 2 (

10.

z = φ1 ( y + x) + φ 2 (

11.

z = φ1 ( y + x) + x φ 2

12.

z = φ1 ( y + x) + φ 2 (

13.

z = φ1 ( x) + φ 2 ( y) + e

14.

z = φ1 ( y − x) + φ 2 ( y − 2 x) + φ 3 ( y + 3 x) +

1 xe x + y 2 5 6 1 5 7 5 y − 2 x) + φ 3 ( y + 3 x) + x + x + x y 72 60 20 1 4 2 1 3 3 1 + x y + x y − x cos ( x − y). 24 6 4 1 y − 2 x) + (2 x + y)5 /2 15 1 ( y + x) + φ 3 ( y + 2 x) + xe y + 2 x − x2 ( y + x)3 /2 3 1 2x − y 1 y + 2 x) + e − xe x + y − cos ( x + 2 y) 12 3 x+ y

1 x 3x + y cos ( x + 2 y) + e 75 20

9.7 A General Method of Finding the Particular Integral Consider the equation ( D − mD ′ ) z = φ ( x, y) or

p − mq = φ ( x, y).

Lagrange’s auxiliary equations are dx dy dz = = ⋅ 1 − m φ ( x, y) Taking the first two members, we get dy + m dx = 0. ∴

y + mx = a (a constant).

Again taking the first and the last members, we get dz = φ ( x, y) dx = φ ( x, a − mx) dx. z= Thus z =

∫ φ ( x, a − mx) dx .

1 φ ( x, y) = D − mD ′

∫ φ ( x, a − mx) dx ,

where after integration the constant a is to be replaced by y + mx, as the particular integral does not contain any arbitrary constant. Now if the equation is F ( D, D ′ ) z = φ ( x, y) where then

F ( D, D ′ ) = ( D − m1 D ′ ) ( D − m2 D ′ )...( D − mn D ′ ), 1 P.I. = φ ( x, y) F ( D, D ′ ) =

1 1 1 ⋅ ... φ ( x, y), D − m1 D ′ D − m2 D ′ D − mn D ′

which can be evaluated by the repeated application of the above method.

D-264

Example 20: Solution:

Solve r + s − 6 t = y cos x.

(Avadh 2009, 10, 11; Purvanchal 07)

The given equation can be written as ( D2 + DD ′ − 6 D ′2 ) z = y cos x.

A.E. is m2 + m − 6 = 0

or (m + 3)(m − 2) = 0 .

∴

m = 2, − 3.

∴

C. F. = φ1 ( y + 2 x) + φ 2 ( y − 3 x). 1 1 P.I. = 2 y cos x = y cos x (D − 2 D ′ ) (D + 3 D ′ ) D + DD ′ − 6 D ′2

Now

=

1 D − 2D ′

=

1 [a sin x + 3 x sin x + 3 cos x ] D − 2D ′

=

1 [( y − 3 x) sin x + 3 x sin x + 3 cos x ] D − 2D ′

=

1 [ y sin x + 3 cos x ] (D − 2 D ′ )

=

∫ [(b − 2 x) sin x + 3 cos x] dx,

∫ (a + 3 x) cos x dx, where

y − 3x = a

where y + 2 x = b

= − b cos x − 2 (− x cos x + sin x) + 3 sin x = − ( y + 2 x) cos x + 2 x cos x + sin x = − y cos x + sin x. Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y + 2 x) + φ 2 ( y − 3 x) − y cos x + sin x. Example 21: Solution:

∴

Solve ( D2 + 2 DD ′ + D ′2 ) z = 2 cos y − x sin y.

A.E. is m2 + 2 m + 1 = 0

(Lucknow 2011)

or (m + 1)2 = 0 .

m = − 1, − 1.

C. F. = φ1 ( y − x) + x φ 2 ( y − x). 1 Now P.I. = 2 (2 cos y − x sin y) D + 2 DD ′ + D ′2

∴

=

1 (2 cos y − x sin y) ( D + D ′ )( D + D ′ )

=

1 D + D′

=

1 [2 sin ( x + a) − {− x cos ( x + a) + sin ( x + a)}] D + D′

=

1 [sin ( x + a) + x cos ( x + a)] D+ D′

∫ [2 cos ( x + a) − x sin ( x + a)] dx,

where y − x = a

D-265

=

1 [sin y + x cos y ] D + D′

=

∫ [sin ( x + b) + x cos ( x + b)] dx,

where y − x = b

= − cos ( x + b) + x sin ( x + b) + cos ( x + b) = x sin ( x + b) = x sin y. Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y − x) + x φ 2 ( y − x) + x sin y. Example 22: Solution:

Solve ( D2 − 2 DD ′ − 15 D ′2 ) z = 12 x y. 2

A.E. is m − 2 m − 15 = 0

(Purvanchal 2010)

or (m − 5)(m + 3) = 0 .

m = 5, − 3. C. F. = φ 1 ( y + 5 x) + φ 2 ( y − 3 x). 1 1 P.I. = 2 12 xy = 12 xy 2 ( D + 3 D ′ ) (D − 5 D ′ ) D − 2 DD ′ − 15 D ′

∴ ∴ Now

=

12 D + 3D ′

=

12  1 2 5 3   ax − x  D + 3D ′  2 3 

∫ x (a − 5 x) dx, where

y + 5x = a

12 5 3   1 2  ( y + 5 x) x − x  D + 3D ′  2 3  2 = {3 yx2 + 5 x3} D + 3D ′ =

= 2 ∫ [3 x2 (3 x + b) + 5 x3 ] dx , where y − 3 x = b =2

3

∫ (14 x

+ 3 x2 b) dx

 x4 x3  = 2  14 ⋅ +3⋅ b  = 7 x 4 + 2 x3 b 4 3   = 7 x4 + 2 x3 ( y − 3 x) = x 4 + 2 x3 y. Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y + 5 x) + φ 2 ( y − 3 x) + x 4 + 2 x3 y. Example 23: Solution:

Solve r − t = tan3 x tan y − tan x tan3 y.

(Lucknow 2006; Avadh 10)

The given equation can be written as ( D2 − D ′2 ) z = tan3 x tan y − tan x tan3 y.

A.E. is m2 − 1 = 0 . ∴

m = 1, − 1.

∴

C. F. = φ 1 ( y + x) + φ 2 ( y − x). 1 P.I. = 2 (tan3 x tan y − tan x tan3 y) D − D ′2

Now

=

1 (tan3 x tan y − tan x tan3 y) ( D + D ′ )( D − D ′ )

D-266

=

1 D + D′

3

∫ [tan

x tan (a − x) − tan x tan3 (a − x)] dx, where y + x = a

1 = D + D′

∫ [{ − 1 + sec

2

x} tan x tan (a − x)] − tan x tan (a − x) {− 1 + sec2 (a − x)}] dx

=

1 D + D′

[ ∫ tan (a − x) tan x sec −

 tan2 x 1 1 = + tan (a − x) ⋅ D + D ′  2 2

1 2 (D + D ′ )

x dx

∫ tan x tan (a − x) sec ∫ sec

2

2

]

(a − x) dx

(a − x) ⋅ tan2 x dx

tan2 (a − x) 1 − 2 2

 x tan2 (a − x) dx  2 2 [tan x tan (a − x) + tan x tan (a − x) + tan x ⋅

=

2

∫ sec

2

− ∫ sec2 x {sec2 (a − x) − 1 − sec2 (a − x) ⋅ (sec2 x − 1)} dx] =

1 [tan2 x tan (a − x) + tan x tan2 (a − x) 2 (D + D ′ ) + ∫ {sec2 x − sec2 (a − x)} dx]

1 [tan2 x tan y + tan x tan2 y + (tan x + tan y)] 2 (D + D ′ ) 1 = [tan x. sec2 y + tan y. sec2 x ] 2 (D + D ′ ) 1 = ∫ [tan x sec2 (b + x) + tan (b + x) sec2 x] dx, where y − x = b 2 1 1 = tan x tan (b + x) − ∫ sec2 x tan (b + x) dx 2 2 1 + tan (b + x) sec2 x dx 2 ∫ 1 1 = tan x tan (b + x) = tan x tan y. 2 2 Hence the general solution of the given equation is 1 z = C. F. + P.I. = φ1 ( y + x) + φ 2 ( y − x) + tan x tan y. 2 =

Comprehensive Exercise 4 1.

Solve

2.

Solve

∂z ∂z + = sin x. ∂x ∂y ∂2 z 2

∂x

+

∂2 z ∂2 z − 6 2 = y sin x. ∂x ∂y ∂y

(Kanpur 2011)

D-267

3. 4. 5.

Solve

∂2 z 2

∂x

−4

∂2 z ∂y

2

=

Solve r − s − 2 t = (2 x

4x y

2

2

−

y x2

⋅ (Rohilkhand 2000; Agra 03) 2

+ xy − y ) sin xy − cos xy.

2

Solve ( D − DD′ − 2 D′2 ) z = ( y − 1) e x .

A nswers 4 1.

z = φ1 ( y − x) − cos x

2.

z = φ1 ( y + 2 x) + φ 2 ( y − 3 x) − ( y sin x + cos x)

3.

z = φ1( y + 2 x) + φ 2 ( y − 2 x) + x log y + y log x + 3 x.

4.

z = φ1 ( y + 2 x) + φ 2 ( y − x) + sin xy.

5.

z = φ1 ( y + 2 x) + φ 2 ( y − x) + ye x .

9.8 Non-Homogeneous Linear Equations with Constant Coefficients A linear partial differential equation which is not homogeneous is called a non-homogeneous linear equation. Consider the differential equation F ( D, D′ ) z = f ( x, y).

…(1)

When F ( D, D′ ) is a homogeneous function in D, D′ it can always be resolved into linear factors. But the result is not always true when F ( D, D′ ) is non-homogeneous. We classify linear differential operators F ( D, D′ ) into two main types. These are : (i)

F ( D, D′ ) is reducible if it can be written as the product of linear factors of the form D + aD′ + b, with a, b constants.

(ii)

F ( D, D′ ) is irreducible if it cannot be so written.

Complementary function of non-homogeneous linear equation: When F ( D, D ′ ) can be resolved into linear factors. The complementary function of non-homogeneous linear equation (1) is the general solution of the equation F ( D, D′ ) z = 0 .

…(2)

Let us consider a simple non-homogeneous equation ( D − mD′ − k ) z = 0 or

p − mq = kz .

Lagrange’s auxiliary equations for it are dy dx dz = = ⋅ 1 − m kz

…(3)

D-268

The first two members give dy + m dx = 0. y + mx = a (a constant).

∴

Again taking the first and the third members, we get dz = k dx. z ∴ log z = k x + log b or z = be kx . ∴

z = e kx φ ( y + mx), which is the solution of (3).

If F ( D, D′ ) can be factorised into non-repeated linear factors ( D − m1 D′ − k1), ( D − m2 D ′ − k2 ), ..., ( D − mn D ′ − k n), then the equation (2) is equivalent to [( D − m1 D′ − k1) ( D − m2 D′ − k2 )... ( D − mn D′ − k n)] z = 0

…(4)

The complete solution of (2) or (4) is made up of the solutions of ( D − m1 D′ − k1) z = 0 , ( D − m2 D′ − k2 ) z = 0 ,..., ( D − mn D′ − k n) z = 0 . Hence the general solution (complete solution) of (2) is z = e k1 x φ 1 ( y + m1 x) + e k2 x φ 2 ( y + m2 x) + ... + e kn

x

φn ( y + mn x).

…(5)

If the operator F ( D, D′ ) is reducible, the order in which the linear factors occur is immaterial. Note 1: Note 2:

It can be shown that if the equation is (α D + βD′ + γ ) z = 0, then z = e(− γ

x / α)

φ ( βx − αy).

F ( D, D ′ ) has repeated factors. Let a factor ( D − mD′ − k ) occur twice in F ( D, D′ ). Consider the equation ( D − mD′ − k )2 z = 0 . Let ( D − mD′ − k ) z = u. Then (6) reduces to ( D − mD ′ − k ) u = 0 . kx This gives u = e φ1 ( y + mx). Hence or

( D − mD′ − k ) z = e kx φ1 ( y + mx) p − mq = kz + e kx φ1 ( y + mx).

Lagrange’s auxiliary equations for this are dy dx dz = = ⋅ kx 1 − m kz + e φ1 ( y + mx) The first two members give, y + mx = a (a constant). From the first and the last members, we get dz − kz = e kx φ1 ( y + mx) = e kx φ1 (a), dx which is an ordinary linear equation.

…(6)

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I. F. = e ∫ − k dx = e − kx . ∴

ze − kx =

∴

z = e kx [ x φ1 ( y + mx) + φ 2 ( y + mx)],

∫e

− kx

e kx φ1 (a) dx + b = x φ 1 (a) + b . …(7)

which is the general solution of (6). Hence if ( D − mD′ − k ) occurs twice in F ( D, D′ ) then the C.F. corresponding to this factor is e kx [ x φ1( y + mx) + φ 2 ( y + mx)]

…(8)

It can be shown that the general solution of ( D − mD′ − k )r z = 0 is

z = e kx [ φ1 ( y + mx) + x φ 2 ( y + mx) + ... + x

r −1

φ r ( y + mx)].

Hence if ( D − mD ′ − k ) occurs r times in F ( D, D ′ ) then the C.F. corresponding to this factor is e kx [φ1 ( y + mx) + x φ2 ( y + mx) + … + x r −1 φ r ( y + mx)] .

…(9)

Case when linear factors of F ( D, D ′ ) are not possible: In case F ( D, D ′ ) is irreducible, i. e., it cannot be resolved into linear factors in D and D ′, the above methods of finding the complementary function fail. In such cases a trial method is used to find solutions.

9.9 Particular Integral The complete solution of F ( D, D ′ ) z = f ( x, y) is where

z = C. F. + P.I. 1 P.I. = f ( x, y). F ( D, D ′ )

The methods of obtaining particular integrals of non-homogeneous partial differential equations are very similar to those of ordinary linear equations with constant coefficients. We give some cases of finding the particular integrals. 1 1 Case I: e ax + by = e ax + by, if F (a, b) ≠ 0 . F ( D, D ′ ) F (a, b) We have Dr (e ax + by) = a r e ax + by, D′ s (e ax + by) = b s e ax + by and

( Dr D ′ s ) (e ax + by) = a r b s e ax + by.

∴

F ( D, D ′ ) (e ax + by) = F (a, b) e ax + by.

Operating both the sides by

1 , we get F ( D, D ′ )

e ax + by = F (a, b)

1 e ax + by F ( D, D ′ )

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1 1 e ax + by = e ax + by, provided F (a, b) ≠ 0 . F ( D, D ′ ) F (a, b)

⇒ Case II:

The value of

1 sin (ax + by) is obtained by putting F ( D, D ′ )

D2 = − a2 , DD ′ = − ab and D ′2 = − b2 , provided the denominator is not zero. Similar is the rule for cos (ax + by). 1 Case III: x m y n = [ F ( D, D ′ )]− 1 x m y n, F ( D, D ′ ) which can be evaluated after expanding [ F ( D, D′ )]−1 in ascending powers of D or D′. 1 1 (e ax + by . V ) = e ax + by V. F ( D, D ′ ) F ( D + a, D ′ + b)

Case IV:

It follows from the fact that F ( D, D ′ ) {e ax + by. V } = e ax + by F ( D + a, D ′ + b) V .

Example 24: Solution:

Solve r + 2 s + t + 2 p + 2 q + z = 0.

The given equation can be written as ( D2 + 2 DD ′ + D ′2 + 2 D + 2 D ′ + 1) z = 0 ( D + D ′ + 1)2 z = 0

or

or {D − (− 1) D ′ − (− 1)}2 z = 0 .

There are repeated linear factors. Hence the solution is z = e − x φ1 ( y − x) + xe − x φ 2 ( y − x). Example 25:

Solve ( D2 − a2 D ′2 + 2 ab D + 2 abD ′ ) z = 0 .

The given equation can be written as ( D + aD ′ ) ( D − aD ′ + 2 ab) z = 0 . There are distinct linear factors. Hence the solution is Solution:

z = φ1 ( y − ax) + e − 2 abx φ 2 ( y + ax). Example 26: Solution:

Solve (2 D4 − 3 D2 D ′ + D ′2 ) z = 0 .

The given equation can be written as (2 D2 − D ′ ) ( D2 − D ′ ) z = 0 .

Let z = Ae hx + ky be the solution corresponding to ( D2 − D ′ ) z = 0 . ∴

( D2 − D ′ ) z = A h2 e hx + ky − Ake hx + ky = A (h2 − k ) e hx + ky = 0

⇒

h2 − k = 0

⇒

k = h2 .

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Hence the general solution of ( D2 − D ′ ) z = 0 is z = ΣAe hx + h

2

y

.

Similarly the general solution of (2 D 2 − D ′ ) z = 0 is 2

z = Σ Be h′ x + 2 h′

y

.

Hence the most general solution of the given equation is z = Σ Ae hx + h Example 27: Solution:

2

y

2

+ Σ Be h′ x + 2 h′

y

.

Solve ( D − D ′ − 1) ( D − D ′ − 2) z = e 2 x − y + x.

C. F. = e x φ1 ( y + x) + e2 x φ 2 ( y + x).

Now P.I. corresponding to e 2 x − y 1 = e2 x − y ( D − D ′ − 1) ( D − D ′ − 2) =

1 1 e2 x − y = e2 x − (2 + 1 − 1) (2 + 1 − 2) 2

y

and P.I. corresponding to x =

1 x ( D − D ′ − 1) ( D − D ′ − 2)

1 (1 − 2 1 = (1 + 2 1 = (1 + 2 1 = (1 + 2 =

1 1 D + D ′ )− 1 x 2 2 1 1 D − D ′ ...) (1 + D − D ′ ...) x 2 2 1 D + D + ...) x 2 3 1 3 D) x = ( x + ). 2 2 2 D + D ′ )−1 (1 −

Hence the general solution of the given equation is z = C. F. + P.I. = e x φ1 ( y + x) + e 2 x φ 2 ( y + x) + Example 28: Solution:

Solve ( D2 − DD ′ + D ′ − 1) z = cos ( x + 2 y) + e y.

The given equation can be written as ( D − 1) ( D − D ′ + 1) z = cos ( x + 2 y) + e y.

∴

C. F. = e x φ1 ( y) + e − x φ 2 ( y + x).

Now P.I. corresponding to cos ( x + 2 y) 1 = 2 cos ( x + 2 y) D − DD ′ + D ′ − 1 =

1 2

− 1 − (− 1 ⋅ 2) + D ′ − 1

cos ( x + 2 y)

1 2x − y x 3 e + + ⋅ 2 2 4 (Avadh 2010, 11)

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D′ 1 cos ( x + 2 y) = cos ( x + 2 y) D′ D ′2

= =

D′ 2

−2

cos ( x + 2 y) = −

1 {− 2 sin ( x + 2 y)} 4

1 = sin ( x + 2 y) 2 and P.I. corresponding to e y 1 1 = 2 ey= ey ( D − 1)( D − D ′ + 1) D − DD ′ + D ′ − 1 =

1 e (0 − 1)( D − D ′ + 1)

=−ey

y

=−e

y

1 1 D − ( D ′ + 1) + 1

D′  1 1  1= − e y 1 −  D − D′ D D 

−1

1= − e y

1 1 = − xe y. D

Hence the general solution of the given equation is z = C. F. + P.I. = e x φ1 ( y) + e − x φ 2 ( y + x) + Example 29:

1 sin ( x + 2 y) − xe y. 2

Solve ( D2 − D ′2 − 3 D + 3 D ′ ) z = xy + e

x +2 y

.

(Rohilkhand 2007; Lucknow 08; Purvanchal 2007)

Solution:

The given equation can be written as ( D − D ′ ) ( D + D ′ − 3) z = xy + e

∴

3x

C. F. = φ1 ( y + x) + e

x +2 y

.

φ2 ( y − x).

Now P.I. corresponding to xy 1 = xy ( D − D ′ )( D + D ′ − 3) −1

−1

=−

D′  1  1 −  3D  D 

=−

D′    1  D D ′ 2 DD ′ + ... 1 + + + + ... xy 1 +    3D  D 3 3 9

1  1 + 3D  1  =−  xy 3D  1 1 = −  x2 3 2 =−

 D D ′ − 1 −   3 3 

xy

 D D ′ D ′ D ′ 2 DD ′ + + + + + ... xy  3 3 D 3 9 y 2 1 2 + + x+ x+  3 3 D 9 1 1 2 1 3 2  y + xy + x + x + x  3 3 6 9 

and P.I. corresponding to e x + 2

y

=

1 ex + 2y ( D − D ′ ) ( D + D ′ − 3)

=

1 e (1 − 2) ( D + D ′ − 3)

x +2 y

=−

1 e ( D + D ′ − 3)

x +2 y

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1 e2 y, putting D = 1 D′ − 2 1 1 = − e x . e2 y 1= − ex +2 y 1 ( D ′ + 2) − 2 D′ = − ye x + 2 y. = − ex

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y + x) + e3 x φ 2 ( y − x) − Example 30:

1 1 2 1 1 2 1 3 2   x y + xy + x + x + x  − ye 3 2 3 3 6 9 

Solve ( D2 − DD ′ − 2 D ′2 + 2 D + 2 D ′ ) z = e2 x + 3

y

x +2 y

.

+ xy + sin (2 x + y). (Avadh 2013)

Solution:

∴

The given equation can be written as ( D + D ′ ) ( D − 2 D ′ + 2) z = e2 x + 3 C. F. = φ 1 ( y − x) + e

−2 x

y

+ xy + sin (2 x + y).

φ 2 ( y + 2 x).

2 x +3 y

Now P.I. corresponding to e

1 e2 x +3 y ( D + D ′ ) ( D − 2 D ′ + 2) 1 1 2 x +3 y = e2 x +3 y = − e , (2 + 3) (2 − 6 + 2) 10 =

P.I. corresponding to xy =

1 xy ( D + D ′ ) ( D − 2 D ′ + 2)

=

D′  1  1 +  2D  D 

=

D′  1  1 1 + ...  {1 − ( D − 2 D ′ ) + ( D − 2 D ′ )2 +...} xy 1 −  2D  D 2 4

−1

{1 +

1 ( D − 2 D ′ )}− 1 xy 2

D′  1  1 + ... ( xy − y + x − 1) 1 −  2D  D 2 1 1 1 1 = ( xy − y + x − 1 − x2 + x) 2D 2 2 2 1  1 1 2 3  = y − x + x − 1  xy −  2D  2 2 2 1 1 2 1 1 3 3 2  =  x y − xy − x + x − x   2 2 2 6 4 =

and P.I. corresponding to sin (2 x + y) 1 = 2 sin (2 x + y) D − DD ′ − 2 D ′2 + 2 D + 2 D ′ 1 = sin (2 x + y) 2 − 2 − (− 2 ⋅ 1) − 2 ⋅ (− 12 ) + 2 D + 2 D ′ =

D − D′ 1 sin (2 x + y) = sin (2 x + y) 2 (D + D ′ ) 2 ( D2 − D ′2 )

D-274

=

1 D − D′ 1 sin (2 x + y) = − ( D − D ′ ) sin (2 x + y) 2 − 22 − (− 12 ) 6

=−

1 1 { 2 cos (2 x + y) − cos (2 x + y)} = − cos (2 x + y). 6 6

Hence the general solution of the given equation is z = C. F. + P.I. Example 31:

C. F. = e 2 x φ1 ( y + 3 x) + xe 2 x φ 2 ( y + 3 x).

Solution:

Now

Solve ( D − 3 D ′ − 2)2 z = 2 e 2 x tan ( y + 3 x).

P.I. =

1

2 e 2 x tan ( y + 3 x)

( D − 3 D ′ − 2)2 = 2e 2 x = 2e 2 x

1

tan ( y + 3 x)

{( D + 2) − 3 D ′ − 2}2 1 ( D − 3 D ′ )2

= 2e 2 x ⋅

tan ( y + 3 x)

x2 tan ( y + 3 x) = x2 e 2 x tan ( y + 3 x). 2!

Hence the general solution of the given equation is z = C. F. + P.I. Example 32: Solution:

Solve ( D3 − 3 DD ′ + D + 1) z = e 2 x + 3 y.

(Kanpur 2008)

Since D3 − 3 DD ′ + D + 1 cannot be resolved into linear factors in D and C. F. = Σ Ae hx + ky

D′ hence

h3 − 3 hk + h + 1 = 0 i. e., k =

where Now

P.I. = =

1 3

D − 3 DD ′ + D + 1 1 23 − 3 ⋅ 2 ⋅ 3 + 2 + 1

h3 + h + 1 ⋅ 3h

e2 x +3

e2 x +3

y

y

=−

1 2 x +3 y e . 7

Hence the general solution of the given equation is 1 z = C. F. + P.I. = Σ Ae hx + ky − e 2 x + 3 y, 7 h3 − 3 hk + h + 1 = 0 .

where Example 33: Solution:

Solve ( D2 − DD ′ − 2 D) z = sin (3 x + 4 y).

Since D2 − DD ′ − 2 D cannot be resolved into linear factors in D and D ′

hence C. F. = Σ Ae hx + ky, where h2 − hk − 2 h = 0 i. e., k = h − 2.

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Now

P.I. =

1 D2 − DD ′ − 2 D

sin (3 x + 4 y)

=

1 sin (3 x + 4 y) − 9 − (−3.4) − 2 D

=

3 + 2D 1 sin (3 x + 4 y) = sin (3 x + 4 y) 3 − 2D 9 − 4 D2

=

3 + 2D sin (3 x + 4 y) 9 − 4 (− 9)

1 {3 sin (3 x + 4 y) + 2 ⋅ 3 cos (3 x + 4 y)} 45 1 = {sin (3 x + 4 y) + 2 cos (3 x + 4 y)}. 15 =

Hence the general solution of the given equation is z = C. F. + P.I. Example 34: Solution:

Solve

∂2 z 2

∂x

+

∂2 z ∂2 z −6 = x2 sin ( x + y). ∂x ∂y ∂ y2

The given equation can be written as ( D2 + DD ′ − 6 D ′2 ) z = x2 sin ( x + y)

or

( D − 2 D ′ ) ( D + 3 D ′ ) z = x2 sin ( x + y).

C. F. = φ1( y + 2 x) + φ 2 ( y − 3 x). 1 Now P.I. = x2 sin ( x + y) 2 D + DD ′ − 6 D ′2

∴

1

= imaginary part of = I.P. of e i ( x + y ) = I.P. of e i ( x + y )

2

D + DD ′ − 6 D ′2

x2 e i( x + y )

1 2

2

( D + i ) + ( D + i ) ( D ′ + i) − 6 ( D ′ + i) 1 2

2

D + 3 iD + DD ′ − 11 D ′ i − 6 D ′ + 4

x2 x2

 D2 3 iD DD ′ 11 D ′ i 6 D ′2  + + − − 4 4 4 4  4

1 = I.P. of e i ( x + y ) ⋅ 4

  1+ 

1 = I.P. of e i ( x + y ). 4

 D2 3 iD DD ′ 11D ′ i 6 − − + + D ′2 1 − 4 4 4 4 4  +

1 = I.P. of e i ( x + y ). 4

 x2 − 1 − 3 i x − 9   2 2 8 

 9 i2 D2 + ...  x2 16 

    

−1

x2

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= I.P. of =

1 13 3 ix  {cos ( x + y) + i sin ( x + y)}  x2 − −  4 8 2  

1  2 13  3  x −  sin ( x + y) − x cos ( x + y).   4 8 8

Hence the general solution of the given equation is z = C. F. + P.I. = φ1 ( y + 2 x) + φ 2 ( y − 3 x) 1 13 3 +  x2 −  sin ( x + y) − x cos ( x + y).   4 8 8

Comprehensive Exercise 5 1.

Solve ( D2 − D ′2 + D − D ′ ) z = 0 .

2.

Solve DD ′ ( D − 2 D ′ − 3) z = 0 .

3.

Solve ( D − 2 D ′ − 1) ( D − 2 D ′2 − 1) z = 0 .

4.

Solve t + s + q = 0.

(Avadh 2007)

5.

Solve r − s + p = 1.

(Avadh 2010)

6.

Solve ( D2 − D ′ ) z = 2 y − x2 .

7.

Solve ( D2 + DD ′ + D ′ − 1) z = sin ( x + 2 y). 2

∂ z

2

∂ z ∂ z ∂z ∂z +4 + −2 = e x + y. 2 ∂x ∂y ∂ ∂ x y ∂y

8.

Solve

9.

Solve ( D2 − DD ′ − 2 D) z = sin (3 x + 4 y) − e 2 x + y .

10.

2

∂x

−4

2

Solve ( D − D ′ ) z = xe 2

ax + a2 y

Solve ( D − D ′ + D − D ′ ) z = e 2 x + 3 y .

12.

Solve ( D2 − 4 DD ′ + D − 1) z = e 3 x − 2 y .

13.

Solve ( D2 − D ′ − 1) z = x2 y.

14.

Solve ( D 2 − DD ′ + D ′ − 1) z = 2 cos ( x + 2 y) − e y .

2

2

Solve

∂ z ∂x2

(Avadh 2010)

.

11.

15.

(Rohilkhand 2008)

2

2

−3

2

∂ z ∂ z ∂z ∂z +2 − +2 = (2 + 4 x) e − y . ∂x ∂y ∂y ∂ y2 ∂ x

A nswers 5 1.

z = φ1 ( y + x) + e − x φ 2 ( y − x)

2.

z = φ1 ( x) + φ 2 ( y) + e3 x φ 3 ( y + 2 x)

(Avadh 2009)

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3.

2 z = e x φ1 ( y + 2 x) + Σ Ae(2 k + 1) x + ky

4.

z = φ1 ( x) + e − x φ 2 ( y − x)

5.

z = φ1 ( y) + e − x φ 2 ( y + x) + x

6.

2 z = Σ Ae hx + h y + yx2

7. 8. 9. 10. 11. 12.

1 z = e − x φ1 ( y) + e x φ 2 ( y − x) − {cos ( x + 2 y) + 2 sin ( x + 2 y)} 10 1 x+ y z = φ1 ( y + 2 x) + e − x φ 2 ( y + 2 x) + ye 2 1 1 z = φ1 ( y) + e 2 x φ 2 ( y + x) + {sin (3 x + 4 y) + 2 cos (3 x + 4 y)} + e 2 x + y . 2 15  x2 2 x  ax + a2 y  e z = Σ Ae hx + h y +  −  4 a 4 a2  1 z = φ1( y + x) + e − x φ 2 ( y − x) − e2 x + 3 y 6 1 3x − 2 y 1 2 , where k = z = Σ Ae hx + ky + e (h + h − 1) 35 4h

13.

2 z = Σ Ae hx + ( h − 1) y + x2 − x2 y − 2 y + 4

14.

z = e x φ1 ( y) + e − x φ 2 ( y + x) + sin ( x + 2 y) + x e y

15.

z = φ1 ( y + 2 x) + e x φ2 ( y + x) + e − y ( x + x2 )

9.10 Equations Reducible to Linear Form with Constant Coefficients A partial differential equation having variable coefficients can sometimes be reduced to an equation with constant coefficients by suitable substitutions. We shall discuss the reduction of an equation of the form n   ∂n z ∂ nz n ∂ z + A1 x n − 1 y + ... + An y n + ...  = f ( x, y)  A0 x n n n 1 − ∂x ∂y ∂x ∂y   …(1)

into a linear equation with constant coefficients. It should be noted that in the equation (1) the term

∂ nz

is multiplied by the ∂x r ∂y n − r

variable expression x r y n − r . To transform the equation (1), we put x = e X , y = eY so that, X = log x and Y = log y.

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Then or ∴

∂z ∂z ∂X 1 ∂z = ⋅ = ∂x ∂X ∂x x ∂X ∂z ∂z x = ∂x ∂X ∂ ∂ x ≡ ≡ D (say). ∂x ∂X

…(2)

2 ∂  ∂z  ∂z 2 ∂ z +x x  = x 2 ∂x  ∂x  ∂x ∂x

Now

x

⇒

x2

∂2 z

 ∂  ∂z = x − 1 x  ∂x  ∂x

∂x2

= ( D − 1) Dz = D ( D − 1) z .

…(3)

n

In general

∂ z xn = D ( D − 1) ( D − 2)...( D − n + 1) z . ∂x n

…(4)

Similarly differentiating w.r.t. y, we get y

∂z ∂z ∂ ∂ = = D ′ z i. e., y ≡ ≡ D ′, ∂y ∂Y ∂y ∂Y

y2

∂2 z ∂ y2

= D ′ ( D ′ − 1) z ,

∂ nz yn = D ′ ( D ′ − 1)...( D′ − n + 1) z . ∂y n Also

xy

and

xm yn

∂2 z = DD ′ z ∂x ∂y ∂ m + nz

= D ( D − 1)...( D − m + 1) D ′ ( D ′ − 1)... ( D′ − n + 1) z . ∂x m ∂y n

These substitutions reduce the equation (1) to an equation having constant coefficients and now it can easily be solved by the methods discussed for homogeneous and non-homogeneous linear equations with constant coefficients.

Example 35:

∂2 z ∂2 z ∂2 z Solve x2 + 2 xy + y2 = 0. ∂x ∂y ∂x2 ∂ y2

Solution: Putting x = e X , y = eY

and denoting

∂ ∂ and by D and D ′ respectively, the given equation transforms to ∂X ∂Y [ D ( D − 1) + 2 DD ′ + D ′ ( D ′ − 1)] z = 0

D-279

or

[ D2 + 2 DD ′ + D ′2 − D − D ′ ] z = 0

or

( D + D ′ ) ( D + D ′ − 1) z = 0 .

Hence the general solution of the given equation is z = φ1 (Y − X ) + e X φ 2 (Y − X ) = φ1 (log y − log x) + x φ2 (log y − log x) y y    y  y = φ1  log  + x φ 2  log  = f1   + x f2   ⋅    x  x x x ∂2 z ∂2 z ∂2 z ∂z Solve x2 − 4 xy + 4 y2 +6y = x3 y 4 . 2 2 ∂x ∂y ∂y ∂x ∂y

Example 36: Solution:

Putting x = e X , y = eY

and denoting

∂ ∂ and by D and D ′ respectively, the given equation transforms to ∂X ∂Y [ D ( D − 1) − 4 DD ′ + 4 D ′ ( D ′ − 1) + 6 D′ ] z = e3 X + 4Y

or

( D2 − D − 4 DD ′ + 4 D ′2 + 2 D ′ ) z = e3 X + 4Y .

or

( D − 2 D ′ ) ( D − 2 D ′ − 1) z = e3 X + 4Y .

∴

C. F. = φ1 (Y + 2 X ) + e X φ 2 (Y + 2 X ) = φ 1 (log y + 2 log x) + x φ2 (log y + 2 log x) = φ 1 (log ( yx2 )) + x φ2 (log ( yx2 )) = f1 ( yx2 ) + x f2 ( yx2 ). 1 P.I. = e3 X + 4Y ( D − 2 D ′ ) ( D − 2 D ′ − 1)

Now

=

1 1 3 4 e3 X + 4Y = x y . (3 − 2.4) (3 − 2.4 − 1) 30

Hence the general solution of the given equation is z = C. F. + P.I. = f1 ( yx2 ) + x f2 ( yx2 ) + Example 37:

Solution:

1 3 4 x y . 30

∂2 z ∂2 z ∂2 z Solve x2 + 2 xy + y2 = ( x2 + y2 )n /2 . 2 2 x y ∂ ∂ ∂x ∂y

∂ ∂ Putting x = e X , y = eY and denoting and by D and D′ respectively, ∂X ∂Y

the given equation transforms to {D ( D − 1) + 2 DD ′ + D ′ ( D ′ − 1)} z = (e2 X + e2Y )n /2 or

{D2 + 2 DD ′ + D ′2 − D − D ′} z = (e2 X + e2Y )n /2

or

( D + D ′ ) ( D + D ′ − 1) z = (e2 X + e2Y )n /2 .

D-280

C. F. = φ1 (Y − X ) + e X φ 2 (Y − X )

∴

= φ1 (log y − log x) + x φ 2 (log y − log x) y y    y  y = φ1  log  + x φ 2  log  = f1   + x f2   ⋅    x  x x  x Now

P.I. =

1 (e 2 X + e 2Y )n /2 ( D + D ′ ) ( D + D ′ − 1)

=

1 e nX {1 + e 2(Y − X )}n /2 ( D + D ′ ) ( D + D ′ − 1)

=

1 ( D + D ′ ) ( D + D ′ − 1)

e nX + 1 ne( n − 2 ) X + 2Y  2 1 1  n ( n − 1)  ( n − 4 ) X + 4Y 2 2 + + ...  e 2!  

=

e nX n (n − 1)

=

e nX [1 + e 2 (Y − X )]n /2 (e 2 X + e 2Y )n /2 ( x2 + y2 )n /2 = = ⋅ n (n − 1) n (n − 1) n (n − 1)

 1 + 1 ne 2 (Y − X ) + ...    2

Hence the general solution of the given equation is  z = C. F. + P.I. = f1   Example 38: Solution:

Solve yt − q = xy.

y  ( x2 + y2 )n /2 ⋅  + x n (n − 1) (Avadh 2009)

The given equation can be written as y

or

y   + x f2   x

∂2 z ∂y

y2

−

2

∂2 z ∂y

2

∂z = xy ∂y − y

∂z = xy2 . ∂y

∂ ∂ Putting x = e X , y = eY and denoting and by D and D′ respectively the given ∂X ∂Y equation transforms to {D ′ ( D ′ − 1) − D ′} z = e X + 2Y or ∴

D ′ ( D ′ − 2) z = e X + 2Y . C. F. = φ1 ( X ) + e 2Y φ 2 ( X ) = φ1 (log x) + y2 φ 2 (log x) = f1 ( x) + y2 f2 ( x).

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Now

1 e X + 2Y D ′ ( D ′ − 2)

P.I. = =

1 e X + 2Y 2 ( D ′ − 2)

=

1 X + 2Y 1 e 1 2 D′ + 2 − 2

=

1 X + 2Y 1 1 e 1 = Ye X + 2Y 2 D′ 2

=

1 2 xy log y. 2

Hence the general solution of the given equation is z = C. F. + P.I. = f1 ( x) + y2 f2 ( x) + Example 39:

Solution:

Solve

Put

1 ∂2 z x2 ∂x2

−

1 ∂z 1 ∂2 z 1 ∂z = − ⋅ 3 ∂x 2 2 y ∂y y3 ∂ y x

1 2 1 x = X and y2 = Y , 2 2

so that

x dx = dX and y dy = dY .

Hence

∂z ∂z ∂x 1 ∂z = = ∂X ∂x ∂X x ∂x

and

∂2 z ∂X

2

∂  ∂z    ∂X  ∂X 

= =

∂  1 ∂z  ∂x  ⋅ ∂x  x ∂x  ∂X

=

1 ∂  1 ∂z  ⋅   x ∂x  x ∂x 

=− i. e., Similarly

1 2 xy log y. 2

1 ∂2 z 2

x

−

2

∂x

1 ∂2 z y

2

∂y

1 ∂z 1 ∂2 z + 3 ∂x x2 ∂x2 x

2

1 ∂z ∂2 z = ⋅ 3 ∂x x ∂X 2

−

1 ∂z ∂2 z = ⋅ 3 ∂y y ∂Y 2

Thus the given equation transforms to ∂2 z ∂X 2

=

∂2 z ∂Y 2

or

or

( D2 − D′ 2 ) z = 0

where

D≡

∂ ∂ , D′ ≡ ∂X ∂Y

∂2 z ∂X 2

−

∂2 z ∂Y 2

=0

D-282

or

( D + D ′ ) ( D − D ′ ) z = 0.

Hence the solution is z = φ1 (Y − X ) + φ 2 (Y + X ) = f1 ( y2 − x2 ) + f2 ( y2 + x2 ).

Comprehensive Exercise 6 1.

∂2 z ∂2 z ∂z ∂z Solve x2 − y2 − y +x = 0. 2 2 y ∂ ∂x ∂x ∂y

2.

∂2 z ∂2 z ∂2 z ∂z ∂z Solve x2 + 2 xy + y2 +x + y − z = 0. ∂x ∂y ∂x ∂y ∂x2 ∂ y2

3.

Solve x2 r − 3 xys + 2 y2 t + px + 2 q y = x + 2 y.

4.

∂2 z ∂2 z Solve x2 − y2 = x2 y. ∂x2 ∂ y2

5.

Solve ( x2 D2 + 2 xyDD ′ + y2 D ′2 ) z = x m y n.

6.

∂2 z ∂2 z ∂z x3 Solve x2 + 2 xy −x = ⋅ ∂x ∂y ∂x ∂x2 y2

7.

Solve x2 r − y2 t + px − qy = log x.

A nswers 6 y  x

1.

 z = f1 ( xy) + f2  

2.

 y 1  y z = x f1   + f    x x 2  x

3. 4. 5. 6. 7.

z = f1 ( xy) + f2 ( x2 y) + x + y 1 z = f1 ( xy) + x f2 ( y / x) + x2 y 2 1  y  y z = f1   + x f2   + xm yn  x  x  (m + n)(m + n − 1) 1 z = f1 ( y) + x2 f2 ( y / x2 ) − ( x3 / y2 ) 9 1 z = f1 ( y / x) + f2 ( xy) + (log x)3 6

(Lucknow 2006)

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Objective Type Questions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

Out of the following four P.D.E., the equation which is linear: (a)

(b)

(c)

∂3 z ∂x3

−3

∂2 z ∂z ∂2 z +8 = sin x ∂x2 ∂y ∂ y2 2

∂2 z

 ∂z  +   + 9z = 0 2  ∂y  ∂x

∂2 z ∂x2

+5

∂2 z ∂ y2

=0

(d) None of these. 2.

(Rohilkhand 2003)

The A.E. of the equation 2 r + 5 s + 2 t = 0 is (a) 2 m2 + 5 m + 2 = 0 (b) 2 m2 − 5 m + 2 = 0 (c) 2 m2 + 5 m − 2 = 0 (d) 2 m2 − 5 m − 2 = 0

3.

The general solution of the differential equation ( D2 − 2 DD ′ + D ′2 ) z = 0 is (a) z = c1 e x + c2 e y (b) z = c e ( x + c2 y ) (c) z = φ1 ( x + y) + φ 2 ( y + x) (d) z = φ1 ( y + x) + x φ 2 ( y + x).

4.

3

(Rohilkhand 2002) 2

3

The C.F. of the equation ( D − 3 DD ′ + 2 D ′ ) z = ( x + 2 y)1 /2 is (a) φ1 ( y + x) + φ 2 ( y − 2 x) (b) φ1 ( y + x) + x φ 2 ( y + x) + φ 3 ( y + 2 x) (c) φ1 ( y + x) + x φ 2 ( y + x) + φ 3 ( y − 2 x) (d) None of these.

5.

The P.I. of the differential equation ( D2 + 3 DD ′ + 2 D ′2 ) z = x + y is (a)

( x + y)3 6

(b)

( x + y)3 12

(c)

( x + y)3 36

(d) None of these.

D-284

6.

The C.F. of the equation log s = x + y is (a) φ1 ( x) + φ 2 ( y) (c) φ1 ( x) + x φ 2 ( y)

7.

(b) φ1 ( y) + x φ 2 ( y) (d) None of these.

The C.F. of the equation ( D2 + 2 DD ′ + D ′2 ) z = 2 cos y − x sin y is (a) φ 1 ( y + x) + φ 2 ( y − x) (b) φ1 ( y + x) + x φ 2 ( y + x) (c) φ1 ( y − x) + x φ 2 ( y − x) (d) None of these.

8.

The solution of non-homogeneous equation ( D − mD ′ − k ) z = 0 is (a) z = e kx φ ( y + mx) (b) z = e kx φ ( y − mx) (c) z = e x φ ( y + mx) (d) z = e x φ ( y − mx).

9.

If z = Ae hx + ky be the solution of ( D − D ′2 ) z = 0 then (b) h = k 2

(a) h = k (c) h = k 3 10.

11.

(d) None of these. 2

The C.F. of ( D − DD ′ − 2 D) z = sin (3 x + 4 y) is Σ Ae hx + ky where (a) h2 − hk − 2 h = 0

(b) h2 + hk − 2 h = 0

(c) h2 − hk + 2 h = 0

(d) None of these.

P.I. of the equation ( D3 − 3 DD ′ + D ′ + 1) z = e4 x + 5 y is 1 4x + 5 y e 4 1 4x + 5 y (c) e 10 (a)

12.

(b)

1 4x + 5 y e 5

(d) None of these.

The equation A0 x n

∂n z

n ∂n z n −1 n ∂ z + A x y + … + A y + … = f ( x, y) n 1 ∂x n ∂y n ∂x n − 1 ∂y

can be reduced into a linear equation with constant coefficients in the variables X and Y by substitutions (a) x = log X , y = log Y (b) x = e X , y = eY (c) x = e2 X , y = e2Y (d) None of these

D-285

Fill In The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

In the homogeneous linear partial differential equation with constant coefficients all the partial derivatives appearing in the equation are of the …… order.

2.

If u is the complementary function and z1 a particular integral of a linear partial differential equation F ( D, D ′ ) z = f ( x, y) then …… is a general solution of the equation.

3.

A linear partial differential equation which is not homogeneous is called a …… linear equation.

4.

The A.E. of a linear homogeneous nth order partial differential equation with constant coefficients will give in general …… roots.

5.

The A.E. of ( D2 + D ′2 ) z = 30 (2 x + y) is …… .

True or False Write ‘ T ’ for true and ‘F’ for false statement. 1.

The A.E. of a linear homogeneous nth order partial differential equation with constant coefficients of the form F ( D, D ′ ) z = f ( x, y) is obtained by putting D = 1, D ′ = m in F ( D, D ′ ) = 0 .

2.

The value of

3.

The solution of ( D2 − 2 aDD ′ + a2 D ′2 ) z = 0 is

1 xr + 1 x r φ (ax + by) = φ (ax + by). (bD − aD ′ ) b (r + 1)

z = φ1 ( y + ax) + x φ 2 ( y + ax). 4.

In case of non-homogeneous linear partial differential equation with constant 1 1 coefficients the value of e ( ax + by ) . V = e ax + by V. F ( D, D ′ ) ( D − a, D ′ − b)

5.

The C.F. of ( D3 − 3 DD ′ + D + 1) z = e2 x + 3 y is Σ Ae hx + ky where h3 − 3 hk + h + 1 = 0 .

6.

In the equation ( D − mD ′ ) z = φ ( x, y) the value of 1 z= φ ( x, y) = ∫ φ ( x, a − mx) dx, where a = y + mx . D − mD ′

A nswers Multiple Choice Questions 1. 4. 7. 10.

(c) (c) (c) (a)

2. 5. 8. 11.

(a) (c) (a) (c)

3. 6. 9. 12.

(d) (a) (b) (b)

D-286

Fill in the Blank(s) 1.

same

3.

u + z1

4.

n

5.

m2 + 1 = 0

3. non-homogeneous.

True or False 1. 4.

F F

2. T 5. T

3. 6.

T T

¨

D-287

10 P artial D ifferential E quations of S econd O rder with V ariable C oefficients

10.1 Partial Differential Equations of Second Order with Variable Coefficients partial differential equation is said to be of second order if it contains at least one of the second order partial differential coefficients r , s and t but none of higher order. The differential coefficients p and q may also appear in the equation. Thus the general form of a second order partial differential equation is

A

F ( x, y, z , p, q, r, s, t) = 0. The complete solutions of these equations will contain two arbitrary functions. In this chapter we deal with the second order partial differential equations with variable coefficients also. Below we give some examples of equations that are readily solvable. It is to be noted that x and y, being independent, are constant with respect to each other in differentiation and integration.

D-288

Example 1:

Solution:

Solve s =

x + a. y

The given equation can be written as

Integrating w.r.t. ‘x’, we get

∂2 z x = + a. ∂x ∂y y

∂z 1 x2 = ⋅ + ax + f ( y). ∂y y 2

Now integrating w.r.t. ‘ y’ we get x2 log y + axy + ∫ f ( y) dy + ψ ( x) 2 1 z = x2 log y + axy + φ ( y) + ψ ( x). 2 z= or Example 2: Solution:

Solve s = 2 x + 2 y. The given equation can be written as

Integrating w.r.t. ‘x’, we get

∂2 z = 2 x + 2 y. ∂x ∂y

∂z = x2 + 2 xy + f ( y). ∂y

Now integrating w.r.t. ‘ y ’, we get z = x2 y + xy2 + or Example 3: Solution:

∫

f ( y) dy + ψ ( x)

z = x2 y + xy2 + φ ( y) + ψ ( x). Solve t = sin xy. The given equation can be written as

Integrating w.r.t. ‘x ’, we get

Solution:

∂ y2

= sin xy.

∂z 1 = − cos xy + f ( x). ∂y x

Again integrating w.r.t. ‘ y’, we get z = − Example 4:

∂2 z

1 x2

sin xy + yf ( x) + ψ ( x).

Solve ys + p = cos ( x + y) − y sin ( x + y). The given equation can be written as ∂q ∂z y + = cos ( x + y) − y sin ( x + y). ∂x ∂x

Integrating w.r.t. ‘x ’, we get

or

yq + z = sin ( x + y) + y cos ( x + y) + f ( y) ∂z y + z = sin ( x + y) + y cos ( x + y) + f ( y). ∂y

D-289

Now integrating w.r.t. ‘ y ’, we get yz = y sin ( x + y) + or

f ( y) dy + ψ ( x)

∫

yz = y sin ( x + y) + φ ( y) + ψ ( x). Solve t − xq = x2 .

Example 5: Solution:

(Lucknow 2011)

The given equation can be written as ∂q − xq = x2 , which is linear in q regarding x as constant. ∂y I. F. = e − ∫ x dy = e − xy.

∴

solution is q e − xy =

2

∫x

e − xy dy + f ( x) = − xe − xy + f ( x)

∂z =−x+e ∂y

xy

or

q=

Integrating,

z = − xy + f ( x) ∫ e

or

z = − xy +

or

z = − xy + φ ( x) e

1 = y

∴

q

or

q=

Integrating,

z=x

+ ψ ( x)

+ ψ ( x). (Avadh 2009)

∫

x⋅

y) dy

= e − log

y

= 1 / y.

1 dy + f ( x) = x log y + f ( x) y

∂z = xy log y + y f ( x). ∂y

∫

y log y dy + f ( x)

∫

y dy + ψ ( x)

1 y2 1 y2  z = x  y2 log y − ∫ ⋅ dy + f ( x) + ψ( x) 2 y 2 2  1 1 1 2 z = xy2 log y − x y2 + y f ( x) + ψ ( x). 2 4 2

or or

Solution:

xy

dy + ψ ( x)

The given equation can be written as ∂q 1 − q = x, which is linear in q regarding x as constant. ∂y y I. F. = e ∫ (− 1 /

Example 7:

xy

xy

Solve yt − q = xy.

Example 6: Solution:

1 f ( x) e x

f ( x).

Solve rx = (n − 1) p. The given equation can be written as

Integrating, log

∂z = (n − 1) log x + log f ( y) ∂x

∂2 z / ∂x2 n − 1 = ⋅ ∂z / ∂x x or

∂z = x n − 1 f ( y). ∂x

Again integrating w.r.t. ‘x’, we get z=

xn f ( y) + ψ ( y) or z = x n φ ( y) + ψ ( y). n

D-290 Example 8: Solution:

Solve p + r + s = 1.

(Avadh 2006)

The given equation can be written as

∂z ∂x

+

∂p ∂x

+

∂q ∂x

= 1.

Integrating w.r.t. ‘x’, we get z + p + q = x + f ( y) or

p + q = x + f ( y) − z . dx dy dz Lagrange’s auxiliary equations for this are = = ⋅ 1 1 x + f ( y) − z The first two members give x − y = a. From the last two members, we get dz dz + z = x + f ( y) or + z = a + y + f ( y), which is linear in z. dy dy I. F. = e ∫ dy = e y. ∴

ze

y

= ∫ {a + y + f ( y)} e = ae

y

y

dy + b = a e

y

+ ∫ { y + f ( y)} e

y

dy + b

+ φ ( y) + b y

or

z = a + e−

φ ( y) + b e −

or

z = x − y + e−

y

y

φ ( y) + e −

y

ψ ( x − y).

Solutions of equations under given conditions. After finding the general solutions by the usual methods, the geometrical conditions given in the problem are used to find the arbitrary functions. Example 9:

Find the surface passing through the parabolas z = 0 , y2 = 4 ax and z = 1, y2 = − 4 ax

and satisfying the equation x r + 2 p = 0. Solution:

The given equation can be written as x

∂p + 2p = 0 ∂x

or

x2

∂p + 2 px = 0 . ∂x

Integrating w.r.t. ‘x’, we get px2 = f ( y) or Again integrating w.r.t. ‘x’, we get z = −

p=

∂z 1 = f ( y). ∂x x2

1 f ( y) + φ ( y). x

…(1)

Now using the geometrical conditions of the problem we are to determine the values of f ( y) and φ ( y). Since the required surface is to pass through the parabola z = 0, y2 = 4 ax hence putting z = 0 and x = y2 / 4 a in (1), we get 0 = −

4a y2

f ( y) + φ ( y).

…(2)

D-291

Again putting z = 1 and x = − y2 / 4 a in (1), we get 1=

4a

f ( y) + φ ( y).

y2

…(3)

Solving (2) and (3) for f ( y) and φ ( y), 1 we have φ ( y) = and f ( y) = y2 / 8 a. 2 Thus we have determined the arbitrary functions. Putting the values of f ( y) and φ ( y) in (1), the required surface is z=−

2 1 y 1 ⋅ + x 8a 2

8 axz − 4 ax + y2 = 0 .

or Example 10:

Find a surface passing through the two lines z = x = 0, z − 1 = x − y = 0, satisfying r − 4 s + 4 t = 0.

Solution:

The given equation can be written as ( D2 − 4 DD′ + 4 D ′2 ) z = 0 . …(1)

A.E. is m2 − 4 m + 4 = 0 or

(m − 2)2 = 0 .

∴

m = 2, 2.

Hence the general solution of (1) is z = φ1 ( y + 2 x) + x φ 2 ( y + 2 x).

…(2)

If the surface (2) passes through the lines z = x = 0 and z − 1 = x − y = 0 , then we have and

0 = φ1 ( y + 2 x)

…(3)

1 = φ1 ( y + 2 x) + x φ 2 ( y + 2 x).

…(4)

From (3) and (4), we get φ 2 ( y + 2 x) =

1 3 3 = = ⋅ x 3x 2 x + y

Putting the values of φ1 and φ 2 in (2), the required surface is 3 z = x⋅ or z (2 x + y) = 3 x. 2x + y

Comprehensive Exercise 1 1.

Solve xr + p = 9 x2 y3 .

2.

Solve r = 2 y2 .

3.

Solve log s = x + y.

4.

Solve r = 6 x.

[∵ y − x = 0 ]

D-292

5.

Solve xr + 2 p = 0.

6.

Solve xr = p.

7.

Solve xs + q = 4 x + 2 y + 2.

8.

Solve t + s + q = 0.

9.

Solve 2 yq + y2 t = 1.

10.

Find a surface satisfying r + s = 0 and touching the elliptic paraboloid z = 4 x2 + y2 along its section by the plane y = 2 x + 1.

11.

Solve the equation r + t = 2 s and determine the arbitrary functions by the conditions that bz = y2 when x = 0 and az = x2 when y = 0.

A nswers 1 1.

z = x3 y3 + f ( y) log x + φ ( y)

2.

z = x2 y2 + x f ( y) + φ ( y)

3.

z=e

4.

z = x3 + x f ( y) + φ ( y)

5.

z=−

6.

z=

7.

z x = 2 x2 y + xy2 + 2 x y + φ ( x) + ψ ( y)

8.

ze

9.

yz = y log y − f ( x) + y φ ( x)

x+ y

+ φ ( y) + ψ ( x)

1 f ( y) + φ ( y) x

1 2 x f ( y) + φ ( y) 2 x

= φ ( x) + ψ ( x − y)

10.

z + 4 x2 + y2 − 8 xy − 4 y + 8 x + 2 = 0

11.

y x z = ( y + x)  +  a b

10.2 Classification of Linear Partial Differential Equation of Second Order The linear partial differential equation of the second order in n independent variables x1, x2 , … , xn can be written as n

n

Σ

Σ

i =1 j =1

aij

n ∂2 u ∂u + Σ bi + c u=0 ∂x i ∂x j i = 1 ∂x i

where aij , bi and c are constants or functions of x1, x2 , … , xn .

…(1)

D-293

Let δ i represent

∂ , for i = 1, 2 , … , n ∂xi

and

δ i δj represent

∂2 , for i = 1, 2 , 3, … , n and ∂x i ∂x j

j = 1, 2 , 3, … , n.

Consider the operator n

n

…(2)

Σ aij δ i δ j

φ= Σ

i =1 j =1

for all real values of δ i positive or negative. Now at a point x1, x2 , … , xn we call the linear partial differential equation given by (1) as : (i)

elliptic if φ is positive for all real values of δ i and it reduces to zero only when all δ i’s are zero.

(ii)

hyperbolic if φ can be both positive or negative.

(iii) parabolic if the determinant ∆ vanishes, where a11 a12 ... a1n a21 a22 ... a2 n ... ...

... ...

... ...

an1

an2

... ann

∆=

... ...

=0.

If aij are functions of x1, x2 , … , xn the same differential equation can be elliptic, hyperbolic and parabolic at different points. If aij are constants, the equation will have the same nature throughout.

10.3 Classification of Linear Partial Differential Equation of Second Order in Two Independent Variables Let us consider the equation of second order in two independent variables x and y A

∂2 u 2

∂x

+B

 ∂2 u ∂2 u ∂u ∂u  + C 2 + f  x, y, u, ,  = 0 , ∂x ∂y ∂x ∂y  ∂y 

where A is positive. Here φ = Aδ12 + B δ1 δ2 + C δ22 . The equation (1) is (i)

elliptic if B 2 − 4 AC < 0 ,

(ii)

hyperbolic if B 2 − 4 AC > 0 ,

(iii) parabolic if B 2 − 4 AC = 0 .

…(1)

D-294

If A , B , C are constants then the nature of the equation (1) will be the same in

Note 1:

the whole region i. e.,for all values of x and y.The nature will depend on B 2 − 4 AC . The equation (1) will be elliptic if B 2 − 4 AC < 0 . The equation (1) will be hyperbolic if B 2 − 4 AC > 0 . The equation (1) will be parabolic if B 2 − 4 AC = 0 . If A , B , C are functions of x and y then the nature of equation (1) will not be

Note 2:

same in the whole region i. e., for all values of x and y. The equation (1) will be elliptic in the region where B 2 − 4 AC < 0 . The equation (1) will be hyperbolic in the region where B 2 − 4 AC > 0 . The equation (1) will be parabolic in the region where B 2 − 4 AC = 0 .

Classify the operator

Example 11: 2

(i)

∂ u 2

∂t

2

+

∂ u ∂2 u + 2 ∂x ∂t ∂x

2

(ii) (iii)

∂ u 2

∂t

∂2 u ∂t2

Solution:

2

(Meerut 2011)

2

−4

∂ u ∂ u + 2 ∂x ∂t ∂x

+4

∂2 u ∂2 u +4 2 ⋅ ∂x . ∂t ∂x

(i) Here A = 1, B = 1, C = 1 and so B2 − 4 AC = 1 − 4 = − 3 < 0 .

Therefore, the given operator is elliptic. (ii)

Here A = 1, B = − 4, C = 1 and so B2 − 4 AC = 16 − 4 = 12 > 0 . Therefore the given operator is hyperbolic.

(iii) Here A = 1, B = 4, C = 4 and so B2 − 4 AC = 16 − 16 = 0 . Therefore the given operator is parabolic. Example 12:

Classify the equation (1 − x2 )

Solution:

∂2 z ∂x

2

− 2 xy

∂z ∂z ∂2 z ∂2 z +x + 3x 2 y − 2z = 0. + (1 − y2 ) ∂x ∂y ∂x ∂y ∂y 2

Consider the operator φ = Aδ 12 + B δ 1 δ 2 + C δ 22 , where δ 1 ≡

∂ ∂ ,δ2 ≡ ⋅ ∂x ∂y

Here

A = 1 − x2 , B = − 2 xy, C = 1 − y2 ,

and so

B2 − 4 AC = 4 x2 y2 − 4 (1 − x2 ) (1 − y2 ) = 4 (− 1 + x2 + y2 ).

D-295

Since A , B , C are functions of x and y, the given differential equation is hyperbolic in the region where B 2 − 4 AC > 0 i. e., x2 + y2 > 1, parabolic in the region where B 2 − 4 AC = 0

i. e., at points on the circle x2 + y2 = 1,

and elliptic in the region where B2 − 4 AC < 0

i. e., x2 + y2 < 1.

Comprehensive Exercise 2 1.

Classify the following equations : (i)

∂2 u

∂2 u

+

∂x 2

∂y 2

2

(ii)

2

∂ u

∂ u

+

∂x2

∂ y2

2

∂ u

(iii) 2.

∂2 u ∂t2

+

(Laplace’s equation) (Meerut 2007, 08, 11)

2

+

∂ u ∂z 2

∂ u ∂y

=0

2

∂z

2

(iii)t

+t

∂2 u

(ii) x2

4.

∂2 u

2

2

=

1 ∂ u C2 ∂t2

2

+

∂ u ∂z

2

=

(Wave equation) (Meerut 2007, 10)

1 ∂u ⋅ C 2 ∂t

(Heat equation) (Meerut 2007)

Find where the following operator is hyperbolic, parabolic and elliptic (i)

3.

2

∂x

+

2

∂t

∂2 u ∂t2

∂2 u ∂2 u +x 2 ∂x ∂t ∂x −

+2

∂2 u ∂x2

+u

∂2 u ∂2 u ∂u +x 2 + ⋅ ∂x ∂t ∂x ∂x

Show that the equation

∂2 u ∂t2

= c2

∂2 u

is hyperbolic.

∂x2

(Rohilkhand 2008, 10)

Classify the following as elliptic, parabolic or hyperbolic : (i)

(iii)

∂2 z 2

∂x

∂2 z ∂x2

=

+

∂z ∂y ∂2 z ∂ y2

(ii)

∂2 z 2

∂x

=

∂2 z ∂ y2

= 0.

A nswers 2 1. 2.

(i) elliptic. hyperbolic if t

(ii) hyperbolic. 2

(iii) parabolic.

>4 x,

parabolic if t 2 = 4 x and elliptic if t 2 < 4 x . 4.

(i) Parabolic

(ii) Hyperbolic

(iii) Elliptic.

D-296

10.4 Canonical Forms (Method of Transformations) Now we shall consider the equation of the type …(1)

Rr + Ss + Tt + F ( x, y, z , p, q) = 0 ,

where R, S, T are continuous functions of x and y possessing continuous partial derivatives of as high an order as necessary. We shall show that any equation of the type (1) can be reduced to one of the three canonical forms by a suitable change of the independent variables. Suppose we change the independent variables from x, y to u, v where …(2)

u = u ( x, y), v = v ( x, y). Then, we have ∂z ∂z ∂u ∂z ∂v = ⋅ + ⋅ , ∂x ∂u ∂x ∂v ∂x ∂z ∂z ∂u ∂z ∂v q= = + ⋅ ∂y ∂u ∂y ∂v ∂y p=

∴

Now

∂ ∂u ∂ ∂v ∂ ∂ ∂u ∂ ∂v ∂ ≡ ⋅ + ⋅ , ≡ ⋅ + ⋅ ⋅ ∂x ∂x ∂u ∂x ∂v ∂y ∂y ∂u ∂y ∂v r=

∂2 z ∂x2

=

∂v ∂   ∂u ∂z ∂v ∂z  ∂  ∂z   ∂u ∂ + +    =  ∂x  ∂x   ∂x ∂u ∂x ∂v   ∂x ∂u ∂x ∂v  2

=

s=

=

∂2 z ∂u ∂v ∂2 z ∂2 z  ∂u + 2 ⋅ +   ∂u ∂v ∂x ∂x ∂v2 ∂u2  ∂x  ∂2 z ∂ = ∂x ∂y ∂x

 ∂z   ∂u ∂ ∂v ∂   ∂z ∂u ∂z ∂v     = + + ⋅   ∂y   ∂x ∂u ∂x ∂v   ∂u ∂y ∂v ∂y 

∂2 z ∂u ∂u ∂2 z  ∂u ∂v ∂u ∂v    + + ⋅ ∂u2 ∂x ∂y ∂u ∂v  ∂x ∂y ∂y ∂x  +

and

t=

2

∂z ∂2 u ∂z ∂2 v  ∂v  + ,   +  ∂x  ∂u ∂x2 ∂v ∂x2

∂2 z ∂y

2

=

∂ ∂y

∂2 z ∂v ∂v ∂z ∂2 u ∂z ∂2 v + + ∂v2 ∂x ∂y ∂u ∂y ∂x ∂v ∂y ∂x

 ∂z   ∂u ∂ ∂v ∂   ∂u ∂z ∂v ∂z    =   + +  ∂y   ∂y ∂u ∂y ∂v   ∂y ∂u ∂y ∂v  2

2

∂2 z  ∂u ∂2 z ∂u ∂v ∂2 z  ∂v  ∂z ∂2 u ∂z ∂2 v = 2   +2 ⋅ + 2   + + ⋅ ∂u ∂v ∂y ∂y ∂v  ∂y  ∂ u ∂ y2 ∂ v ∂ y2 ∂u  ∂y  Substituting these values of p, q, r, s and t in (1), it takes the form A

∂2 z ∂u2

+2 B

∂2 z ∂2 z ∂z ∂z   , + C 2 + F  u, v, z ,  =0  ∂u ∂v ∂u ∂v  ∂v

…(3)

2

where

2  ∂u  ∂u ∂u  ∂u  +T  , A= R  + S  ∂x  ∂x ∂y  ∂y 

…(4)

D-297

B= R

∂u ∂v 1  ∂u ∂v ∂u ∂v  ∂u ∂v  +T ⋅ + S + , ∂x ∂x 2  ∂x ∂y ∂y ∂x  ∂y ∂y

2  ∂v  ∂v ∂v  ∂v   +T  C= R  +S  ∂x  ∂x ∂y  ∂y 

…(5)

2

…(6)

and the function F is the transformed form of the function f . Now the problem is to determine u and v so that the equation (3) takes the simplest possible form. The procedure is simple when the discriminant S2 − 4 RT of the quadratic equation Rλ2 + Sλ + T = 0

…(7)

is everywhere either positive, negative or zero, and we shall discuss these three cases separately. Case I:

S2 − 4 RT > 0 . If this condition is satisfied then the roots λ1, λ 2 of the

equation (7) are real and distinct. The coefficients of

∂2 z 2

∂u

and

will vanish if we choose u and v such that ∂u ∂u = λ1 , ∂x ∂y and

∂v ∂v = λ2 ⋅ ∂x ∂y

∂2 z ∂v 2

in the equation (3)

…(8) …(9)

The differential equations (8) and (9) will determine the form of u and v as functions of x and y. For this, from (8), Lagrange’s auxiliary equations are dx dy du = = ⋅ 1 − λ1 0 The last member gives du = 0 i. e., u = constant. The first two members give dy + λ1 = 0. dx

…(10)

Let f1 ( x, y) = constant be the solution of the equation (10). Then the solution of the equation (8) can be taken as u = f1 ( x, y).

…(11)

Similarly, if f2 ( x, y) = constant is a solution of dy + λ 2 = 0, dx then the solution of the equation (9) can be taken as v = f2 ( x, y).

…(12)

D-298

Also it can be easily seen that, in general, 2

 ∂u ∂v ∂u ∂v   , AC − B2 = (4 RT − S2 )  −  ∂x ∂y ∂y ∂x  so that when A and C are zero 2

 ∂u ∂v ∂u ∂v   . B = (S − 4 RT )  −  ∂x ∂y ∂y ∂x  2

2

…(13)

It follows that B2 > 0 since S2 − 4 RT > 0 and hence we can divide both sides of the equation by it. Thus making the substitutions defined by the equations (11) and (12), the equation (1) transforms to the form ∂2 z ∂z ∂z  , = φ  u, v, z ,  ∂u ∂v ∂u ∂v

 , 

…(14)

which is the canonical form in this case. Case II:

S2 − 4 RT = 0 . In this case the roots of the equation (7) are equal. We define

the function u as in Case I and take v to be any function of x and y, which is independent of u. Then, we have, as before, A = 0. S2 − 4 RT = 0 ,

Since

hence from (13), B2 = 0 i. e., B = 0 . On the other hand, in this case, C ≠ 0, otherwise v would be a function of u. Putting A = 0 , B = 0 and dividing by C, we see that in this case the canonical form of the equation (1) is, ∂2 z ∂v 2 Case III:

∂z ∂z   , = φ  u, v, z , ⋅  ∂u ∂v 

…(15)

S2 − 4 RT < 0 . Formally it is the same as Case I except that now the roots of

the equation (7) are complex. Proceeding as in Case I, we find that the equation (1) reduces to the form (14) but that the variables u, v are not real but are in fact complex conjugates. To find a real canonical form let u = α + iβ, v = α − iβ 1 1 so that α = (u + v), β = i (v − u). 2 2 Now

∂z ∂z ∂α ∂z ∂β = + ⋅ ∂u ∂α ∂u ∂β ∂u =

1  ∂z ∂z  −i  ⋅ 2  ∂α ∂β 

D-299

Similarly

∂z 1  ∂z ∂z  =  +i ⋅ ∂v 2  ∂α ∂β 

∴

∂2 z ∂  ∂z  =   ∂u ∂v ∂u  ∂v  =

1 ∂ ∂   ∂z ∂z  −i  +i   4  ∂α ∂β   ∂α ∂β 

=

1 4

 ∂2 z ∂2 z  + 2 ∂ β2  ∂α

  ⋅ 

Thus, transforming the independent variables u, v to α, β the desired canonical form is ∂2 z ∂α

2

+

∂2 z

 ∂z ∂z  = ψ  α, β , z , , ⋅ ∂α ∂β   ∂β

…(16)

2

Second order partial differential equations of the type (1) are classified by their canonical forms ; we say that an equation of this type is : (i) Hyperbolic if S2 − 4 RT > 0 , (ii) Parabolic if S2 − 4 RT = 0 , S2 − 4 RT < 0 .

(iii) Elliptic if

Example 13:

(Rohilkhand 2009)

Reduce the equation ∂2 z ∂x2

= x2

∂2 z ∂ y2

, (Lucknow 2007)

to canonical form. Solution:

The given equation can be written as r − x2 t = 0.

…(1)

Comparing the equation (1) with Rr + Ss + Tt + f ( x, y, z , p, q) = 0 , we have R = 1, S = 0 , T = − x2 . The quadratic equation Rλ2 + Sλ + T = 0 therefore becomes λ2 − x2 = 0 ⇒ λ = x, − x (real and distinct roots). The equations

dy

+ λ1 = 0 and

dx dy

dx

+ x = 0 and

dy dx dy dx

+ λ 2 = 0 become − x = 0.

D-300

These on integration give 1 1 y + x2 = constant and y − x2 = constant, 2 2 so that to change the independent variables from x, y to u, v, we take 1 1 u = y + x2 and v = y − x2 . 2 2 ∂z ∂z ∂u ∂z ∂v ∂z ∂z  ∂z ∂z  ∴ p= = ⋅ + ⋅ =x −x =x − ,  ∂u ∂v  ∂x ∂u ∂x ∂v ∂x ∂u ∂v ∂z ∂z ∂u ∂z ∂v ∂z ∂z = ⋅ + ⋅ = + , ∂y ∂u ∂y ∂v ∂y ∂u ∂v

q=

∂2 z

r=

2

∂x

=x

=

∂ ∂x

∂   ∂z ∂z    ∂z  −    = x   ∂x  ∂x   ∂u ∂v  

∂  ∂z ∂z   ∂z ∂z  −  + 1⋅  −    ∂u ∂v  ∂x  ∂u ∂v 

∂ =x  ∂u

 ∂z ∂z  ∂u ∂ + −    ∂u ∂v  ∂x ∂v

 ∂z ∂z  ∂v  ∂z ∂z − ⋅ + −   ∂u ∂v  ∂x  ∂u ∂v

 ∂2 z ∂2 z ∂2 z  ∂z ∂z = x2  2 − 2 − + 2  + ∂u ∂v ∂v  ∂u ∂v  ∂u and

t=

∂2 z ∂ y2

=

∂2 z ∂u2

=

∂ ∂y

+2

 ∂z   ∂ ∂   ∂z ∂z    = +   +   y u v   ∂u ∂v  ∂ ∂ ∂  

∂2 z ∂2 z + 2 ⋅ ∂u ∂v ∂v

Substituting these values in (1), it reduces to ∂2 z 1  ∂z ∂z  = −   ∂u ∂v 4 x2  ∂u ∂v  ∂2 z 1 = ∂u ∂v 4 (u − v)

or

 ∂z ∂z  − ,   ∂u ∂v 

which is the required canonical form of the given equation. Example 14:

Reduce the equation ∂2 z 2

∂x

+2

∂2 z ∂2 z + 2 =0 ∂x ∂y ∂y

to canonical form and hence solve it. Solution:

(Lucknow 2010)

The given equation can be written as r + 2 s + t = 0.

Comparing the equation (1) with Rr + Ss + Tt + f ( x, y, z , p, q) = 0 , we have R = 1, S = 2, T = 1.

…(1)

D-301

The quadratic equation Rλ2 + Sλ + T = 0 is therefore given by λ2 + 2 λ + 1 = 0 , or (λ + 1)2 = 0 . (equal roots). dy dy The equation + λ = 0 becomes − 1 = 0, dx dx

∴

λ = − 1, − 1.

which on integration gives x − y = constant. To change the independent variables x, y to u, v we take u = x − y. We have to take v as some function of x and y independent of u. Let v = x + y. Then

p=

∂z ∂z ∂u ∂z ∂v ∂z ∂z = ⋅ + ⋅ = + , ∂x ∂u ∂x ∂v ∂x ∂u ∂v

q=

∂z ∂z ∂u ∂z ∂v ∂z ∂z = ⋅ + ⋅ =− + , ∂y ∂u ∂y ∂v ∂y ∂u ∂v

r=

∂  ∂z   ∂ ∂   ∂z ∂z  +   +    = ∂x  ∂x   ∂u ∂v   ∂u ∂v 

=

and

∂2 z 2

∂u

+2

∂2 z ∂2 z + 2 , ∂u ∂v ∂v

 ∂z   ∂ ∂2 z ∂2 z ∂   ∂z ∂z    = +  − +  =− 2 + 2 ∂u ∂v  ∂y   ∂u ∂v   ∂u ∂v 

s=

∂ ∂x

t=

∂  ∂z   ∂ ∂   ∂z ∂z    = − +  − +   ∂u ∂v   ∂u ∂v  ∂y  ∂y 

=

∂2 z ∂u2

−2

∂2 z ∂2 z + 2 ⋅ ∂u ∂v ∂v

Substituting these values in (1), it reduces to ∂2 z ∂v2

=0

which is the required canonical form. Integrating it w.r.t. v, we get ∂z = φ1 (u). ∂v Again integrating w.r.t. v, we get z = v φ1 (u) + φ 2 (u), where φ1 and φ2 are arbitrary functions of u. Hence the solution is z = ( x + y) φ1 ( x − y) + φ 2 ( x − y).

D-302 Example 15:

Reduce the equation ∂2 z ∂x2

+ x2

∂2 z ∂ y2

=0,

to canonical form. Solution:

(Lucknow 2011; Meerut 11)

The given equation can be written as r + x2 t = 0.

…(1)

Comparing the equation (1) with Rr + Ss + Tt + f ( x, y, z , p, q) = 0 , we have R = 1, S = 0 , T = x2 . The quadratic equation Rλ2 + Sλ + T = 0 is therefore given by λ2 + x2 = 0 ⇒ λ = i x, − i x. (Complex roots) The equations

dy dy + λ1 = 0 and + λ 2 = 0 become dx dx dy dy + i x = 0 and − i x = 0. dx dx

These on integration give 1 1 y + i x2 = constant and y − i x2 = constant, 2 2 so that to change the independent variables from x, y to u, v, we take 1 u = y + i x2 = α + iβ (say) 2 1 and v = y − i x2 = α − iβ. 2 1 Then α = y, β = x2 . 2 Now we shall transform the independent variables x and y to α and β. We have ∂z ∂z ∂α ∂z ∂β ∂z p= = ⋅ + ⋅ =x , ∂x ∂α ∂x ∂β ∂x ∂β q=

∂z ∂z ∂α ∂z ∂β ∂z = ⋅ + ⋅ = , ∂y ∂α ∂y ∂β ∂y ∂α

r=

∂  ∂z  ∂  ∂z   x    = ∂x  ∂x  ∂x  ∂β 

= 1⋅ =

∂z ∂  ∂z  + x⋅   ∂β ∂x  ∂β 

 ∂  ∂z  ∂α ∂z ∂  ∂z  ∂β  +x +    ⋅  ∂β  ∂α  ∂β  ∂x ∂β  ∂β  ∂x 

D-303

= and

∂z ∂2 z + x2 ∂β ∂ β2 ∂ ∂y

t=

 ∂z  ∂  ∂z  ∂2 z   = ⋅   =  ∂ y  ∂ y  ∂ α  ∂ α2

Substituting these values in (1), it reduces to ∂2 z ∂α

2

+

∂2 z ∂ β2

=−

1 ∂z , 2α ∂α

which is the required canonical form of (1). Example 16:

Reduce the equation (n − 1)2

∂2 z 2

∂x

− y2 n

∂2 z ∂y

2

= ny2 n−1

∂z , ∂y

to canonical form and find its general solution.

(Lucknow 2009)

Solution: The given equation can be written as

(n − 1)2 r − y2 n t − ny2 n−1 q = 0 .

…(1)

Comparing the equation (1) with Rr + Ss + Tt + f ( x, y, z , p, q) = 0 , we have R = (n − 1)2 , S = 0 , T = − y2 n. The quadratic equation Rλ2 + Sλ + T = 0 is therefore given by (n − 1)2 λ2 − y2 n = 0 or λ2 = ∴ The equations

λ=

1 1 yn , − y n. (n − 1) (n − 1)

1 (n − 1)2

y2 n.

(Real and distinct roots).

dy dy + λ1 = 0 and + λ 2 = 0 become dx dx dy 1 + yn = 0 dx (n − 1)

and

dy 1 − yn = 0 dx (n − 1)

or

(n − 1) y − n dy + dx = 0

and

(n − 1) y − n dy − dx = 0 .

These on integration give x − y − n+1 = constant and

x + y − n+1 = constant,

so that to change the independent variables from x, y to u, v, we take u = x − y − n+1 and v = x + y − n+1.

D-304

∴

p=

∂z ∂z ∂u ∂z ∂v ∂z ∂z = ⋅ + ⋅ = + , ∂x ∂u ∂x ∂v ∂x ∂u ∂v

q=

∂z ∂z ∂u ∂z ∂v  ∂z ∂z  = ⋅ + ⋅ = (n − 1) y − n  − ,  ∂u ∂v  ∂y ∂u ∂y ∂v ∂y

r=

∂  ∂z   ∂ ∂   ∂z ∂z  +  +    = ∂x  ∂x   ∂u ∂v   ∂u ∂v 

= and

t=

∂2 z ∂u2

+2

∂2 z ∂2 z + 2 ∂u ∂v ∂v

 ∂z  ∂  ∂z   − n  ∂z   = (n − 1) y  ∂u − ∂v    ∂y  ∂y  

∂ ∂y

∂  ∂z ∂z   ∂z ∂z  = − n (n − 1) y − n−1  −  + (n − 1) y − n −    ∂u ∂v  ∂y  ∂u ∂v   ∂z ∂z  = − n (n − 1) y − n−1  −   ∂u ∂v   ∂  ∂z ∂z  ∂u ∂  ∂z ∂z  ∂v  + (n − 1) y − n   −  + −     ∂u  ∂u ∂v  ∂y ∂v  ∂u ∂v  ∂y   ∂z ∂z  = − n (n − 1) y − n−1  −   ∂u ∂v   ∂2 z ∂2 z ∂2 z + (n − 1)2 y −2 n  2 − 2 + 2 ∂u ∂v ∂v  ∂u

 ⋅ 

Substituting these values in (1), it reduces to ∂2 z = 0 , which is the required canonical form. ∂u ∂v Integrating it, w.r.t. v, we get ∂z = φ 1 (u), where φ1 is an arbitrary function. ∂u Again integrating w.r.t. u, we get z = ψ1 (u) + ψ2 (v), where ψ1 and ψ2 are arbitrary functions. Hence the required general solution is z = ψ1 ( x − y − n +1) + ψ2 ( x + y − n +1).

Comprehensive Exercise 3 1.

Reduce the equation y2

∂2 z 2

∂x

− 2 xy

y2 ∂z x2 ∂z ∂2 z ∂2 z + x2 2 = + ∂x ∂y x ∂x y ∂y ∂y

to canonical form and hence solve it.

(Meerut 2006, 07)

D-305

2.

Reduce the equation xyr − ( x2 − y2 ) s − xyt + py − qx = 2 ( x2 − y2 ) to canonical form and hence solve it.

3.

Reduce the equation x2 ( y − 1) r − x ( y2 − 1) s + y ( y − 1) t + xyp − q = 0 to canonical form and hence solve it.

4.

Reduce the equation x ( xy − 1) r − ( x2 y2 − 1) s + y ( xy − 1) t + ( x − 1) p + ( y − 1) q = 0 to canonical form and hence solve it.

5.

Reduce the equation x2 r − 2 xy s + y2 t − xp + 3 yq = 8 y / x to canonical form and hence solve it.

A nswers 3 1.

∂2 z ∂v2

= 0 ; z = ( x2 − y2 ) φ1 ( x2 + y2 ) + φ 2 ( x2 + y2 )

2.

v2 − 1 ∂2 z = 2 ; z = − xy + ψ1 ( x2 + y2 ) + ψ2 ( y / x) ∂u ∂v (v + 1)2

3.

∂2 z = 0 ; z = φ1 ( xy) + φ 2 ( xe y) ∂u ∂v

4.

∂2 z = 0 ; z = φ1 ( xe y ) + φ 2 ( ye x ) ∂u ∂v

5.

v

∂2 z ∂v2

+2

y ∂z 1 + x2 ψ1 ( xy) + φ 2 ( xy), where ψ1 ( xy) = − = 2; z = φ1 ( xy). x ∂v xy

Objective Type Questions

Fill In The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

The solution of the equation s = 0 is z = …… .

2.

The solution of the equation r = 6 x is z = …… .

3.

The solution of the equation ar = xy is az = …… .

D-306

4.

The equation A

∂2 u ∂x2

+B

 ∂2 u ∂2 u ∂u ∂u  + C 2 + f  x, y, u, ,  =0 ∂x ∂y ∂x ∂y  ∂y 

is elliptic, hyperbolic or parabolic according as B2 − 4 AC …… .

A nswers Fill in the Blank(s) 1.

φ ( y) + ψ ( x)

2. x3 + x f ( y) + φ ( y)

3.

1 3 x y + x f ( y) + φ ( y) 6

4. < , >, = 0

¨

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11 M onge's M ethod

11.1 Introduction

I

t is only in special cases that a partial differential equation F ( x, y, z , p, q, r, s, t) = 0

of the second order can be integrated. The most important method of solution, due to Monge, is applicable to a wide class of such equations but by no means to all. Monge’s method depends on establishing one or two intermediate integrals (first integrals) of the form u = f (v) where u and v are functions of x, y, z , p, q and f is some arbitrary function.

11.2 Monge’s Method of Integrating Rr + Ss + Tt = V ; where r, s, t have their usual meanings and R, S, T , V are functions of x, y, z , p and q. (Avadh 2006, 12; Rohilkhand 07; Lucknow 09, 11)

The given equation is

Rr + Ss + Tt = V .

…(1)

D-308

We have

dp =

∂p ∂p dx + dy = r dx + s dy ∂x ∂y

and

dq =

∂q ∂q dx + dy = s dx + t dy. ∂x ∂y

∴

r=

dp − s dy dq − s dx and t = ⋅ dx dy

Substituting these values of r and t in (1), we get  dp − s dy  R  + Ss + T  dx 

 dq − s dx    =V dy  

or

R dy (dp − s dy) + Ss dx dy + T dx (dq − s dx) = V dx dy

or

( R dp dy + T dq dx − V dx dy) − s ( R dy2 − S dx dy + T dx2 ) = 0 .

If any relation between x, y, z , p, q makes each of the bracketed expressions vanish, this relation will satisfy the differential equation (1). From

R dy2 − S dx dy + T dx2 = 0   R dp dy + T dq dx − V dx dy = 0 

and

dz = p dx + q dy,

…(2)

it may be possible to obtain either one or two relations between x, y, z , p, q called intermediate integrals, and therefrom to deduce the general solution of (1). The equations (2) are called Monge’s Subsidiary Equations. The method of solution is explained in the following procedure: Resolve the equation R dy2 − S dx dy + T dx2 = 0

…(3)

into the two equations dy − m1 dx = 0 , dy − m2 dx = 0 .

…(4)

Now from the first of the equations (4) and from the equation R dp dy + T dq dx − V dx dy = 0

…(5)

combining if necessary with dz = p dx + q dy, find two integrals u1 = a, v1 = b. Then u1 = f1 (v1) is an intermediate integral, f1 being an arbitrary function. In the same way, taking the second of the equations (4), find another pair of integrals, u2 = a, v2 = b. Then u2 = f2 (v2 ), where f2 is an arbitrary function, is another intermediate integral. To obtain the final integral, either of these intermediate integrals may be integrated and this must be done when m1 = m2 . If m1 and m2 are unequal then solve the two intermediate integrals for p and q and substitute the values of p and q thus found in dz = p dx + q dy. Integrating it obtain the complete integral of (1).

D-309

Example 1:

Solve r = a2 t.

…(1)

(Avadh 2007, 10, 11, 13; Lucknow 07, 08, 10; Rohilkhand 08; Purvanchal 10, 14)

Solution:

We have dp =

∂p ∂p dx + dy = r dx + s dy ∂x ∂y

and

dq =

∂q ∂q dx + dy = s dx + t dy. ∂x ∂y

∴

r=

dp − s dy dq − s dx and t = ⋅ dx dy

Putting these values of r and t in (1), we get dp − s dy dq − s dx = a2 dx dy (dp dy − a2 dq dx) − s (dy2 − a2 dx2 ) = 0 .

or

Monge’s subsidiary equations are dp dy − a2 dq dx = 0 2

and

2

…(2)

2

dy − a dx = 0 .

…(3)

The equation (3) resolves into the two equations: and

dy − a dx = 0

…(4)

dy + a dx = 0.

…(5)

From (2) and (4), we get dp (a dx) − a2 dq dx = 0 From (4) and (6), we get ∴

or dp − a dq = 0.

…(6)

y − ax = a1 and p − aq = b1.

p − aq = f1 ( y − ax)

…(7)

is an intermediate integral, f1 being an arbitrary function. Similarly from (2) and (5) we find another intermediate integral p + aq = f2 ( y + ax), where f2 is an arbitrary function. Solving (7) and (8) for p and q, we have 1 p = { f1 ( y − ax) + f2 ( y + ax)} 2 1 and q= { f2 ( y + ax) − f1 ( y − ax)}. 2a Substituting these values of p and q in dz = p dx + q dy, we get 1 1 dz = [ f1 ( y − ax) + f2 ( y + ax)] dx + [ f2 ( y + ax) − f1 ( y − ax)] dy 2 2a

…(8)

D-310

1 1 f2 ( y + ax)(dy + a dx) − f1 ( y − ax)(dy − a dx). 2a 2a 1 1 z= φ 2 ( y + ax) − φ1 ( y − ax) 2a 2a =

Integrating, or

z = F2 ( y + ax) + F1 ( y − ax),

which is the complete integral of the equation (1). Solve r + (a + b) s + abt = xy.

Example 2:

…(1) (Avadh 2006)

Solution:

We have

∴

dp = r dx + s dy and dq = s dx + t dy. dp − s dy dq − s dx r= and t = ⋅ dx dy

Putting these values of r and t in (1), we get dp − s dy dq − s dx + (a + b) s + ab = xy dx dy or

{ dp dy + ab dq dx − xy dx dy} − s {dy2 − (a + b) dx dy + ab dx2} = 0 .

Monge’s subsidiary equations are dp dy + ab dq dx − xy dx dy = 0 and

2

2

dy − (a + b) dx dy + ab dx = 0 .

…(2) …(3)

The equation (3) resolves into the two equations: and

dy − b dx = 0

…(4)

dy − a dx = 0.

…(5)

From (4), we get y − bx = A. From (2) and (4), we have

Integrating, ∴ or

dp + a dq − xy dx = 0 or dp + a dq − x ( A + bx) dx = 0 . 1 1 p + aq − b . x3 − A . x2 = B. 3 2 1 1 3 p + aq − b . x − ( y − bx) x2 = f1 ( y − bx) 3 2 1 3 1 2 p + aq + bx − x y = f1 ( y − bx), 6 2

…(6)

is an intermediate integral, f1 being an arbitrary function. Similarly from (2) and (5) we find another intermediate integral 1 1 p + bq + ax3 − x2 y = f2 ( y − ax), 6 2 …(7) where f2 is an arbitrary function. Solving (6) and (7) for p and q, we have 1 1 1 p = x2 y − (a + b) x3 + {a f2 ( y − ax) − b f1 ( y − bx)} 2 6 a−b

D-311

and

q=

1 3 1 x + { f1 ( y − bx) − f2 ( y − ax)}. 6 a−b

Substituting these values of p and q in dz = p dx + q dy, we get 1 1 1 dz = x2 y dx + x3 dy − (a + b) x3 dx 2 6 6 1 + {a f2 ( y − ax) − b f1 ( y − bx)} dx a−b + or

dz =

1 { f1 ( y − bx) − f2 ( y − ax)} dy a−b

1 1 (3 x2 y dx + x3 dy) − (a + b) x3 dx 6 6 1 + [ f2 ( y − ax) (dy − a dx) − f1 ( y − bx) (dy − b dx)]. (b − a)

Integrating,

or

z=

1 3 1 1 x y− (a + b) x4 + [ φ 2 ( y − ax) − φ1 ( y − bx)] 6 24 b−a

z=

1 3 1 x y− (a + b) x4 + ψ2 ( y − ax) + ψ1 ( y − bx), 6 24

which is the complete integral of the equation (1). Example 3:

Solve q2 r − 2 pqs + p2 t = 0 .

…(1)

Show that the integral represents a surface generated by straight lines which are parallel to a fixed plane. (Agra 2003; Kanpur 14) Solution:

∴

We have

dp = r dx + s dy and dq = s dx + t dy. dp − s dy dq − s dx r= and t = ⋅ dx dy

Putting these values of r and t in (1), we get dp − s dy dq − s dx q2 − 2 p qs + p2 =0 dx dy or

(q2 dp dy + p2 dq dx) − s (q2 dy2 + 2 pq dx dy + p2 dx2 ) = 0 .

Monge’s subsidiary equations are q2 dp dy + p2 dq dx = 0 and

2

2

…(2) 2

2

q dy + 2 pq dx dy + p dx = 0

…(3)

2

The equation (3) gives (q dy + p dx) = 0 or

q dy + p dx = 0.

…(4)

∴

dz = p dx + q dy ⇒ dz = 0 ⇒ z = a.

…(5)

From (2) and (4), we get q dp − p dq = 0 or ⇒

p = qb.

dp p

=

dq q …(6)

D-312

From (5) and (6), we get p − q f (z ) = 0 ,

…(7)

as an intermediate integral, f being an arbitrary function. For the differential equation (7), Lagrange’s auxiliary equations are dy dx dz = = ⋅ 1 − f (z ) 0 The last two members give dz = 0 i. e., z = A. Taking the first two members, we get dy + f (z ) dx = 0 or dy + f ( A) dx = 0 . ∴

y + x f ( A) = B

or

y + x f (z ) = φ (z ),

…(8)

which is the complete integral of the equation (1). The integral of the given differential equation is the surface (8) which is the locus of the straight lines given by the intersection of planes y + x f (c ) = φ (c ) and z = c . These lines are all parallel to the planes z = 0 as they lie on the plane z = c for varying values of c. Example 4:

Solve t − r sec 4 y = 2 q tan y.

…(1)

(Rohilkhand 2002, 04, 09; Lucknow 07; Purvanchal 10; Avadh 10, 14; Meerut 13B)

Solution:

∴

We have dp = r dx + s dy and dq = s dx + t dy. dp − s dy dq − s dx and t = r= . dx dy

Putting these values of r and t in (1), we get dq − s dx dp − s dy − sec4 y = 2 q tan y dy dx or

{dq dx − dp dy sec4 y − 2 q tan y dx dy } − s {dx2 − dy2 sec4 y} = 0 .

Monge’s subsidiary equations are dq dx − dp dy sec4 y − 2 q tan y dx dy = 0 . and

2

2

4

dx − dy sec y = 0 .

…(2) …(3)

The equation (3) resolves into the two equations dx − dy sec2 y = 0 and

2

dx + dy sec y = 0.

Integrating (4), we get x − tan y = a. From (2) and (4), we get dq − dp sec2 y − 2 q tan y dy = 0 or

dp − (cos2 y dq − 2 q sin y cos y dy) = 0 .

…(4) …(5)

D-313

Integrating, we have p − q cos2 y = b. ∴

p − q cos2 y = f1 ( x − tan y),

…(6)

which is an intermediate integral, f1 being an arbitrary function. Similarly from (2) and (5) we find another intermediate integral p + q cos2 y = f2 ( x + tan y),

…(7)

where f2 is an arbitrary function. Solving (6) and (7) for p and q , we have 1 p = { f1 ( x − tan y) + f2 ( x + tan y)} 2 1 and q = sec2 y { f2 ( x + tan y) − f1 ( x − tan y)}. 2 Substituting these values of p and q in dz = p dx + q dy, we get 1 dz = { f1 ( x − tan y) + f2 ( x + tan y)} dx 2 1 + sec2 y { f2 ( x + tan y) − f1 ( x − tan y)} dy 2 1 1 = f1 ( x − tan y) (dx − sec2 y dy) + f2 ( x + tan y) (dx + sec2 y dy). 2 2 Integrating, we get z = φ1 ( x − tan y) + φ 2 ( x + tan y), which is the complete integral of the equation (1). Example 5: Solve r − t cos2 x + p tan x = 0.

…(1) (Lucknow 2006)

Solution:

∴

We have

dp = r dx + s dy and dq = s dx + t dy. dp − s dy dq − s dx and t = r= . dx dy

Putting these values of r and t in (1), we get dp − s dy dq − s dx − cos2 x + p tan x = 0 dx dy or

(dp dy − dq dx cos2 x + p tan x dx dy) − s (dy2 − cos2 x dx2 ) = 0 .

Monge’s subsidiary equations are dp dy − dq dx cos2 x + p tan x dx dy = 0 and

2

2

2

dy − cos x dx = 0 .

…(2) …(3)

The equation (3) resolves into the two equations dy − cos x dx = 0

…(4)

and

…(5)

dy + cos x dx = 0 .

Integrating (4), we get y − sin x = a. Also from (2) and (4), we get dp − dq cos x + p tan x dx = 0

D-314

or

sec x dp + p sec x tan x dx − dq = 0 .

Integrating, we get ∴

p sec x − q = b. …(6)

p sec x − q = f1 ( y − sin x)

is an intermediate integral, f1 being an arbitrary function. Similarly from (2) and (5) we find another intermediate integral …(7)

p sec x + q = f2 ( y + sin x), where f2 is an arbitrary function. Solving (6) and (7) for p and q, we have 1 p = cos x { f1 ( y − sin x) + f2 ( y + sin x)} 2 1 and q = { f2 ( y + sin x) − f1 ( y − sin x)}. 2

Substituting these values of p and q in dz = p dx + q dy, we get 1 dz = cos x { f1 ( y − sin x) + f2 ( y + sin x)} dx 2 1 + { f2 ( y + sin x) − f1 ( y − sin x)} dy 2 1 1 =− f1 ( y − sin x) (dy − cos x dx) + f2 ( y + sin x) (dy + cos x dx) 2 2 1 1 or dz = − f1 ( y − sin x) d ( y − sin x) + f2 ( y + sin x) d ( y + sin x). 2 2 Integrating, we get z = φ1 ( y − sin x) + φ 2 ( y + sin x), which is the complete integral of the equation (1). Example 6:

Solve pt − qs = q3 .

…(1)

(Rohilkhand 2003, 14; Lucknow 07; Purvanchal 09, 11; Meerut 13)

Solution:

∴

We have dq = s dx + t dy. dq − s dx t= . dy

Putting this value of t in (1), we get dq − s dx p − qs = q3 dy

or ( p dq − q3 dy) − s ( p dx + q dy) = 0 .

Monge’s subsidiary equations are p dq − q3 dy = 0,

…(2)

and

…(3)

p dx + q dy = 0.

[ ∵ dz = p dx + q dy ]

From (3), we have dz = 0. ∴ z = a. Again combining (3) with (2), we get dq + q2 dx = 0

or

dq q2

+ dx = 0 .

D-315

Integrating, we get − ∴

−

1 + x = b. q

1 1 + x = f (z ), or − + x = f (z ), q (∂z / ∂y)

…(4)

which is an intermediate integral, f being an arbitrary function. Now we shall integrate the equation (4) regarding x as constant. dy Regarding x as constant, the equation (4) can be written as − x = − f (z ). dz Integrating w.r.t. z, treating x as constant, we get y−xz=−

∫

f (z ) dz + c = f1 (z ) + f2 ( x),

taking the constant of integration c as a function of x or

y = x z + f1 (z ) + f2 ( x),

which is the complete integral of the equation (1). Example 7:

Solve y2 r − 2 ys + t = p + 6 y.

…(1) (Rohilkhand 2006)

Solution:

∴

We have dp = r dx + s dy and dq = s dx + t dy. dp − s dy dq − s dx and t = r= . dx dy

Putting these values of r and t in (1), we get dp − s dy dq − s dx y2 − 2 ys + = p+6 y dx dy or

{ y2 dp dy + dq dx − ( p + 6 y) dx dy } − s { y2 dy2 + 2 y dx dy + dx2} = 0.

Monge’s subsidiary equations are y2 dp dy + dq dx − ( p + 6 y) dx dy = 0 and

2

2

2

y dy + 2 y dx dy + dx = 0 . 2

The equation (3) gives ( y dy + dx) = 0

or

…(3) y dy + dx = 0.

2

Integrating, we get y + 2 x = a. From (2) and (4), we get or

…(2)

…(4) …(5)

y dp − dq + p dy + 6 y dy = 0

( y dp + p dy) − dq + 6 y dy = 0 .

Integrating, we get py − q + 3 y2 = b.

…(6) 2

2

From (5) and (6), we have py − q + 3 y = f ( y + 2 x) which is an intermediate integral, f being an arbitrary function. Lagrange’s subsidiary equations for the equation (7) are dx dy dz = = . 2 y − 1 − 3 y + f ( y2 + 2 x)

…(7)

D-316

Taking the first two members, we get

y dy + dx = 0.

y2 + 2 x = A.

∴

Again from the last two members, we get dz = {3 y2 − f ( y2 + 2 x)} dy dz = 3 y2 dy − f ( A) dy.

or Integrating,

z = y3 − y f ( A) + B

or

z = y3 − y f ( y2 + 2 x) + φ ( y2 + 2 x),

which is the complete integral of the equation (1).

Comprehensive Exercise 1 Solve the following equations by Monge’s method : 1.

r = t.

2.

(q + 1) s = ( p + 1) t.

3.

x2 r + 2 xys + y2 t = 0 .

(Purvanchal 2007)

2

4.

x r − 2 x s + t + q = 0.

5.

r − 2 s + t = sin (2 x + 3 y).

6.

(r − t) x y − s ( x2 − y2 ) = qx − py.

7.

r x2 − 3 s xy + 2 ty2 + px + 2 q y = x + 2 y.

8.

2 x r − ( x + 2 y) s + yt =

9.

(r − s) y + (s − t) x + q − p = 0 .

10. 11.

x +2 y (2 p − q). x −2 y

r + ka2 t − 2 as = 0. x

−2

r− y

−2

t= x

−3

(Kanpur 2014)

p− y

−3

q.

A nswers 1 1.

z = f1 ( x + y) + f2 ( y − x)

2.

z = φ ( x) + ψ ( x + y + z )

3.

z = y f1 ( y / x) + f2 ( y / x)

4.

z = f1 ( y + log x) + x f2 ( y + log x)

5. 6.

z = f ( x + y) + x φ ( x + y) − sin (2 x + 3 y) z = f1 ( x2 + y2 ) + f2 ( y / x)

D-317

7.

z = x + y + f ( x y) + φ ( x2 y)

8.

z = f1 ( x + 2 y) + f2 ( x y)

9.

z = f1 ( x + y) + f2 ( x2 − y2 )

10.

z = f1 { y + a (1 + t) x} + f2 { y + a (1 − t) x}, t 2 = 1 − k

11.

z = f1 ( x2 + y2 ) + f2 ( x2 − y2 )

11.3 Monge’s Method of Integrating Rr + Ss + Tt + U (rt − s2 ) = V ,

…(1)

where r, s, t have their usual meanings and R, S, T, U, V are functions of x, y, z, p and q. We have dp = r dx + s dy, dq = s dx + t dy. dp − s dy dq − s dx and t = ∴ r= ⋅ dx dy Putting these values of r and t in (1), we get dp − s dy dq − s dx R⋅ + Ss + T ⋅ +U dx dy or

 (dp − s dy) (dq − s dx) 2  − s  =V  dx dy  

( R dp dy + T dq dx + U dp dq − V dx dy) − s ( R dy2 − S dx dy + T dx2 + U dp dx + U dq dy) = 0 .

∴ Monge’s subsidiary equations are L ≡ R dp dy + T dq dx + U dp dq − V dx dy = 0 and

…(2)

M ≡ R dy2 − S dx dy + T dx2 + U dp dx + U dq dy = 0.

…(3) Here the equation (3) cannot be resolved into two linear equations on account of the presence of the term U dp dx + U dq dy. However we try to resolve M + λL = 0 into two linear equations, where λ is some multiplier to be determined. Now

M + λL = R dy2 + T dx2 − (S + λV ) dx dy + U dp dx + U dq dy + λ R dp dy + λ T dq dx + λ U dp dq. …(4)

Also let M + λL ≡ (α dy + β dx + γ dp)(α ′ dy + β ′ dx + γ ′ dq).

…(5)

Equating the coefficients of dy2 , dx2 and dp dq in (4) and (5), we have R = αα ′ , T = ββ ′ , λU = γγ ′ 1 λ If we choose α ′ = 1, β ′ = , γ ′ = then α = R, β = kT , γ = mU . k m Now equating the coefficients of the remaining five terms in (4) and (5), we have R kT + = − (S + λV ) k …(6) kT λ mU = λT , U = , …(7) m k

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λR = U , λR = mU . m λR From (7), m = k and from (8), m = ⋅ U Putting k = λ R / U in (6), we have λR U T +R = − (S + λV ) U λR or λ2 ( RT + UV ) + λUS + U 2 = 0 .

…(8)

…(9) Apart from the special case when S2 = 4 ( RT + UV ), this equation will have two

distinct roots λ1, λ 2 . For λ = λ1, m = k = R λ1 / U . Then from (5), M + λL = 0 gives RT U U    dx + λ1 R dp  dy + dx + dq = 0  R dy + λ1   U R λ1 R  or

(U dy + T λ1 dx + U λ1 dp) ( Rλ1 dy + U dx + U λ1 dq) = 0

Similarly for λ = λ

2

…(10)

, M + λL = 0 gives

(U dy + T λ 2 dx + U λ 2 dp) ( R λ 2 dy + U dx + U λ 2 dq) = 0 .

…(11)

Now one factor of (10) is combined with one factor of (11) to give an intermediate integral and similarly other pair gives another intermediate integral. These cannot be obtained if we combine the first factor of (10) with the first factor of (11) or the second factor of (10) with the second factor of (11). Thus the problem reduces to the solution of the pairs U dy + λ1 T dx + λ1 U dp = 0   λ 2 R dy + U dx + λ 2 U dq = 0  and

U dy + λ 2 T dx + λ 2 U dp = 0   λ1 R dy + U dx + λ1 U dq = 0 

…(12) …(13)

From each of these pairs we shall derive two integrals of the form u = a, v = b. Let u1 = a1, v1 = b1 be the integrals obtained from the equations (12) and u2 = a2 , v2 = b2 be the integrals obtained from the equations (13). Then the two intermediate integrals are u1 = f1 (v1) and u2 = f2 (v2 ), which can often be solved to find the values of p and q as functions of x, y and z. Substituting these values of p and q in dz = p dx + q dy and integrating it we obtain the solution of the original equation. Note 1:

If the two values of λ are equal, then it is possible to find only one intermediate

integral. If it is not possible to solve the equations u1 = f1 (v1) and u2 = f2 (v2 ) for p and q, then we may take one of the intermediate integrals say u1 = f1 (v1) and one of the integrals from u2 = a2 and v2 = b2 to determine p and q. Note 2:

An integral of a more general form can be obtained by taking the arbitrary function occurring in the intermediate integral to be linear. Note 3:

Let u1 = mv1 + n, where m and n are some constants.

D-319

Then integrating it by Lagrange’s method we find the solution of the given equation.

Example 8: Solution:

Solve r + 3 s + t + (rt − s2 ) = 1.

…(1)

Comparing the given equation (1) with Rr + Ss + Tt + U (rt − s2 ) = V ,

we have R = 1, S = 3, T = 1, U = 1, V = 1. The λ-quadratic equation is λ2 (UV + RT ) + λSU + U 2 = 0 2 λ2 + 3 λ + 1 = 0 or (2 λ + 1) (λ + 1) = 0 . 1 ∴ λ1 = − 1, λ 2 = − ⋅ 2 One intermediate integral is given by the equations U dy + λ1 T dx + λ1 U dp = 0 and λ 2 R dy + U dx + λ 2 U dq = 0 1 1 or dy − dx − dp = 0 and − dy + dx − dq = 0 2 2 or dy − dx − dp = 0 and dy − 2 dx + dq = 0. Integrating, y − x − p = a1 and y − 2 x + q = b1. or

Hence one intermediate integral is y − x − p = f1 ′ ( y − 2 x + q) = f1 ′ (α),

…(2)

where α = y − 2 x + q. Another intermediate integral is given by the equations U dy + λ 2 T dx + λ 2 U dp = 0 and λ1 R dy + U dx + λ1 U dq = 0 1 1 or dy − dx − dp = 0 and − dy + dx − dq = 0. 2 2 Integrating, 2 y − x − p = a2 and − y + x − q = b2 . Hence another intermediate integral is 2 y − x − p = f2 ′ (− y + x − q) = f2 ′ ( β),

…(3)

where β = − y + x − q. From the above relations , we have x = − (α + β) so that dx = − (dα + dβ), y = f2 ′ (β) − f1 ′ (α) so that dy = f2 ′ ′ ( β) dβ − f1 ′ ′ (α) dα and p = y − x − f1 ′ (α), q = x − y − β. Putting these values in dz = p dx + q dy, we get dz = { y − x − f1 ′ (α)} dx + {x − y − β} dy = − ( x − y) (dx − dy) − f1 ′ (α) {− dα − dβ } − β { f2 ′ ′ (β) dβ − f1 ′ ′ (α) dα} = − ( x − y) (dx − dy) + f1 ′ (α) dα + { f1 ′ (α) dβ + β f1 ′ ′ (α) dα} − β f2 ′ ′ (β) dβ. 1 2 Integrating , z = − ( x − y) + f1(α) + βf1 ′ (α) − {βf2 ′ ( β) − f2 ( β)} 2

D-320

1 ( x − y)2 + f1(α) + f2 (β) + β { f1 ′ (α) − f2 ′ ( β)} 2 1 = − ( x − y)2 + f1(α) + f2 (β) − βy, 2 which is the required solution of the equation (1). or

Example 9:

z=−

Solve 2 s + (rt − s2 ) = 1.

…(1) (Lucknow 2011; Purvanchal 11)

Solution:

Comparing the given equation (1) with Rr + Ss + Tt + U (rt − s2 ) = V ,

we have R = 0 , S = 2, T = 0 , U = 1 and V = 1. The λ-quadratic equation is λ2 (UV + RT ) + λSU + U 2 = 0 or

λ2 + 2 λ + 1 = 0 .

∴ λ1 = − 1, λ 2 = − 1. In this case we can find only one intermediate integral, which is given by the equations U dy + λ1 T dx + λ1 U dp = 0 and λ 2 R dy + U dx + λ 2 U dq = 0 or …(2) dy − dp = 0 and dx − dq = 0. Integrating, y − p = a and x − q = b. ∴ The intermediate integral is y − p = f ( x − q).

…(3)

From (2), p = y − a and q = x − b. Putting these values of p and q in dz = p dx + q dy, we get dz = ( y − a) dx + ( x − b) dy or dz = ( y dx + x dy) − a dx − b dy. Integrating, we get z = xy − ax − by + c , which is the complete integral of (1). Alite:. To find the more general solution , let y − p = m ( x − q) + n, where m and n are constants or p − mq = y − mx − n. Lagrange’s auxiliary equations are dy dx dz = = . 1 −m y − mx − n From the first two members , we have dy + m dx = 0. ∴ y + mx = a. Again from the first and the last members, we have dz = ( y − mx − n) dx = (a − 2 mx − n) dx. Integrating, z = ax − mx2 − nx + b or

z = ( y + mx) x − mx2 − nx + b

or

z = xy − nx + φ ( y + mx), which is the required general solution of (1).

D-321 Example 10:

Solve z (1 + q2 )r − 2 pqzs + z (1 + p2 )t + z 2 (rt − s2 ) + (1 + p2 + q2 ) = 0 . …(1)

Solution:

Comparing the equation (1) with Rr + Ss + Tt + U (rt − s2 ) = V ,

we have

R = z (1 + q2 ), S = − 2 pqz , T = z (1 + p2 ), U = z 2

and

V = − (1 + p2 + q2 ).

The λ-quadratic equation is λ2 (UV + RT ) + λSU + U 2 = 0 or

{− z 2 (1 + p2 + q2 ) + z 2 (1 + p2 ) (1 + q2 )} λ2 − 2 pqz 3 λ + z 4 = 0

or

p2 q2 λ2 − 2 pqzλ + z 2 = 0

or ( pqλ − z )2 = 0 .

∴ λ1 = λ 2 = z /( pq). The intermediate integral is given by the equations U dy + λ1 T dx + λ1 U dp = 0 and λ 2 R dy + U dx + λ 2U dq = 0 or pq dy + (1 + p2 ) dx + z dp = 0 and pq dx + (1 + q2 ) dy + z dq = 0 or p ( p dx + q dy) + dx + z dp = 0 and q ( p dx + q dy) + dy + z dq = 0 or p dz + z dp + dx = 0 and q dz + z dq + dy = 0. Integrating, zp + x = a and zq + y = b. …(2) ∴ The intermediate integral is …(3) zp + x = f (zq + y). a− x b− y From (2), p = and q = ⋅ z z Putting these values of p and q in dz = p dx + q dy, we get a− x b− y dz = dx + dy z z or z dz = (a − x) dx + (b − y) dy. Integrating, we have z 2 = − (a − x)2 − (b − y)2 + c , which is the required complete integral of (1). Aliter. To find the more general solution, let z p + x = m (zq + y) + n, where m and n are constants or z p − mzq = − x + my + n. Lagrange’s auxiliary equations are dy dx dz = = . z − mz − x + my + n Taking the first two members , we get y + mx = a. x dx + y dy + z dz Also each fraction = nz dx x dx + y dy + z dz ∴ = ⇒ x dx + y dy + z dz − n dx = 0. z nz Integrating, x2 + y2 + z 2 − 2 nx = b or

x2 + y2 + z 2 − 2 nx = φ ( y + mx),

which is the required general solution of (1).

D-322 Example 11: Solution:

Solve (rt − s2 ) − s (sin x + sin y) = sin x sin y.

…(1)

Comparing the equation (1) with Rr + Ss + Tt + U (rt − s2 ) = V , we have

R = 0 , S = − (sin x + sin y), T = 0 , U = 1, V = sin x sin y. The λ-quadratic equation is λ2 (UV + RT ) + λSU + U 2 = 0 or

λ2 sin x sin y − λ (sin x + sin y) + 1 = 0

or

(λ sin x − 1) (λ sin y − 1) = 0 .

∴

λ1 = cosec y

and λ 2 = cosec x.

One intermediate integtral is given by the equations U dy + λ1 T dx + λ1 U dp = 0 and

λ 2 R dy + U dx + λ 2U dq = 0

or

dy + cosec y dp = 0 and dx + cosec x dq = 0

or

sin y dy + dp = 0 and sin x dx + dq = 0

Integrating, p − cos y = a, q − cos x = b. ∴ Intermediate integral is p − cos y = f (q − cos x). Let

p − cos y = m (q − cos x) + n, where m and n are constants

or

p − mq = cos y − m cos x + n.

Lagrange’s auxiliary equations are dy dx dz = = . 1 − m cos y − m cos x + n The first two members give y + mx = A. Again taking the first and the last members, we get dz = (cos y − m cos x + n) dx = {cos ( A − mx) − m cos x + n} dx. 1 Integrating, z = − sin ( A − mx) − m sin x + nx + B m or mz + sin y + m2 sin x − mn x = mφ ( y + mx). Example 12: Solution:

Solve qr + ( p + x) s + yt + y (rt − s2 ) = − q.

Comparing the equation (1) with Rr + Ss + Tt + U (rt − s2 ) = V ,

we have R = q, S = p + x, T = y, U = y, V = − q. The λ-quadratic equation is λ2 (UV + RT ) + λ SU + U 2 = 0 or

λ2 (− yq + qy) + λy ( p + x) + y2 = 0 .

∴ λ1 = − y / ( p + x), λ 2 = ∞. One intermediate integral is given by the equations U dy + λ1 T dx + λ1 U dp = 0 and λ 2 R dy + U dx + λ 2 U dq = 0

…(1)

D-323

y2 y2 dx − dp = 0. p+ x p+ x

or

y dy −

and

q dy + 0 . dx + y dq = 0 dp + dx dy − 2 ( p + x) + = 0 and q dy + y dq = 0. y y

or

Integrating, we get ( p + x) / y = a and q y = b. Hence one intermediate integral is  p + x ⋅ qy = f   y  Charpit’s auxiliary equations are dy dp dx = = ⋅ p+ x   p + x y 1 1   − f′ f′ y y  y   y  From the first and the last members, we have dx + dp = 0. ∴ p + x = c. Also from dz = p dx + q dy, we get dz = (c − x) dx + (1/ y) f (c / y) dy. 1 Integrating, z = cx − x2 + φ (c / y) + ψ (c ), which is the required solution. 2

Comprehensive Exercise 2 1.

Solve r t − s2 + a2 = 0.

2.

Solve rt − s2 + 1 = 0.

3.

Solve 2 pr + 2 qt − 4 pq (rt − s2 ) = 1.

4.

Solve 3 r + 4 s + t + (rt − s2 ) = 1.

5.

Solve 3 s + (rt − s2 ) = 2.

6.

Solve r + t − (rt − s2 ) = 1.

(Rohilkhand 2011)

A nswers 2 1. 3. 4. 5. 6.

1 { f2 ( β) − f1(α)} + βy 2a 3 z = 3 c ± 2 (b + x)3 /2 ± 2 (a + y)3 /2 1 3 2 z = 2 xy − x2 − y + ax + by + c 2 2 z = xy − f1(α) + f2 ( β) + βy 1 z = ( x2 + y2 ) + ax + by + c 2 z = − ax y +

2. z = − xy −

1 { f1(α) − f2 ( β)} + βy 2

D-324

Objective Type Questions

Multiple Choice Questions 1.

2.

Monge’s subsidiary equations for r = t are (a) dy2 + dx2 = 0 , dp dy − dx dq = 0

(b) dy2 − dx2 = 0 , dp dy − dx dq = 0

(c) dy2 − dx2 = 0 , dp dy + dx dq = 0

(d) dy2 + dx2 = 0 , dp dy + dx dq = 0 .

The λ-quadratic equation for the equation rt − s2 + 1 = 0 is (a) λ2 + 1 = 0

(b) λ2 − 1 = 0

(c) λ2 + 2 = 0

(d) λ2 − 2 = 0 .

Fill In The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

Monge’s method is usually used to integrate the …… order partial differential equations.

2.

The equations R dy 2 − S dx dy + T dx2 = 0 , R dp dy + T dq dx − V dx dy = 0

3. 4.

are called …… in integrating the equation of the form R r + S s + Tt = V . dp − … … We have r = and t = ⋅ dx dy In integrating the equation of the type Rr + Ss + Tt + U (rt − s2 ) = V by Monge’s method the λ-quadratic equation is …… .

A nswers Multiple Choice Questions 1.

(b)

2.

(b)

Fill in the Blank(s) 1.

second

2.

Monge’s subsidiary equations

3.

s dy, dq − s dx

4.

λ2 (UV + RT ) + λ SU + U 2 = 0

¨

D-325

12 S eries S olutions

Of

D ifferential E quations

12.1 Introduction If a homogeneous linear differential equation has constant coefficients, it can be solved by algebraic methods, and its solutions are elementary functions known from calculus. However, if such an equation has variable coefficients (functions of x), it must usually be solved by other methods. Legendre’s equation, the hypergeometric equation, and Bessel’s equation are very important equations of this type, since these equations and their solutions play a basic role in applied mathematics. In the present chapter we shall discuss the solutions of some linear differential equations of second order with variable coefficients in the form of series. Consider the linear differential equation of the second order d2 y dx 2

+ P ( x)

dy + Q ( x) y = 0 , dx

…(1)

where P ( x) and Q ( x) are functions of x . A point x = a may have a special nature with respect to the differential equation (1).

D-326

Without loss of generality, we can restrict our study to the nature of the origin with respect to (1), because the form of the differential equation (1) remains unchanged by shifting the origin. Hence to check the nature of a point a with respect to (1), it is convenient to change the differential equation by shifting the origin to the point a and then to determine the nature of the origin with respect to the transformed equation. Ordinary and singular points: The point x = 0 (i. e., the origin) is called an ordinary point of the differential equation if P ( x) and Q ( x) do not become infinite in a neighbourhood of the origin and they can be expanded in the form of a power series. The point x = 0 is called a singular point of the differential equation (1) if it is not an ordinary point of the differential equation. The singular points are of the two types : (i)

Regular singular points,

(ii) Irregular singular points.

The origin is a regular singular point if both xP ( x) and x 2 Q ( x) can be expanded in a power series of x in the neighbourhood of x = 0. Otherwise x = 0 is called an irregular singular point. If we consider the differential equation P0 ( x)

d2 y 2

dx

+ P1 ( x)

dy dx

+ P2 ( x) y = 0 ,

…(2)

then x = a is called an ordinary point of (2) if P0 (a) ≠ 0 , otherwise it is a singular point. A singular point x = a of (2) is called regular, if when (2) is put in the form d2 y 2

dx

+

Q1 ( x) dy x − a dx

+

Q 2 ( x) ( x − a)2

= 0,

then Q1 ( x) and Q 2 ( x) possess derivatives of all orders in the neighbourhood of a.

12.2 Frobenius Method The method of finding a solution in a series for the linear differential equation (1) near a regular singular point or near an ordinary point is called Frobenius method.

12.3 Solution Near an Ordinary Point We begin by taking a trial solution of the given differential equation d2 y dx

2

+ P ( x)

dy + Q ( x) y = 0 dx

…(1)

in the form of the series ∞

y = Σ c n x n. n=0

∴

dy = Σ n cn x dx

n −1

…(2)

D-327

and

d2 y dx 2

= Σ n (n − 1) c n x

n−2

…(3)

.

If P ( x) and Q ( x) are not polynomials in x , then they can be expanded in the form of power series given by ∞

P ( x) = Σ pn x

n

n=0

and

∞

Q ( x) = Σ qn x n.

…(4)

n=0

Putting the respective values from (2), (3), (4) in (1) and then equating to zero the coefficients of various powers of x we determine the values of different coefficients of (2). Putting these values in (2) we get the general solution of (1) because in general it contains two arbitrary constants.

Example 1 :

Solve

d2 y dx2

− 2 x2

dy + 4 xy = x 2 + 2 x + 2 in powers of x . dx (Avadh 2008, 14)

Here x = 0 is an ordinary point. Let a trial solution in the form of the series of the given differential equation be Solution :

y = c0 + c1 x + c2 x 2 + c3 x 3 + … + c n x Differentiating (1), we get dy = c1 + 2 c2 x + 3 c3 x 2 + … + n c n x dx and

d2 y dx 2

n

+ ….

n −1

= 2 c2 + 6 c3 x + … + n (n − 1) c n x

…(1)

+…

n−2

…(2) …(3)

+…

Putting these values in the given equation, we get (2 c2 + 6 c3 x + …) − 2 x 2 (c1 + 2 c2 x + 3 c3 x 2 + …) + 4 x (c0 + c1 x + c2 x 2 + c3 x 3 + …) − x 2 − 2 x − 2 = 0 or

(2 c2 − 2) + (6 c3 + 4 c0 − 2) x + (12 c4 + 2 c1 − 1) x 2 + 20 c5 x 3 + … + {(n + 2) (n + 1) c n + 2 − 2 (n − 1) c n − 1 + 4 c n − 1} x

n

+ … = 0,

which is an identity in x .We can equate to zero the coefficients of various powers of x . Equating to zero the coefficients of various powers of x , we get 2 c2 − 2 = 0

i. e., c2 = 1 1 2 6 c3 + 4 c0 − 2 = 0 i. e., c3 = − c0 3 3 1 1 12 c4 + 2 c1 − 1 = 0 i. e., c4 = − c1 . 12 6 All other coefficients are given by the relation 2 (n − 3) cn + 2 = c n−1 , n ≥ 3. (n + 1) (n + 2)

D-328

Hence the required complete solution in series is 2 2 6 1 1 7     x − …… y = c0 1 − x 3 − x − …… + c1  x − x 4 −     3 45 6 63 1 1 4 1 6 + x2 + x3 + x + x + …… , 3 12 45 where c0 and c1 are arbitrary constants.

12.4 Solution Near a Regular Singular Point We assume a trial series solution y = x m (c0 + c1 x + c2 x 2 + …) = x

m

∞

Σ

n=0

cn x

n

where all the c’s are constants and c0 ≠ 0 . Here we have to determine m and c’s. dy d2 y and and substitute their values in the given differential dx dx 2 equation. Then the equation (1) reduces to an identity in x . For this, we find

By equating to zero the coefficient of the lowest power of x in this identity, we get a quadratic equation in m. It is called the indicial equation. It will determine m. Equating to zero the coefficients of other various powers of x , we can determine the values of the constants c1 , c2 ,…… etc. in terms of c0 . The following cases arise according to the nature of the roots of the indicial equation : (i) The roots of the indicial equation equal. (ii) The roots of the indicial equation unequal and differing by a quantity not an integer. (iii) The roots of the indicial equation unequal, differing by an integer and making the coefficients of some powers of x in the series for y infinity. (iv) The roots of the indicial equation unequal, differing by an integer and making the coefficient of some power of x in the series for y indeterminate. Case I: Roots of the indicial equation equal: Let m = α be two equal roots of the indicial equation. Then putting m = α in y and in ∂y / ∂m, we can obtain the two independent solutions. The second solution always consists of a numerical multiple of the product of the first solution and log x added to another series.

Example 2:

Solve x

d2 y dx 2

+

dy + xy = 0 . (Bessel’s equation when n = 0). dx

…(1)

(Avadh 2008; Purvanchal 11)

D-329 Solution :

m

Putting y = x

in the left hand side of the given equation (1), we get

x . m (m − 1) x or

x

m +1

+ m2 x

m −2 m −1

+mx

m −1

+ x.x

m

.

Obviously the common difference of the powers is (m + 1) − (m − 1) i. e., 2 . ∴

Let the series solution of the given equation (1) in ascending powers of x be ∞

y = Σ cr x

m +2 r

r =0

∴ and

= c0 x

∞ dy = Σ c r (m + 2 r) x dx r = 0

d2 y 2

dx

m

+ c1 x

m +2

+ c2 x

m +4

+…

…(2)

m + 2 r −1

∞

= Σ c r (m + 2 r) (m + 2 r − 1) x

m +2 r −2

r =0

.

Since (2) is a solution of (1) so substituting the values of y, dy / dx and d2 y / dx2 in (1), we have ∞

Σ c r [(m + 2 r) (m + 2 r − 1) x

m + 2 r −1

r =0

+ (m + 2 r) x or

∞

Σ cr [x

m + 2 r +1

r =0

+ (m + 2 r)2 x

m + 2 r −1

m + 2 r −1

+x

m + 2 r +1

]=0

] = 0,

which is an identity in x . Equating to zero the coefficient of the lowest power of x i. e., of x m −1, we have the indicial equation as c0 m2 = 0 . Now c0 ≠ 0 , as it is the coefficient of the first term with which we start to write the series. ∴

m = 0, 0.

Thus the indicial equation has equal roots. Now equating to zero the coefficient of the general term i. e., of x

m + 2 p +1

, we get

2

c p + (m + 2 p + 2) c p + 1 = 0 . ∴

c p +1 = −

1 (m + 2 p + 2)2

…(3)

cp .

Putting p = 0 , 1, 2 , …… in (3), we get 1 1 1 c1 = − c0 , c2 = − c1 = (− 1)2 c0 , (m + 2)2 (m + 4)2 (m + 2)2 (m + 4)2 c3 = −

1 2

(m + 6)

c2 = (− 1)3

1 2

(m + 2) (m + 4)2 (m + 6)2

c0 , etc.

Substituting these values in (2), we get y = c0 x

m

  x2 x4 + − … 1 − 2 2 2 (m + 2) (m + 4)  (m + 2) 

…(4)

D-330

Putting m = 0 in (4), we get   x2 x4 x6 y = c0 1 − 2 + 2 2 − 2 2 2 + … 2 2 .4 2 .4 .6   or

…(5)

y = au (say),

which is one solution of the given equation (1). Here

u = 1−

x2 22

x4

+

22 . 42

−

x6

+ …,

22 . 42 . 62

and a is an arbitrary constant which replaces c0 . Since the two values of m are equal so a second solution cannot be obtained from (4). Differentiating (4), we get  dy = c0 m x dx  and

d2 y 2

dx

m −1

−

m +1

(m + 2) x

(m + 2)

 = c0 m (m − 1) x 

m −2

−

m +3

 − …… (m + 2) (m + 4)  (m + 4) x

+

2

2

(m + 2) (m + 1)

m

x

(m + 2)2 +

2

(m + 4) (m + 3) x

m +2

2

2

(m + 2) (m + 4)

 − … ⋅ 

Substituting these values in the given equation (1), we have L.H.S.

 = x c0 m (m − 1) x 

m −2

−

(m + 2) (m + 1) 2

(m + 2)

 + c0 m x 

m −1

−

 + xc0  x  = c0 m2 x ∴

m −1

x

m

(m + 4) (m + 3) x

+

(m + 2) x

2

m +1

(m + 2)

−

2

(m + 2)

 − … 

m +3

+

 − … (m + 2) (m + 4) 

+

 − … (m + 2) (m + 4) 

m +2

x

2

(m + 2) (m + 4) 2

m

m +2

(m + 4) x 2

x

2

m +4

2

2

[∵ all other terms cancel]

.

 d2  d + x y = c0 m2 x  x 2 + dx  dx 

m −1

.

Differentiating both sides partially w.r.t. m, we get  d ∂ ∂  d2 + + x y = (c0 m2 x  x ∂ m ∂m  dx 2 dx  or

 d2  d + + x x 2 dx  dx 

 ∂y   = c0 . 2 mx   ∂m 

  ∵ the operators 

m −1

m −1

)

+ c0 m2 x

m −1

log x .

  d2 d ∂   x + + x and commute 2 dx m ∂    dx

D-331

Putting m = 0, we get  d2  d + + x x 2 dx  dx 

 ∂y   = 0 .   ∂m  0

 ∂y    satisfies the equation (1). Hence it is also a solution of (1).  ∂m  m = 0

Thus

Differentiating (4) partially w.r.t. m, we get ∂y ∂m

m

= c0 x

+ c0 x

  x2 x4 log x 1 − + − … 2 2 2 (m + 2) (m + 4)  (m + 2) 

 2 x2    4 −2 −2 + + x + …  3 3 2 2 3 (m + 2) (m + 4)   (m + 2) (m + 4) (m + 2) 

m

Putting m = 0, we obtain    ∂y  x2 x4   = c0 log x 1 − 2 + 2 2 − … 2 2 .4  ∂m  m = 0    x 2  −2  −2  + c0  2 +  3 2 + 2 3  x 4 + … 2 . 4   2 . 4  2  x 2  3 = bu log x + b  2 − 3 2 x 4 + …… = bv, (say) 2 .4  2  which is another independent solution of (1). x 2  3 v = u log x +  2 − 3 2 x 4 + …… 2 .4  2 

Here

and b is an arbitrary constant which replaces c0 . Hence the required general solution of (1) is y = au + bv, where a and b are arbitrary constants. Note.

The equation (1) is called the Bessel’s equation of order zero.

Example 3 :

Solve x

d2 y dx

2

+ (1 + x)

dy + 2 y = 0. dx

…(1) (Purvanchal 2010)

Solution :

Putting y = x

m

in the left hand side of the given equation (1), we get

x . m (m − 1) x or

(m + 2) x

m

m −2

+ m2 x

+ (1 + x) m x m −1

m −1

+2 x

m

.

Obviously the common difference of the powers is 1. ∴

Let the solution of (1) in ascending powers of x be y = c0 x

m

+ c1 x

m +1

+ c2 x

m +2

+…=

∞

Σ

r =0

cr x

m+r

.

…(2)

D-332

dy

∴

dx

=

d2 y

and

2

dx

∞

c r (m + r) x

Σ

r =0 ∞

= Σ

r =0

m + r −1

c r (m + r) (m + r − 1) x

m + r −2

.

Putting these values in the equation (1), we get ∞

Σ c r [(m + r) (m + r − 1) x

m + r −1

r =0

or

∞

Σ

r =0

c r [(m + r + 2) x

m+r

+ (1 + x) (m + r) x

+ (m + r)2 x

m + r −1

m + r −1

+ 2x

m+r

]=0

] = 0,

which is an identity in x . Equating to zero the coefficient of the lowest power of x i. e., of x m −1, we have the indicial equation as c0 m2 = 0 . Now c0 ≠ 0 , as it is the coefficient of the first term with which we start to write the series. ∴

m = 0, 0.

Thus the indicial equation has equal roots. Now equating to zero the coefficient of the general term i. e., of x

m+p

, we get

c p (m + p + 2) + (m + p + 1)2 c p + 1 = 0 . ∴

c p +1 = −

(m + p + 2) (m + p + 1)2

…(3)

cp

Putting p = 0 , 1, 2 , …… in (3), we get (m + 2) (m + 3) (m + 2) (m + 3) c1 = − c0 , c2 = − c1 = (− 1)2 c0 , (m + 1)2 (m + 2)2 (m + 1)2 (m + 2)2 c3 = −

(m + 4) 2

(m + 3)

c2 = (− 1)3

(m + 2) (m + 3) (m + 4) (m + 1)2 (m + 2)2 (m + 3)2

c0 , etc.

Substituting these values in (2), we get y = c0 x

m

  (m + 2) (m + 2) (m + 3) 2 x+ x − … ⋅ 1 − 2 2 2 (m + 1) (m + 2)  (m + 1) 

…(4)

Putting m = 0 in (4), we get   3 2 4 3 y = c0 1 − 2 x + x − x + …… 2! 3!  

…(5)

= au (say), which is one solution of (1). 3 2 4 3 Here u = 1 − 2 x + x − x + …… and a is an arbitrary constant which replaces c0 . 2! 3!  ∂y   Proceeding as in Ex. 2, we find that the second solution is  ⋅  ∂m  m = 0 Differentiating (4) partially w.r.t. m and then putting m = 0, we get

D-333

 ∂y   1 3  1 1 2     = bu log x + b 2 2 −  x −  − + 2 +  x + … 2 2! 3 2    ∂m  m = 0  = bv (say), which is another independent solution of (1). Hence the required general solution of (1) is y = au + bv, where a and b are arbitrary constants. Case II:

Roots of the indicial equation unequal and not differing by an integer:

If the indicial equation has two unequal roots m = α and β which do not differ by an integer, then the two independent solutions can be obtained by putting m = α and β in the series for y.

Example 4 :

Solve in series the Bessel’s equation d2 y

x2

2

dx

taking 2n as non-integral. Solution :

dy + ( x 2 − n2 ) y = 0 dx

+x

(Avadh 2009, 10; Rohilkhand 10; Kanpur 09; Lucknow 08)

Putting y = x

m

in the left hand side of the given equation (1), we get

x 2 m (m − 1) x or

…(1)

m −2

+xmx

{ m (m − 1) + m − n2 } x

m

+x

m −1

+ ( x 2 − n2 ) x

m +2

m

.

Obviously the common difference of the powers is (m + 2) − m i. e., 2 . Let the solution of (1) be ∞

y = Σ cr x

m + 2r

…(2)

r =0

∴ and

∞ dy = Σ c r (m + 2 r) x dx r = 0

d2 y 2

dx

m + 2 r −1

∞

= Σ c r (m + 2 r) (m + 2 r − 1) x

m + 2r − 2

r =0

.

Substituting the values of y, dy / dx , d2 y / dx 2 in (1), we get ∞

Σ c r [(m + 2 r) (m + 2 r − 1) x

m + 2r

r =0

+ (m + 2 r) x or or

∞

Σ c r [{(m + 2 r)2 − n2 } x

r =0 ∞

m + 2r

+x

Σ c r [(m + 2 r + n) (m + 2 r − n) x

r =0

m +2 r

m + 2r + 2

m + 2r

+x

+ ( x 2 − n2 ) x

]=0 m +2 r +2

] = 0,

m + 2r

]=0

D-334

which is an identity in x . Equating to zero the coefficient of the lowest power of x i. e., of x m , we have the indicial equation as c0 (m + n) (m − n) = 0 . Now c0 ≠ 0 , as it is the coefficient of the first term with which we start to write the series. ∴

m = n, − n.

Here the roots of the indicial equation are unequal and differ by 2n which is not an integer. Now equating to zero the coefficient of the general term i. e., of x

∴

m + 2p

, we get

c p (m + 2 p + n) (m + 2 p − n) + c p − 1 = 0 . 1 cp = − c p −1 . (m + 2 p + n) (m + 2 p − n)

Putting p = 1, 2 , …… , we get c1 = −

1 c0 , (m + 2 + n) (m + 2 − n)

c2 = −

1 c1 (m + 4 + n) (m + 4 − n)

= (−1)2

1 c0 , etc. (m + 4 + n) (m + 4 − n) (m + 2 + n) (m + 2 − n)

Substituting these values in (2), we get y = c0 x

 1 x2 1 − m + + n m + − n ( 2 ) ( 2 ) 

m

+

 x4 − … ⋅ (m + 4 + n) (m + 4 − n) (m + 2 + n) (m + 2 − n) 

Putting m = n and −n successively, we get y = c0 x

n

= c0 x

n

  x4 x2 + − … 1 −  2 (2 n + 2) 2 . 4 . (2 n + 2) (2 n + 4)    x2 x4 + 4 − … 1 − 2   2 . 1 !(n + 1) 2 . 2 !(n + 1) (n + 2)

= au (say), which is one solution of (1) and

y = c0 x

−n

= c0 x

−n

  x4 x2 + − … 1 −  2 (−2 n + 2) 2 . 4 . (−2 n + 2) (−2 n + 4)    x2 x4 + 4 − … 1 − 2   2 . 1 !(− n + 1) 2 . 2 !(− n + 1) (− n + 2)

= bv (say), which is another independent solution of (1). Hence the required general solution of (1) is y = au + bv, where a and b are arbitrary constants.

D-335 Example 5:

Solve in descending powers of x the Legendre’s equation (1 − x 2 ) y ′ ′ − 2 xy ′ + p ( p + 1) y = 0 .

Solution :

Putting y = x

m

in the left hand side of the given equation (1), we get

(1 − x 2 ) m (m − 1) x or

…(1)

m −2

(− m2 − m + p2 + p) x

m

− 2 x . mx

m −1

+ p ( p + 1) x

m −2

+ m (m − 1) x

m

.

Obviously the common difference of the powers is m − (m − 2) i. e., 2 . ∴

Let the solution of (1) in a series in descending powers of x be m

y = c0 x ∴

y′=

and

y ′′ =

+ c1 x

∞

m −2

+ c2 x

Σ c r (m − 2 r) x

m −4

∞

+…= Σ

r =0

cr x

m − 2r

…(2)

.

m − 2 r −1

r =0 ∞

Σ c r (m − 2 r) (m − 2 r − 1) x

m − 2r − 2

r =0

.

Substituting the values of y, y ′ , y ′ ′ in (1), we get ∞

Σ c r [(1 − x2 ) (m − 2 r) (m − 2 r − 1) x m − 2 r − 2 − 2 x (m − 2 r) x

m − 2 r −1

r =0

+ p ( p + 1) x or

∞

Σ c r [{− (m − 2 r) (m − 2 r − 1) − 2 (m − 2 r) + p ( p + 1)} x

m − 2r

∞

Σ c r [{ p2 − (m − 2 r)2 + ( p − m + 2 r)} x

m − 2r − 2

]=0

m − 2r

r =0

+ (m − 2 r) (m − 2 r − 1) x or

]=0

r =0

+ (m − 2 r) (m − 2 r − 1) x or

m − 2r

∞

Σ c r [( p − m + 2 r) ( p + m − 2 r + 1) x

m − 2r − 2

]=0

m − 2r − 2

]=0

m − 2r

r =0

+ (m − 2 r) (m − 2 r − 1) x

which is an identity in x. Equating to zero the coefficient of the highest power of x i. e., of x m , we have the indicial equation as c0 ( p − m) ( p + m + 1) = 0 . Now c0 ≠ 0 ,as it is the coefficient of the first term with which we start to write the series. ∴

m = p, − ( p + 1).

Now equating to zero the coefficient of x

∴

m − 2r

, we get

c r ( p − m + 2 r) ( p + m − 2 r + 1) + (m − 2 r + 2) (m − 2 r + 1) c r − 1 = 0 . (m − 2 r + 2) (m − 2 r + 1) cr = − c r −1 . ( p − m + 2 r) ( p + m − 2 r + 1)

Putting r = 1, 2 , … , we get

D-336

c1 = −

m (m − 1) c0 , ( p − m + 2) ( p + m − 1)

c2 = −

(m − 2) (m − 3) c1 ( p − m + 4) ( p + m − 3)

= (− 1)2

m (m − 1) (m − 2) (m − 3) c0 , etc. ( p − m + 2) ( p − m + 4) ( p + m − 1) ( p + m − 3)

Substituting these values in (2), we get  y = c0  x 

m

+ Putting

and

−

m (m − 1) x ( p − m + 2) ( p + m − 1)

m −2

m (m − 1) (m − 2) (m − 3) x ( p − m + 2) ( p − m + 4) ( p + m − 1) ( p + m − 3)

m = p, − ( p + 1) successively, we get p ( p − 1) p −2 p ( p − 1) ( p − 2) ( p − 3)  y = c0  x p − x x + 2 (2 p − 1) 2 . 4 . (2 p − 1) (2 p − 3) 

m −4

p −4

 − … 

 − … 

= au (say), which is one solution of (1) ( p + 1) ( p + 2) − p − 3  y = c0  x − p − 1 + x 2 (2 p + 3)  +

( p + 1) ( p + 2) ( p + 3) ( p + 4) 2 . 4 . (2 p + 3) (2 p + 5)

 x − p − 5 + … 

= bv (say), which is another independent solution of (1). Hence the required general solution of (1) is y = au + bv, where a and b are arbitrary constants. Case III: Roots of the indicial equation differing by an integer and making the coefficients of some powers of x in the series for y infinity: Let m = α and β be two roots of the indicial equation which differ by an integer and some of the coefficients of powers of x in the series for y become infinite for m = β. In this case substitute c (m − β) for c0 . It will give two independent solutions for m = β, namely the modified y and ∂y / ∂m as in case I. Hence, in all we find three solutions : (i) the solution obtained by putting m = α in y, (ii) the solution obtained by putting m = β in the modified y, (iii) the solution obtained by putting m = β in the partial differential coefficient w.r.t. m of the modified y i. e., in ∂y / ∂m. But the solution (i) is a numerical multiple of (ii). Thus only two of these three solutions are independent.

D-337

Example 6 :

Obtain a general solution in series of powers of x of the equation (Bessel’s equation of

order one) d2 y

x2 Solution :

+x

dx 2

Putting y = x

dy + ( x 2 − 1) y = 0 . dx

m

…(1) (Avadh 2006)

in the left hand side of the given equation (1), we get

x2 . m (m − 1) x m − 2 + x . m x m − 1 + ( x2 − 1) x or

(m2 − 1) x

m

+x

m +2

m

.

Obviously the common difference of the powers is 2 . ∴

Let the solution of (1) in a series of ascending powers of x be y = c0 x

and

dy

Σ cr x

m +2

+ c2 x

m +4

+…

m + 2r

…(2)

r =0

∞

= Σ c r (m + 2 r) x

dx

m + 2 r −1

r =0

d2 y dx

+ c1 x

∞

= ∴

m

2

=

∞

Σ c r (m + 2 r) (m + 2 r − 1) x

m + 2r − 2

r =0

.

Putting these values in the given equation (1), we get ∞

Σ c r [ x 2 (m + 2 r) (m + 2 r − 1) x

m + 2r − 2

r =0

+ x . (m + 2 r) x or or

∞

Σ c r [{(m + 2 r)2 − 1} x

r =0 ∞

m + 2r

+x

Σ c r [(m + 2 r + 1) (m + 2 r − 1) x

m + 2 r −1

m + 2r + 2

m + 2r

r =0

+x

+ ( x 2 − 1) x

m + 2r

]=0

]=0 m + 2r + 2

]=0

which is an identity in x . Equating to zero the coefficient of the lowest power of x i. e., of x m , we have the indicial equation as c0 (m + 1) (m − 1) = 0 . Now c0 ≠ 0 ,as it is the coefficient of the first term with which we start to write the series. ∴

m = 1, − 1.

Here the roots of the indicial equation are unequal and the difference of the roots is 2 which is an integer. Now equating to zero the coefficient of the general term i. e., of x or

c p (m + 2 p + 1) (m + 2 p − 1) + c p − 1 = 0 1 cp = − c p −1 . (m + 2 p + 1) (m + 2 p − 1)

m + 2p

, we get

D-338

Putting p = 1, 2 , 3, …… , we get 1 c1 = − c0 , (m + 3) (m + 1) 1 1 c2 = − c1 = (− 1)2 c0 , (m + 5) (m + 3) (m + 5) (m + 3)2 (m + 1) 1 1 c3 = − c2 = (− 1)3 c0 , etc. (m + 7) (m + 5) (m + 7) (m + 5)2 (m + 3)2 (m + 1) Substituting these values in (2), we get  x2 x4 y = c0 x m 1 − + 2  (m + 1) (m + 3) (m + 1) (m + 3) (m + 5) −

x6

 + … (m + 1) (m + 3) (m + 5) (m + 7)  2

2

…(3)

Now if we put m = − 1 in the above series, the coefficients of x 2 and onwards in the bracketed expression become infinite because of the factor (m + 1) in the denominator. To overcome this difficulty, put c (m + 1) for c0 in (3).  1 1 Thus y = c x m (m + 1) − x2 + x4 2 ( m + ) 3 (m + 3) (m + 5)  − Putting

 x 6 + … (m + 3) (m + 5) (m + 7)  1

2

2

m = − 1 in (4), we get  1  1 1 y = c x −1 − x 2 + 2 x4 − 2 2 x 6 + … 2 2 .4 2 .4 .6  

…(4)

…(5)

= au (say), which is one solution of the equation (1). Putting m = 1in (3), we get a series which is a numerical multiple of the series in (5). So this solution is not independent of the solution obtained in (5). Hence so far, by all means we have obtained only one series solution. As in case I, to find the second series solution, students may show that (∂y / ∂m)m = − 1 also satisfies the given differential equation while y is given by (4). Thus (∂y / ∂m)m = − 1 will give the second series solution. Note that substituting (4) in the given differential equation (1) and on simplification, d2 y dy we have x 2 +x + ( x 2 − 1) y = cx m (m + 1)2 (m − 1). dx dx There is a factor (m + 1)2 present on the R.H.S. showing that y as well as ∂y / ∂m satisfy the differential equation when m = − 1. Differentiating (4) partially w.r.t. m, we get ∂y ∂m

= cx

m

  x2 x4 log x . (m + 1) − + − … 2 + 3 m (m + 3) (m + 5)  

D-339 m

+ cx

  x2 2 1   4 − + + …… x 1 +    2  (m + 3)3 (m + 5) (m + 3)2 (m + 5)2   (m + 3) 

Putting m = − 1 in this, we get  ∂y    = cu log x + c x −1  ∂m  m = − 1

  x2 5 4 x + …… = bv (say), − 1 + 4 64  

which is another independent solution of (1). Hence the required general solution of the given equation (1) is y = au + bv, where a and b are arbitrary constants. Case IV: Roots of the indicial equation differing by an integer and making a coefficient of the series for y indeterminate: Let m = α and β (say α > β) be two roots of the indicial equation which differ by an integer. If one of the coefficients of the series for y becomes indeterminate when m = β the complete primitive is obtained by putting m = β in y, which contains two arbitrary constants. The series obained by putting m = α in y is merely a numerical multiple of one of the series contained in the first solution.

Example 7 : Solution :

Solve (1 − x 2 )

and

dx

2

+ 2x

dy dx

+ y = 0.

…(1) (Rohilkhand 2008)

Let the series solution of the given equation (1) be y = c0 x

∴

d2 y

dy

+ c1 x

∞

m +1

= Σ c r (m + r) x

dx

+ c2 x

m +2

∞

+ … = Σ cr x

m+r

…(2)

r =0

m + r −1

r =0

d2 y dx

m

2

∞

= Σ c r (m + r) (m + r − 1) x

m + r −2

r =0

.

Putting these values in the given equation (1), we get ∞

Σ c r [(1 − x 2 ) (m + r) (m + r − 1) x

m + r −2

r =0

+ 2 x (m + r) x or

∞

Σ c r [{− (m + r) (m + r − 1) + 2 (m + r) + 1} x

m + r −1

]=0

m + r −2

]=0

m + r −2

]=0

m+r

r =0

+ (m + r) (m + r − 1) x or

m+r

+x

∞

Σ c r [− {(m + r) (m + r − 3) − 1} x

m+r

r =0

+ (m + r) (m + r − 1) x

D-340

which is an identity in x . Equating to zero the coefficient of the lowest power of x i. e., of x m − 2 , we have the indicial equation as c0 m (m − 1) = 0 . Now c0 ≠ 0 ,as it is the coefficient of the first term with which we start to write the series. m = 0 , 1.

∴

Now equating to zero the coefficient of the next higher power of x i. e.,of x

m −1

, we get …(3)

c1 (m + 1) . m = 0 , which makes c1 indeterminate (0 / 0 ) when m = 0. But if m = 0, from (3) we have the identity c . 0 = 0,

which is true for every value of c1. Hence, in this case c1 can be taken as an arbitrary constant. Further equating to zero the coefficient of the general term i. e., of x

m+p

, we get

− c p {(m + p) (m + p − 3) − 1} + c p + 2 (m + p + 2) (m + p + 1) = 0 (m + p) (m + p − 3) − 1 cp + 2 = cp . (m + p + 1) (m + p + 2)

or

Putting p = 0 , 1, 2 , …… , we get m (m − 3) − 1 (m + 1) (m − 2) − 1 c2 = c0 , c3 = c1 , (m + 1) (m + 2) (m + 2) (m + 3) c4 =

{(m + 2) (m − 1) − 1}{ m (m − 3) − 1} (m + 2) (m − 1) − 1 c2 = c0 , (m + 3) (m + 4) (m + 1) (m + 2) (m + 3) (m + 4)

c5 =

(m + 3) m − 1 {m (m + 3) − 1}{(m + 1) (m − 2) − 1} c3 = c1 , etc. (m + 4) (m + 5) (m + 2) (m + 3) (m + 4) (m + 5)

Substituting these values in (2), we get y = c0 x + c1 x

m

m

  m (m − 3) − 1 2 {(m + 2) (m − 1) − 1} { m (m − 3) − 1} 4 x + x + … 1 + (m + 1) (m + 2) (m + 3) (m + 4)  (m + 1) (m + 2) 

  (m + 1) (m − 2) − 1 3 { m (m + 3) − 1}{(m + 1) (m − 2) − 1} 5 x + x + … ⋅ x + (m + 2) (m + 3) (m + 2) (m + 3) (m + 4) (m + 5)  

Putting m = 0 in this, we get 1 1 1 5  1    y = c0 1 − x 2 + x 4 + … + c1  x − x 3 + x + … 8 2 40  2    1 3 1 5  1 2 1 4    = a 1 − x + x + … + b  x − x + x + … , 8 2 40  2   

…(4)

where a and b are arbitrary constants. This is the required complete primitive of the given equation (1) because it contains two arbitrary constants.

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If we take m = 1, we get from (3) c1 = 0 and hence c3 , c5 ,…… all vanish. In this case we shall get a series which is merely a numerical multiple of the second series in the general solution (4).

12.5 Some Cases of Failure of the Method of Frobenius It is not always necessary that every equation has a convergent series solution in ascending or descending powers of x. There may be cases in which the indicial equation may not have any root or may have one root, which may lead to a divergent series. These cases lead to the failure of the method of Frobenius.

Example 8 :

Transform the Bessel’s equation x2

d2 y 2

dx

+x

dy + ( x 2 − n2 ) y = 0 dx

…(1)

by the substitution x = 1 / z . Hence show that it has no integrals that are regular in descending powers of x . 1 dx 1 Solution : We have x = , so that =− 2 ⋅ z dz z dy dy dz dy Now = ⋅ = − z2 dx dz dx dz and

d2 y dx2

d  2 dy  d  2 dy  dz − z  = − z ⋅ dx  dz  dz  dz  dx

=

 d2 y dy  d2 y dy  (− z 2 ) = z 4 2 z + 2 z3 ⋅ =  − z 2 − 2 2  dz  dz dz dz  Putting these values in the given equation (1), we get the transformed equation as dy  1 1  4 d2 y + z + 2z 3 2  2 dz  z z  dz or

z4

d2 y dz

2

+ z3

 2 dy   1  − n2  y = 0 − z  +   dz   z 2

dy + (1 − n2 z 2 ) y = 0 . dz

…(2)

Let the solution of (2) in ascending powers of z be y= Then and

dy = dz d2 y dz

2

∞

Σ

r =0 ∞

Σ

r =0 ∞

cr z

m+r

.

c r (m + r) z

…(3) m + r −1

= Σ c r (m + r) (m + r − 1) z r =0

m + r −2

.

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Putting these values in (2), we get ∞

Σ c r [(m + r) (m + r − 1) z

m + r +2

r =0

or

∞

Σ c r [{(m + r)2 − n2 } z

m + r +2

r =0

+z

+ (m + r) z m+r

m + r +2

+ (1 − n2 z 2 ) z

m+r

]=0

]=0

which is an identity in z . Equating to zero the coefficient of the lowest power of z i. e., of z m , we have the indicial equation as c0 = 0 , which has no roots as c0 ≠ 0 , it being the coefficient of the first term with which we start to write the series. Hence the differential equation (2) has no regular integrals in ascending powers of z i. e., the given equation (1) has no regular integrals in descending powers of x .

Example 9 :

Transform the equation d2 y dx2

…(1)

− y = 0,

by the substitution x = 1 / z and show that it has no integrals that are regular in descending powers of x . Solution : Proceeding exactly as in Ex. 8 after article 12.5, we get the transformed equation

as

z4

d2 y dz

2

+ 2z 3

dy − y =0 dz

…(2)

Let the solution of (2) in ascending powers of z be ∞

y = Σ cr z

m+r

…(3)

r =0

∴ and

∞ dy = Σ c r (m + r) z dz r = 0

d2 y dz

2

m + r −1

∞

= Σ c r (m + r) (m + r − 1) z

m + r −2

r =0

.

Putting these values in (2), we get ∞

Σ c r [(m + r) (m + r − 1) z

m + r +2

+ 2 (m + r) z

m + r +2

−z

r =0

or

∞

Σ c r [(m + r) (m + r + 1) z

r =0

m+r

m + r +2

−z

m+r

]=0

] = 0,

which is an identity in z. Equating to zero the coefficient of the lowest power of z i. e., of z m , we have the indicial equation as − c0 = 0, which has no roots as c0 ≠ 0 , it being the coefficient of the first term with which we start to write the series. Hence the differential equation (2) has no regular integrals in ascending powers of z i. e., the given equation (1) has no regular integrals in descending powers of x.

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12.6 Method of Differentiation

Example 10 : Solution :

Solve (1 − x 2 )

d2 y 2

dx

−x

dy + m 2 y = 0 , where x = 0 , y = 0 , dy / dx = m. dx

The given equation can be put in the form (1 − x 2 ) y2 − x y1 + m 2 y = 0 .

…(1)

Differentiating the equation (1) n times by Leibnitz’s theorem w.r.t. x, we get (1 − x 2 ) yn + 2 − (2 n + 1) xyn + 1 + (m 2 − n 2 ) yn = 0

…(2)

Putting x = 0 in (1) and (2), we get ( y2 )0 = − m 2 ( y)0 = 0 ( yn + 2 )0 = (n 2 − m 2 ) ( yn)0

and

…(3)

Putting n = 2 , 4, … in (3), we get ( y4 )0 = ( y6 )0 = … = 0 . Putting n = 1, 3, 5, … in (3), we get ( y3 )0 = (12 − m 2 ) ( y1)0 = (12 − m 2 ) m ( y5 )0 = (32 − m 2 ) ( y3 )0 = (32 − m 2 ) (12 − m 2 ) m, etc. By Maclaurin’s Theorem, we have y = ( y)0 +

x x2 x3 ( y1)0 + ( y2 )0 + ( y3 )0 + … 1! 2! 3!

Putting the values of ( y)0 , ( y1)0 , … , we get y = mx + (12 − m2 ) m

x3 x5 + (32 − m2 ) (12 − m 2 ) m + …. 3! 5!

12.7 The Particular Integral If in a linear equation, right hand side is of the form x m , then the usual method to determine the particular integral corresponding to x m , is followed.

Example 11 : Solution :

Solve x4

d2 y 2

dx

+x

Substituting y = x

m

dy + y = x −1. dx in the left hand side of the given equation, we have

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x 4 . m (m − 1) x or

m (m − 1) x

m +2

m −2

+ x.m x

+ (m + 1) x

m −1

+x

m

m

i. e.,

the common difference of the powers is (m + 2) − m = 2 .

∴

Let us assume

∞

y = Σ Ar x

m − 2r

…(1)

r =0

as the C.F. of the given equation i. e., the solution of the equation x4

d2 y dx2

+x

dy + y = 0. dx

…(2)

Then substituting the value of y from (1) in (2), we get ∞

Σ [(m − 2 r) (m − 2 r − 1) Ar x

m − 2r + 2

r =0

+ (m − 2 r + 1) Ar x

m − 2r

]=0

which is an identity. ∴

Equating to zero the coefficient of the highest power of x i. e., of x m (m − 1) A0 = 0 .

[∵

m = 0 , 1.

∴

Now equating to zero the coefficient of the general term i. e., of x

or Case I:

m −2 r

, we get A0 ≠ 0 ]

, we get

(m − 2 r − 2) (m − 2 r − 3) Ar + 1 + (m − 2 r + 1) Ar = 0 (m − 2 r + 1) Ar + 1 = − Ar . (m − 2 r − 3) (m − 2 r − 2) When m = 0 , Ar + 1 =

m +2

…(3)

2r − 1 Ar . (2 r + 2) (2 r + 3)

…(4)

Now putting r = 0 , 1, 2 , 3, … in (4), we get 1 1 1 1 A1 = − A0 = − A0 , A2 = A1 = − A0 , 2 .3 3! 4 .5 5!

∴

A3 =

1. 3 3 A2 = − A0 , etc. 6 .7 7!

y=

Σ Ar x

∞

m − 2r

r =0

=

∞

Σ Ar x

= A0 + A1 x −2 + A2 x  1 −2 1 = A 1 − x − x 5!  3! Case II:

− 2r

[ ∵ m = 0]

r =0

−4

+ A3 x

−4

−

−6

+…

1. 3 − 6  x … ⋅ 7! 

When m = 1, from (3), we have Ar +1 =

[Taking A0 = A]

r −1 Ar . (r + 1) (2 r + 1)

…(5)

Now putting r = 0, we have A1 = − A0 , putting r = 1, we have A2 = 0 , putting r = 2 , 3 , … , we have A3 = A4 = … = 0 (each). ∴

y=

∞

Σ Ar x

m − 2r

r =0

= A0 x + A1 x

−1

=

∞

Σ Ar x1 − 2 r

r =0

+ A2 x

−3

+…

[ ∵ m = 1]

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1 1   = A0  x −  = B  x −  , taking   x x ∴

the C.F. of the equation is  1 −2 1 A 1 − x − x 3! 5! 

−4

−

1. 3 x 7!

A0 = B.

 1  … + B  x −  ⋅  x 

−6

To find P.I. To find the P.I., put y = C0 x m . Thus we must have m (m − 1) C 0 x m= −3

⇒ ∴

Let

y=

m +2

and

=x

C0 =

∞

m − 2r

Σ Cr x

r =0

−1

m (m − 1) C 0 = 1 and

⇒

m + 2 = −1

1 ⋅ 12 ∞

= Σ Cr x

− 3 − 2r

r =0

be the P.I. of the given equation. ∞

Substituting y = Σ Cr x − 3 − 2 r in (2) and proceeding as above, we get r =0

Cr + 1 =

2 (r + 1) Cr . (2 r + 5) (2 r + 6)

Putting r = 0 , 1, 2 , … C1 = ∴

2 .4 2 4 C 0 , C2 = C1 = C 0 , etc. 5 .6 7 .8 5 .6 .7 .8 ∞

P.I. = Σ Cr x − 3 − 2 r = C 0 x

−3

r =0

= C0 x

−3

1 x 12

−3

=

= 2x

−3

[1 +

1 2 x  + 4 ! 6 ! 

−5

−2

+

2 .4 x 5 .6 .7 .8

−2

+

2 .4 x 5 .6 .7 .8

2 x 5 .6

 2 x 1 + 5 .6 

+ C1 x

−2

+

2 .4 x 8!

−4

+ C2 x −4

−4

−7

+…

+…  + … 

 + … ⋅ 

Hence the solution of the given equation is  1 y = A 1 − x ! 3 

−2

−

1 x 5!

−4

−

1. 3 x 7!

−6

 1  … + B  x −   x 

+ 2x

−3

1 2  + x 4 ! 6 !

Comprehensive Exercise 1 Solve completely in series the following equations : 1.

(1 − x 2 ) y ′ ′ − 2 xy ′ + p ( p + 1) y = 0 .

−2

+

2 .4 x 8!

−4

 + … ⋅ 

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2. 3.

y ′ ′ + ( x − 1)2 y ′ − 4 ( x − 1) y = 0 about x = 1. d2 y

(x − x 2)

+ (1 − 5 x)

d2 y

dy

dx

4.

(2 x + x 3 )

5.

2x 2

d2 y dx

2

2

−x

2

6.

x2

d y 2

dx

−

dx 2

dy dx

dy

+x

dx

d y

(2 + x 2 )

8.

(x − x 2)

9.

9 x (1 − x)

10.

d2 y dx 2

+ (1 − x 2 ) y = x 2 .

dy

+x

2

+ (1 − x)

dx d2 y dx

2

− 4 y = 0.

dx

− 12

(Agra 2008)

− 6 xy = 0 .

dx

2

dx d2 y

dx

+ ( x2 − 4) y = 0 .

2

7.

dy

(Avadh 2006)

(Purvanchal 2010)

+ (1 + x) y = 0 .

dy dx

dy dx

− y = 0.

+ 4 y = 0.

− y = x.

A nswers 1 1.

2. 3.

p ( p + 1) 2 ( p − 2) p ( p + 1) ( p + 3) 4   x − … y = c0 1 − x + 4! 2!   ( p − 1) ( p + 2) 3 ( p − 3) ( p − 1) ( p + 2) ( p + 4) 5   + c1  x − x + x − … ⋅ 3! 5!   1 1 1  2   3 6 9 4 ( x − 1) − ( x − 1) + … + b ( x − 1) + ( x − 1)  ⋅ y = a 1 + ( x − 1) + 45 1620 4  3    y = au + bv, where u = 1 + 22 x + 32 x2 + 42 x3 + … and v = u log x − 2 [1 . 2 x + 2 . 3 x2 + 3 . 4 x3 + …].

4.

3 1 6   y = a 1 + 3 x2 + x4 − x + … 5 15    3  1. 3 4 1. 3 . 5 6 + bx3 /2 1 + x2 − x + x − … ⋅ 8 . 16 8 . 16 . 24  8 

5.

y = au + bv + f ( x), where   x2 x4 x6 u = x 1 + + + + …, 2 . 5 2 . 4 . 5 . 9 2 . 4 . 6 . 5 . 9 . 13     x2 x4 x6 v = x1 /2 1 + + + + …,  2 . 3 2 . 4 . 3 . 7 2 . 4 . 6 . 3 . 7 . 11 

D-347

f ( x) = 6.

1 2 1 1 4 1 1 x + ⋅ x + ⋅ x6 + … . 3 3 3 .7 3 3 . 5 . 7 . 11

y = au + bv, where   x4 x6 x8 u = x −2 − 2 + 3 − 3 2 + …,  2 . 4 2 . 4 . 6 2 . 4 . 6 . 8 

7. 8.

  x2 x4 v = u log x + x −2 1 + 2 + 2 2 + … ⋅ 2 .4  2  1 3 5 4  1 1 4  1   y = a 1 − x2 − x + x … + b  x − x3 − x + … ⋅ 4 12 96 6 24     y = au + bv, where

9.

u = 1+ x +

y = au + bv, where

10.

14 3 2 2 2 .5 3   x + x + … , v = u log x +  −2 x − x2 − x − … ⋅   27 4 4 .9

u = 1+

1. 4 2 1. 4 . 7 3 1 x+ x + x + …, 3 3 .6 3 .6 .9

  8 . 11 2 8 . 11 . 14 3 8 v = x7 /3 1 + x+ x + x + … . 10 10 . 13 10 . 13 . 16   1 1 4 1 1 5     y = a 1 + x2 + x + … + b  x + x3 + x + …     2 24 6 120 1 5  1 x + … ⋅ +  x3 +  6 120

Objective Type Questions Fill in the Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

The point x = 0 is called an …… of the differential equation d2 y dy + P ( x) + Q ( x) y = 0 , 2 dx dx if P ( x) and Q ( x) do not become infinite in a neighbourhood of the origin and they can be expanded in the form of a power series.

2. The origin is a …… point if both xP ( x) and x2 Q ( x) can be expanded in a power series of x in the neighbourhood of x = 0. 3. The method of finding a solution in a series for the linear differential equation d2 y 2

+ P ( x)

dy + Q ( x) y = 0 dx

dx near a regular singular point or near an ordinary point is called …… .

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4. 5.

6. 7.

In Question 1 if x = 0 is not an ordinary point, it is called a …… point. The singular points are of two types : …… points and …… points d2 y dy The indicial equation of x 2 + + xy = 0 is … dx dx The indicial equation of (2 x + x3 )

d2 y dx2

−

dy − 6 xy = 0 is … dx

True or False Write ‘ T ’ for true and ‘F’ for false statement. 1.

x = 0 is an ordinary point of (1 − x 2 ) y ′ ′ − 2 xy ′ + p ( p + 1) y = 0 .

2.

x = 0 is a singular point of

d2 y 2

dx

− 2x 2

dy + 4 xy = x 2 + 2 x + 2. dx

A nswers Fill in the Blank(s) 1. 3. 5. 6.

ordinary point. Frobenius method. regular singular, irregular singular. c0 m2 = 0 .

2. 4.

regular singular. singular.

7.

c0 m (2 m − 3) = 0 .

2.

F.

True or False 1.

T.

o

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13 L egendre's F unctions

13.1 Legendre’s Equation. (Avadh 2006, 08)

T

he differential equation of the form dy d2 y (1 − x 2 ) 2 − 2 x + n (n + 1) y = 0 dx dx

is called Legendre’s differential equation (or Legendre’s equation), where n is a constant. dy  d  2 This equation can also be written as (1 − x )  + n (n + 1) y = 0 . dx  dx 

13.2 Solution of Legendre’s Equation (Avadh 2013; Bundelkhand 13)

The Legendre’s equation is (1 − x 2 )

d2 y 2

dx

− 2x

dy + n (n + 1) y = 0 . dx

...(1)

D-350

It can be solved in a series of ascending or descending powers of x. The solution in descending powers of x is more important than the one in ascending powers. Let us assume,

∞

y = Σ ar x

k−r

r =0

, a0 ≠ 0 .

∞ dy = Σ ar (k − r) x dx r = 0

∴

d2 y

and

dx

2

k − r −1

∞

= Σ ar (k − r) (k − r − 1) x

k − r −2

r =0

.

Substituting in (1), we have ∞

(1 − x 2 ) Σ ar (k − r) (k − r − 1) x

k − r −2

r =0

∞

−2 x Σ

r =0

ar (k − r) x

k − r −1

∞

+ n (n + 1) Σ

ar x

r =0

∞

or

Σ ar [(k − r) (k − r − 1) x

k−r

=0

k − r −2

r =0

k−r

]=0

+ { n (n + 1) − (k − r) (k − r + 1)}x

k−r

]=0

+ { n2 − (k − r)2 + n − (k − r)} x

k−r

]=0

+ { n (n + 1) − (k − r) (k − r − 1) − 2 (k − r)} x ∞

or

Σ ar [(k − r) (k − r − 1) x

k − r −2

r =0 ∞

or

Σ

r =0

ar [(k − r)(k − r − 1) x

∞

or

Σ ar [(k − r) (k − r − 1) x

k − r −2

k − r −2

r =0

+ (n − k + r) (n + k − r + 1) x

k−r

] = 0. ...(2)

Now (2) being an identity, we can equate to zero the coefficients of various powers of x. ∴

Equating to zero the coefficient of the highest power of x, i. e. of x k , we have a0 (n − k ) (n + k + 1) = 0 .

Now a0 ≠ 0 , as it is the coefficient of the first term with which we start to write the series. ∴ or

k=n   k = − (n − 1) 

...(3)

Equating to zero the coefficient of the next lower power of x i. e. of x

k −1

, we have

a1 (n − k + 1) (n + k ) = 0 . ∴ a1 = 0, since neither (n − k + 1) nor (n + k ) is zero by virtue of (3). Again equating to zero the coefficient of the general term i. e. of x

∴

k−r

, we have

ar − 2 (k − r + 2) (k − r + 1) + (n − k + r) (n + k − r + 1) ar = 0 . (k − r + 2) (k − r + 1) ar = − ar − 2 . (n − k + r) (n + k − r + 1)

...(4)

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(k − 1) (k − 2) a1 = 0 , since a1 = 0 . (n − k + 3) (n + k − 2)

Putting r = 3, a3 = − ∴

we have, a1 = a3 = a5 = … = 0 (each).

Now two cases arise : Case I:

When k = n, from (4), we have (n − r + 2) (n − r + 1) ar = − ar − 2 . r . (2 n − r + 1)

Putting r = 2, 4,… etc. a2 = −

n (n − 1) . a0 , 2 (2 n − 1)

a4 = −

(n − 2) (n − 3) n (n − 1) (n − 2) (n − 3) . a2 = . a0 , etc. 4 . (2 n − 3) 2. 4 . (2 n − 1) (2 n − 3) n

∴

y = a0 x

or

 y = a0  x 

+ a2 x

n

−

n−2

+ a4 x

n (n − 1) x 2 . (2 n − 1)

n−4

n−2

+

+ .... n (n − 1) (n − 2) (n − 3) x 2 . 4 (2 n − 1) (2 n − 3)

n−4

 − … ,  ...(5)

which is one solution of Legendre’s equation. Case II:

When k = − (n + 1), from (4), we have (n + r − 1) (n + r) ar = ar − 2 . r (2 n + r + 1)

Putting r = 2, 4,… etc. (n + 1) (n + 2) a2 = . a0 2 (2 n + 3) a4 = ∴

y=

(n + 3) (n + 4) (n + 1) (n + 2) (n + 3) (n + 4) . a2 = . a0 etc. 4 . (2 n + 5) 2. 4 . (2 n + 3) (2 n + 5) ∞

Σ

r =0

ar x

− n −1 − r

= a0 x − n − 1 + a2 x − n − 3 + a4 x − n − 5 + . . .

 (n + 1) (n + 2) − n − 3 = a0  x − n − 1 + x 2. (2 n + 3)   (n + 1) (n + 2) (n + 3) (n + 4) − n − 5  x + … + 2.4. (2 n + 3) (2 n + 5)  

...(6)

which is other solution of Legendre’s equation.

13.3 Definition of Pn (x) (Legendre’s Polynomial) and Q n (x) The solution of Legendre’s equation is called Legendre’s function. 1. 3 . 5. … (2 n − 1) When n is a positive integer and a0 = , n!

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the solution (5) of article 13.2 is denoted by Pn ( x) and is called Legendre’s function of the first kind. ∴

1.3.5 … (2 n − 1) n!

Pn ( x) =

 x 

n

−

n (n − 1) x 2. (2 n − 1)

n−2

 n (n − 1) (n − 2) (n − 3) x +  2.4. (2 n − 1) (2 n − 3)

n−4

 − … . 

Pn ( x) is a terminating series and gives what are called Legendre’s Polynomials for different values of n. (n / 2 )

We can write Pn ( x) = Σ

r =0

(− 1)r

(2 n − 2 r) ! n

2 r !(n − 2 r) !(n − r) !

x

n − 2r

(Avadh 2006; Rohilkhand 10)

where

 n / 2, if n is even (n / 2) =  (n − 1) / 2, if n is odd

Again when n is a positive integer and a0 =

n! , 1.3.5 … (2 n + 1)

the solution (6) of 13.2 is denoted by Qn ( x) and is called the Legendre’s function of the second kind. ∴

Qn ( x) =

n! 1.3.5. … (2 n + 1)

 − n − 1 (n + 1) (n + 2) − n − 3 + x x 2 (2 n + 3)  +

(n + 1) (n + 2) (n + 3) (n + 4) x 2.4 (2 n + 3) (2 n + 5)

− n−5

 + .... . 

Qn ( x) is an infinite or non-terminating series as n is positive.

13.4 General Solution of Legendre’s Equation The most general solution of the Legendre’s equation is y = APn ( x) + BQn ( x), where A and B are arbitrary constants.

13.5 Pn (x) is the Coefficient of h n in the Expansion of (1 − 2 xh + h 2 ) − 1/2 in Ascending Powers of h (Rohilkhand 2002; Avadh 07, 09; Lucknow 11; Purvanchal 10)

We have

(1 − 2 x h + h2 )−1 /2 = {(1 − h (2 x − h)}−1 /2 1 1.3 2 h (2 x − h) + h (2 x − h)2 + … 2 2.4 1.3. .... (2 n − 3) n − 1 1.3 … (2 n − 1) n + h (2 x − h)n − 1 + h (2 x − h)n + … 2.4 … (2 n − 2) 2.4. … (2 n)

= 1+

D-353

∴

Coefficient of h n =

1.3 … (2 n − 1) 1.3 … (2 n − 3) . (2 x)n − 2.4. … 2 n 2.4. … (2 n − 2) +

=

1.3. … (2 n − 1) n  2 x 2.4. … 2 n 

n

n −1

C1 (2 x)n − 2

1.3 . . . (2 n − 5) 2.4. … (2 n − 4)

n−2

C2 (2 x)n − 4 − …

2n x n−2 (n − 1) 2 2n − 1 2

−

  2 n (2 n − 2) (n − 2) (n − 3) x n − 4 . . 4 − … + 2! 2  (2 n − 1) (2 n − 3)  =

1.3. … (2 n − 1) n!

 x 

n

−

2n x n−2 . (n − 1) . 2 2n − 1 2   2 n (2 n − 2) (n − 2) (n − 3) x n − 4 . . 4 − … + ( 2 n 1 ) ( 2 n 3 ) 2 ! − − 2  

=

1.3. … (2 n − 1)  x n! 

n

−

n (n − 1) x 2. (2 n − 1)

n−2

+

n (n − 1) (n − 2) (n − 3) .x 2.4. (2 n − 1) (2 n − 3)

n−4

 − … 

= Pn ( x). Thus we can say that ∞

Σ

n=0

h n Pn ( x) = (1 − 2 xh + h2 ) − 1 /2 , where P0 ( x) = 1.

2 − 1 /2

Note. (1 − 2 xh + h )

(Meerut 2009)

is called the generating function of the Legendre polynomials.

13.6 Laplace’s Definite Integrals For Pn (x) (I) Laplace's First Integral for Pn (x): When n is a positive integer, 1 π Pn ( x) = ∫ [ x ± √ ( x 2 − 1) cos φ]n dφ. π 0 (Rohilkhand 2004, 14; Meerut 06, 08; Agra 10; Lucknow 07; Avadh 14)

Proof:

Putting

From integral calculus, we have π dφ π 2 2 ∫0 a ± b cos φ = √ (a 2 − b 2 ) , where a > b . a = 1 − hx and b = h √ ( x 2 − 1) so that a 2 − b 2 = (1 − hx)2 − h 2 ( x 2 − 1) = 1 − 2 xh + h 2 ,

we have

π (1 − 2 xh + h2 ) − 1 /2 =

π

∫0

[1 − hx ± h √ ( x 2 − 1) cos φ] − 1 dφ

D-354

or

π

[1 − h {x + √ ( x 2 − 1) cos φ)}] − 1 dφ

π

[1 − ht] − 1 dφ , where t = x + √ ( x 2 − 1) cos φ

=

∫0

=

∫0

π

∞

h n Pn ( x) =

Σ

n=0

π

(1 + ht + h 2 t 2 + … + h nt

∫0

n

+ ...) dφ.

Equating the coefficients of h n on both sides, we have πPn ( x) = Pn ( x) =

∴

t

n

dφ =

π

[ x + √ ( x 2 − 1) cos φ] n dφ.

∫0

1 π [ x ± √ ( x 2 − 1) cos φ] n dφ. π ∫0

Putting x = cos θ, we have 1 π Pn (cos θ) = ∫ (cos θ ± i sin θ cos φ) n dφ. π 0

Note:

(II)

π

∫0

Laplace’s Second Integral for Pn (x) : When n is a positive integer, 1 π dφ Pn ( x) = ∫ . 2 0 π [ x ± √ ( x − 1) cos φ]n + 1 (Meerut 2007; Avadh 08; Kanpur 07)

Proof:

Putting

From integral calculus, we have π dφ π 2 2 ∫0 a ± b cos φ = √ (a2 − b2 ) , where a > b . a = xh − 1 and b = h √ ( x 2 − 1) so that a 2 − b 2 = 1 − 2 xh + h 2 ,

we have

π (1 − 2 xh + h 2 ) − 1 /2 =

or

π h

or

π h =

1 1  1 − 2 x h + h2  ∞

Σ

n=0 π

∫0

1 ht

1 h

n

− 1 /2

Pn ( x) =

1  1 − ht 

π

[− 1 + xh ± h √ ( x 2 − 1) cos φ] − 1 dφ

∫0

π

∫0

π

(ht − 1) − 1 dφ , where t = x ± √ ( x 2 − 1) cos φ

∫0

−1

dφ =

π

∫0

1 1 1  + 2 2 + 3 3 ht h t h t  ∞  1 π 1  = Σ  n +1 ∫ dφ n + 1 0 n = 0 h t  =

π

∫0

∞  1 = Σ  n +1 n = 0 h 

∴

{ x ± √ (x

Equating the coefficients of π Pn( x) =

π

∫0

 1 1 1 + 2 2 + ... + n 1 + ht h t h t   1 + . . . + n + 1 n + 1 + . . . dφ h t  1 ht

dφ

π

∫0

[h{ x ± √ ( x 2 − 1) cos φ} − 1] − 1 dφ

=

2

− 1) cos φ}

1 , we have h n +1 dφ

{x ± √ ( x 2 − 1) cos φ} n + 1



n +1 



.

n

 + . . .  dφ 

D-355

or

Pn( x) =

Note.

1 π

dφ

π

∫0

{x ± √ ( x 2 − 1) cos φ} n + 1

.

Replacing n by − (n + 1) in Laplace’s second integral, we have P− (n + 1) x =

1 π

π

∫0

{x ± √ ( x 2 − 1) cos φ} n dφ

= Pn( x), from Laplace’s first integral. Hence

Example 1 :

(ii)

P− (n + 1) = Pn .

Show that (i)

Pn(1) = 1,

Pn(− x) = (− 1)n Pn( x).

(Meerut 2007, 11; Rohilkhand 11; Kanpur 12; Purvanchal 10) (Meerut 2009; Agra 03, 10; Lucknow 10)

n

Hence deduce that Pn(− 1) = (− 1) .

(Kanpur 2007, 12)

(iii) Prove that Pn( x) is an even or odd function of x according as n is even or odd respectively. Solution :

(i) We know that ∞

Σ h n Pn( x) = (1 − 2 xh + h 2 ) − 1 /2 .

n=0

∞

Putting x = 1 on both sides, we have Σ h n Pn(1) = (1 − 2 h + h 2 ) − 1 /2 = (1 − h) − 1 n=0

= 1 + h + h2 + . . . + h n + . . . =

∞

Σ

n=0

h n.

Equating the coefficients of h n on both sides, we have Pn(1) = 1. ∞

(ii) We have, (1 − 2 xh + h2 ) − 1 /2 = Σ h n Pn ( x).

...(1)

n=0

Replacing h by − h on both sides of (1), we have ∞

∞

n=0

n=0

(1 + 2 xh + h2 )−1 /2 = Σ (− h) n Pn ( x) = Σ (− 1) n h n Pn ( x).

...(2)

Again replacing x by − x on both sides of (1), we have ∞

(1 + 2 xh + h 2 ) − 1 /2 = Σ h n Pn (− x). n=0

...(3)

From (2) and (3), we have ∞

∞

Σ h n Pn(− x) = Σ (− 1) n h n Pn ( x).

n=0

n=0

n

Equating the coefficients of h on both sides, we have Pn ( − x) = ( − 1) n Pn ( x). Deduction: Putting x = 1, we have Pn (− 1) = (− 1)n Pn (1) = (− 1) n.

[ ∵ Pn(1) = 1. See Ex. 1]

D-356

(iii) From part (ii) of this question, we have Pn(− x) = (− 1)n Pn( x).

...(4)

If n is even then (− 1) n = 1. ∴

From (4), if n is even, we have Pn(− x) = Pn( x).

Hence Pn( x) is an even function of x, if n is even. Again if n is odd, then (− 1) n = − 1. ∴

From (4), if n is odd, we have Pn(− x) = − Pn( x).

Hence Pn( x) is an odd function of x, if n is odd. Example 2 :

Prove that Pn(0 ) = 0 , for n odd n /2

and

Pn(0 ) =

Solution :

(− 1)

n!

n

2 {(n / 2) !}2

(Agra 2006)

, for n even.

∞

We know that Σ h n Pn( x) = (1 − 2 xh + h2 ) − 1 /2 .

...(1)

n=0

Putting x = 0 on both sides of (1), we have ∞

Σ h n Pn(0 ) = (1 + h2 ) − 1 /2 = {1 − (− h2 )} − 1 /2

n=0

= 1+

1 1.3 1.3.5 ⋅ (− h2 ) + (− h2 )2 + (− h2 )3 2 2.4 2.4.6 13 . .5 . . . (2 r − 1) + ... + (− h2 )r + . . . . 2.4 . . . 2 r

. . . (2)

We observe that all the powers of h on the R.H.S. of (2) are even. Therefore equating the coefficients of h n on both sides of (2), we have Pn (0 ) = 0 , if n is odd. Again equating the coefficients of h2 m on both sides of (2), we have 1.3.5 . . . (2 m − 1) (2 m) ! P2 m (0 ) = (− 1) m = (− 1) m 2 m 2.4.6 . . . 2 m 2 (m !)2 i. e.

when n = 2 m, we have Pn(0 ) =

Example 3 : Solution :

Show that

(−1)n /2 n !

1 − z2

Proved.

.

2 n{(n / 2) !}2 ∞

2 3 /2

(1 − 2 xz + z )

= Σ (2 n + 1) Pn( x) z n. n=0

(Agra 2003; Meerut 10)

∞

We have (1 − 2 x z + z 2 ) − 1 /2 = Σ z

n

n=0

...(1)

Pn( x).

Differentiating w.r.t. z, we have ∞

( x − z )(1 − 2 xz + z 2 ) − 3 /2 = Σ n z n=0

∴

∞

n −1

Pn ( x).

2 ( x − z ) z (1 − 2 xz + z 2 ) − 3 /2 = Σ 2 n z n=0

n

Pn ( x).

...(2)

D-357

Adding (1) and (2), we have 1 − 2 xz + z

2

+ 2 ( x − z )z 2 3 /2

(1 − 2 xz + z ) 1− z

or

2

(1 − 2 xz + z )

n

n=0

∞

2 3 /2

Example 4 :

∞

= Σ (2 n + 1) z

= Σ (2 n + 1) z n=0

n

Pn ( x)

Pn ( x).

Proved.

Express Pn (cos θ) as a series in cosines of multiples of θ. ∞

We have Σ h n Pn ( x) = (1 − 2 xh + h2 ) − 1 /2 .

Solution :

n=0

…(1)

Putting x = cos θ in (1), we get ∞

Σ h n Pn (cos θ) = (1 − 2 h cos θ + h2 ) − 1 /2

n=0

= {1 − h (e iθ + e − iθ ) + h2 } − 1 /2 = (1 − he iθ )−1 /2 (1 − he − iθ ) − 1 /2   1 . 3 2 2 iθ 1 . 3 . 5 … (2 n − 1) n niθ 1 = 1 + he iθ + +…+ h e h e + … 2 .4 2 . 4 . 6 … 2n  2    1 . 3 2 − 2 iθ 1 . 3 . 5 … (2 n − 1) n − niθ 1 × 1 + he − iθ + +…+ h e h e + … ⋅ 2 .4 2 . 4 . 6 … 2n  2  Equating the coefficients of h n from both sides, we get 1 . 3 . 5 … (2 n − 1)  niθ 1 2n − niθ Pn (cos θ) = { e(n − 2) iθ + e − (n − 2) iθ } )+ ⋅ (e + e 2 . 4 . 6 … 2n  2 2n − 1 + =

 1. 3 2 n (2 n − 2) ⋅ { e(n − 4) iθ + e − (n − 4) iθ } + … 2 . 4 (2 n − 1) (2 n − 3) 

1 . 3 . 5 … (2 n − 1)  1 2n cos (n − 2) θ 2 cos nθ + 2 ⋅ 2 . 4 . 6 … 2n  2 2n − 1 +2⋅

 1. 3 2 n (2 n − 2) ⋅ cos (n − 4) θ + … ⋅ 2 . 4 (2 n − 1) (2 n − 3) 

13.7 Orthogonal Properties of Legendre’s Polynomials (i)

(ii)

+1

∫ −1

+1

∫ −1

Proof: (i)

Pm ( x) Pn ( x ) dx = 0 if m ≠ n.

(Rohilkhand 2009; Agra 06, 08, 09, 10; Meerut 09, 10, 11; Avadh 10; Kanpur 07, 08, 10; Lucknow 10, 11; Purvanchal 10)

[ Pn ( x )]2 dx =

2 . 2 n+1

(Agra 2009, 10; Rohilkhand 09, 10; Kanpur 10; Purvanchal 10)

Legendre’s equation may be written as d  2 dy  (1 − x )  + n(n + 1) y = 0 . dx  dx 

Since Pn ( x) is a solution of Legendre’s equation, therefore

D-358

d dx d dx

Similarly,

 2 dPn  (1 − x )  + n(n + 1) Pn = 0 . dx    2 dPm  (1 − x )  + m (m + 1) Pm = 0 . dx  

...(1) ...(2)

Multiplying (1) by Pm and (2) by Pn and then subtracting, we have d  d  2 dPn  2 dPm  Pm (1 − x )  − Pn (1 − x )  dx  dx  dx  dx  +{ n(n + 1) − m(m + 1)}Pn Pm = 0. Integrating between the limits –1 to 1, we have 1  1  d  d 2 dP   ∫ −1  Pm dx (1 − x ) dxn   dx − ∫ −1  Pn dx Integrating by parts, we have  2 dPn   Pm (1 − x ) dx 

 2 dPm   (1 − x )  dx dx   

+ { n(n + 1) − m(m + 1)}∫ +1

− −1

+1

+ −1

+1

∫ −1

+ [n (n + 1) − m (m + 1)] ∫

Hence,

{ n (n + 1) − m (m + 1)} ∫ +1

∫ −1

+1 −1

−1

Pm Pn dx = 0 .

dPm  2 dPn  (1 − x )  dx dx  dx 

+1

∫ −1

dP   −  Pn(1 − x 2 ) m  dx  

∴

1

+1 −1

dPn dx

 2 dPm  (1 − x )  dx dx  

Pm Pn dx = 0 .

Pm Pn dx = 0 .

Pm ( x) Pn ( x) dx = 0 since m ≠ n.

(ii) We have

∞

(1 − 2 xh + h2 ) − 1 /2 = Σ h n Pn( x). n=0

Squaring both sides, we have ∞

(1 − 2 xh + h2 ) − 1 = Σ h2 n {Pn ( x)}2 + n=0

∞

Σ

m, n = 0

h m + n Pm ( x) Pn( x).

m≠n

Integrating between the limits –1 to +1, we have ∞

+1

h2 n [ Pn ( x)]2 dx +

n = 0 ∫ −1

Σ

∞

+1

m, n = 0 ∫ −1

Σ

h m + n Pm ( x) Pn ( x) dx

m≠n

= or

∞

Σ

n=0

+1

∫ −1

h2 n [ Pn( x)]2 dx =

+1

∫ −1

dx (1 − 2 xh + h2 )

+1

∫ −1

dx (1 − 2 xh + h2 )

,

since other integrals on the L.H.S. are zero by (i) as m ≠ n +1 1 1 =− log (1 − 2 xh + h2 ) =− { log (1 − h)2 − log (1 + h)2 } −1 2h 2h

[

]

D-359

=

1 + 1  log  2h 1 − 

2 1 + h  1   = log  h  h 1 − 

 h 2  h3 h5 + + ...   = h + 3 5 h h  

  h2 h4 h2 n = 2 1 + + + ... + + ...  = 3 5 2 n +1  

∞

Σ

n=0

2 h2 n . 2 n +1

Equating the coefficients of h2 n, we have +1 2 2 ∫ −1 [ Pn( x)] dx = 2 n + 1 ⋅ Remark :

Making use of the Kronecker delta, the above results (i) and (ii) can be

written in compact form as 1

∫ −1

Pm ( x) Pn( x) dx =

2 δm n , 2 n +1

(Avadh 2008; Lucknow 09)

where Kronecker delta, δ m n , is defined as 0 , if m ≠ n δm n =  1, if m = n.

13.8 Recurrence Formulae (I) (2 n + 1) x Pn = ( n + 1) Pn + 1 + n Pn − 1.

(Meerut 2006, 08; Agra 02, 06; Rohilkhand 06; Avadh 06, 13; Lucknow 09, 11; Purvanchal 09, 11)

Proof:

∞

We have (1 − 2 xh + h2 ) − 1 /2 = Σ h n Pn( x). n=0

Differentiating both sides w.r.t. ‘h’, we have ∞ 1 − (1 − 2 xh + h2 ) − 3 /2 (−2 x + 2 h) = Σ nh n − 1 Pn( x) n=0 2 ∞

or

( x − h)(1 − 2 xh + h2 ) − 1 /2 = (1 − 2 xh + h2 ) Σ nh n − 1 Pn( x)

or

( x − h) Σ h n Pn( x) = (1 − 2 xh + h2 ) Σ nh n − 1 Pn( x)

or

n=0

∞

∞

n=0

n=0

( x − h)[ P0 ( x) + hP1( x) + . . . + h

n −1

Pn − 1( x) + h n Pn( x) + . . . ]

= (1 − 2 xh + h2 )[ P1( x) + 2 hP2 ( x) + . . . + (n − 1)h n − 2 Pn − 1( x) + nh n − 1 Pn( x) + (n + 1)h n Pn + 1( x) + . . . ] Equating the coefficients of h n on both sides of (1), we have xPn( x) − Pn − 1( x) = (n + 1) Pn + 1( x) − 2 xn Pn( x) + (n − 1) Pn − 1( x) or

(2 n + 1) x Pn( x) = (n + 1) Pn + 1( x) + nPn − 1( x).

In short (2 n + 1) x Pn = ( n + 1) Pn + 1 + nPn − 1 .

... (1)

D-360

Equating the coefficients of h n −1 on both sides of (1), we get

Note.

xPn − 1( x) − Pn − 2 ( x) = nPn( x) − 2 x (n − 1) Pn − 1( x) + (n − 2) Pn − 2 ( x) or (II)

nPn = (2 n − 1) x Pn − 1 − ( n − 1) Pn − 2 . n Pn = x Pn ′ − Pn − 1 ′ , where dashes denote differentiation w.r.t. ‘x’. (Agra 2003; Avadh 07; Lucknow 06; Kanpur 08, 11, 12)

Proof:

We have (1 − 2 xh + h 2 )

− 1 /2

∞

= Σ

n=0

h

n

...(1)

Pn( x)

Differentiating (1) w.r.t. ‘h’, we have ∞

( x − h)(1 − 2 xh + h 2 ) − 3 /2 = Σ nh n − 1 Pn( x). n=0

...(2)

Again differentiating (1) w.r.t. ‘x’, we have ∞

h (1 − 2 xh + h 2 ) − 3 /2 = Σ h n Pn ′ ( x) n=0

or

∞

h ( x − h)(1 − 2 xh + h2 ) − 3 /2 = ( x − h) Σ h n Pn ′ ( x). n=0

...(3)

From (2) and (3) , we have h or

∞

Σ

n=0

nh n − 1 Pn( x) = ( x − h)

∞

Σ

n=0

h n Pn ′ ( x)

h [h0 P1 ( x) + 2 hP2 ( x) + . . . + nh n − 1 Pn ( x) + . . . ] = ( x − h)[ P0 ′ ( x) + hP1 ′ ( x) + . . . + h n − 1 Pn − 1 ′ ( x) + h n Pn ′ ( x) + . . . ].

Equating the coefficients of h n on both sides, we have nPn( x) = xP ′ n ( x) − P ′ n −1 ( x). In short nPn = x P ′ n − P ′ n − 1. (III) Proof:

(2 n + 1) Pn = P ′ n + 1 − P ′ n − 1 .

(Lucknow 2008; Avadh 12)

From recurrence formula I, we have (2 n + 1) x Pn = (n + 1) Pn + 1 + nPn − 1.

Differentiating w.r.t. x, we have (2 n + 1) x P ′ n + (2 n + 1) Pn = (n + 1) P ′ n + 1 + nPn − 1 ′ .

...(1)

From recurrence formula II, we have xP ′ n = nPn + P ′ n −1.

...(2)

Eliminating xPn ′ from (1) and (2), we have (2 n + 1)(nPn + P ′ n − 1 ) + (2 n + 1) Pn = (n + 1) P ′ n + 1 + nP ′ n − 1 or

(2 n + 1)(n + 1) Pn = (n + 1) P ′ n + 1 + nP ′ n − 1 − (2 n + 1) P ′ n − 1

or

(2 n + 1)(n + 1) Pn = (n + 1) P ′ n + 1 − (n + 1) P ′ n − 1 .

∴

(2 n + 1) Pn = P ′ n + 1 − P ′ n − 1.

(IV) Proof:

( n + 1) Pn = P ′ n + 1 − x P ′ n .

(Rohilkhand 2006, 11; Kanpur 10; Purvanchal 14)

Writing recurrence formulae II and III, we have nPn = xP ′ n − P ′ n −1

...(1)

D-361

and

...(2)

(2 n + 1) Pn = P ′ n + 1 − P ′ n − 1 .

Subtracting (1) from (2), we have ( n + 1) Pn = P ′ n + 1 − x P ′ n . (V)

2

(1 − x ) P ′ n = n ( Pn − 1 − x Pn ).

Proof:

(Roholkhand 2006, 11; Kanpur 10)

Replacing n by (n − 1) in recurrence formula IV, we have nPn − 1 = P ′ n − xP ′ n − 1.

...(1)

Writing recurrence formula II, we have ...(2)

nPn = xP ′ n − P ′ n −1 . Multiplying (2) by x and then subtracting from (1), we have n( Pn − 1 − xPn) = (1 − x2 ) P ′ n (1 − x 2 ) P ′ n = n ( Pn − 1 − x Pn ).

i. e. Aliter:

From Laplace’s first integral, we have 1 π Pn( x) = ∫ {x + √ ( x2 − 1) cos φ}n dφ. π 0

Replacing n by (n − 1), we have 1 π Pn − 1 ( x) = ∫ {x + √ ( x2 − 1) cos φ} n − 1 dφ. π 0 π 1 ∴ Pn − 1 − xPn = ∫ [{x + √ ( x2 − 1) cos φ} n − 1 − x{x + √ ( x2 − 1) cos φ}n ] dφ π 0 1 π = ∫ {x + √ ( x2 − 1) cos φ} n − 1[1 − x{x + √ ( x2 − 1) cos φ}] dφ π 0 1 π = ∫ {x + √ ( x2 − 1) cos φ} n − 1[(1 − x2 ) − x √ ( x2 − 1) cos φ] dφ π 0 =−

( x2 − 1) π 2 n −1 ∫0 {x + √ ( x − 1) cos φ} π

=−

( x2 − 1) π 2 ∫0 { x + √ ( x − 1) cos φ} π

=−

( x2 − 1) π  1 d n 2 ∫0  n dx {x + √ ( x − 1) cos φ}  dφ π

=

[

(1 − x2 ) d πn dx

π

∫0

(1 − x2 ) d {πPn( x)} πn dx (1 − x2 ) = ⋅ Pn ′ ( x). n Hence (1 − x 2 ) Pn ′ = n ( Pn − 1 − x Pn ).

Proof:

n −1

×

 φ dφ 

d { x + √ ( x2 − 1) cos φ}] dφ dx

{x + √ ( x2 − 1) cos φ} n dφ

=

(VI)

x  cos 1 + 2 √ ( x − 1) 

(1 − x 2 ) Pn ′ = ( n + 1) ( x Pn − Pn + 1 ) Writing recurrence formula I, we have

[from Laplace’s first integral]

(Agra 2011)

D-362

...(1)

(2 n + 1) x Pn = (n + 1) Pn + 1 + n Pn − 1 which may be written as (n + 1) x Pn + n x Pn = (n + 1) Pn + 1 + n Pn − 1 or

(n + 1)( x Pn − Pn + 1) = n ( Pn − 1 − x Pn).

Writing recurrence formula V, we have (1 − x 2 ) Pn ′ = n( Pn − 1 − xPn).

...(2)

From (1) and (2), we have (1 − x 2 ) Pn ′ = ( n + 1) ( x Pn − Pn + 1 ). Aliter:

∴

From Laplace’s second integral, we have 1 π dφ Pn( x) = ∫ . π 0 [ x + √ ( x2 − 1) cos φ]n + 1

xPn − Pn + 1 =

x π

1 π 1 = π =

dφ

π

∫0

2

[ x + √ ( x − 1) cos φ]

−

1 π

dφ

π

∫0

2

[ x + √ ( x − 1) cos φ]n + 2

π

{x + √ ( x2 − 1) cos φ} − n − 2 [ x{x + √ ( x2 − 1) cos φ} − 1] dφ

π

{x + √ ( x2 − 1) cos φ} − n − 2{( x2 − 1) + x √ ( x2 − 1) cos φ} dφ

∫0 ∫0

=

( x2 − 1) π

∫0

=

( x2 − 1) π

∫0

=−

n +1

( x2 − 1) π

π

 x {x + √ ( x2 − 1) cos φ} − n − 2 1 + cos 2 √ ( x − 1) 

 φ dφ 

π

[{x + √ ( x2 − 1) cos φ} − n − 2 ⋅ dxd {x + √ ( x2 − 1) cos φ}] dφ  1 d  {x + √ ( x2 − 1) cos φ} − n − 1 dφ  1 n + dx  

π

∫0

=

(1 − x2 ) d π(n + 1) dx

=

(1 − x2 ) d (1 − x2 ) Pn ′ {π Pn( x)} = . π(n + 1) dx n +1

π

∫0

{x + √ ( x2 − 1) cos φ} − n − 1 dφ

Hence (1 − x 2 ) Pn ′ = ( n + 1) ( x Pn − Pn + 1 ).

13.9 Beltrami’s Result To prove that Proof:

and

(2 n + 1)( x2 − 1) Pn ′ = n(n + 1)( Pn + 1 − Pn − 1).

From recurrence formulae V and VI, we have (1 − x 2 ) Pn ′ = n( Pn − 1 − xPn)

...(1)

(1 − x 2 ) Pn ′ = (n + 1)( xPn − Pn + 1).

...(2)

Multiplying (1) by n + 1, (2) by n and adding, we get

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[(n + 1) + n](1 − x 2 ) Pn ′ = n(n + 1)( Pn − 1 − Pn + 1) or

(2 n + 1)( x 2 − 1) Pn ′ = n(n + 1)( Pn + 1 − Pn − 1).

This result is known as Beltrami’s result.

13.10 Christoffel’s Expansion To prove that Pn ′ = (2 n − 1) Pn − 1 + (2 n − 5) Pn − 3 + (2 n − 9) Pn − 5 + . . . , the last term of the series being 3 P1 or P0 according as n is even or odd. Proof. From Recurrence formula III, we have ...(A)

P ′ n + 1 = (2 n + 1) Pn + P ′ n − 1 . Replacing n by (n − 1), we have Pn ′ = (2 n − 1) Pn − 1 + P ′ n − 2 .

...(1)

Replacing n by (n − 2), (n − 4), ... in (i), we have P ′ n − 2 = (2 n − 5) Pn − 3 + P ′ n − 4

...(2)

P ′ n − 4 = (2 n − 9) Pn − 5 + P ′ n − 6

...(3)

.... .... .... .... .... .... .... P2 ′ = 3 P1 + P0 ′ , when n is even. Adding (1), (2), (3), etc., we have when n is even Pn ′ = (2 n − 1) Pn − 1 + (2 n − 5) Pn − 3 + (2 n − 9) Pn − 5 + . . . + 3 P1 + P0 ′ = (2 n − 1) Pn − 1 + (2 n − 5) Pn − 3 + (2 n − 9) Pn − 5 + . . . + 3 P1

[∵

P0 ′ = 0 ].

Again when n is odd, the last of the above relations is [∵ P1 ′ = 1 = P0 ]

P3 ′ = 5 P2 + P1 ′ = 5 P2 + P0 . Adding as before, we have when n is odd Pn ′ = (2 n − 1) Pn − 1 + (2 n − 5) Pn − 3 + . . . + 5 P2 + P0 . Hence

Pn ′ = (2 n − 1) Pn − 1 + (2 n − 5) Pn − 3 + (2 n − 9) Pn − 5 + . . .

the last term of the series being 3 P1 or P0 according as n is even or odd.

13.11 Christoffel’s Summation Formula n

To prove that Σ

r =0

Proof: and

(2 r + 1) Pr ( x) Pr ( y) = (n + 1)

Pn + 1 ( x) Pn ( y) − Pn + 1 ( y) Pn (x) ( x − y)

.

From Recurrence formula I, we have (2 r + 1) x Pr ( x) = (r + 1) Pr + 1 ( x) + rPr − 1 ( x)

...(1)

(2 r + 1) y Pr ( y) = (r + 1) Pr + 1 ( y) + r Pr − 1 ( y).

...(2)

Multiplying (1) by Pr ( y) and (2) by Pr ( x) and then subtracting, we have (2 r + 1)( x − y) Pr ( x) Pr ( y) = (r + 1) [ Pr + 1 ( x) Pr ( y) − Pr + 1 ( y) Pr ( x)] − r [ Pr − 1 ( y) Pr ( x) − Pr − 1 ( x) Pr ( y)]. ...(3) Now

( x − y) P0 ( x) P0 ( y) = P1 ( x) P0 ( y) − P1 ( y) P0 ( x)

. . . (A0 )

[∵ P0 ( x) = 1 = P0 ( y), and P1 ( x) = x, P1 ( y) = y]

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Again putting r = 1, 2 , 3, . . . , (n − 1), n in (3), we have 3( x − y) P1 ( x) P1 ( y) = 2 [ P2 ( x) P1 ( y) − P2 ( y) P1 ( x)] −1 . [ P0 ( y) P1 ( x) − P0 ( x) P1 ( y)]

. . . (A1)

5 ( x − y) P2 ( x) P2 ( y) = 3 [ P3 ( x) P2 ( y) − P3 ( y) P2 ( x)] − 2 . [ P1 ( y) P2 ( x) − P1 ( x) P2 ( y)] .... ....

.... ....

.... ....

.... ....

.... ....

.... ....

.... ....

.... ....

.... ....

.... ....

.... ....

.... ....

.... ....

. . . (A2 ) .... ....

.... ....

(2 n − 1)( x − y) Pn − 1 ( x) Pn − 1 ( y) = n . [ Pn ( x) . Pn − 1 ( y) − Pn ( y) Pn − 1 ( x)] − (n − 1) [ Pn − 2 ( y) Pn − 1 ( x) − Pn − 2 ( x) Pn − 1 ( y)]

. . . (A n − 1)

(2 n + 1)( x − y) Pn ( x) Pn ( y) = (n + 1 )[ Pn + 1 ( x) Pn ( y) − Pn + 1 ( y) Pn ( x)] − n [ Pn − 1 ( y) Pn ( x) − Pn − 1 ( x) Pn ( y)]

. . . (A n )

Adding (A 0 ), (A1 ), (A 2 ), . . . , (A n -1 ) and (A n ), we have n

( x − y) Σ (2 r + 1) Pr ( x) Pr ( y) r =0

Hence

n

Σ

r =0

= (n + 1) [ Pn + 1 ( x) Pn ( y) − Pn + 1 ( y) Pn ( x)]. Pn + 1( x) Pn( y) − Pn + 1( y) Pn(x) (2 r + 1) Pr ( x) Pr ( y) = (n + 1) . ( x − y)

This is Christoffel’s Summation Formula.

Example 5 :

Show that xPn ′ ( x) = nPn ( x) + (2 n − 3) Pn − 2 ( x) + (2 n − 7) Pn − 4 ( x) + …

and hence deduce the following : 1 2n (i) ∫ x Pn Pn ′ dx = ⋅ −1 2n + 1 Solution :

(ii)

1

∫ −1

x Pn Pm ′ dx = either 0 , 2 or

2n ⋅ 2n + 1

From recurrence formula II, we have …(1)

xPn ′ = nPn + P ′ n −1 Also from recurrence formula III, we have P ′ n + 1 = (2 n + 1) Pn + P ′ n − 1 .

…(2)

In (2) replacing n by (n − 2), (n − 4), (n − 6) and so on, we get P ′ n − 1 = (2 n − 3) Pn − 2 + P ′ n − 3

…(3)

P ′ n − 3 = (2 n − 7) Pn − 4 + P ′ n − 5

…(4)

P ′ n − 5 = (2 n − 11) Pn − 6 + P ′ n − 7

…(5)

…

…

…

…

…

Adding (1), (3), (4), (5) etc., we get xPn ′ = nPn + (2 n − 3) Pn − 2 + (2 n − 7) Pn − 4 + …

…(6)

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Deduction (i): Multiplying both sides of (6) by Pn and then integrating w.r.t. 'x' between the limits − 1 to 1, we get 1

∫ −1

xPn Pn ′ dx = n∫

1 −1

Pn2 dx, the other integrals being zero as 1

Pm Pn dx = 0 , if m ≠ n

∫ −1 = n⋅ Deduction (ii):

2 2n = ⋅ 2n + 1 2n + 1

From (6), we have

xPm ′ = mPm + (2 m − 3) Pm − 2 + (2 m − 7) Pm − 4 + … 1

∫ −1

⇒

xPn Pm ′ dx = m ∫

1 −1

Pn Pm dx + (2 m − 3)∫

1 −1

Pn Pm − 2 dx

+ (2 m − 7)∫

1 −1

Pn Pm − 4 dx + …

Now if m and n are distinct positive integers, then 1

∫ −1

xPn Pm ′ dx = 0 , since all the integrals on the R.H.S. vanish. 1

Again if n = m, then

∫ −1

xPn Pm ′ dx = n∫

being zero

1 −1

= n⋅

If n = m − 2, then

1

∫ −1

Pn2 dx, the other integrals on the R.H.S. 2 2n = 2n + 1 2n + 1

xPn Pm ′ dx = 0 + (2 m − 3)∫

1 −1

= {2 (n + 2) − 3} ⋅ If n = m − 4, then

1

∫ −1

2 = 2. 2n + 1

xPn Pm ′ dx = 0 + 0 + (2 m − 7)∫ = {2 (n + 4) − 7}

Pn2 dx + 0 + …

1 −1

Pn2 dx + 0 + …

2 = 2. 2n + 1

In the same way, if n = (m − 6), (m − 8) etc., the value of the integral is 2. Example 6 :

Prove that P02 ( x) + 3 P12 ( x) + 5 P22 ( x) + . . . + (2 n + 1) Pn2 ( x) = (n + 1)[ Pn( x) Pn + 1 ′ ( x) − Pn + 1( x) Pn ′ ( x)] = (n + 1)2 Pn2 ( x) + (1 − x2 ){Pn ′ ( x)}2 .

Solution :

From Christoffel’s summation formula, we have ( x − y)

n

Σ

r =0

(2 r + 1) Pr ( x) Pr ( y)

= (n + 1) [ Pn + 1 ( x) Pn ( y) − Pn + 1 ( y) Pn ( x)]. Putting y = x + h, where h is a small quantity, we have n

− h Σ (2 r + 1) Pr ( x) Pr ( x + h) = (n + 1)[ Pn + 1( x) Pn( x + h) − Pn + 1( x + h) Pn( x)]. r =0

D-366

Expanding by Taylor’s theorem, we have n

−h Σ

r =0

(2 r + 1) Pr ( x){Pr ( x) + hPr ′ ( x) + . . . }

   h2 Pn ′ ′ ( x) + . . .  = (n + 1)  Pn + 1( x) Pn( x) + hPn ′ ( x) + 2 !       h2 −  Pn + 1( x) + hP ′ n + 1 ( x) + Pn + 1 ′ ′ ( x) + . . .  Pn( x) 2!     = − h(n + 1)[{Pn( x) P ′ n + 1 ( x) − Pn ′ ( x) Pn + 1( x)} + h(. . .) + . . . ] or

n

Σ (2 r + 1) Pr ( x){Pr ( x) + hPr ′ ( x) + . . . }

r =0

= (n + 1)[ Pn( x) P ′ n + 1 ( x) − Pn ′ ( x) Pn + 1( x) + h(. . .) + . . . ]. Taking limit as h → 0, we have n

2

Σ

r =0

or

(2 r + 1) Pr ( x) = (n + 1)[ Pn( x) P ′ n + 1 ( x) − Pn ′ ( x) Pn + 1( x)]

2

2

2

2

P0 ( x) + 3 P1 ( x) + 5 P2 ( x) + . . . + (2 n + 1) Pn ( x) = (n + 1)[ Pn( x) P ′ n + 1 ( x) − Pn + 1( x) Pn ′ ( x)].

Again

2

2

2

Proved.

2

(n + 1) Pn ( x) + (1 − x ){Pn ′ ( x)}

= (n + 1) Pn( x)[(n + 1) Pn( x)] + Pn ′ ( x)[(1 − x2 ) Pn ′ ( x)] = (n + 1) Pn( x)[ P ′ n + 1 ( x) − xPn ′ ( x)] + Pn ′ ( x)[(n + 1){ xPn( x) − Pn +1( x)}], from recurrence formula IV and VI = (n + 1)[ Pn( x) P ′ n + 1 ( x) − Pn + 1( x) Pn ′ ( x)]. +1 2 2 n(n + 1) Example 7 : Prove that ∫ x Pn + 1 Pn − 1 dx = . −1 (2 n − 1)(2 n + 1)(2 n + 3) Solution :

From Recurrence formula I, we have (Rohilkhand 2011; Purvanchal 10) (2 n + 1) x Pn = (n + 1) Pn + 1 + nPn − 1.

Replacing n by (n − 1) and (n + 1) successively, we have and

(2 n − 1) xPn − 1 = nPn + (n − 1) Pn − 2

...(1)

(2 n + 3) xPn + 1 = (n + 2) Pn + 2 + (n + 1) Pn .

...(2)

Multiplying (1) and (2), we get (2 n − 1)(2 n + 3) x2 Pn + 1 Pn − 1 = n(n + 1) Pn2 + n(n + 2) Pn Pn + 2 + (n − 1)(n + 2) Pn − 2 Pn + 2 + (n − 1)(n + 1) Pn − 2 Pn. Integrating between the limits –1 to +1, we have (2 n − 1)(2 n + 3)∫

+1 −1

x2 Pn + 1 Pn − 1 dx = n(n + 1)∫

+1 −1

Pn

2

dx,

all other integrals being zero by article 13.7 (i) 2 [By article 13.7] = n (n + 1) ⋅ (2 n + 1)

D-367 +1

∫ −1

∴ Example 8 :

x2 Pn + 1 Pn − 1 dx =

Prove that

+1

∫−1

2 n(n + 1) . (2 n − 1)(2 n + 1)(2 n + 3)

( Pn ′ )2 dx = n(n + 1).

From Christoffel’s expansion, we have

Solution :

...(1)

Pn ′ = (2 n − 1) Pn − 1 + (2 n − 5) Pn − 3 + (2 n − 9) Pn − 5 + . . . the last term is P0 i. e. 1 or 3 P1 i. e. 3 x according as n is odd or even. When n is odd, let m be the number of terms on the R.H.S. of (1). Then the last coeff. 1 = (2 n − 1) + (m − 1) ⋅ (−4); n +1 ∴ m= ⋅ 2 Again when n is even, let m′ be the number of terms on the R.H.S. of (1). Then the last coeff. 3 = (2 n − 1) + (m ′ − 1)(−4). ∴

m ′ = n / 2.

Now

( Pn ′ )2 = (2 n − 1)2 Pn − 1 + (2 n − 5)2 Pn − 3 + . . .

2

2

+ 2(2 n − 1)(2 n − 5) Pn − 1 Pn − 3 + . . .. +1

∫ −1

∴

( Pn ′ )2 dx = (2 n − 1)2

+1

∫ −1

Pn2− 1 dx + (2 n − 5)2

+1

∫ −1

2

Pn − 3 dx + … ,

the other integrals are zero by article 13.7 (i) 2 2 = (2 n − 1)2 + (2 n − 5)2 ⋅ 2(n − 1) + 1 2(n − 3) + 1 + (2 n − 9)2 = 2 [(2 n − 1) + (2 n − 5) + (2 n − 9) + . . . ],

2 + ... 2(n − 5) + 1 ...(2)

the last term of the A.P. on the R.H.S. of (2) being 1 or 3 according as n is odd or even. Case I: When n is even, the number of terms in the arithmetic progression on the R.H.S. of (2) is n / 2. +1

∫ −1

∴

( Pn ′ )2 dx = 2 ⋅ =

Case II:

n/2 ⋅ [first term + last term], summing the A.P. 2

n [(2 n − 1) + 3] = n (n + 1). 2

When n is odd, the number of terms in the A.P. on the R.H.S. of (2) is

( n + 1) / 2. +1

∫ −1

∴

Hence

+1

∫ −1

(n + 1) / 2 ⋅ [first term + last term] 2 n +1 = ⋅ [(2 n − 1) + 1] = n (n + 1). 2

( Pn ′ )2 dx = 2 ⋅

( Pn ′ )2 dx = n(n + 1).

Proved.

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13.12 Rodrigue’s Formula To prove that

Pn ( x) =

1 n !2

⋅ n

dn dx n

( x2 − 1)n.

(Meerut 2002, 07, 09, 11, 13; Avadh 11; Rohilkhand 02, 04, 07, 11; Lucknow 06, 08, 10; Agra 03, 11; Kanpur 08; Purvanchal 09, 11, 14)

Proof:

Let

y = ( x2 − 1)n.

Differentiating ,

dy dx

= n( x2 − 1)n − 1 . 2 x.

( x2 − 1)

∴

dy dx

= 2 nxy.

Differentiating both sides (n + 1) times by Leibnitz’s Theorem, we have ( x2 − 1)

dn + 2 y

+ (n + 1) .

dx n + 2

dn +1 y dx n + 1

⋅2x +

n (n + 1)n d y ⋅ ⋅2 2! dx n

 dn +1 y dn y  = 2 n x ⋅ + + 1 ⋅ 1 n ( ) dx n + 1 dx n   or

( x2 − 1)

or

(1 − x2 )

Putting

dn y dx n

dn + 2 y

+2 x

dx n + 2 dn + 2 y

−2 x

dx n + 2

dn +1 y dx n + 1 dn +1 y dx n + 1

− n(n + 1) + n(n + 1)

dn y dx n dn y dx n

=0 = 0.

= z , the above equation becomes (1 − x2 )

d2 z dx2

− 2x

dz + n(n + 1) z = 0 , dx

which is Legendre’s equation. Hence its solution is z = cPn( x), where c is a constant. ∴

dn y dx n

Putting x = 1, we have Now

...(1)

= c Pn ( x ).  dn y  , since Pn (1) = 1. c =  n   dx  x =1

y = ( x2 − 1)n = ( x + 1)n ⋅ ( x − 1)n.

Differentiating n times by Leibnitz’s theorem, we have dn y dx

n

= ( x − 1)n ⋅

 dn −1  ( x + 1)n + n ⋅  n − 1 ( x + 1)n ⋅ n( x − 1)n − 1 + . . . dx  dx  dn

n

n −1 dn d  d + n  ( x + 1)n n − 1 ( x − 1)n + ( x + 1)n ⋅ n ( x − 1)n  dx  dx dx

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n! ( x + 1)n( x − 1)n − 1 + . . . 1!

= ( x − 1)n n ! + n ⋅

+ n ⋅ n( x + 1)n − 1

Putting

 dn y x = 1,  n  dx 

n! ( x − 1) + ( x + 1)n ⋅ n !. 1!

  = (1 + 1)n ⋅ n ! = 2 n ⋅ n ! = c .   x =1

∴

From (1), we have Pn( x) =

or

Pn ( x ) =

n 1d y c dx n

dn

1

( x 2 − 1) n .

2 n ⋅ n ! dx n

This is Rodrigue’s formula. Corollary:

Show that

1

∫0

x

m

Pn ( x) dx =

m (m − 1) (m − 2) … (m − n + 2) , (m + n + 1) (m + n − 1) … (m − n + 3)

when m > n − 1 and n is a positive integer. By Rodrigue’s formula, we have Pn ( x) = ∴

1

x

∫0

m

2 . n ! dx

n

Pn ( x) dx =

 x n 2 .n!  1

=

dn

1 n

m

( x2 − 1)n. 1

2

n

dn −1 dx

n −1

1

x

. n ! ∫0

dn

m

dx

n

( x2 − 1)n dx

1

 1 ( x2 − 1)n − m ∫ x 0 0

d n −1

m −1

dx

n −1

 ( x2 − 1)n dx , 

integrating by parts 0=

(− 1) m 2

n

1

. n ! ∫0

x

m −1

d

n −1

dx

n −1

( x2 − 1)n dx .

[The first term vanishes in the given limits.] Again integrating the R.H.S. by parts, we get 1

∫0

x

m

Pn ( x) dx =

(− 1)2 m (m − 1) 2 n.n!

1

∫0

x

m −2

d n−2 dx

n−2

( x2 − 1)n dx .

Proceeding in the same way, we get 1

∫0 = =

x

m

Pn ( x) dx =

(− 1)n . m (m − 1) … (m − n + 1)

m (m − 1)…(m − n + 1) 2 n.n!

n

2 .n!

m (m − 1) … (m − n + 1) 2

n +1

.n!

1

∫0

1

∫0

x

m−n

1

∫0

x

m−n

( x2 − 1)n dx

(1 − x 2 )n dx

( x2 )(m − n − 1) /2 (1 − x 2 ) n 2 x dx ,

D-370

=

m (m − 1) … (m − n + 1) 2n +1 . n !

1

∫0

z

(m − n − 1) /2

(1 − z )n dz , putting x2 = z , 2 x dx = dz

= =

m (m − 1) … (m − n + 1) 2

n +1

.n!

m (m − 1) … (m − n + 1) 2

n +1

.n!

m − n − 1  ⋅B  + 1, n + 1 2   m − n + 1  ⋅B  , n + 1 2  

 m − n + 1 Γ  Γ (n + 1)   2 m (m − 1) … (m − n + 1) = . 2n +1 . n !  m + n + 3 Γ    2

=

= =

m (m − 1) … (m − n + 1) 2n +1 . n ! m (m − 1) … (m − n + 1) 2n +1 . n !

 Γ (m) Γ (n)  ∵ B (m, n) =  Γ (m + n)  

 m − n + 1 Γ  n!   2 ⋅  m + n + 1  m + n − 1  m − n + 3   m − n + 1     ...             2 2 2 2 ⋅

2 n +1 . n ! (m + n + 1) (m + n − 1) … (m − n + 3) (m − n + 1)

m (m − 1) (m − 2) … (m − n + 2) ⋅ (m + n + 1) (m + n − 1) … (m − n + 3)

Case I:

…(1)

When n is even. 1

∫0

x

m

Pn ( x) dx =

m (m − 2) (m − 4) … (m − n + 2) ⋅ (m + n + 1) (m + n − 1) … (m + 1)

…(2)

Since for n = 2 , the last factor i. e.,(m − n + 3) in the Dr. is (m + 1) and its preceding terms are (m − 1), (m − 3) etc., which cancel out with the corresponding factors in the Nr. Case II:

When n is odd. 1

∫0

x

m

Pn ( x) dx =

(m − 1) (m − 3) … (m − n + 2) ⋅ (m + n + 1) (m + n − 1) … (m + 2)

…(3)

Since for n = 1, the last factor i. e.,(m − n + 3) in the Dr. is (m + 2) and its preceding terms are m, (m − 2) etc. which cancel out with the corresponding factors in the Nr.

Example 9 :

(i) and (ii)

+1

∫ −1

+1

∫ −1

Prove that Pn ( x) dx = 0 , n ≠ 0 P0 ( x) dx = 2 .

(Kanpur 2012; Rohilkhand 14)

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(i) From Rodrigue’s formula, we have

Solution :

1

Pn ( x) = +1

2

d n −1 dx

n −1

( x2 − 1)n .

1

+1

n ! ∫ −1

n

dx

n

( x2 − 1)n dx

 d n −1 2  ( x − 1)n   n n −1 2 n !  dx  d n −1

( x2 − 1)n =

dn

1

=

Now

n

2 n n ! dx

Pn ( x) dx =

∫ −1

∴

dn

⋅

dx

n −1

= ( x + 1)n

+1

...(1) −1

{( x + 1)n ( x − 1)n } d n −1 dx

n −1

( x − 1)n + (n − 1) n ( x + 1)n − 1 + … + ( x − 1)n

d n−2 dx

n−2

d n −1 dx

n −1

( x − 1)n ( x + 1)n,

by Leibnitz’s theorem of differential calculus n! n! = ( x + 1)n ( x − 1) + n (n − 1) ( x + 1)n − 1 ( x − 1)2 + … 1! 2! … + ( x − 1)n n !( x + 1)   n! r n ( x + a)n − r   ∵ D ( x + a) = (n − r) !   when x = − 1 or 1,

=0

since each term contains both x + 1 and x − 1 as factors. ∴

from (1),

+1

∫ −1

Pn ( x) dx = 0 .

(ii) We know that P0 ( x) = 1. +1

∫ −1

∴

P0 ( x) dx =

+1

∫ −1

dx

+1

= [ x]− 1 = 2 . Example 10 :

Let

1

∫ −1

x 4 P6 ( x) dx .

(Meerut 2006, 13B)

We know that Pn ( x) is a polynomial in x of degree n . Therefore we can

Solution :

express x

Evaluate

4

as a linear combination of P0 , P1 , P2 , P3 and P4 . x 4 = c0 P0 + c1 P1 + c2 P2 + c3 P3 + c4 P4 , where c0 , c1 , c2 , c3 , c4 are some constants.

Then

1

I =

∫ −1

=

∫ −1

1

x 4 P6 ( x) dx (c0 P0 + c1 P1 + c2 P2 + c3 P3 + c4 P4 ) P6 dx

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= c0

1

P0 P6 dx + c1

∫ −1

+ c3 = 0, since

1

∫ −1

1

∫ −1 1

∫ −1

P1 P6 dx + c2

P3 P6 dx + c4

1

∫ −1 1

∫ −1

P2 P6 dx

P4 P6 dx

Pm Pn dx = 0 if m ≠ n .

Comprehensive Exercise 1 1. Show that P0 ( x) = 1, P1( x) = x, P2 ( x) = (3 x2 − 1) / 2, P3 ( x) = (5 x3 − 3 x) / 2, P4 ( x) = (35 x4 − 30 x2 + 3) / 8. 4

3

(Kanpur 2008; Lucknow 08; Avadh 14)

2

2.

Express P( x) = x + 2 x + 2 x − x − 3 in terms of Legendre’s polynomials.

3.

(i)

Prove that Pn ′ − P ′ n − 2 = (2 n − 1) Pn − 1 . [Hint : Replace n by (n − 1) in recurrence formula (III)].

(ii) Prove that x P9 ′ = P8 ′ + 9 P9 .

(Meerut 2008)

[Hint : Put n = 9 in recurrence formula II]. 4.

Prove that (1 − 2 xz + z 2 )−1 /2 is a solution of the equation

5.

(i)

∂2 (zv)

 2 ∂v  (1 − x )  = 0 . ∂x   ∞ 1+ z 1 Prove that − = Σ [ Pn ( x) + Pn + 1( x)] z n. 2 z √ (1 − 2 x z + z ) z n = 0

z

∂z 2

+

∂ ∂x

(ii) Prove that n

P ′ n + 1 + Pn ′ = P0 + 3 P1 + 5 P2 + … + (2 n + 1) Pn = Σ

(2 r + 1) Pr ( x).

r =0

6.

Prove that (i) and (ii)

1

∫x

Pn( x) dx = [ Pn + 1( x) − Pn − 1( x)] / (2 n + 1) + C,

∫

Pn( x) dx = [ Pn − 1( x) − Pn + 1( x)] / (2 n + 1).

[Hint : Integrate recurrence formula III]. 7.

(i)

Prove that ∫

(ii) Show that ∫ [Hint: 1

∫ −1

+1 −1 1 −1

( x2 − 1) Pn + 1 Pn ′ dx =

2 n(n + 1) . (2 n + 1)(2 n + 3)

xPn( x) Pn − 1( x) dx = 2 n / (4 n2 − 1).

(Agra 2007) (Rohilkhand 2004; Purvanchal 07)

Using recurrence formula I, we have n +1 1 xPn( x) Pn − 1( x) dx = Pn + 1( x) Pn − 1( x) dx 2 n + 1 ∫ −1 1 n + [ Pn − 1( x)]2 dx. 2 n + 1 ∫ −1

D-373

8.

9.

(i)

∫ −1

(ii) Prove that

∫ −1

(i)

Pn( x)

1

Prove that

√ (1 − 2 xh + h2 )

+1

(1 − x2 )( Pn ′ )2 dx =

+1

Prove that

dx =

∫ −1

2hn . 2n + 1

2 n(n + 1) . 2n + 1

(Agra 2010)

(1 − x2 ) Pm ′ Pn ′ dx = 0

where m and n are distinct positive integers. 1 2 n(n + 1) (ii) Prove that ∫ (1 − x2 ) Pm ′ Pn ′ dx = δ mn , −1 2n + 1 where δ mn is Kronecker delta. 10.

1

Prove that (i)

x

∫ −1

m

(Lucknow 2006)

Pn( x) dx = 0

where m and n are integers and m < n. and 11.

1

(ii)

∫ −1

x n Pn( x) dx =

Prove that (i) Pn ′ (1) =

2 n + 1.(n !)2 ⋅ (2 n + 1) !

1 n(n + 1) 2

(Lucknow 2008)

1 (ii) Pn ′ (−1) = (−1)n − 1 ⋅ n (n + 1). 2

and

12. Prove that  1  1  1  1  1  1  1 Pn  −  = P0  −  P2 n   + P1  −  P2 n − 1   + … + P2 n  −  P0   ⋅  2  2  2  2  2  2  2 1 1 + sin θ 1 1 1 2 ⋅ 13. Prove that 1 + P1 (cos θ) + P2 (cos θ) + P3 (cos θ) + … = log 1 2 3 4 sin θ 2 π 1 . 3 . 5 … (2 n − 1) 14. Show that ∫ Pn (cos θ) cos nθ dθ = (Avadh 2012) π. 0 2 . 4 . 6 … 2n π

1 1  Pn (cos θ) cos nθ dθ = B  n + ,  ⋅  2 2

15.

Show that

16.

Prove that if n is odd 1

∫0 17.

∫0

Pn ( x) dx =

(− 1)(n − 1) /2 (n − 1) ! n

2 . {(n + 1) / 2} ! {(n − 1) / 2}!

(Avadh 2012)

⋅ (Avadh 2008)

Prove that 1

∫ −1

x2 Pn2 ( x) dx =

1 3 1 + + ⋅ 8 (2 n − 1) 4 (2 n + 1) 8 (2 n + 3)

18.

Show that all the roots of Pn ( x) = 0 are real and lie between − 1 and 1.

19.

Show that all the roots of Pn ( x) = 0 are distinct.

D-374

A nswers 1 2.

P( x) =

8 4 40 1 224 P4 ( x) + P3 ( x) + P2 ( x) + P1( x) − P0 ( x). 35 5 21 5 105

Objective Type Questions Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1. The value of the integral ∫

2.

3.

4.

5.

6.

1 −1

[ Pn ( x)]2 dx is

(a) 1

(b) 0

2 (c) 2 n +1

(d) 2 n2 . (Rohilkhand 2011)

The value of xPn ′ ( x) − P ′ n − 1 ( x) is (a) nPn ( x)

(b) n2

(c) Pn − 1 ′ ( x)

(d) None of these.

The value of P ′ n + 1 ( x) − P ′ n − 1 ( x) is (a) Pn ( x)

(b) (2 n + 1) Pn ( x)

(c) Pn ′ ( x)

(d) None of these.

The value of n ( Pn − 1 − xPn) is (a) (1 − x2 ) Pn ( x)

(b) (1 − x2 ) Pn ′ ( x)

(c) Pn − 1 ′ ( x)

(d) None of these.

The value of (n + 1) ( xPn − Pn + 1) is (a) (1 − x2 ) Pn ′

(b) xPn

(c) (n + 1) Pn ′

(d) None of these.

The value of Pn (1) is (a) 0

(b) 1

(c) − 1

(d) None of these. (Rohilkhand 2010, 11; Agra 10)

7.

The value of Pn (− 1) is (a) 1

(b) 0

(c) − 1

(d) (− 1)n. (Rohilkhand 2008, 09; Agra 08)

D-375

8. If Pn ( x) is a solution of Legendre’s equation, then value of P0 ( x) will be

9.

(a) 1

(b) 0

(c) − 1

(d) x.

The value of

dn

1 n

2 . n ! dx

n

(Rohilkhand 2004)

( x2 − 1) n is

(a) Pn ′ ( x)

(b) Qn ( x)

(c) Pn −1 ( x)

(d) Pn ( x).

(Agra 2006, 10)

Fill In The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

The differential equation (1 − x2 )

2.

The polynomial solution 1 . 3 . 5 … (2 n − 1) n!

 x 

n

−

d2 y 2

dx

n (n − 1) x 2 . (2 n − 1)

−2 x

n−2

+

dy dx

+ n (n + 1) y = 0 is called … .

n (n − 1)(n − 2)(n − 3) x 2 . 4 . (2 n − 1)(2 n − 3)

n−4

 − … 

of Legendre’s equation is called Legendre’s function of … 3.

… is called the generating function of Legendre’s polynomials.

4.

If m ≠ n, ∫

5.

Laplace’s first integral for Pn ( x) gives Pn ( x) = … .

6.

Qn ( x) is called Legendre’s function of … .

7.

The general solution of Legendre’s equation is given by …

1 −1

Pm ( x) Pn ( x) dx = … .

True or False Write ‘ T ’ for true and ‘F’ for false statement. 1.

Laplace’s second integral for Pn ( x) gives dφ 1 π Pn ( x) = ∫ ⋅ 2 0 π [ x ± √ ( x − 1) cos φ]n + 1

2.

Pn ( x) is the coefficient of h n + 1 in the expansion of (1 − 2 xh + h2 )−1 /2 in ascending powers of h.

3.

0 , if m ≠ n δ mn =  1, if m = n.

4.

(2 n + 1) ( x2 − 1) Pn ′ = n (n − 1) ( Pn + 1 − Pn − 1) is known as Beltrami’s result.

5.

(2 n + 1) x Pn = (n + 1) Pn + 1 + nPn − 1.

D-376

A nswers Multiple Choice Questions 1. 6.

(c). (b).

2. 7.

(a). (d).

3. 8.

(b). (a).

4. 9.

(b). (d).

5.

(a).

Fill in the Blank(s) 1.

Legendre’s equation. 2 −1 / 2

2. first kind.

3.

(1 − 2 xh + h )

4. 0.

5.

1 [ x ± √ ( x2 − 1) cos φ]n dφ. π ∫0

7.

APn ( x) + BQn ( x).

1.

T.

.

π

6. second kind.

True or False 2.

F.

3.

T.

4.

F.

5. T .

o

D-377

14 B essel’s F unctions

14.1 Bessel’s Equation (Avadh 2007, 10; Lucknow 07)

T

he differential equation of the form d2 y 2

dx

+

1 dy  n2  + 1 − 2  y = 0 x dx  x 

is called Bessel’s differential equation (or Bessel’s equation).

14.2 Solution of Bessel’s Differential Equation The Bessel’s differential equation is d2 y 2

dx

+

1 dy  n2  + 1 − 2  y = 0 . x dx  x 

...(1)

We can integrate (1) in a series of ascending powers of x. Assume that its series solution is ∞ ∞ dy y = Σ ar x k + r . ∴ = Σ ar (k + r) x k + r − 1 r=0 r=0 dx

D-378

d2 y

and

2

dx

=

∞

Σ

r=0

ar (k + r)(k + r − 1) x k + r − 2 .

Substituting these values in (1), we have ∞

Σ

r=0

or

∞

Σ

r=0

 ar (k + r)(k + r − 1) x 

k + r −2

+

ar [{(k + r)2 − n2} x k + r − 2 + x

1 (k + r) x x

k+r

k + r −1

 n2  + 1 − 2  x x  

k+r

  =0  ...(2)

] = 0.

Since the relation (2) is an identity, the coefficients of various powers of x must be zero. ∴ Equating to zero the coefficient of the lowest power of x, i. e.of x 2

k −2

in (2), we have

2

a0 (k − n ) = 0 . Now a0 ≠ 0 as it is the coefficient of the first term with which we begin to write the series. k 2 − n2 = 0 or k = ± n.

∴

Now equating to zero the coefficient of x

...(3) k −1

in (2), we have

a1{(k + 1)2 − n2} = 0 . But (k + 1)2 − n2 ≠ 0 for k = ± n given by (3);

∴

a1 = 0 .

Again equating to zero the coefficient of the general term i. e.of x 2

k+r

in (2), we have

2

ar + 2 {(k + r + 2) − n } + ar = 0 or

ar + 2 (k + r + n + 2)(k + r − n + 2) = − ar . ar ar + 2 = − . (k + r + n + 2)(k + r − n + 2)

∴

Putting r = 1 in (4), we have a3 =

...(4)

a1 = 0 , since a1 = 0 . (k + n + 3)(k − n + 3)

Similarly putting r = 3, 5, 7 etc. in (4), we have a1 = a3 = a5 = . . . = 0 (each). Now two cases arise. ar . (2 n + r + 2)(r + 2) a0 a Putting r = 0 , 2, 4 etc., we have a2 = − =− 2 0 (2 n + 2)(2) 2 .1 !(n + 1) a2 a2 a0 , etc. a4 = − =− 2 = (2 n + 4)(4) 2 . 2 . (n + 2) 24 . 2 !(n + 1)(n + 2) Case I:

When k = n, from (4), we have ar + 2 = −

  x n+2 x n+4 y = a0  x n − 2 + 4 − . . . 2 .1 !(n + 1) 2 . 2 !(n + 1)(n + 2)     x2 x4 = a0 x n. 1 + (− 1) 2 + (− 1)2 4 + . . . ⋅ 2 .1 !(n + 1) 2 . 2 !(n + 1)(n + 2)  

∴

If a0 =

1 n

2 Γ (n + 1)

kind of order n.

, this solution is called Jn( x), known as Bessel’s function of the first

D-379

  x4 x2 2 1 + ( − 1 ) + ( − 1 ) + . . .  n 2 4 2 Γ(n + 1)  2 .1 !(n + 1) 2 . 2 !(n + 1)(n + 2) 

Jn( x) =

∴

=

x

n

x

n

2 n Γ(n + 1) ∞

= Σ

r =0

Case II:

∞

x2 r

(− 1)r

Σ

r =0

x  ( − 1) r   2 

22 r r !(n + 1)(n + 2) . . . (n + r)

n + 2r

1 . r ! Γ( n + r + 1)

When k = − n.

The series solution is obtained by replacing n by −n in the value of Jn. J− n( x) =

∴

∞

 x (− 1)r    2

Σ

r=0

− n + 2r

1 . r ! Γ(− n + r + 1)

When n is not an integer J− n( x) is distinct from Jn( x). Note.

The solutions of Bessel’s equation are called Bessel’s functions.

14.3 General Solution of Bessel’s Equation (Lucknow 2007)

When n is not an integer, the most general solution of Bessel’s equation is y = A J− n ( x) + B Jn ( x), where A, B are two arbitrary constants.

14.4 Integration of Bessel’s Equation in Series for n = 0 (Avadh 2008) 2

The differential equation

d y 2

dx

+

1 dy + y =0 x dx

...(1)

is known as Bessel’s equation for n = 0. Let us assume that its solution be y = ∴

dy = dx

∞

ar (k + r) x

Σ

r=0

k + r −1

and

∞

ar x k + r .

Σ

r=0

d2 y 2

dx

=

∞

Σ

r=0

ar (k + r)(k + r − 1) x

k + r −2

.

Substituting these values in (1), we get ∞

Σ

or

r=0 ∞

Σ

r=0

1   ar (k + r)(k + r − 1) x k + r − 2 + (k + r) x k + r − 1 + x k + r  = 0 x   ar [(k + r)2 x k + r − 2 + x k + r ] = 0 ,

...(2)

which is an identity. Equating to zero the coefficient of the lowest power of x i. e., of x k − 2 , we have a0 k 2 = 0 . Since a0 ≠ 0 for the same reason, as in 14.2, therefore k 2 = 0 .

∴ k = 0.

Now equating to zero the coefficient of the next lowest power of x, i. e., of x

k −1

, we have

D-380

a1 (k + 1)2 = 0 . Since k + 1 ≠ 0 by virtue of (3), we have a1 = 0 . Again equating to zero the coefficient of the general term, i. e., of x ar + 2 (k + r + 2)2 + ar = 0 . When k = 0, we have ar + 2 = −

ar (r + 2)2

∴ ar + 2 = −

ar (k + r + 2)2

k+r

, we have

.

.

Putting r = 1, 3, 5, etc., we have a1 = a3 = a5 = . . . = 0 (each). Again putting r = 0 , 2, 4, etc., we have a a a a2 = − 02 , a4 = − 22 = 2 0 2 etc. 2 4 2 ⋅4 ∞

ar x r , when k = 0

Since

y=

∴

  x2 x4 x6 y = a0 1 − 2 + 2 2 − 2 2 2 + . . .  ⋅ 2 2 ⋅4 2 ⋅4 ⋅6  

Σ

r =0

If a0 = 1 , this solution is denoted by J0 ( x). J0 ( x) = 1 −

∴

x2 22

+

x4 22 ⋅ 4 2

−

x6 22 ⋅ 4 2 ⋅ 62

+…

where J0 ( x) is called Bessel’s function of zeroeth order.

14.5 Definition of J 0 ( x ) J0 ( x) is the solution of Bessel’s equation for n = 0 i.e., of d2 y 2

+

1 dy + y = 0. x dx

dx which is equal to 1 for x = 0.

Example 1:

Show that when n is a positive integer

(i)

J− n ( x) = (− 1)n Jn( x);

(ii)

Jn(− x) = (− 1)n Jn( x) for + ive or − ive integers ; (Meerut 2006, 11; Purvanchal 10)

(Meerut 2006, 08, 11; Kanpur 07, 08; Agra 06, 11; Rohilkhand 14; Purvanchal 10)

and (iii) Jn( x) is an even or an odd function of x according as n is even or odd respectively. Solution:

(i) We have J− n ( x) =

∞

Σ

r=0

 x (− 1)r    2

− n + 2r

⋅

1 . r ! Γ(− n + r + 1)

If p is an integer, then Γ(− p) is infinity for p ≥ 0. Therefore we get terms in J− n equal to zero till − n + r + 1 < 1 i. e., r < n. Hence we can write

D-381

J− n ( x) = =

− n + 2r (− 1)r  x   r ! Γ(− n + r + 1)  2 

∞

Σ

r=n

n + 2s (− 1)n + s  x ,   (n + s) ! Γ(s + 1)  2 

∞

Σ

s=0

= (− 1)n (ii) We know that Jn( x) = Case I:

Σ

s= 0 ∞

Σ

r=0

putting r = n + s

n + 2s (− 1)s  x = (− 1)n Jn( x).   Γ (n + s + 1) ⋅ s !  2 

(−1)r

1  x   r ! Γ(n + r + 1)  2 

n + 2r

.

Let n be a +ive integer. Replacing x by − x, we have ∞

Jn(− x) =

Σ

r=0

(−1)r ∞

= (− 1)n

Σ

r=0

1  x −  r ! Γ(n + r + 1)  2  (− 1)r

n + 2r

1  x   r ! Γ(n + r + 1)  2 

n + 2r

[ ∵ (− 1)2 r = 1]

= (− 1)n Jn( x). Case II:

Let n be a –ive integer, say n = − m where m is a +ive integer.

Then

Jn( x) = J− m ( x) = (− 1)m Jm ( x), as proved in part (i) of this question.

Replacing x by − x, we have Jn (− x) = (− 1)m Jm (− x) = (− 1)m (− 1)m Jm ( x)

[as proved in case I]

= (− 1)m J− m ( x), by part (i) of this question = (− 1)2 m (− 1)− m . J− m ( x) = (− 1)− m J− m ( x)

[∵ (− 1)2 m = 1]

= (− 1)n Jn( x). Hence Jn(− x) = (− 1)n Jn( x) for +ive or –ive integers. n

(iii) We have Jn(− x) = (− 1) Jn( x).

Proved ...(1)

n

When n is even, (− 1) = 1, ∴ from (1), Jn (− x) = Jn( x). ∴ Jn( x) is an even function of x if n is even. When n is odd, (− 1)n = − 1, ∴ from (1), Jn (− x) = − Jn ( x). ∴ Jn( x) is an odd function of x, if n is odd. Example 2:

Prove that

Jn ( x) = (− 2)n x

n

dn d ( x2 )n

J0 ( x).

Solution: J0 ( x) is the solution of Bessel’s equation of zeroeth order

d2 y 2

dx

+

1 dy + y = 0. x dx

…(1)

Let us change the independent variable from x to X, by the relation x2 = X , so that dy dy dX dy dy = ⋅ = 2x =2√ X dx dX dx dX dX

D-382

d2 y

and

dx

dy  d  2 √ X  =  dX  dX

d dx

=

2

dy  dX  2 √ X ⋅  dX  dx

 d2 y d2 y dy 1 dy   ⋅2 √ X = 4X +2 ⋅ = 2 √ X + dX √ X dX  dx2 dX 2  Putting these values in (1), we get  d2 y dy  dy 1  + 4 X + y =0 + 2 2√X 2   dX dX √ X dX   or

4X

d2 y dX

2

+4

dy + y = 0. dX

…(2)

Now differentiating (2) n times w.r.t. X, by Leibnitz’s theorem, we get  dn+2 y d n +1 y  d n +1 y dn y 4 X + n . 1 ⋅ + 4 + =0  n+2 dX n +1  dX n +1 dX n  dX or

dn+2 y

4X

Let Y =

dn y dX n

=

dX

dn dX n

n+2

d n +1 y

+ 4 (n + 1)

dX

n +1

dn y

+

dX n

…(3)

= 0.

J0 ( x).

…(4)

Using this substitution the equation (3) reduces to d2Y

4X

dX 2

+ 4 (n + 1)

dY + Y = 0. dX

…(5)

Again, Jn ( x) is the solution of the Bessel’s equation d2 y dx2

+

1 dy  n2  + 1 − 2  y = 0 . x dx  x 

y = xn Z

Let so that

or

dy dZ = xn + nx dx dx

…(6)

Jn ( x) = x n Z

n −1

2

d y

…(7) d Z

+ 2 nx

n −1

dZ 1  dZ + n (n − 1) x n − 2 Z +  x n + nx dx x dx

n −1

Z and

dx2

= xn

2

dx2

dZ + n (n − 1) x dx

n−2

Z.

Putting these values in (6), we get xn

d2 Z dx2

+ 2 nx n − 1

or or

xn

d2 Z 2

+ (2 n + 1) x n − 1

dZ +x dx

n

dx 1 dZ d2 Z + (2 n + 1) + Z = 0. 2 x dx dx

n2    Z  + 1 − 2  x n Z = 0   x 

Z =0 …(8)

Changing the independent variable from x to X, by the relation x2 = X , the equation (8) becomes  d2 Z dZ  (2 n + 1) dZ  + 4 X + Z =0 +2 2√X 2 dX  dX √X dX  or

4X

d2 Z dX

2

+ 4 (n + 1)

dZ + Z = 0. dX

…(9)

D-383

Comparing (5) and (9), we get Z = cY , where c is a constant to be determined. ∴ x n Z = c x n Y or y = c x n Y . Hence

dn

Jn ( x) = c x n ⋅

d ( x2 )n

…(10)

J0 ( x).

Now it remains to find the value of c. x2 x4 x6 We have J0 ( x) = 1 − 2 + 2 2 − 2 2 2 + … = 2 2 .4 2 .4 .6 dn

∴

d ( x2 )n

∞

(− 1)r x2 r

r =0

(2 r r !)2

dn

J0 ( x) =

Σ

d ( x2 )n

=

∞

(− 1)r x2 r

r =0

(22 r !)2

Σ

∞

(− 1)n + r ( x2 )n + r

r =0

{2 n + r (n + r) !}2

dn

Σ

d ( x2 )n

⋅

[∵ On differentiation all those terms on R.H.S. will vanish in which the power of x2 is less than n] =

∞

(− 1)n + r (n + r) (n + r − 1) … (r + 1) ( x2 )r

r =0

{2 n + r (n + r) !}2

Σ

=

∞

(− 1)n + r (n + r) ! x2 r

r =0

r ! 22 n + 2 r {(n + r) !}2

Σ

= (− 1)n

∞

r =0

Jn ( x) = cx n (−1)n

∴

=

c (− 1)n 2

n

(− 1)r

Σ

∞

Σ

r =0 ∞

Σ

r =0

1 2 n+2 r

r !2

(− 1)r (− 1)r

(n + r) !

x2 r .

1 r ! 22 n + 2 r (n + r) !

1  x   r ! Γ (n + r + 1)  2 

x2 r

n + 2r

=

c (− 2)n

⋅ Jn ( x).

c = (− 2)n.

∴

Putting this value of c in (10), we get Jn ( x) = (− 2)n x n

dn d ( x2 )n

J0 ( x).

14.6 Recurrence Formulae (I)

x J n ′ ( x) = n ⋅ J n ( x) − x ⋅ J n + 1 ( x). (Meerut 2006, 07, 09, 11; Lucknow 07, 11; Rohilkhand 04, 14)

Proof:

We know that Jn( x) =

∞

Σ

r=0

 x (− 1)r    2

n+2 r

1 ⋅ r ! Γ(n + r + 1)

Differentiating w.r.t. x, we have n + 2 r −1 ∞ (n + 2 r) 1  x Jn ′ ( x) = Σ (− 1)r   r=0 r ! Γ(n + r + 1) 2  2  ∴

∞

x Jn ′ ( x) = Σ

r=0

(− 1)r

(n + 2 r)  x  n + 2 r =   r ! Γ(n + r + 1)  2  +

∞

Σ

r =0

∞

Σ

r=0

(− 1)r

(− 1)r

n  x   r ! Γ(n + r + 1)  2 

x x 2r ⋅   r ! Γ (n + r + 1) 2  2 

n + 2r

n + 2 r −1

D-384

= n J n( x) + x = n J n( x) − x

∞

Σ (− 1)r

r =1 ∞

(− 1)s

Σ

s=0

1  x   (r − 1) ! Γ(n + r + 1)  2  1  x   s ! Γ(n + 1 + s + 1)  2 

n + 2 r −1

n +1 + 2 s

,

putting r − 1 = s = n Jn( x) − x Jn + 1( x). Hence x J n ′ ( x) = n J n ( x) − x J n + 1 ( x). Multiplying both sides by x − n −1, we have x − n Jn ′ = nx − n − 1 Jn − x − n Jn + 1 or − n x − n − 1 Jn + x − n Jn ′ = − x − n Jn + 1 d ( x − n J n) = − x − n J n + 1 , dx which is another form of recurrence formula (I). (II) x J n ′ ( x) = − n J n ( x) + x J n − 1 ( x). (Meerut 2008, 13, 10; Agra 10) or

Proof:

As in I, we have x Jn ′ ( x) = =

∞

Σ

r=0

= −n

∞

(− 1)r

Σ

r=0

(2 n + 2 r − n)  x  n + 2 r   r !. Γ(n + r + 1)  2 

(− 1)r ∞

Σ

r=0

(n + 2 r)  x  n + 2 r   r ! Γ(n + r + 1)  2 

(− 1)r

1  x   r ! Γ(n + r + 1)  2  +

∞

= − n Jn( x) +

Σ

r =0

= − n Jn( x) + x = − n Jn( x) + x

∞

Σ

r =0 ∞

Σ

r =0

(− 1)r

n + 2r

∞

(− 1)r

Σ

r=0

(2 n + 2 r)  x  n + 2 r   r !. Γ(n + r + 1)  2 

n + 2 r −1 2(n + r)  x  x      2 r !. Γ(n + r + 1)  2 

1 r ! Γ(n + r)

(− 1)r (− 1)r

 x    2

n + 2 r −1

1  x   r ! Γ{(n − 1) + r + 1}  2 

(n − 1) + 2 r

= − n Jn( x) + x Jn − 1( x). Hence x J n ′ ( x) = − n J n ( x) + x J n − 1 ( x) Multiplying both sides by x x

n

Jn ′ = − nx

n −1

n −1 n

, we have

Jn + x Jn − 1

or

nx

n −1

Jn + x

n

Jn ′ = x n Jn − 1

d which is another form of recurrence formula (II). ( x n J n) = x n J n − 1, dx (III) 2 J n ′ ( x) = J n − 1 ( x) − J n + 1 ( x). (Rohilkhand 2002, 07; Avadh 07, 10; or

Kanpur 07)

Proof:

Recurrence formulae I and II, are x Jn ′ ( x) = n Jn( x) − x Jn + 1( x) and

x Jn ′ ( x) = − n Jn( x) + x Jn − 1( x).

Adding, we have 2 x Jn ′ ( x) = x [ Jn − 1( x) − Jn + 1( x)]. Hence

2 J n ′ ( x) = J n − 1 ( x) − J n + 1 ( x).

D-385

Aliter. Jn( x) = ∴

2 Jn ′ ( x) = =

= =

∞

Σ

r=0 ∞

Σ

r=0 ∞

Σ

r=0

∞

Σ

r=0 ∞

Σ

r=0

1  x   r ! Γ(n + r + 1)  2 

(− 1)r

2 1 (n + 2 r) r ! Γ(n + r + 1) 2

(n + r) r ! Γ(n + r + 1)

(− 1)r

=

Σ

r=0

(− 1)r

 x    2

1  x   r ! Γ(n + r)  2 

 x    2

n + 2 r −1

n + 2 r −1

+

∞

Σ

r=0

(− 1)r

r r ! Γ(n + r + 1)

∞

(− 1)r − 1

r =1

1  x   r ! Γ [(n − 1) + r + 1]  2  ∞

− Σ (− 1)s s =0

1 (r − 1) ! Γ (n + r + 1)

n + 2 r −1

 x    2

n + 2 r −1

(n − 1) + 2 r

1  x   s ! Γ [(n + 1) + s + 1]  2 

= Jn − 1( x) − Jn + 1( x). Hence 2 J n ′ ( x) = J n − 1 ( x) − J n + 1 ( x). (IV) 2 n J n ( x) = x [ J n − 1 ( x) + J n + 1 ( x)] Proof:

 x    2

n + 2 r −1

Σ

− ∞

.

[(n + r) + r]  x  n + 2 r − 1   r ! Γ(n + r + 1)  2 

(− 1)r (− 1)r

n + 2r

(− 1)r

(n + 1)+ 2 s

, putting r − 1 = s

(Avadh 2009; Kanpur 11)

Writing Recurrence formulae I and II, we have x Jn ′ ( x) = n Jn( x) − x Jn + 1( x) and

x Jn ′ ( x) = − n Jn( x) + x Jn − 1( x).

Subtracting, we have 0 = 2 n Jn( x) − x [ Jn + 1( x) + Jn − 1( x)]. Hence 2 n J n ( x) = x [ J n + 1 ( x) + J n − 1 ( x)] ∞

Aliter:

Jn( x) =

∴

2 n Jn( x) =

Σ

r=0

=

(− 1)r ∞

Σ

r=0 ∞

1  x   r ! Γ(n + r + 1)  2 

(− 1)r

Σ (− 1)r

r=0

∞

Σ

r=0

.

(2 n + 2 r − 2 r) r ! Γ(n + r + 1)

 x    2

n + 2r

n + 2r 2(n + r)  x   r ! Γ(n + r + 1)  2 

− =x

n + 2r

(− 1)r +x

∞

Σ

r=0

(− 1)r

2r  x ⋅  r ! Γ (n + r + 1)  2 

1  x   r ! Γ [(n − 1) + r + 1]  2  ∞

Σ

r =1

(− 1)r − 1

n + 2r

(n − 1) + 2 r

1  x   (r − 1) ! Γ (n + r + 1)  2 

n + 2 r −1

D-386

= x J n − 1( x) + x

∞

Σ

s=0

(− 1)s

1  x   s ! Γ [(n + 1) + s + 1]  2 

(n + 1) + 2 s

, putting r − 1 = s

= x Jn − 1( x) + x Jn + 1( x). Hence 2 n J n ( x) = x [ J n − 1( x) + J n +1( x)] d (V) [ x − n J n ( x)] = − x − n J n + 1 ( x). dx (Avadh 2006, 10; Agra 08; Lucknow 06, 08; Meerut 13B; Rohilkhand 14)

Proof:

d −n [x Jn ( x)] = − n x − n − 1 Jn( x) + x − n Jn ′ ( x) dx = x − n − 1 [− n Jn ( x) + x Jn ′ ( x)] = x − n − 1 [− n Jn ( x) + {n J n ( x) − x J n + 1 ( x)}], from recurrence formula I = x − n − 1 [− x Jn + 1 ( x)] = − x − n Jn + 1 ( x).

Hence Aliter:

d [ x − n J n ( x)] = − x − n J n + 1 ( x). dx We have

=

n + 2r (− 1)r d −n d  −n ∞  x [ x Jn ( x)] = Σ   x  r = 0 r ! Γ (n + r + 1)  2  dx dx  

n 2r (− 1)r d  ∞  1  x   ⋅  ⋅    Σ dx r = 0 r !Γ (n + r + 1)  2   2  

n 2 r −1 ∞  (−1)r 1  1  x = Σ  ⋅    ⋅ 2r    2 r = 0  r !Γ (n + r + 1)  2  2  

= x− n = x− n = − x− n (VI)

n −1 + 2 r (− 1)r  x ⋅  (r − 1) ! Γ(n + r + 1)  2 

∞

Σ

r =1

n + 2 s +1 (− 1)s + 1  x , putting r − 1 = s ⋅  s ! Γ(n + s + 2)  2 

∞

Σ

s=0 ∞

Σ

s=0

(− 1)s s ! Γ(n + 1 + s + 1)

d [ x n J n ( x)] = x n J n − 1 ( x). dx

 x    2

n +1 + 2 s

= − x − n Jn + 1( x).

(Agra 2003; Avadh 09, 11; Kanpur 10; Purvanchal 10; Bundelkhand 13)

Proof:

d [ x n Jn ( x)] = n x n −1 Jn ( x) + x n Jn ′ ( x) = x n − 1[n Jn ( x) + x Jn ′ ( x)] dx = x n − 1[n Jn ( x) + {− n Jn ( x) + x Jn − 1 ( x)}], from recurrence formula II =x

n −1

[ x Jn − 1 ( x)] = x

n

Jn − 1 ( x).

d [ x n J n ( x)] = x n J n − 1 ( x). dx Aliter: We have n + 2r (− 1)r d n d  n ∞  x [ x Jn( x)] =   x Σ  dx dx  r = 0 r ! Γ (n + r + 1)  2  

Hence

D-387

=

d dx

2 (n + r )  ∞ (−1)r  x ⋅ 2n ⋅    Σ   2 r = 0 r !Γ (n + r + 1) 

∞ (− 1)r ⋅ 2 n ⋅ 2 (n + r) x 2 n + 2 r − 1 1   = Σ  ⋅     2 r = 0  r ! Γ (n + r + 1) 2  

= xn = xn

∞

Σ

r=0 ∞

Σ

r=0

n + 2 r −1 (− 1)r  x ⋅  r ! Γ(n + r)  2 

(− 1)r r ! Γ(n − 1 + r + 1)

 x    2

n −1 + 2 r

n

= x ⋅ Jn − 1( x).

Example 3:

Prove that

n +1 2  d n 2 2 2 [ J n + J n + 1] = 2  J n − J n + 1 . x  dx x

or

Jn Jn ′ + Jn + 1 Jn + 1 ′ =

Solution:

We have

1 2 2 [ n Jn − (n + 1) Jn + 1] . x

d 2 2 ( J n + J n + 1) = 2 J n J n ′ + 2 J n + 1 J n + 1 ′ . dx

From recurrence formula I, we have

Jn ′ =

Also from recurrence formula II, we have

n Jn − Jn +1 . x Jn ′ = −

...(1) ...(2)

n Jn + Jn −1. x

Replacing n by n + 1 in this result, we have n +1 Jn + 1 ′ = − Jn + 1 + Jn . x

...(3)

Substituting the values of Jn ′ and Jn + 1 ′ from (2) and (3) in (1), we have d  n +1  2 2  n Jn + 1 + Jn ( J n + J n + 1) = 2 J n  J n − J n + 1 + 2 J n + 1  −  x   x dx n +1 2  n 2 = 2  Jn − J n + 1 ⋅ x  x Example 4:

(i)

(ii)

Prove that

 2 J1 /2 ( x) =   ⋅ sin x  π x

(Avadh 2007, 12, 13; Lucknow 08, 10; Meerut 09; Kanpur 09; Rohilkhand 10; Purvanchal 08)

 2   ⋅ cos x. J− 1 /2 ( x) =   π x

(iii) [ J1 /2 ( x)]2 + [ J− 1 /2 ( x)]2 = 2 / πx.

(Avadh 2007, 11; Purvanchal 10) (Avadh 2007, 08, 14; Purvanchal 09, 14)

D-388

(iv)

(v)

 2  1    cox x + sin x . J− 3 /2 ( x) = −    π x  x   3 − x2  3 ⋅  2  cos x + sin x. x  x  

 2   J− 5 /2 ( x) =   π x

Solution:

(Avadh 2008)

We know that Jn( x) =

(i) Putting n =

  x2 x4 + − . . . 1 − 2 Γ(n + 1)  2.(2 n + 2) 2.4.(2 n + 2)(2 n + 4)  x

n

n

...(1)

1 in (1), we have 2   x2 x4 + − … 1 − 2 Γ (3 / 2)  2 . 3 2 . 4 . 3 . 5  3 5   x x 2 2 =   x − + − . . . =   . sin x  πx   πx   3! 5! 

J1 /2 ( x) =

x1 /2

1 /2

(ii) Again putting n = − J− 1 /2 ( x) =

...(2)

1 in (1) , we have 2

x − 1 /2  1 2 −1 / 2 Γ    2

  x2 x4 + − ... 1 −  2 .1 2 . 4 .1. 3 

 x2 x4 2 2  =   1 − + − . . . =   cos x.  πx   πx   2! 4! 

...(3)

(iii) Squaring and adding (2) and (3), we have 2 [ J1 /2 ( x)]2 + [ J−1 /2 ( x)]2 = ⋅ πx (iv) From recurrence formula IV, we have 2 n Jn( x) = x [ Jn − 1( x) + Jn + 1( x)] or Jn − 1( x) = (2 n / x) Jn( x) − Jn+1( x). 1 Putting n = − in (4), we have 2 J−3 /2 ( x) = − (1 / x) J−1 /2 ( x) − J1 /2 ( x)

...(4)

...(5)

Substituting for J1 /2 ( x) and J−1 /2 ( x) from (2) and (3) in (5), we have J−3 /2 ( x) = −

 2  1  2  sin x ⋅   cos x −  x  π x  π x

 2  1  = −    cos x + sin x ⋅  πx   x  Aliter:

We know that Jn( x) =

  x2 x4 1 − + − . . .  n 2 Γ (n + 1)  2.(2 n + 2) 2.4(2 n + 2)(2 n + 4)  x

n

...(6)

D-389

=

n

  x2 x4 1 − + − . . . ,  n 2 .( 2 n + 2 ) 2 . 4 .( 2 n + 2 )( 2 n + 4 ) 2 Γ (n + 2)   (n + 1) x

multiplying the numerator and the denominator by n + 1. Now putting n = − 3 / 2, we have  1 −3 /2 −  x  2 J−3 /2 ( x) =  1 2 −3 / 2 Γ    2

  x6 x2 x4 + − + . . . 1 −  2 ⋅ (− 1) 2 ⋅ 4 (− 1) (+ 1) 2.4.6(− 1) (1) (3) 

 x2 x4 x6 2 1 = −   ⋅ 1 + − + − ...   πx  x  2 2 .4 2 .3 .4 .6   1 2 3 4 5 6 2 1 = −   ⋅ 1 + x − x + x − . . .  πx  x  2 ! 4! 6!    2  1  2 −1 2 4 −1 4 6 −1 6 x − x + x − . . . = −   ⋅ 1 +  πx  x  2! 4! 6!   2  1 3 4 3 6 5  2   1 1 = −   ⋅  − x+ x − . . . +  x− x + x − . . .   πx   x 2 ! 4! 4! 6!  2 !       x2 x4 x3 x5  2  1  + − . . .  = −    1 − + − . . . +  x −  πx   x  2! 4! 3! 5!       2  1  = −    cos x + sin x ⋅  πx   x  (v) Putting n = − 3 / 2 in (4), we have J−5 /2 ( x) = (−3 / x) J−3 /2 ( x) − J−1 /2 ( x)  2  2  1  3  =  −  ⋅ −   ⋅  cos x + sin x  −   cos x,   πx   x  x   πx  substituting for J−3 /2 ( x) and J−1 /2 ( x) from (6) and (3)

Aliter:

2 =    πx 

3  1 ⋅   cos x + sin x  x

2 =    πx 

 3 − x2  3  2  cos x + sin x  x 

 x − cos 

 x 

 x. 

We know that Jn( x) =

=

xn

  x2 x4 1 − + − . . .  n 2 Γ(n + 1)  2(2 n + 2) 2.4(2 n + 2)(2 n + 4) 

(n + 2)(n + 1) x

n

n

2 Γ(n + 3)

  x2 x4 + − . . . , 1 −  2.(2 n + 2) 2.4.(2 n + 2)(2 n + 4) 

multiplying the numerator and the denominator by (n + 2)(n + 1). Now putting n = − 5 / 2, we have

D-390

J−5 /2 ( x) =

(−1 / 2)(−3 / 2) x −5 /2  x2 x4 ⋅ 1 − +  2 −5 /2 Γ (1 / 2)  2 ⋅ (− 3) 2 ⋅ 4 ⋅ (− 3) ⋅ (− 1) −

 x6 + . . . 2 . 4 . 6 . (− 3) (− 1) (1) 

 x2 x 4 5 x6 2 3  =   ⋅ 2 1 + + − + . . .  πx  x  3! 4! 6!   1 3 x2 15 x 2 3 =    2 + + − + . . .  πx   x 2! 4! 6!    2   3 3 − 4 3 + 0 2 3 + 12 4 + x − x + . . . =    2 −  πx   x 2! 4! 6!  2 =    πx  2 =    πx 

     3 − 3 + 3 x2 − 3 x4 − . . . +  4 + 0 x2 − 12 x 4 − . . .  2  x 2! 4! 6! 6!    2 ! 4 !  3   x2 x4 x6 + − − . . .  2 1 − 2! 4! 6!   x  −

 2(1 − 3) 4(3 − 3) 2 6(5 − 3) 4 + x − x − . . . 2! 4! 6! 

21 . 4.3 2 6.5 4 2 3 + x − x + ... =    2 cos x −  πx   x 2! 4! 6! 2  4 2 6 4 +3 − x + x − . . .  6! 2 ! 4 !    3   x2 x4 x3 x5 2 3 . . .  +  x − + =    2 cos x − 1 − + . . .    πx   x 2! 4! 3 ! 5 !    x   3 2 3 =    2 cos x − cos x + sin  πx   x x

2 3   2  3 − x x =    2 cos x + sin   πx x   x

 x ⋅ 

14.7 Generating Function for J n ( x ) Prove that when n is a positive integer, Jn( x) is the coefficient of z n in the expansion of e x [z − (1 / z)] /2 in ascending and descending powers of z . Also prove that Jn( x) is the coefficient of z − n multiplied by (− 1)n in the expansion of the above expression. (Rohilkhand 2002, 06; Lucknow 06, 09, 11; Meerut 08; Avadh 06, 08)

Proof:

e

x [ z − (1 / z )] /2

= e xz /2 ⋅ e −

x /2 z

n +1 n+2 2 n   1  xz  1 xz 1  xz  1  xz   xz  = 1 + + + + ...       + ... +   + (n + 1)!  2  (n + 2)!  2  2 2!  2  n!  2    n + 1 n + 2 2 n n + 1 n + 2 n   (− 1) (−1) (−1)  x  x 1  x  x  x × 1 − + + + …       − ... +   +         (n + 1)! 2 z (n + 2) ! 2 z n ! 2z   2 z 2 ! 2 z

Coefficient of z n in this product

D-391 n

n +1

2

1  x 1  x  x       − n !  2 (n + 1) !  2   2 

=

1  x 1  x     − n !  2 (n + 1) !  2 

=

n+2 n+4 (− 1)0  x  n (− 1) (− 1)2  x  x ⋅  + ... + ⋅    + Γ(n + 1)  2  1 ! Γ(n + 2)  2  2 ! Γ(n + 3)  2 

n

∞

Σ

=

r=0

(− 1)r

1 1  x  x ⋅     (n + 2) ! 2 !  2   2 

n+2

=

+

n+2

+

1  x   2 !(n + 2) !  2 

1  x ⋅  r ! Γ(n + r + 1)  2 

− ...

n+4

− ...

n+2 r

= Jn( x).

Similarly, the coefficient of z − n in the above product (− 1)n  x  n (− 1)n + 1 x  x  n + 1 (− 1)n + 2 1  x  2  x  n + 2 +… ⋅ ⋅  = + ⋅ ⋅      + n !  2 (n + 1) ! 2  2  (n + 2) ! 2 !  2   2  n+4  (−1)0  x  n  (−1)  x  n + 2 (− 1)2  x = (−1)n  + + …   +     Γ (n + 2)  2  2 ! Γ (n + 3)  2  Γ (n + 1)  2   n = (− 1) Jn( x).

In the above product the term independent of z is x2 x4 x6 1 − 2 + 2 2 − 2 2 2 + . . . = J0 ( x). 2 2 ⋅4 2 ⋅4 ⋅6 1 1 1    x (z − 1 / z) /2 Hence e = J0 +  z −  J1 +  z 2 + 2  J2 + … + z n + (− 1)n n  Jn + …   z z   z  Note.

=

∞

Σ

−∞

z

n

Jn ( x),

since J− n ( x) = (− 1)n Jn( x). See Example 1, after article 14.5.

14.8 Orthogonal Property of Bessel’s Functions 1

∫0

, α ≠β 0   xJ n (α x) J n ( β x) dx =  1 2  2 [ Jn + 1 (α)] , α = β,

where α , β are the roots of Jn ( x) = 0 . Proof:

(Lucknow 2008, 11; Meerut 10; Agra 11)

The solutions of the equations x2 u′ ′ + xu ′ + (α2 x2 − n2 ) u = 0

…(1)

and

x2 v ′ ′ + xv ′ + ( β2 x2 − n2 ) v = 0

…(2)

are

u = Jn (αx) and v = Jn ( βx) respectively.

Multiplying (1) by v / x and (2) by u / x and subtracting, we get x (u ′ ′ v − uv ′ ′ ) + (u ′ v − uv ′ ) + (α2 − β2 ) xuv = 0 d or { x (u ′ v − uv ′ ) } = ( β2 − α2 ) xuv. dx Integrating both sides of (3) from 0 to 1, we get

…(3)

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( β 2 − α2 ) ∫ Now

1 0

1

xuv dx = [ x (u ′ v − uv ′ ] 0 = (u′ v − uv ′ ) x = 1

…(4)

u = Jn (α x) d (α x) d d u′ = [ Jn (α x)] = [ Jn (α x)] ⋅ = α Jn ′ (α x) dx d (α x) dx

⇒

Similarly, v = J n ( βx) ⇒ v ′ = β J n ′ ( βx). Substituting these values in (4), we get 1

∫0

x Jn (α x) Jn ( βx) dx =

α Jn ′ (α) Jn ( β) − β Jn (α) Jn ′ ( β) β 2 − α2

⋅

…(5)

If α and β are distinct roots of Jn ( x) = 0 ,then Jn (α) = Jn ( β) = 0 and thus (5) reduces to 1

∫0

…(6)

x Jn (α x) Jn ( βx) dx = 0 .

This is known as the orthogonality relation of Bessel’s functions. If β = α, then the right hand side of (5) is of the form 0/0. Its value can be found by considering α as a root of Jn ( x) = 0 and β as a variable approaching α . Then, from (5) we have lim lim α Jn ′ (α) Jn ( β) 1 x Jn (α x) Jn ( βx) dx = β → α ∫0 β→α β 2 − α2 lim 1 α Jn ′ (α) Jn ′ ( β) 2 or ∫0 x Jn (α x) dx = β → α 2β 1 = { Jn ′ (α)}2 , by L’Hospital’s rule 2 1 …(7) = [ Jn+1 (α)]2 , by recurrence formula. 2 Corollary:

If the interval be from 0 to a, then a

∫0

x Jn2 (α x) dx =

a2 2 J (aα), 2 n+1

…(8)

where α’s are the roots of Jn (ax) = 0 .

14.9 Some Trignometric Expansions Involving Bessel’s Functions Here we give some trigonometric expansions involving Bessel’s functions. (i) (ii) (iii) (iv) (v) (vi)

cos ( x sin θ) = J0 + 2 cos 2θ J2 + 2 cos 4θ J4 + . . . sin ( x sin θ) = 2 sin θ J1 + 2 sin 3θ J3 + . . . cos ( x cos θ) = J0 − 2 cos 2θ J2 + 2 cos 4θ J4 − . . . sin ( x cos θ) = 2 cos θ J1 − 2 cos 3θ J3 + 2 cos 5θ J5 − . . . cos x = J0 − 2 J2 + 2 J4 − 2 J6 + . . . sin x = 2 J1 − 2 J3 + 2 J5 − 2 J7 + . . ..

Proof:

We know that

(Kanpur 2011) (Kanpur 2011)

D-393

1 1 1   e x (z − 1 / z) /2 = J0 +  z −  J1 + (z 2 + 2 ) J2 +  z 3 − 3  J3 + … …(1)   z z z  iθ − iθ iθ − iθ p Putting z = e , so that 1 / z = e , z − 1 / z = e − e = 2 isin θ, z = e ipθ , 1 / z p = e − ipθ , z p + 1/ z p = e ipθ + e − ipθ = 2 cos pθ, z p − 1 / z p = e ipθ − e − ipθ = 2 i sin pθ in (1), we have e i( x sin θ) = J0 + (2 i sin θ) J1 + (2 cos 2θ) J2 + (2 i sin 3θ) J3 + (2 cos 4θ) J4 + . . . cos ( x sin θ) + i sin ( x sin θ) = ( J0 + 2 cos 2θ J2 + 2 cos 4θ J4 + . . .) + i(2 sin θ J1 + 2 sin 3θ J3 + . . .) ...(2) (i) Equating the real parts on both sides of (2), we have cos ( x sin θ) = J0 + 2 cos 2θJ2 + 2 cos 4θJ4 + . . . (ii) Equating the imaginary parts on both sides of (2), we have sin ( x sin θ) = 2 sin θ J1 + 2 sin 3θ J3 + . . . (iii) Replacing θ by π / 2 − θ in (i), we have cos ( x cos θ) = J0 − 2 cos 2θ J2 + 2 cos 4θ J4 − . . . (iv) Replacing θ by π / 2 − θ in (ii), we have sin ( x cos θ) = 2 cos θ J1 − 2 cos 3θ J3 + 2 cos 5θ J5 − . . . (v) Putting θ = π / 2 in (i), we have cos x = J0 − 2 J2 + 2 J4 − 2 J6 + . . . (vi) Putting θ = π / 2 in (ii), we have sin x = 2 J1 − 2 J3 + 2 J5 − 2 J7 + . . .

or

Using generating function, prove that

Example 5:

J n ( x + y) =

∞

Σ

r=−∞

We know that ∞ 1  1  Σ exp  ( x + y)  z −   = J n ( x + y) z n   n= − ∞ z  2 ∞ 1  1  1  1  Σ J n ( x + y) z n exp  x  z −   . exp  y  z −   =     n= − ∞ z  z  2 2

Solution:

or

∞

Jr ( x) z

r

Equating the coefficients of z

n

or

Σ

r=−∞ ∞

Σ

r=−∞

Example 6:

(a) π Jn = (b)

(i)

Jr ( x) Jn− r ( y).

.

∞

Σ

k=−∞

J k ( y) z

k

=

∞

Σ

n= − ∞

J n ( x + y) z n

on both the sides, we get

Jr ( x) Jn − r ( y) = Jn ( x + y).

Show that when n is integral π

∫0

cos (nθ − x sin θ) dθ.

π J0 =

π

∫0

cos ( x cos φ) dφ

(Purvanchal 2007, 08; Avadh 12) (Avadh 2012)

D-394

(ii) π J0 =

π

cos ( x sin φ) dφ

∫0

(Agra 2006; Purvanchal 09)

2

and hence deduce that

J0 ( x) = 1 −

x

2

2

4

x

+

2

6

2

2 .4

−

x 2

2

2

2 .4 .6

+ ... =

∞

(− 1)r x2 r

r=0

(2 r r !)2

Σ

.

From article 14.9, we have

Solution:

cos ( x sin θ) = J0 + 2 J2 cos 2θ + . . . + 2 J2 m cos 2 mθ + . . . and sin ( x sin θ) = 2 sin θ. J1 + 2 sin 3θ. J3 + . . . + 2 J2 m + 1 sin (2 m + 1)θ + . . .

. . . (1) . . . (2)

(a) Multiplying both sides of (1) by cos 2mθ and then integrating between the limits 0 to π, π

∫0

cos ( x sin θ) cos 2 mθ dθ

= J0 ∫

π 0

π

cos 2 mθ dθ + 2 J2

∫0

cos 2θ cos 2 mθ dθ + . . . + 2 J2 m

= 0 + . . . + J2 m

π

∫0

π

∫0

cos2 2 mθ dθ + . . .

(1 + cos 4 mθ) dθ + 0 + . . . = π J2 m .

Similarly we can prove that π

∫0

cos ( x sin θ) cos (2 m + 1)θ dθ = 0 .

Again multiplying both sides of (2) by sin (2 m + 1)θ and then integrating between the limits 0 to π, we have π

∫0

sin ( x sin θ) sin (2 m + 1)θ dθ

= 2 J1 ∫

π 0

π

sin θ.sin (2 m + 1)θ dθ + 2 J3

∫0

+ 2 J2 m + 1

∫0

= 0 + 0 + . . . + J2 m + 1

π

∫0

π

sin 3θ sin(2 m + 1)θ dθ + . . . sin2 (2 m + 1) θ dθ + . . .

{1 − cos 2(2 m + 1) θ} dθ + . . .

= J2 m + 1 [ θ ]0π = π J2 m + 1. Similarly,

π

∫0

Therefore, ∫

π 0

sin ( x sin θ) sin 2 mθ dθ = 0 . cos (2 mθ − x sin θ) dθ =

Also ∫

π 0

π

∫0

cos 2 mθ.cos ( x sin θ) dθ +

π

∫0

sin 2 mθ.sin ( x sin θ) dθ = π J2 m .

cos [(2 m + 1) θ − x sin θ] dθ =

π

∫0

cos (2 m + 1) θ . cos ( x sin θ) dθ +

= π J2 m + 1.

π

∫0

sin (2 m + 1)θ sin ( x sin θ) dθ

D-395

Hence for all positive integral values of n, we have π

∫0 cos ( nθ −

x sin θ) dθ = π J n.

If n is negative, say n = − m, where m is +ive, then π

∫0

cos (nθ − x sin θ) dθ =

=−

∫π

0

π

=

∫0

=

∫0

π

π

∫0

cos (− mθ − x sin θ) dθ

cos {− m(π − φ) − x sin (π − φ)} dφ, putting θ = π − φ

cos {− mπ + (mφ − x sin φ)} dφ {cos mπ.cos (mφ − x sin φ) + sin mπ sin (mφ − x sin φ)} dφ

= (− 1)m

π

∫0

cos (mφ − x sin φ) dφ

= (− 1)m π Jm ( x), since we have proved the result for positive integers = π J− m ( x), since J− m ( x) = (− 1)m Jm ( x) = π Jn( x). Hence for all integral values of n, π

∫0 (b)

Proved

cos (nθ − x sin θ) dθ = π Jn .

Putting θ = π /2 + φ in the value of cos ( x sin θ) from (1), we have cos ( x cos φ) = J0 − 2 J2 cos 2 φ + 2 J4 cos 4 φ − . . .

∴

π

∫0

cos( x cos φ) dφ = J0

π

∫0

dφ − 2 J2

π

∫0

cos 2 φ dφ + . . . Proved

= π J0 . From (1) we have cos ( x sin φ) = J0 + 2 J2 cos 2 φ + 2 J4 cos 4 φ + . . . ∴

π

∫0

cos( x sin φ) dφ = J0

π

∫0

dφ + 2 J2

π

∫0

cos 2 φ dφ + . . . Proved

= π J0 . Deduction:

We have proved that 1 π

=

π

∫0

1 J0 ( x) = π

π

∫0

cos( x cos φ) dφ

  x 2 cos2 φ x 4 cos4 φ x 6 cos6 φ 1 − + − + . . . dφ.  2! 4! 6!  

But from definite integrals, we have π

∫0

cos2 r φ dφ = 2 ∫ = 2.

π /2 0

cos2 r φ dφ

(2 r − 1)(2 r − 3) . . . 5.31 . π 13 . .5. . . . (2 r − 1) ⋅ = π. 2 r (2 r − 2) . . . 6.4.2 2 2.4.6. . . . (2 r)

∴ from (1), we have J0 ( x) =

 1 x2 1 x4 1. 3 x6 1. 3 . 5 ⋅ π+ ⋅ π− ⋅ π + . . . π − π 2! 2 4 ! 2 .4 6 ! 2 .4 .6 

...(1)

D-396

x2

= 1−

2

x2

= 1−

2

2

2 .4

−

x4

+

22

x4

+

2

24 .(2 !)2

x6

+ ...

2

2 .42 .62 −

x6

+ ... =

26 .(3 !)2

∞

(− 1)r x2 r

r =0

(2 r . r !)2

Σ

.

Comprehensive Exercise 1 Jn( x)

1

1.

(i) Prove that lim

2.

(ii) Prove that Jn + 3 + Jn + 5 = (2 / x) (n + 4) Jn + 4 . (Kanpur 2009; Rohilkhand 11) (iii) Prove that 4 Jn ′ ′ ( x) = Jn − 2 ( x) − 2 Jn( x) + Jn + 2 ( x). Prove that (i) J0 ′ = − J1 (Kanpur 2008, 12; Lucknow 08; Agra 10;

x

x→0

n

=

n

2 . Γ(n + 1)

; n > − 1. (Kanpur 2007; Purvanchal 11)

Rohilkhand 10, 11; Agra 2007)

(ii)

J2 = J0 ′ ′ − x

−1

J0 ′

and (iii)

J2 − J0 = 2 J0 ′ ′ .

(Agra 2007)

3.

Establish the differential formula x2 Jn ′ ′ ( x) = (n2 − n − x2 ) Jn( x) + x Jn + 1( x), (n = 0 , 1, 2, . . .).

4.

(i)

2 [n Jn − (n + 2) Jn + 2 + (n + 4) Jn + 4 − . . .] x and hence deduce that 1 x Jn = (n + 1) Jn + 1 − (n + 3) Jn + 3 + (n + 5) Jn + 5 − . . . 2

Prove that Jn − 1 =

(Avadh 2008; Purvanchal 14)

5.

(ii) Prove that

2 n  Jn ′ =  Jn − (n + 2) Jn + 2 + (n + 4) Jn + 4 − . . . . x 2 

(i)

J0

Prove that

2

+ 2( J1

2

+ J2

2

+ J3

2

+ . . .) = 1.

(Agra 2010, 11)

−1 / 2

6.

Deduce that | J0 ( x)|≤ 1,| Jn ( x)| ≤ 2 , n ≥ 1. (Agra 2009) d 2 2 (ii) Prove that ( x Jn Jn + 1) = x ( Jn − Jn + 1). dx (Purvanchal 2011) (i) Prove that x = 2 J0 J1 + 6 J1 J2 + . . . + 2(2 n + 1) Jn Jn + 1 + . . . (ii) Prove that

7.

∫

Jn +1 ( x) dx =

2 Prove that (i) J3 /2 ( x) =    πx 

∫

Jn − 1( x) dx − 2 Jn( x) + C.

1  x sin x − cos

 x 

(Meerut 2007, 09; Agra 03;

Rohilkhand 08, 11; Kanpur 11; Purvanchal 11)

2 (ii) J5 /2 ( x) =    πx  8.

 x . 

Show that (i)

9.

 3 − x2  3  2  sin x − cos x  x 

x

∫0

x

x n Jn − 1( x) dx = x n Jn( x) and (ii)

If n > − 1, show that ∫

x 0

x − n Jn + 1( x) dx =

∫0

x

n +1

1 n

2 Γ(n + 1)

Jn( x) dx = x

n +1

Jn + 1( x).

− x − n Jn( x). (Purvanchal 2011)

D-397 π /2

10.

Prove that ∫

0

11.

Prove that ∫

0

π

√ (πx) J1 /2 (2 x) dx = 1.

e− a

x

J0 (bx) dx =

1 2

√ (a + b2 )

, a > 0.

(Agra 2002; Lucknow 10)

12. From the recurrence formula 2 Jn ′ = Jn − 1 − Jn + 1 r (r − 1) deduce the result 2 r Jnr = J n − r − rJn − r + 2 + Jn − r + 4 + … + (1)r Jn + r . 2! 1

n− 1 ( x /2)n 1 13. Prove that for n > − , Jn ( x) = (1 − t2 ) 2 e i x t dt. ∫ 1 − 1 2 √ π √ (n + ) 2 2 x x 14. Prove that ∫ t { Jn (t)}2 dt = { Jn2 ( x) − Jn − 1 ( x) Jn + 1 ( x)}. 0 2

15.

(i) (ii)

Express

∫ Express ∫

J3 ( x) dx in terms of J0 and J1 . x

−3

(Lucknow 2011)

J4 ( x) dx in terms of J0 and J1 .

Objective Type Questions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

2.

3.

4.

5.

n Jn ( x) − x Jn +1 ( x) = (a) x Jn ( x)

(b) (n + 1) Jn ′ ( x)

(c) x Jn ′ ( x)

(d) None of these.

− n Jn ( x) + x Jn − 1 ( x) = … (a) J ′ n −1 ( x)

(b) x Jn ′ ( x)

(c) (n − 1) Jn ( x)

(d) None of these.

Jn − 1 ( x) − Jn + 1 ( x) = … (a) 2 Jn ′ ( x)

(b) 2 Jn ( x)

(c) n Jn ( x)

(d) None of these.

x [ Jn − 1 ( x) + Jn + 1 ( x)] = … (a) 2 Jn ( x)

(b) 2 Jn ′ ( x)

(c) 2 n Jn ( x)

(d) None of these.

The value of Jn ( x) is (a)

∞

Σ

r =0 ∞

(b) Σ

r =0

 x (− 1)r    2

n+2 r

1 r ! Γ (n + r + 1)

 x (− 1)r    2

n+2 r

1 r ! Γ (n + r)

(Agra 2006)

D-398

(c)

∞

Σ

r =0

 x (− 1)r    2

n+ r

1 r ! Γ(n + r + 1)

(d) None of these.

(Rohilkhand 2004)

Fill In The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

The differential equation

d2 y dx2

+

 1 dy n2  + 1 − 2  y = 0 is called … . x dx  x 

2. The most general solution of Bessel’s equation is … . 3. When n is a positive integer, Jn ( x) is the coefficient of z n in the expansion of … . 4.

J− n ( x) = (−1)n … .

5.

d −n [x Jn ( x)] = … . dx

6. 7.

J0 ( x) is the solution of the differential equation … . d [ x n Jn ( x)] = … dx

True or False Write ‘T’ for true and ‘F’ for false statement. 1. 2. 3. 4.

J0 ( x) is called Bessel’s function of zeroeth order. The most general solution of Bessel’s equation is y = A Jn ( x) + B J− n ( x). Jn ( x) is called Bessel’s function of the first kind of order n. J− n ( x) is called Bessel’s function of second kind of order −n.

A nswers 1 Multiple Choice Questions 1.

(c).

2.

(b).

3.

(a).

4.

(c).

5. (a)

Fill in the Blank(s) 1.

Bessel’s equation.

2. A Jn ( x) + B J− n ( x). 3. e 2

5. − x − n Jn + 1 ( x).

6.

d y 2

dx

+

x [z − (1 / z)] /2

1 dy + y = 0 . 7. x x dx

n

.

4. Jn ( x).

Jn −1 ( x).

True or False 1.

T.

2.

T.

3.

T.

4.

F.

¨

SECTION

B INTEGRAL TRANSFORMS C hapters 1.

The Laplace Transform

2.

The Inverse Laplace Transform

3.

Applications of Laplace Transform

4.

Fourier Transforms

5.

Finite Fourier Transforms

6.

Applications of Fourier Transforms in Initial and Boundary Value Problems

7.

Fourier Series

T-3

1 T he L aplace T ransform

aplace transform or Laplace transformation reduces the problem of solving a differential equation to an algebraic problem. It is a method for solving linear differential equations arising in Physics and Engineering.

L

1.1 Integral Transform Let K ( p, t) be a function of two variables p and t, where p is a parameter (may be real or complex) independent of t. The function f ( p) defined by the integral (assumed to be convergent) f ( p) =

∞

∫− ∞

K ( p, t) F (t) dt

is called the integral transform of the function F (t) and is denoted by T {F (t)}. The function K ( p, t) is called the kernel of the transformation. Remark: Some authors use the letter ‘s’ in place of ‘p’.

T-4

1.2 Laplace Transformation If the kernel K ( p, t) is defined as for t < 0 0 K ( p, t) =  − pt for t ≥ 0 e then

f ( p) =

∞

∫0

e − pt F (t) dt.

…(1)

The function f ( p) defined by the integral (1) is called the Laplace transform of the function F (t) and is also denoted by L {F (t)}

or

F ( p).

Thus Laplace transform is a function of a new variable (or parameter) p given by (1). Note: The Laplace transform of F (t) is said to exist if the integral (1) converges for some values of p, otherwise it does not exist.

1.3 Linearity Property of Laplace Transformation (Bundelkhand 2014)

A transformation T is said to be linear if for every pair of functions F1 (t) and F2 (t) and for every pair of constants a1 and a2 , we have T { a1 F1 (t) + a2 F2 (t)} = a1 T {F1 (t)} + a2 T {F2 (t)}. Theorem: The Laplace transformation is a linear transformation, i.e. L { a1 F1 (t) + a2 F2 (t)} = a1 L {F1 (t)} + a2 L {F2 (t)} where a1 , a2 are constants. Proof:

(Avadh 2010)

We have L {F (t)} =

∴

∞

∫0

e − pt F (t) dt .

L { a1 F1 (t) + a2 F2 (t)} = = a1 ∫

∞ 0

∞ − pt

e

∫0

e − pt F1 (t) dt + a2

∞

∫0

{ a1 F1(t) + a2 F2 (t)} dt

e − pt F2 (t) dt

= a1 L {F1 (t)} + a2 L {F2 (t)}.

1.4 Piecewise (or Sectionally) Continuous Function A function F (t) is said to be piecewise (or sectionally) continuous on a closed interval a ≤ t ≤ b, if it is defined on that interval and is such that the interval can be subdivided into a finite number of intervals, in each of which F (t) is continuous and has finite right and left hand limits.

T-5

1.5 Functions of Exponential Order A function F (t) is said to be of exponential order α as t → ∞ if there exists a positive constant (real) M, a number α and a finite number t0 such that | F (t)| < Me αt or | e − α t F (t)| < M , for all t ≥ t0 . If a function F (t) is of exponential order α, it is also of β, β > α.

1.6 A Function of Class A A function which is piecewise (or sectionally) continuous on every finite interval in the range t ≥ 0 and is of exponential order as t → ∞ is known as ‘a function of class A’.

1.7 Existence of Laplace Transform Theorem: If F (t) is a function which is piecewise continuous on every finite interval in the range t ≥ 0 and satisfies| F (t)| ≤ Me at for all t ≥ 0 and for some constants a and M, then the Laplace transform of F (t) exists for all p > a. Proof:

We have

The integral ∫

t0 0

∞

L {F (t)} =

∫0

=

∫0

t0

e − pt F (t) dt e − pt F (t) dt +

∞

∫ t0

e − pt F (t) dt

…(1)

e − pt F (t) dt exists since F (t) is piecewise continuous on every finite

interval 0 ≤ t ≤ t0 . Now

∞

∫ t0

e − pt F (t) dt ≤ ∫

∞ t0

| e − pt F (t)| dt ≤

∞

∫ t0

e − pt Me at dt , since | F (t)| ≤ Me at

= ∴ But

∞

∫ t0 ∞

∫ t0

e −(p − a) t M dt = e − pt F (t) dt ≤

Me − ( p − a) t0 , p > a. p− a

Me −(p − a)t0 , p > a. p− a

Me − ( p − a) t0 can be made as small as we please by taking t0 sufficiently large. p− a

Thus from (1), we conclude that L {F (t)} exists for all p > a. Note 1: The above theorem of existence of Laplace transform can also be stated

as :

T-6

“If F (t) is a function which is piece-wise continuous on every finite interval in the range t ≥ 0 and is of exponential order a as t → ∞, the Laplace transform of F (t) exists for all p > a”.

Or “If F (t) is a function of class A, the Laplace transform of F (t) exists for p > a”. Note 2: Conditions in the theorem are sufficient but not necessary for the existence of

Laplace transform. If these conditions are satisfied, the Laplace transform must exist. If these conditions are not satisfied, the Laplace transform may or may not exist. We can show this by the following example. Example:

Consider the function F (t) = 1 / √ t.

Here F (t) → ∞ as t → 0, from the right. Thus the function F (t) is not piece-wise continuous on every finite interval in the range t ≥ 0. But F (t) is integrable from 0 to any positive value t0 . | F (t)| < Me at for all t > 1 with

Also Now

L {F (t)} = ∞

∞

∫0

e − pt ⋅ F (t) dt 1 dt, which converges for p > 0 √t

=

∫0

e − pt ⋅

=

2 √p

∫0

=

2 √π ⋅ , since √p 2

∞

M = 1 and a = 0.

2

e − x dx, putting √ ( pt) = x so that

= √ (π / p),

∞

∫0

2

e − x dx =

dt 2 = dx √t √ p

√π 2

p > 0.

Thus L {1/ √ t} exists for p > 0 even if 1/ √ t is not piecewise continuous in the range t ≥ 0.

1.8 Laplace Transforms of Some Elementary Functions (i)

Laplace transform of the function F (t) = 1.

Solution: We have L { F (t)} =

∴

L {1} =

∞

∫0

∞

∫0

e − pt F (t) dt. ∞

 e − pt  1 e − pt ⋅ 1 dt = −  = , p> 0 . p p  0

Here the condition p > 0 is necessary, since the integral is convergent for p > 0 and divergent for p ≤ 0.

T-7 (ii) Laplace transform of the function F (t) = t n, n is any real number greater than –1. Solution:

We have,

∴

∞

L {F (t)} =

∫0

L { t n} =

∞

e − pt F (t) dt.

e − pt t ndt =

∫0

∞

e − pt t(n + 1) − 1dt

∫0

…(1)

Now from Eulerian integral of the second kind known as ‘Gamma function’, we have ∞ − ax m −1 Γ(m) ∫0 e x dx = am , if a > 0 and m > 0. Γ (n + 1) ∴ from (1), L { t n} = , if p > 0 and n + 1 > 0 i.e., n > − 1. p n+1 Thus if n is any real number greater than –1, we have Γ (n + 1) L { t n} = , p > 0. p n+1 Here the condition p > 0 is necessary for the convergence of the integral (1). (iii) Laplace transform of the function F (t) = t n, n is a positive integer. (Gorakhpur 2008)

Solution:

We have

∴

∞

L {F (t)} =

∫0

L { t n} =

∞

=

∫0

Γ (n + 1) p n+1

e − pt F (t) dt

e − pt t n dt =

∞

∫0

e − pt t(n + 1) −1 dt

if p > 0

,

∞ − ax m −1 Γ(m)  ∵ e x dx = m ,  ∫ 0 a  if a > 0 and m > 0. Here n + 1 > 0, n being a positive integer.   n! = n +1 , p > 0. p [ ∵ Γ (n + 1) = n !, n being a positive integer] n! Thus if n is any positive integer, we have L { t n} = n + 1 , p > 0 . p

Here the condition p > 0 is necessary for the convergence of the integral defining the Laplace transform of t n. (iv)

Laplace transform of the function F (t) = e at .

Solution:

(Rohilkhand 2009)

Here L { e at} =

∞

∫0

e − pt ⋅ e at dt =

∞

∫0

e − (p − a) t dt

T-8 ∞

 e −(p − a) t  = −  , p≠ a p− a   0 =

1 , p > a. p− a

Here the condition p > a is necessary, since the integral is convergent for p > a and divergent for p ≤ a. (v) Laplace transform of the function F (t) = sin at. (Avadh 2010, 11)

Solution: L {sin at} =

∞

e − pt sin at dt

∫0

∞

 e − pt  (− p sin at − a cos at) = 2 2 p + a 0

  e ax ax (a sin bx − b cos bx)  ∵ ∫ e sin bx dx = 2 2 a +b   =

a 2

p + a2

, p > 0.

Here the condition p > 0 is necessary for the convergence of the integral (1). (vi) Laplace transform of the function F (t) = cos at. Solution:

L {cos at } =

∞

(Rohilkhand 2006, 07, 10)

e− pt cos at dt

∫0

∞

 e − pt  (− p cos at + a sin at) = 2 2 p + a 0  ∵  =

p 2

p + a2

∫e

ax

cos bx dx =

e ax

 (a cos bx + b sin bx) a +b  2

, p > 0.

(vii) Laplace transform of the function F (t) = cosh at. Solution:

1 L {cosh a t} = L { (e at + e − at )} 2 1 1 = L { e at} + L { e − at} 2 2 1 1 1 1 = ⋅ + ⋅ , p > a and p > − a 2 p− a 2 p+ a =

p 2

p −a2

, p > | a |.

2

T-9 (viii) Laplace transform of the function F (t) = sinh at.

1 Solution: L {sinh at} = L { (e at − e − at )}

2 1 = [ L { e at} − L { e − at}] 2

= =

1 1 1  −   , p > a and p > − a 2  p − a p + a a 2

p − a2

, p > | a |.

1.9 Laplace Transforms of Some Elementary Functions Laplace Transforms of some elementary functions obtained in 1.8 are given here in the form of a table. Laplace Transforms of Some Elementary Functions F (t)

L {F (t)}

1.

1

1 , p> 0 p

2.

tn (n is a positive integer)

3.

ta (a > − 1)

4.

e at

5.

sin at

6.

cos at

7.

sinh at

8.

cosh at

n! p n +1

, p> 0

Γ (a + 1) p a +1

, p> 0

1 , p> a p− a a 2

p + a2 p 2

p + a2 a p2 − a2 p 2

p − a2

, p> 0

, p> 0

, p > | a|

, p > | a|

If we know the Laplace transforms in the above table then nearly all the transforms can be obtained by using the general theorems which we shall consider later on.

T-10

Example 1:

Find L {(t 2 + 1)2} .

Solution: We have

L { (t2 + 1)2} = L { t 4 + 2 t2 + 1} = L { t 4} + 2 L {t2} + L {1} 4!

= Example 2:

p5

+2⋅

2! p3

+

2 4 1 24 + 4 p + p = , p > 0. p p5

Find the L.T. of the function F (t) = (sin t − cos t)2 .

Solution:

2

2

We have L { (sin t − cos t) } = L {sin t + cos t − 2 sin t cos t} = L {1} − L {sin 2 t} = =

Example 3: (i) Solution:

(Gorakhpur 2006, 10) 2

p2 − 2 p + 4

1 2 − , p> 0 p p2 + 22

, p > 0.

p ( p2 + 4)

Evaluate L {4 cos2 2 t} .

We have L {4 cos2 2 t} = L {2 (1 + cos 4 t)} = 2 [ L{1} + L {cos 4 t}] 1 p  =2 + 2 , p> 0 2 p p + 4  =

Example 3: (ii) Solution:

4 ( p2 + 8) p ( p2 + 16)

, p > 0.

Evaluate L { sin2 at}.

(Rohilkhand 2014)

We have 1 1 L {sin2 at} = L { (1 − cos 2 at)} = [ L {1} − L {cos 2 at}] 2 2 =

= Example 4: Solution:

 p 1 1  − 2  , p> 0 2 2  p p + (2 a)  2 a2 p ( p2 + 4 a2 )

, p > 0.

Find L {3 t 4 − 2 t3 + 4 e −3 t − 2 sin 5 t + 3 cos 2 t}. We have L {3 t 4 − 2 t 3 + 4 e −3 t − 2 sin 5 t + 3 cos 2 t} = 3 L { t 4} − 2 L { t 3} + 4 L { e −3 t} − 2 L{sin 5 t} + 3 L {cos 2 t}

T-11

4!

=3⋅

5

p

−2⋅

3!

+4⋅

4

p

p 1 5 −2⋅ 2 +3⋅ 2 , 2 p+3 p +5 p + 22 p > 0 and p > − 3 i. e. p > 0

=

Example 5: Solution:

72 p5

−

12

3p

4 10 − + , p > 0. p + 3 p2 + 25 p2 + 4

+

p4

(t − 1)2 , t > 1 Find L { F (t)} if F (t) =  0 , 0 < t < 1.  We have

L { F (t)} = =

=

∞

1

∫0

F (t) e − pt dt = ∫

∞

(t − 1)2 e − pt dt =

∫1

∞

∫0

e

− p − px

e

0

2

x dx = e

0 ⋅ e − pt dt +

Example 6: Solution:

2! 3

p

=

2e−p p3

(t − 1)2 e − pt dt

∞ 2

∫0 −p

x e − p ( x +1) dx, putting t − 1 = x ,

∞

∫0

so that dt = dx and changing the limits Γ(3) e − px x(3 −1)dx = e − p ⋅ 3 , p > 0 p  ∵ 

= e−p .

∞

∫1

∞ − ax

∫0

e

x m −1dx =

Γ(m) am

, a > 0, m > 0

 

, p > 0. 2

Show that the function e t is not of exponential order as t → ∞. We have 2

lim { e − at F (t)} = lim { e − at e t } = lim e t(t − a)

t→ ∞

t→ ∞

t→ ∞

= ∞ for all values of a. Hence whatever be the value of a, we cannot find a number M such that 2

e t < M e at . ∴ the given function is not of exponential order as t → ∞ . Example 7: Solution:

Find L { sin √ t }.

(Purvanchal 2007; Avadh 12)

We have   (√ t)3 (√ t)5 (√ t)7 L {sin √ t} = L  √ t − + − + ... 3! 5! 7!    1 /2 t3 /2 t5 /2 t7 /2  = L t − + − + .... 3! 5! 7!   1 1 1 1 /2 3 /2 5 /2 = L {t } − L {t } + L {t } − L { t7 /2} + ... 3! 5! 7! 3 5 7 9 Γ( ) Γ( ) Γ( ) Γ( ) 1 1 1 2 2 2 = 3 /2 − ⋅ + ⋅ − ⋅ 2 + ... 3 ! p5 /2 5 ! p7 /2 7 ! p9 /2 p

T-12

1 3 1 5 3 1 √π ⋅ √π ⋅ ⋅ √π 1 1 2 2 2 − ... = 2 3 /2 − ⋅2 2 + ⋅ 1⋅ 2 ⋅ 3 1⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 p p5 /2 p7 /2 2 3  √π  1 1 1  1 1  1  − + − 1 + ...       3 /2 3 !  4 p 2p  1!  4 p 2 !  4 p    √π = 3 / 2 ⋅ e −1 / 4 p . 2p

=

Show that the Laplace transform of the function

Example 8:

F (t) = t n, − 1 < n < 0 , exists, although it is not a function of class A. Here F (t) → ∞ as t → 0 from the right i.e., the function is not piecewise continuous on every finite interval in the range t ≥ 0 . Solution:

 tn  We have lim { e − at F (t)} = lim  at  t→ ∞ t→ ∞ e  1

= lim

t→ ∞ t − n e at

= lim

1

t→ ∞ t m e at

,

where 0 < m < 1

= 0 , if a > 0 . n

∴ F (t) = t is of exponential order. Since F (t) = t n is not sectionally continuous over every finite interval in the range t ≥ 0, hence it is not a function of class A. But t n is integrable from 0 to any positive number t0 . Now

∞ − pt

L { F (t)} = =

∫0

e

Γ(n + 1) p(n + 1)

F (t) dt =

∞ − pt n

∫0

e

t dt = ∫

∞ − pt (n+1)−1 0

e

t

dt

, if p > 0 and n + 1 > 0 i. e., n > − 1.

Hence the Laplace transform of t n, − 1 < n < 0, exists, although it is not a function of class A.

Comprehensive Exercise 1

1.

Find the L.T. of the following functions. (i) 2 t3 − 6 t + 8 (ii) sin t cos t (iii) cosh2 2t.

2.

(iv) 6 sin 2 t − 5 cos 2 t

(i) 3 cosh 5 t − 4 sin 5 t (ii) 7 e2 t + 9 e −2 t + 5 cos t + 7 t3 + 5 sin 3 t + 2 . (iii) 2 e3 t − e −3 t . (iv) F (t) = (e

at

− 1) / a.

(Rohilkhand 2006)

T-13

3.

4.

 et , 0 < t ≤ 1 (i) Find L {F (t)}, if F (t) =  t >1  0, 0 , 0 < t < 2 (ii) Find L {F (t)}, where F (t) =  t>2  4, (i) Find the Laplace transform of the following function :  x / a, 0 < x < a f ( x) =  1, x > a.  0 , 0 < t < 1  (ii) Find L {F (t)}, where F (t) =  t, 1 < t < 2 0 , t > 2. 

5.

6.

(Lucknow 2010)

(i) Find Laplace Transform of the function F (t), where sin t, 0 < t < π . F (t) =  t> π  0,

(Lucknow 2007)

e t, 0 < t < 5 (ii) Find L {F (t)}, if F (t) =  t > 5. 3, 1, 0 < t < 2 (i) Find L {F (t)}, if F (t) =  t > 2.  t,  t, 0 < t < 4 (ii) Find L { F (t)} if F (t) =  t > 4. 5,

7.

(i) Prove that L {H (t)} =

2 (1 − e − π p ) 2

p +4

sin 2 t, , where H (t) =   0,

 1  1 (ii) Prove that L  ⋅ =  √ (πt) √ p 8. 9.

(Avadh 2014)

 cos √ t   π  −1 / 4 p Show that L  .  =   e  p  √t 

(Purvanchal 2007; Kashi 14)

2

Show that t is of exponential order 3.

A nswers 1 1.

(i) 12 / p4 − 6 / p2 + 8 / p, p > 0 (iii)

2.

(i)

p2 − 8 2

p ( p − 16) 3 p − 20 p2 − 25

, p> 4

, p> 5

0< t< π t > π.

(ii) (iv)

1 ( p2 + 4) 12 − 5 p p2 + 4

, p> 0

, p> 0

T-14

(ii)

(iii) (iv) 3.

(i)

4.

(i)

16 p − 4

+

p2 − 4 p+9 p2 − 9

5p

+

p2 + 1

42 + 2 p3

+

p4

15 p2 + 9

, p> 2

, p> 3

1 , p > 0 if a < 0 and p > a if a > 0 p ( p − a) 1 [1 − e − ( p − 1)], p ≠ 1 ( p − 1) 1 2

ap

(ii)

4 −2 p e , p> 0 p

(1 − e − ap ), p > 0

2 1 1  1 (ii) −  + 2  e −2 p +  + 2  e − p , p ≠ 0 . p p  p p  5.

(i)

6.

(i)

e− p π + 1

(ii)

2

p +1 1 1 [1 + e −2 p ] + 2 e −2 p , p > 0 p p

(ii)

1 − e −5( p −1) 3 −5 p + e , p> 0 p−1 p 1 + ( p − 1) e −4 p

, p> 0

p2

1.10 First Translation or Shifting Theorem If L {F (t)} = f ( p) when p > α, then L { e at F (t)} = f ( p − a),

p > α + a; i.e., if f ( p) is the

Laplace transform of F (t) then f ( p − a) is the Laplace transform of e at F (t). (Gorakhpur 2008, 11)

Proof. By definition, we have f ( p) = L { F (t)} =

∞ − pt

∫0

e

F (t) dt,

…(1)

the integral (1) being given to be convergent if p > α. ∴

∞ −( p − a) t

f ( p − a) =

∫0

=

∫0

∞

e

F (t) dt

…(2)

e − pt ⋅ e at F (t) dt = L { e at F (t)}.

Obviously if the integral (1) converges when p > α, the integral (2) converges when p − a > α i. e., p > α + a.

1.11 Second Translation or Shifting Theorem  F (t − a) , t > a If L {F (t)} = f ( p) and G (t) =  then L { G (t)} = e − ap f ( p). 0 , t < a ,  (Avadh 2007; Rohilkhand 07; Lucknow 09)

T-15

Proof:

By definition, we have a

∞

e − pt G (t) dt =

∞

e − pt F (t − a) dt =

L { G (t)} =

∫0

=

∫a

∫0

e − pt ⋅ 0 dt + ∞

∫0

∞

e − pt ⋅ F (t − a) dt

∫a

e − p (a + x) ⋅ F ( x) dx, putting t − a = x so that dt = dx

= e − pa

∞

∫0

e − px F ( x) dx = e − pa

∞

∫0

e − pt F (t) dt, by a property of definite integrals

=e

− pa

L{F (t)} = e

− pa

f ( p).

1.12 Change of Scale Property If L {F (t)} = f ( p), then L { F (at)} =

1  p f  ⋅ a  a

(Avadh 2006, 11; Kanpur 07; 10; Lucknow 07; Rohilkhand 08, 11)

Proof: By definition, we have L {F (at)} =

∞

∫0

1 a 1 = a =

=

Example 9:

e − pt F (at) dt ∞

e − p ( x / a) F ( x) dx, putting at = x, so that dt = (1 / a) dx

∞

e −( p / a)t F (t) dt

∫0 ∫0

1  p f   , since f ( p) = a  a

∵  ∞ − pt

∫0

e

b

∫a

f ( x) dx =

f (t) dt 

F (t) dt.

Find (i) L { e − t sin2 t}

(Rohilkhand 2004)

(ii) L { e t sin2 t}. Solution:

b

∫a

(Kanpur 2012)

1 We have L {sin2 t} = L { (1 − cos 2 t)} 2 =

p  1 1 2 = f ( p) , say.  − 2 = 2 2 2  p p + 2  p ( p + 4)

∴ From first shifting theorem, we have (i)

L { e − t sin2 t} = f ( p + 1) =

(ii)

L { e t sin2 t} = f ( p − 1) =

2 ( p + 1) {( p + 1)2 + 4} 2 2

( p − 1) {( p − 1) + 4}

=

=

2 ( p + 1) ( p2 + 2 p + 5) 2

( p − 1) ( p2 − 2 p + 5)

.

⋅

T-16

Find L { e − t (3 sinh 2 t − 5 cosh 2 t)}.

Example 10:

We have L {3 sinh 2 t − 5 cosh 2 t} p 6 − 5p 2 = f ( p), say. =3⋅ 2 −5⋅ 2 = 2 2 2 p −2 p −2 p −4

Solution:

∴

From first shifting theorem, we have L { e − t (3 sinh 2 t − 5 cosh 2 t)} = f ( p + 1) =

6 − 5 ( p + 1) 2

( p + 1) − 4

=

1− 5p 2

p + 2p − 3

.

Example 11: Using first shifting theorem evaluate

L { e6 t (t + 2)2}. Solution:

We have, L {(t + 2)2} = L { t2 + 4 t + 4} 2!

= ∴

p3

1!

+4⋅

p2

+

4 2 + 4 p + 4 p2 = = f ( p), say. p p3

From first shifting theorem, we have L { e6 t (t + 2)2} = f ( p − 6) =

Example 12:

Given L {F (t)} =

2 + 4 ( p − 6) + 4 ( p − 6)2 ( p − 6)3 p2 − p + 1 (2 p + 1)2 ( p − 1)

=

4 p2 − 44 p + 122 ( p − 6)3

,

applying the change of scale property show that L {F (2 t)} = Solution:

p2 − 2 p + 4 4 ( p + 1)2 ( p − 2)

.

We have L {F (t)} =

p2 − p + 1 (2 p + 1)2 ( p − 1)

= f ( p), say.

∴ by the change of scale property, we have 1 L {F (2 t)} = f ( p / 2) 2 p2 − 2 p + 4 ( p / 2)2 − ( p / 2) + 1 1 ⋅ = ⋅ = 2 [2 ⋅ ( p / 2) + 1]2 [( p / 2) − 1] 4 ( p + 1)2 ( p − 2) Example 13:.

Solution:

Then

 cos (t − 2 π), t > 2 π / 3  Find L {F (t)}, where F (t) =  3  0 , t < 2 π / 3.

Let φ (t) = cos t.  φ (t − 2 π / 3), t > 2 π / 3 F (t) =  0 , t < 2 π / 3. 

.

T-17

p

We have L { φ (t)} = L {cos t} =

2

p +1

= f ( p), say.

∴ From second shifting theorem, we have L {F (t)} = e(−2 π /3)p . f ( p) = e −2 πp /3 ⋅ Aliter:

We have L {F (t)} =

∞

∞

2π  e − pt ⋅ cos  t −  dt  3

=

∫ 2 π /3

=

∫0

∞

.

2π  e − pt ⋅ cos  t −  dt  3

e − pt ⋅ 0 dt +

∫0

p +1

e − pt F (t) dt

∫0

2 π /3

=

p 2

∞

∫ 2 π /3

e − p [ x + (2 π /3)] cos x dx, putting t − 2 π / 3 = x and dt = dx ∞

e − px cos x dx

∞

e − pt cos t dt

= e − p (2 π /3)

∫0

= e − p (2 π /3)

∫0

= e − p (2 π /3) L{cos t} = e −2 p π /3 [ p / ( p2 + 1)], p > 0 .

Comprehensive Exercise 2 Find the Laplace transform of the following functions. 1.

(i) t3 e −3 t

(Kanpur 2008)

(ii) e − at t

n−1

/ (n − 1) !

(iii) t n e at . 2.

3t

(i) e

t

sin 4 t. (Lucknow 2010) 2

(Avadh 2010) 3t

(ii) e

cos 5 t.

− 4t

3.

(i) e cos t.

(ii) e

4.

(i) e − t (3 sin 2 t − 5 cosh 2 t).

(ii) e − 2 t (3 cos 6 t − 5 sin 6 t).

5.

(t + 3)2 . e t .

6.

If L { F (t) } = f ( p), find L { F (t) cos ωt }.

7.

Applying change of scale property, find (i) L {sinh 3 t } and

8. 9. 10.

cosh 2 t.

(ii) L {cos 5 t }.

 et − a , t > a Find L {G (t)}, where G (t) =  , t < a. 0 sin (t − π / 3), t > π / 3 Find L {F (t)}, where F (t) =  0 , t < π / 3.  If L {F (t)} = (1 / p) e −1 / p , prove that L { e − t F (3 t)} =

e −3 /( p + 1) . p+1

T-18

A nswers 2 1.

(i) 6 /( p + 3)4 .

(ii) 1 / ( p + a)n

(iii) n !/ ( p − a)n +1. 2.

(i) 4 /( p2 − 6 p + 25).

3.

(i)

4.

(i)

5.

(ii) ( p − 3) /( p2 − 6 p + 34).

( p2 − 2 p + 3) 2

( p − 1) ( p − 2 p + 5) 6 p2 + 2 p + 5

9 p2 − 12 p + 5 3

( p − 1) 3

7.

(i)

8.

e − ap , p > 1. p−1

2

p −9

−

(ii)

⋅

5 ( p + 1) p2 + 2 p − 3

(ii)

⋅

6.

⋅

(ii)

.

9.

p+4 2

p + 8 p + 12 3 p − 24 p2 + 4 p + 40

⋅ ⋅

1 [ f ( p − iω) + f ( p + iω)]. 2 p 2

p + 25 e − π p /3 p2 + 1

, p > 0. , p > 0.

1.13 Laplace Transform of the Derivative of F ( t ) Theorem: Let F (t) be continuous for all t ≥ 0 and be of exponential order a as t → ∞ and if F ′ (t) is of class A, then Laplace transform of the derivative F ′ (t) exists when p > a, and L {F ′ (t)} = pL {F (t)} − F (0 ). Proof: Case I:

In case F ′ (t) is continuous for all t ≥ 0, then L {F ′ (t)} =

∞

∫0

e − pt F ′ (t) dt ∞

= [e − pt F (t)] 0 + p ∫ =

…(1) ∞ 0

e − pt F (t) dt

[Integrating by parts]

lim − pt e F (t) − F (0 ) + pL {F (t)}. t→ ∞

Now | F (t)| ≤ Me at for all t ≥ 0 and for some constants a and M. We have | e − pt F (t)| = e − pt | F (t)| ≤ e − pt Me at = Me −( p − a)t → 0 as t → ∞ if p > a. ∴

lim − pt e F (t) = 0 for p > a. t→ ∞

Therefore from (2) we conclude that L {F ′ (t)} exists and L {F ′ (t)} = pL {F (t)} − F (0 ).

…(2)

T-19

Case II: In case F ′ (t) is merely piece-wise continuous, the integtral (1) may be broken as the sum of integrals in different ranges from 0 to ∞ such that F ′ (t) is continuous in each of such parts. Then proceeding as in case I, we shall have L {F ′ (t)} = pL {F (t)} − F (0 ). Note 1:

If F (t) fails to be continuous at t = 0 but lim [F (0 + 0 ) is not equal to F (0 ), F (t) = F (0 + 0 ) exists, t→0 which may or may not exist] L {F ′ (t)} = pL {F (t)} − F (0 + 0 ).

then Note 2:

If F (t) fails to be continuous at t = a, then L { F ′ (t)} = pL {F (t)} − F (0 ) − e − ap [ F (a + 0 ) − F (a − 0 )]

where F (a + 0 ) and F (a − 0 ) are the limits of F at t = a, as t approaches a from the right and from the left respectively. The quantity F (a + 0 ) − F (a − 0 ) is called the jump at the discontinuity t = a. Proof:

L {F ′ (t)} =

∞

∫0

e − pt F ′ (t) dt = a

= [e − pt F (t)] 0 + p ∫

a 0

a

∫0

e − pt F ′ (t) dt +

∞

∫a

e − pt F ′ (t) dt ∞

e − pt F (t) dt + [e − pt F (t)] a

= e − ap F (a − 0 ) − F (0 ) +

∞

+ p∫

a

− e − ap F (a + 0 ) + p ∫

0

e − pt F (t) dt

lim − pt e F (t) t→ ∞ ∞

e − pt F (t) dt

= pL {F (t)} − F (0 ) − e − ap [ F (a + 0 ) − F (a − 0 )]. lim   − pt F (t) = 0 , as shows in the theorem ∵ t → ∞ e   For more than one discontinuity of the function F (t), appropriate modification can be made.

Note 3:

1.14

Laplace Transform of the nth Order Derivative of F (t)

Theorem:

Let F (t) and its derivatives F ′ (t), F ′ ′ (t),..., F

n −1

(t) be continuous functions

n

for all t ≥ 0 and be of exponential orders as t → ∞ and if F (t) is of class A, then Laplace transform of F n (t) exists when p > a, and is given by L {F n (t)} = p n L {F (t)} − p n −1 F (0 ) − pn − 2 F ′ (0 ) − ... − F

n −1

(0 ).

T-20

Proof:

From the theorem of 1.13, we have L {F ′ (t)} = pL {F (t)} − F (0 )

…(1)

Applying the result (1) to the second order derivative F ′ ′ (t), we have L {F ′ ′ (t)} = pL {F ′ (t)} − F ′ (0 ) = p [ pL {F (t)} − F (0 )] − F ′ (0 ) = p2 L {F (t)} − pF (0 ) − F ′ (0 ).

…(2) (Gorakhpur 2006, 10)

Again applying (1) to the third order derivative F ′ ′ ′ (t), we have L {F ′ ′ ′ (t)} = pL {F ′ ′ (t)} − F ′ ′ (0 ) = p [ p2 L {F (t)} − pF (0 ) − F ′ (0 )] − F ′ ′ (0 ) = p3 L {F (t)} − p2 F (0 ) − pF ′ (0 ) − F ′ ′ (0 ). Proceeding similarly, we have L {F n (t)} = p n L {F (t)} − p n − 1 F (0 ) − p n−2 F ′ (0 ) − ... − F or

L {F n (t)} = p n L {F (t)} −

n −1

∑

n −1

(0 )

p n − 1 − r F r (0 ).

r =0

1.15 Initial-Value Theorem Let F (t) be continuous for all t ≥ 0 and be of exponential order as t → ∞ and if F ′ (t) is of class A, lim lim then F (t) = p L {F (t)}. t→0 p→ ∞ Proof: By the theorem of 1.13, we have L {F ′ (t)} =

∞

∫0

e − pt F ′ (t) dt = pL{F (t)} − F (0 )

…(1)

Since F ′ (t) is sectionally continuous and of exponential order, lim ∞ − pt ∴ e F ′ (t) dt = 0 . p→ ∞ ∫0 Taking limit as p → ∞ in (1), we have lim 0= pL {F (t)} − F (0 ) or p→ ∞ or Note:

F (0 ) =

lim p L {F (t)} p→ ∞

lim lim F (t) = p L {F (t)}. t→0 p→ ∞ If F (t) fails to be continuous at t = 0, but

by using Note (1), 1.13 theorem.

lim F (t) exists, the result still holds t→0

T-21

1.16 Final-Value Theorem Theorem: Let F (t) be continuous for all t ≥ 0 and be of exponential order as t → ∞ and if F ′ (t) is of class A, then lim lim F (t) = p L {F (t)}. t→ ∞ p→ 0 Proof. By the theorem of 1.13, we have L {F ′ (t)} =

∞

∫0

e − pt F ′ (t) dt

= pL { F (t)} − F (0 ).

…(1)

Taking limit as p → 0 in (1), we have lim lim ∞ − pt e F ′ (t) dt = pL {F (t)} − F (0 ) p→ 0 ∫0 p→ 0 ∞

or

∫0

or

[ F(t)] 0∞ = p → 0

or

lim lim F (t) − F (0 ) = pL {F (t)} − F (0 ) t→ ∞ p→ 0

or

lim lim F (t) = pL {F (t)}. t→ ∞ p→ 0

Note:

F ′ (t) dt = lim pL { F (t) − F (0 )} p→0

lim

p L {F (t)} − F (0 )

If F (t) fails to be continuous at t = 0, but

lim F (t) exists, the result still holds t→0

by using Note (1), 1.13 theorem.

1.17 Laplace Transform of Integrals Theorem:

If F (t) is piecewise continuous and satisfies| F (t)| ≤ Me at for all t ≥ 0 for some

constants a and M, then t 1 L  ∫ F ( x) dx = L { F (t)}, ( p > 0 , p > a) 0  p 

Proof:

Let F (t) be piece-wise continuous such that | F (t)| ≤ Me a t ,

(Lucknow 2010)

…(1)

for some constants a and M. If (1) holds for some negative value of a then it also holds for positive value of a. Therefore suppose that a is positive. Let

G (t) =

t

∫0

F ( x) dx.

Then G (t) is continuous because the integral of an integrable function is continuous.

T-22

Also

t

| G (t)| ≤

t

∫ 0 | F ( x)| dx ≤ ∫ 0

Me ax dx.

M at (e − 1), a > 0 . a Further G ′ (t) = F (t), except for points at which F (t) is discontinuous. | G (t)| ≤

∴

…(2)

Therefore G ′ (t) is piece-wise continuous on each finite interval. Hence from the theorem of 1.13, we have L { G ′ (t)} = pL { G (t)} − G (0 ) = pL {G (t)} [Since G (0 ) = 0 from (2)] 1 ∴ L { G (t)} = L { G ′ (t)} p t 1 L  ∫ F ( x) dx = L { F (t)} 0  p 

1.18 Multiplication By t Theorem: If F (t) is a function of class A and if L {t F (t)} = − f ′ ( p).

L {F (t)} = f ( p), then

(Rohilkhand 2009, Gorakhpur 05, 09)

Proof: We have f ( p) = L {F (t)} = f ′ ( p) =

∴

=

∞

∫0

e

− pt

F (t) dt.

d ∞ − pt e F (t) dt dp ∫ 0 ∞

∫0

∂ − pt {e F (t)} dt, ∂p

by Leibnitz’s rule for differentiating under the sign of integral ∞

te − pt F (t) dt

∞

e − pt { t F (t)} dt = − L { t F (t)}.

=−

∫0

=−

∫0

Thus L { t F ( t)} = − f ′ ( p).

1.19 Multiplication by t n Theorem:

If F (t) is a function of class A and if L {F (t)} = f ( p) then L {t

n

F (t)} = (− 1) n

dn dp n

f ( p), where n = 1, 2, 3,......

(Rohilkhand 2000; Lucknow 06, 09, 11; Kanpur 08; Avadh 10; Agra 01; Gorakhpur 11)

Proof:

We shall prove this theorem by mathematical induction. By 1.18, we have d L { t F (t)} = (− 1)1 f ( p) dp

T-23

i.e. the theorem is true for n = 1. Now assume that the theorem is true for a particular value of n say m. Then, we have m

L {t ∞

or

F (t)} = (− 1)m

e − pt t

∫0

m

dm dp m

f ( p)

F (t) dt = (− 1)m

dm

f ( p).

dp m

Now differentiating both sides w. r. t. p, we have d ∞ − pt e t dp ∫ 0

∂ − pt t {e ∂p

∞

or

m

∫0

F (t) dt = (− 1)m m

d m +1 dp m + 1

F (t)} dt = ( − 1)m

f ( p)

d m +1 dp m + 1

f ( p),

by Leibnitz’s rule for differentiating under the sign of integral or

−∫

∞ 0

∞

e − pt ⋅ t

or

∫0

e − pt ⋅ { t

or

L {t

m +1

m +1

m +1

F (t) dt = ( − 1)m

d m +1 dp m + 1

F (t)} dt = ( − 1) m + 1

F (t)} = ( − 1)m + 1

f ( p)

d m +1 f ( p ) dp m +1

d m + 1 f ( p) dp m +1

which shows that if the theorem is true for any particular value on n, it is true for the next value of n. But we have already seen that the theorem is true for n = 1. Hence it is true for n = 1 + 1 = 2 and n = 2 + 1 = 3, etc. Therefore the theorem is true for every positive integral value of n.

1.20 Division by t 1 Theorem: If L {F (t)} = f ( p), then L  F (t) =  t

∞

∫p

f ( x) dx

1 provided lim  F (t) exists. t→0  t  Proof: ∴

Let

G (t) =

1 F (t) i. e., F (t) = t G (t). t

L {F (t)} = L { t G (t)} d =− L { G (t)}, by the theorem of 1. 18 dp

T-24

or

f ( p) = −

d L { G (t)}. dp

Now integrating both sides with respect to p from p to ∞, we have ∞

− [ L { G(t)}]p = or

−

lim p→ ∞

∞

∫p

f ( p) dp

L { G (t)} + L { G (t)} =

∞

∫p

f ( p) dp

[Note that L {G (t)} is a function of p] 0 + L { G(t)} =

or

∞

∫p

f ( p) dp  L { G (t)} = lim ∵ plim →∞ p→ ∞ 

1 L  F (t) =  t

or

Example 14:

(i)

∞

∫p

∞ − pt

∫0

e

 G(t) dt = 0  

f ( x) dx

Evaluate (ii) L { e at}

L { t}

(Rohilkhand 2011)

(iii) L {− a sin at } Solution:

We have L {F ′ (t)} = pL {F (t)} − F (0 )

(i) Here let F (t) = t, then F ′ (t) = 1 and F (0 ) = 0 . ∴ or

from (1), we have L {1} = pL { t } − 0 1 1 1 1 L { t } = L {1} = ⋅ = 2 , p > 0 . p p p p

(ii) Here let F (t) = e at , then F ′ (t) = a e at and F (0 ) = 1. ∴

from (1), we have L { a e at} = pL { e at} − 1

or

a L {e at} = pL { e at} − 1

or

( p − a) L { e at} = 1.

∴

L { e at} =

1 ⋅ p− a

(iii) Here let F (t) = − a sin at, then F ′ (t) = − a2 cos at and F ′ ′ (t) = a3 sin at so that

F ′ (0 ) = − a2

and

F (0 ) = 0 .

2

∴ from

L {F ′ ′ (t)} = p L {F (t)} − pF (0 ) − F ′ (0 ),

we have

L { a3 sin at} = p2 L { − a sin at} − 0 − (− a2 )

or

( p2 + a2 ) L {− a sin at} = − a2 .

…(1)

T-25

L {− a sin at} = −

∴

Example 15: Show that L { t

2

a2

⋅

p2 + a2

cos at} =

2 p ( p2 − 3 a2 ) ( p2 + a2 )3

, p > 0. (Rohilkhand 2011; Kanpur 07)

Solution:

2

Since L {cos at} = p /( p + a ), p > 0, L { t2 cos at} = (− 1)2

∴

= Example 16: Solution:

2

 p d2  d  − p2 + a2  =     dp2  ( p2 + a2 ) dp  ( p2 + a2 )2 

2 p ( p2 − 3 a2 ) ( p2 + a2 )3

⋅

Find L {(sin at − at cos at)}.

Let F (t) = sin at − at cos at, then F ′ (t) = a2 t sin at and F (0 ) = 0 .

∴ From L {F ′ (t)} = pL {F (t)} − F (0 ), we have L { a2 t sin at} = pL{(sin at − at cos at)} − 0 . ∴

L {(sin at − at cos at)} = =

a2 a2 d L { t sin at} = − L {sin at} p p dp

=− Aliter.

1 L { a2 t sin at} p

a2 d  a  2 a3 ⋅  2  = 2 2 p dp  p + a  ( p + a2 )2

L {sin at − at cos at} = L {sin at} − a L { t cos at} a d = 2 − a ⋅ ( − 1) [ L {cos at}] 2 dp p +a =

=

a 2

2

p +a a

p2 + a2

+a

+

p  d   2  dp  p + a2 

a (a2 − p2 ) ( p2 + a2 )2

=

2 a3 ( p2 + a2 )2

.

Example 17: Show that

L {(t2 − 3 t + 2) sin 3 t} = Solution:

6 p 4 − 18 p 3 + 126 p 2 − 162 p + 432 ( p 2 + 9)3

We have L {(t2 − 3 t + 2) sin 3 t} = L { t2 sin 3 t} − 3 L { t sin 3 t} + 2 L {sin 3 t} = ( − 1)2

d2 dp2

 d  L {sin 3 t} − 3 − L {sin 3 t} + 2 L {sin 3 t} dp  

⋅

T-26

=

=

=

d2  3  d  3  3  2  +3   +2⋅ 2 2 dp  p2 + 9  p +9 dp  p + 9  18 p2 − 54 2

 − 6p  6 + 2 +3⋅ 2 2  ( p + 9)  p + 9

3

( p + 9)

6 p4 − 18 p3 + 126 p2 − 162 p + 432 ( p2 + 9)3

.

Example 18: Find L { t e − t sin t} .

(Lucknow 2007)

1

Solution: Since L {sin t} = 2 p +1

∴

2p d  1   2 = 2 dp  p + 1 ( p + 1)2

L { t sin t} = −

∴

L { t e − t sin t} =

2 ( p + 1) 2

2

[( p + 1) + 1]

=

2p + 2 2

( p + 2 p + 2)2

⋅

sin t  sin at  −1 1 and hence find L   = tan  ⋅ Does the Laplace p  t   t 

Example 19: Prove that L  

transform of Solution:

Now

cos at exist? t

Let F (t) = sin t. lim F (t) lim sin t = = 1. t→0 t t→0 t

We have L {sin t} = ∴

(Lucknow 2009; Gorakhpur 05; Rohilkhand 10; Avadh 13)

1 p2 + 1

= f ( p), say.

sin t  L  =  t 

∞

∫p

f ( x) =

∞

∫p

dx 2

x +1

= (tan−1 x)p∞ =

π − tan−1 p 2

= cot −1 p = tan−1 (1/ p). Now

1  sin at   sin at  = a ⋅ 1 tan−1 L ,   = aL a ( p / a)  at   t 

since

L { F (at)} =

Again, since L {cos at} = L  

1  p f   = tan−1(a / p). p  a

p 2

p + a2

sin at  = t 

∫p

= which does not exist since

∞

= f ( p), we have ∞

1 dx =  log ( x2 + a2 ) 2 2   p 2 x +a x

1 lim 1 log ( x 2 + a2 ) − log ( p2 + a2 ), 2 x→ ∞ 2

lim log ( x2 + a2 ) is infinite. x→ ∞

T-27

cos at  Hence L   does not exist.  t  If L {F (t), t → p} = f ( p),  t F (u)  1 ∞ show that L  ∫ du, t → p  = ∫ f ( y) dy . 0 u   p p

Example 20:

cot −1 p du, t → p = ⋅ u p 

t sin u

Hence show that L  ∫  0 Solution:

From 1.17, we have t 1 L  ∫ F (u) du = f ( p)  p  0

f ( p) = L {F (t)}. F (t) G (t) = . t  F (t) L { G(t)} = L  =  t 

where Let Then

∞

∫p

…(1)

f ( y) dy

[from article 1.20]

= g ( p), say. ∴ From (1), we have t 1 L  ∫ G (u) du = g ( p) 0  p  t 1 ∞ L  ∫ F (u) du = ∫ f ( y) dy 0  p p 

or Deduction. so that

…(2)

Let F (t) = sin t f ( p) = L {sin t} =

1 2

p +1

.

∴ From (2), we have 1 1 ∞ dy = [tan−1 y]∞ du = ∫ p 2 p u  p y +1 p

t sin u

L  ∫  0

=

1 π  1  − tan−1 p = cot −1 p.  p p 2

Comprehensive Exercise 3 Find the Laplace transforms of the following functions. 1.

(i) t cos at. (Lucknow 2008)

(ii) t cosh 3 t.

2.

(i) t 2 sin at. (Lucknow 2006)

(ii) t 2 e 2 t .

3.

(i) Show that L { t3 cos t} =

6 p4 − 36 p2 + 6 ( p2 + 1)4

.

T-28

(ii) Show that L { t n . e at} = 4.

5. 6.

n! ( p − a)n +1

, p > a.

(i) Find L { t (3 sin 2 t − 2 cos 2 t)}. (ii) Find L.T. of f (t) = sin α t + t cos α t. 2 a ( p − a) Prove that L { t e at sin at} = 2 ⋅ ( p − 2 ap + 2 a2 )2 √π

Given L {sin √ t } =

2 p3 /2

(Kanpur 2010)

e −1 /4 p , show that

 cos √ t   π  −1 / 4 p L . =   e  p  √t  7. 9.

10.

 t sin x  8. Find L  ∫ dx .  0 x 

 sinh t  Find L  ⋅  t 

Prove that if L {F (t)} = f ( p), then ∞ F (t) ∞ ∫ 0 t dt = ∫ 0 f ( x) dx provided that the integral converges. Find the Laplace transform of F (t) defined as  t + 1, 0 ≤ t ≤ 2 F (t) =   3, t>2. Also determine L { F ′ (t)}.

A nswers 3 1.

(i)

p2 − a2 ( p2 + a2 )2

2 a (3 p2 − a2 )

2.

(i)

4.

(i)

7.

p+1 1 log ⋅ 2 p−1

10.

1 p2

, p > 0.

2

2 3

(p +a )

8 + 12 p − 2 p2 ( p2 + 4)2

[1 + p − e − 2 p ],

(ii)

, p > 0.

(ii)

.

(ii)

p2 + 9 ( p2 − 9)2 2 ( p − 2)3

, p > 0.

.

(α + 1) p2 + (α − 1) α2 ( p2 + α2 )2

.

8. (1 / p) cot −1 p. 1 − e− 2 p ⋅ p

1.21 Evaluation of Integrals If L {F (t)} = f ( p) i.e., ∫

∞ 0

e − pt F (t) dt = f ( p), taking limit as p → 0, we have

T-29 ∞

∫0

F (t) dt = f (0 )

assuming that the integral is convergent.

Solution:

(e − at − e − bt) dt. t

∞

Example 21: Evaluate

∫0

Let F (t) = e − at − e − bt . L {F (t)} = L { e − at} − L { e − bt} =

Then

 F (t) L =  t 

∴

∞

∫p

f ( x) dx =

∫p

∞

 1 − 1  dx = log ( x + a)      x + a x + b ( x + b) p 

=

p+ a lim x+a log − log x→ ∞ x+b p+ b

=

p+ a lim 1 + (a / x) log − log x→ ∞ 1 + (b / x) p+ b p+ a

= 0 − log  F (t) L =  t 

Thus

∞

1 1 − = f ( p), say. p+ a p+ b

∞ − pt

∫0

e

p+ b ⋅

= log

p+ b p+ a

⋅

p+ b e − at − e − bt dt = log t p+ a

∴ Taking limit as p → 0, we have ∞

∫0 Example 22: Solution:

∴

e − at − e − bt b dt = log ⋅ t a

Prove that

t3 e − t sin t dt = 0 .

(Gorakphur 2006)

3

Let F (t) = t sin t.

L {F (t)} = L { t3 sin t} = (−1)3

=−

or

∞

∫0

∞

∫0

d3 dp 3

2p  d3  1  d2  d  2 − 6 p2  = = − −       dp3  p2 + 1 dp2  ( p2 + 1)2  dp  ( p2 + 1)3 

e − pt . t3 sin t dt =

24 ( p2 − 1) p

Taking limit as p → 1, we have ∞

∫0

L {sin t}

t3 e − t sin t dt = 0 .

( p2 + 1)4

⋅

T-30

Comprehensive Exercise 4 1.

Show that

2.

Evaluate

e− t − e− 3 t dt. t

∞

∫0

Show that

∫0

4.

Show that

∫0

5.

Prove that ∫ Evaluate

− 3t

∞

e − x sin x π dx = ⋅ x 4

∞ 0

∞

∫0

(Rohilkhand 2010)

3 te sin t dt = ⋅ 50 3 t e − 2 t cos t dt = ⋅ 25

∞

3.

6.

sin t π dt = ⋅ t 2 (Purvanchal 2007; Meerut 13, 13B; Rohilkhand 10, 14)

∞

∫0

te

−3 t

(Rohilkhand 2009, 11)

cos 4 t dt. (Gorakhpur 2007, 09)

A nswers 4 3.

6.

log 3.

5 ⋅ 169

1.22 Periodic Functions Fundamental Theorem: Let F (t) be a periodic function with period T > 0, that is F (u + T ) = F (u), F (u + 2T ) = F (u), etc., then T

L {F (t)} = Proof:

∫0

e − pt F (t) dt .

1 − e − pT

We have L {F (t)} =

∞

∫0

e − pt F (t) dt

=

∫0

T

e − pt F (t) dt +

∫T

2T

=

∫0

T

e − pt F (t) dt +

∫0

T

e − pt F (t) dt +

3T

∫ 2T

e − pt F (t) dt + ...

e − p (u + T ) F (u + T ) du +∫

T 0

e − p (u + 2T ) F (u + 2T ) du + ... ,

putting t = u + T , t = u + 2T , etc. in 2nd, 3rd, … integrals respectively =

T

∫0

e − pu F (u) du + e − pT

T

∫0

e − pu F (u) du + e −2 pT ∫

T 0

e − pu F (u) du + ...

T-31

= (1 + e − pT + e −2 pT + ...) ∫

T 0

e − pu F (u) du T

=

1

T

e

∫0

1 − e − pT

− pu

F (u) du =

∫0

e − pt F (t) dt 1 − e − pT

.

1.23 Some Special Functions 1.

The Sine and Cosine Integrals: The sine and cosine integrals, denoted by Si (t) and Ci (t) respectively are defined by t sin u Si (t) = ∫ du, 0 u

and

cos u du. u

∞

Ci (t) =

∫t

2.

The Error Function: The error function denoted by erf (t), is defined by t − u2 2 erf (t) = e du. ∫ 0 √π

3.

The Gamma Function: If n > 0, the gamma function is defined by Γ (n) =

∞

un − 1 e − u du.

∫0

4.

The Unit Step Function (also called Heaviside’s Unit function): 0 , t < a The unit step function, denoted by H (t − a), is defined by H (t − a) =  1, t ≥ a.

5.

The Bessel Function: Bessel function of order n is defined by Jn (t) =

  t2 t4 1 − + −.. .  n 2 .( 2 n + 2 ) 2 ⋅ 4 . ( 2 n + 2 ) ( 2 n + 4 ) 2 Γ (n + 1)   t

n +2 r (− 1)r t ⋅   r ! Γ (n + r + 1)  2 

∞

=

n

∑

r =0

The Bessel function of order zero i.e., J0 (t) is given by J0 (t) = 1 −

Example 23:

(i)

and (ii)

t2 2

2

+

t4 2

2

2 .4

−

Show that

L { sinh at cos at} =

a ( p2 − 2 a2 )

L { sinh at sin at } =

p4 + 4 a4 2 a2 p p4 + 4 a4

⋅

t6 2

2 .42 .62

+ ... .

T-32 Solution:

∴

We have L {sinh at} =

a p2 − a2

= f ( p), say.

L { e iat sinh at} = f ( p − ia) =

=

a 2

2

( p − ia) − a

=

a 2

( p − 2 a2 ) − 2 iap

a {( p2 − 2 a2 ) + 2 iap} ( p2 − 2 a2 )2 − (2 ipa)2

L {sinh at (cos at + i sin at)} =

or

a ( p2 − 2 a2 ) + 2 ia2 p p4 + 4 a4

or

L{sinh at cos at} + i L{sinh at sin at} =

Hence

L {sinh at cos at} =

and

L {sinh at sin at} =

Example 24: Solution:

+i

p4 + 4 a4

2 a2 p p4 + 4 a4

⋅

a ( p2 − 2 a2 ) p4 + 4 a4 2 a2 p p4 + 4 a4

Show that L {(1 + t e − t )3} =

.

1 3 6 6 + + + ⋅ p ( p + 1)2 ( p + 2)3 ( p + 3)4

We have L {(1 + t e − t )3} = L {1 + 3 t e − t + 3 t2 e −2 t + t3 e −3 t} = L {1} + 3 (−1)

Aliter:

a ( p2 − 2 a2 )

d d2 d3 L { e − t} + 3 (−1)2 2 L { e −2 t} + (−1)3 3 L {e −3 t} dp dp dp

=

1! 2! 3! 1 + 3. + 3. + 2 3 p ( p + 1) ( p + 2) ( p + 3)4

=

1 3 6 6 + + + . 2 3 p ( p + 1) ( p + 2) ( p + 3)4

L {(1 + t e − t )3} =

∞

∫0

(1 + t e − t )3 . e − pt dt

∞

[e − pt + 3 t e − ( p + 1)t + 3 t2 e − ( p + 2)t + t3 e − ( p + 3)t ] dt

∞

e − pt dt + 3 ∫

=

∫0

=

∫0

∞ 0

te − ( p + 1)t dt + 3 ∫

∞ 0

t2 e − ( p + 2)t dt +

=

∞

∫0

e − pt t1−1 dt + 3 ∫

∞ 0

∞

∫0

t 3 e − ( p +3)t dt

e − ( p + 1)t t2 − 1 dt

+3∫

∞ 0

e − ( p + 2)t t3 − 1 dt +

∞

∫0

e − ( p + 3)t t 4 − 1 dt

T-33

=

Example 25:

1 3 6 6 + + + , p > 0. 2 3 p ( p + 1) ( p + 2) ( p + 3)4 ∞ − at n−1 Γ(n)   ∵ e t dt = n , if a > 0 and n > 0  ∫ 0  a

Prove that L { J0 (t)} =

1 √ (1 + p2 )

and hence deduce that

(Avadh 2007)

1

(i)

L { J0 (at)} =

(ii)

L { t J0 (at)} =

(iii)

L { e − at J0 (at)} =

(iv)

∫0

∞

Solution:

2

√ ( p + a2 ) p 2

( p + a2 )3 /2

(Rohilkhand 2002)

1 2

√ ( p + 2 ap + 2 a2 )

J0 (t) dt = 1. We know that J0 (t) = 1 −

t2

t4

t6

+ ... 22 .42 22 .42 .62 1 1 1 L { J0 (t)} = L {1} − 2 L { t 2} + 2 2 L { t 4} − 2 2 2 L { t 6} + ... 2 2 .4 2 .4 .6 4! 6! 1 1 2! 1 1 = − 2 ⋅ 3 + 2 2 ⋅ 5 − 2 2 2 7 + ... p 2 p 2 .4 p 2 .4 .6 p

∴

=

=

22

+

−

2 3   1  1  1  1⋅ 3  1  1⋅ 3 ⋅ 5  1  1−  2  +  2 −  2  + .... p  2  p  2 ⋅4  p  2 ⋅4 ⋅6  p    

1 1 1 + 2  p p 

−1 / 2

=

1 √ (1 + p2 )

⋅

Deductions: (i) Since ∴

1  p f   where f ( p) = L { F (t)}, a  a 1 1 1 L { J0 (at)} = ⋅ = . 2 2 a √ [1 + ( p / a) ] √ ( p + a2 )

L { F (at)} =

(ii) L { t J0 (at)} = −

 p 1 d d  L{ J0 (at)} = −  = 2 2 2 dp  √ ( p + a ) ( p + a2 )3 /2 dp

(iii) Since L { e − at F (t)} = f ( p + a) where f ( p) = L {F (t)}, ∴

L { e − at J0 (at)} = =

  1 ∵ L { J0 (at)} = 2 2  √ ( p + a ) 

1 2

2

√ [( p + a) + a ] 1 2

√ ( p + 2 ap + 2 a2 )

.

T-34

(iv)

We have L { J0 (t)} =

∴

Putting p = 0, we have

∞

e − pt J0 (t) dt =

∫0

∞

∫0

1 √ (1 + p2 )

.

J0 (t) dt = 1.

Example 26: Prove that L { J1 (t)} = 1 −

p √ ( p 2 + 1)

where J1 (t) is the Bessel function of order

one

(Kanpur 2009)

and hence deduce that

L {t J1 (t)} =

1 2

( p + 1)3 /2

.

Solution: We know that J0 ′ (t) = − J1 (t).

∴ From L {F ′ (t)} = pL {F (t)} − F (0 ), we have L { J1 (t)} = L { − J0 ′ (t)} = − L { J0 ′ (t)} = − [ pL { J0 (t)} − J0 (0 )]   1 = −p ⋅  2  √ ( p + 1) − 1   1 and J0 (0 ) = 1 ∵ from Ex . 26, L { J0 (t)} = 2 √ (1 + p )   p = 1− ⋅ √ ( p2 + 1) Aliter: We have ∞

Jn (t) =

∑

r =0

n +2 r (− 1)r t . ⋅   r ! Γ (n + r + 1)  2 

2 r +1 (− 1)r t 1 t3 1 t5  t ⋅ = − + ⋅ − ...   ∑ r !(r + 1) !  2 2 22 4 22 .42 6 r =0 ∞

∴

J1(t) =

∴

L { J1 (t)} =

1 1 1 ⋅ L { t } − 2 L { t3} + 2 2 L { t5} − ... 2 2 .4 2 .4 .6 5! 1 1! 1 3! 1 = ⋅ 2 − 2 ⋅ 4 + 2 2 ⋅ 6 − ... 2 p 2 .4 p 2 .4 .6 p 2 3   1  1  1⋅ 3  1  1⋅ 3 ⋅ 5  1  = 1 − 1 −  2  +  2 −  2  + ... 2 ⋅4 ⋅6  p   2  p  2 ⋅4  p    

 1 = 1 − 1 + 2  p   Deduction:

L { t J1 (t)} = − =−

−1 / 2

= 1−

p 2

√ ( p + 1)

d L { J1 (t)} dp  p 1 d  . 1 −  = 2 2 dp  √ ( p + 1) ( p + 1)3 /2

T-35

Find L { erf √ t } and hence prove that 3p + 8 . L { t ⋅ erf (2 √ t)} = 2 p ( p + 4)3 /2

Example 27:

Solution:

(Avadh 2012)

We know that erf (√ t) = =

2 √π

2 √π

√t

∫0

2

e − u du

 u4 u6 − + … du 1 − u2 + 2! 3!  

√t 

∫0

√t

 2  u3 u5 u7 = + − + ... u − 3 5 (2 !) 7 (3 !) √π  0 =

 2  1 /2 t3 /2 t5 /2 t7 /2 − + − + ... t 3 5 (2 !) 7 (3 !) √π  

∴ L {erf √ t} =

2 √π

  1 1 1 1 /2 3 /2 L { t5 /2 } − L { t7 /2 } + ...  L {t } − L {t } − 3 5 (2 !) 7 (3 !)  

=

2 √π

 Γ( 32 )  1 Γ( 52 )  Γ( 7 ) Γ( 9 ) 1 1 ⋅ 92/2 + ... ⋅ 72/2 −  3 /2 −   5 /2 + 7 (3 !) p 3 p 5 (2 !) p  p 

=

 2 √ π 1 √π 1 1 √ π 1⋅ 3 1 √ π 1⋅ 3 ⋅ 5 1 ⋅ 3 /2 − + ⋅ − ⋅ + ...  5 2 7 2 9 / 2 / / 2 2 p 2 2 ⋅4 p 2 2 ⋅4 ⋅6 p √π  2 p 

 1  1 1 1⋅ 3 1 1⋅ 3 ⋅ 5 1 1 1− ⋅ + − + ... = 3 /2 3 /2  2 3 2 p 2 ⋅ 4 2 ⋅ 4 ⋅ 6 p p p p   1 = ⋅ p √ ( p + 1) =

Since L { F (at)} = ∴

 1 1 +  p 

−1 / 2

(Rohilkhand 2004)

1  p f   , where f ( p) = L { F (t)}, a  a

L {erf (2 √ t)} = L {erf √ (4 t)} 1 1 2 = ⋅ = ⋅ 4 p p  p √ ( p + 4) 1 +   4 4  Hence L { t. erf (2 √ t)} = −

Example 28: Solution:

 3p + 8 d d  2 ⋅ L {erf (2 √ t)} = −  = dp dp  p √ ( p + 4) p2 ( p + 4)3 /2

Find the Laplace transform of Si (t).

We know that Si (t) =

t

∫0

sin u du = u

t

∫0

  u2 u4 u6 + − + ... du 1 − 3! 5! 7!  

(Rohilkhand 2009)

T-36

=t−

L { Si (t)} = L { t } −

∴

=

Example 29: Solution:

Let

t3 t5 t7 + − + ... 3 (3 !) 5 (5 !) 7 (7 !)

1! p2

−

1 1 1 L { t 3} + L { t 5} − ⋅ L { t7} + ... 3 (3 !) 5 (5 !) 7 (7 !)

3! 5! 7! 1 1 1 ⋅ + ⋅ − ⋅ + ... 3 (3 !) p4 5 (5 !) p6 7 (7 !) p8

1 1 1 1 1 1 1   − ⋅ 3 + ⋅ 5 − ⋅ 7 + ... 5 p 7 p p 3 p 

=

1 p

=

1 1 tan−1 , by Gregory’s series. p p

Find L { Ci (t)}.

(Rohilkhand 2003, 08)

cos u   L { Ci (t)} = L  ∫ du ⋅ u  t  ∞ cos u t cos u F (t) = ∫ du = − ∫ du t ∞ u u ∞

so that

F ′ (t) = −

cos t t

or

t F ′ (t) = − cos t.

L { t F ′ (t)} = L { − cos t} p d − L{F ′ (t)} = − 2 dp p +1

∴ or

or

p d where f ( p) = L {F (t)} [ p f ( p) − F (0 )] = 2 dp p +1

or

p d [ p f ( p)] = 2 , since F (0 ) is constant. dp p +1

Integrating, p f ( p) =

1 log ( p2 + 1) + C (constant) 2

But from the final-value theorem 1.16, lim lim p f ( p) = F (t) = 0 . p→ 0 t→ ∞ from (1) as p → 0 we have 0 = 0 + C or C = 0 . 1 ∴ from (1), p f ( p) = log ( p2 + 1) 2 log ( p2 + 1) or f ( p) = L {F (t)} = L { Ci (t)} = . 2p ∴

Example 30:

3 t, 0 < t < 2 If F (t) =  6, 2 < t < 4,

find L {F (t)} where F (t) has period 4. Solution:

Here F (t) is a periodic function with period T = 4.

…(1)

T-37

∴ from 1.22 theorem, we have T

L {F (t)} =

=

=

=

∫0

2

e − pt F (t) dt =

1 − e − pT 1 1 − e −4 p 1 1 − e −4 p

∫0

3 t e − pt dt +

4

∫2

6 . e − pt dt

1 − e− 4 p 2

4

 e − pt   e − pt   6 e − pt  1  ⋅ 3 t −  2  ⋅ 3 +  −4 p  − p   p  0 1 − e  − p   2  6 −2 p 3 e −2 p  3 6 6 − + 2 − e −4 p + e −2 p  − e 2 p p p p  p 

3 − 3 e− 2 p − 6 p e− 4 p p2 (1 − e − 4 p )

.

Comprehensive Exercise 5 1.

  1  t  1 1 Given L 2    = 3 /2 , show that 1 /2 = L  ⋅  π p  √ (πt)   p

2.

Find (i) L {F (t)} and (ii) L {F ′ (t)}, for the function given by 2 t , 0 ≤ t ≤ 1 F (t) =  t > 1.  t,

3.

Show that L {(5 e2 t − 3)2} =

4.

 p2 + 4   sin2 t  1 Show that L   = log  2  ⋅  p   t  4

5. 6.

25 30 9 − + , p > 4. p−4 p−2 p (Rohilkhand 2010; Kashi 14)

sin2 t

π dt = . 2 t2 ∞ cos 6 t − cos 4 t 2 Show that ∫ dt = log   ⋅  3 0 t Show that ∫

∞

0

2

7.

Prove that L { J0 (a √ t)} = (1 / p) e − (a

8.

If F (t) = t2 , 0 < t < 2 and F (t + 2) = F (t), find L {F (t)}.

9.

/4 p)

(Rohilkhand 2003)

.

[Hint. Here F (t) is a periodic function with period T = 2] sin t, 0 < t < π Compute L {F (t)}, if F (t) =  where F (t) has period 2π. π < t < 2 π, 0 ,

10.

Find the Laplace transform of the Heaviside’s unit step function H (t − a).

11.

t 1 − e −2 x  Find Laplace transform of ∫   dx. 0 x 

T-38 ∞

log ( p + 1) e− u du, show that L { E (t)} = ⋅ u p

∞

e − t erf √ t dt.

12.

If E (t) =

∫t

13.

Evaluate

∫0

A nswers 5 2. 8. 10.

(i) 2 / p2 − (1/ p + 1/ p2 ) e − p

(ii)

− (4 p2 + 4 p + 2) e −2 p + 2 3

p (1 − e

−2 p

9.

)

e − ap p

11.

2 e −p − p p 1 2

( p + 1) . (1 − e − pπ )  1 2 log 1 +  p p 

12. 1/ √ 2

1.24 Table of Laplace Transform Theorems Laplace Transform Theorems No.

Operation

F (t )

L { F (t )} = f ( p)

1.

Linearity property

a1 F1 (t) + a2 F2 (t)

a1 L{F1(t)} + a2 L{F2 (t)}

2.

First translation or shifting theorem

e at F (t)

f ( p − a)

3.

Second translation or shifting theorem

 F (t − a), t > a G(t) =  , t< a  0

e − ap f ( p)

4.

Change of scale property

F (at)

1  p f   a  a

5.

Differentiation theorems

F ′ (t)

pf ( p) − F (0 )

F n (t)

pn f ( p) −

n−1

∑

r =0

p n −1− r F r (0 )

T-39

t F (t) 6.

Multiplication theorems

t

n

− f ′ ( p)

F (t)

(− 1)n

1 F (t) t

∞

∫p

dn dp

n

f ( p)

f ( x) dx

7.

Division theorem

8.

Integral theorem

9.

Initial value theorem

lim lim F (t) = p L {F (t)} t→0 p→ ∞

10.

Final-value theorem

lim lim F (t) = p L {F (t)} t→ ∞ p→ 0

t

∫0

1 f ( p) p

F ( x) dx

T

11.

Fundamental theorem for Periodic Functions

L {F (t)} =

∫0

e − pt F (t) dt

, 1 − e − pT F (t) is periodic function of period T

Objective Type Questions

Multiple Choice Questions

1.

2.

3.

Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). The Laplace transform of F (t) exists for all p > a, if (a) F (t) is a continuous function (b) F (t) is a differentiable function (c) F (t) is a function of class A (d) None of these. (Rohilkhand 2002) The Laplace transform of F (t) = 1 is (a) p (b) 1 1 1 (c) (d) , p> 0 , p > 1. p p−1 (Rohilkhand 2008)  F (t − a), t > a If L { F (t)} = f ( p) and G (t) =  then L { G (t)} is 0 , t< a  (a) f ( p − a) (b) e − ap f ( p) 1  p (c) (d) f (ap). f   (Rohilkhand 2003) a  a

T-40

4.

If L { F (t)} = f ( p) and F ′ (t) is of class A, then L { F ′ (t)} is (a) pf ( p) − F (0 ) (b) p2 f ( p) (c)

(d)

f ( p) − F (0 )

None of these. (Rohilkhand 2003)

5.

6.

The Laplace transform of cosh at is p (a) 2 p − a2 1 (c) 2 p − a2 If L { F (t)} = f ( p) then L { t F (t)} is (a) f ′ ( p) (c)

7.

∞

f ( x) dx

∫p

The value of Laplace transform of

8.

9.

10.

11.

p2 + a2

⋅

(Rohilkhand 2003)

(b)

− f ′ ( p) 1 (d) f ( p). p

(Rohilkhand 2002)

(b)

tan−1

tan−1 p

(d)

None of these.

(b)

1 f ( p) a

(d)

None of these.

1  p f   a  a

The Laplace transform of e t 1 (a) p−2 (c)

p + a2 1

sin t is t

If L { F (t)} = f ( p) then L { F (at)} is  p (a) f    a (c)

(d)

p 2

1 p

(a) tan−1 (c)

(b)

2

2 p

is

1 p2

(b)

1 ( p − 2)2

(d)

not existing. (Rohilkhand 2003)

 3 tan−1    p

sin t   sin 3 t  −1 1 If L  , then L   = tan  is p  t   t  (a)

 3 1 tan−1   3  p

(b)

(c)

tan−1 (3 p)

(d) 3 tan−1 (3 p).

The value of L { t sin at} is (a) (c)

a 2

2

(p +a ) 2 ap 2

( p + a2 )2

(b) (d)

2 ap 2

( p + a2 ) None of these.

T-41

12.

13.

14.

The value of ∫ (a)

1

(c)

−1

The value of ∫

∞

J0 (t) dt is

0

(b)

0

(d)

None of these.

sin t dt is t

∞ 0

(a)

0

(b)

(c)

1

(d)

If L {erf √ t} =

π 2 None of these.

1 then L { e3 t erf √ t} is p √ ( p + 1)

(a)

1 ( p − 3) √ ( p − 2)

(b)

1 ( p − 3) √ ( p + 2)

(c)

3 ( p − 3) √ ( p − 2)

(d)

None of these.

Fill In The Blanks Fill in the blanks “……” so that the following statements are complete and correct. 1.

If the integral f ( p) =

∞

∫−∞

K ( p, t) F (t) dt is convergent, then f ( p) is called

the …… of the function F (t) and the function K ( p, t) is called the …… of the transformation. ∞ − pt

2.

The integral

3.

The Laplace transform of the function F (t) = cos at is …… .

4.

If f ( p) is the Laplace transform of F (t) then f ( p − a) is the Laplace transform of …… .

5.

If F (t) is a function of class A and if L { F (t)} = f ( p), then L { t

e

∫0

F (t) dt is called the …… transform of the function F (t).

n

F (t)} = ……

where n = 1, 2 , 3, …… . 6.

The value of L { J0 (t)} is …… .

7.

The value of integral ∫

8.

If F (t) is a periodic function with period T > 0 then F (u + T ) = …… .

∞

sin2 t

0

t2

dt is …… .

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

The Laplace transformation is a linear transformation.

2.

The function e t is not of exponential order as t → ∞.

2

T-42

3.

If the Laplace transform of a function F (t) exists then it must be of class A.

4.

The Laplace transform of the function F (t) = t n, n being any real number greater Γn than −1 is n + 1 , p > 0 . p

5.

If F (t) is a function of class A and if L { F (t)} = f ( p), then

6.

L { t F (t)} = f ′ ( p). 1 The value of L { e at} = ⋅ p+ a

7.

The value of L {sin λt . cos λt} =

8.

The value of L { t3 e −3 t } =

λ 2

p + 4 λ2

6 ( p + 3)4

, p > 0.

⋅

A nswers Multiple Choice Questions 1. 4. 7. 10. 13.

(c) (a) (a) (b) (b)

2. (c) 5. (a) 8. (c) 11. (c) 14. (a)

3. 6. 9. 12.

(b) (b) (d) (a)

Fill in the Blank(s) 1. 3.

integral transform ; kernel. p , p> 0 2 p + a2

5.

(− 1)n

7.

π 2

dn n

dp

f ( p)

2. Laplace. 4. e at F (t) 6. 8.

1 √ (1 + p2 ) F (u)

True or False 1. 3. 5. 7.

T F F T

2. T 4. F 6. F 8. T

¨

T-43

2 T he I nverse L aplace T ransform

2.1 Inverse Laplace Transform If f ( p) is the Laplace Transform of a function F (t), i. e. L {F (t)} = f ( p), then F (t) is called the inverse Laplace transform of the function f ( p) and is written as F (t) = L−1 { f ( p)}. L−1 is called the inverse Laplace transformation operator.

2.2 Null Function If N (t) is a function of t such that ∫

t 0

N (t) dt = 0 for all t > 0 then N (t) is called a null

function.

2.3 Lerch’s Theorem If F1 (t) and F2 (t) are two functions having the same Laplace transform f ( p), then F1 (t) − F2 (t) = N (t) where N (t) is a null function for all t > 0. From this it follows that, an inverse Laplace transform is unique except for the addition of a null function.

T-44

If we restrict ourselves to functions F (t) which are sectionally continuous in every finite interval 0 ≤ t ≤ t0 and of exponential order for t > t0 , then the inverse Laplace transform of f ( p) i.e., L−1 { f ( p)} = F (t) is unique.

2.4 Linearity Property Theorem: Let f1 ( p) and f2 ( p) be the Laplace transforms of functions F1 (t) and F2 (t) respectively and c1, c2 be two constants, then L−1 { c1 f1 ( p) + c2 f2 ( p)} = c1 L−1 { f1 ( p)} + c2 L−1 { f2 ( p)} = c1 F1 (t) + c2 F2 (t).(Avadh 2014) Proof: We have

∴

L { c1 F1 (t) + c2 F2 (t)} = c1 L {F1 (t)} + c2 L {F2 (t)} = c1 f1( p) + c2 f2 ( p). L−1 { c1 f1 ( p) + c2 f2 ( p)} = c1 F1 (t) + c2 F2 (t) = c1 L−1 { f1 ( p)} + c2 L−1 { f2 ( p)}.

2.5 Inverse L.T. of Some Special Functions (Rohilkhand 2006)

(i) Inverse L.T. of 1/p, p > 0. ∵

L {1} = 1/ p, p > 0 ,

∴

L−1 {1/ p} = 1, p > 0 .

(ii) Inverse L.T. of 1 / pn + 1, p > 0, n > − 1. Γ (n + 1)

∵

L { t n} =

∴

L−1 {1/ p n+1} = t n / Γ (n + 1), n > − 1.

p n+1

, p > 0, n > − 1

If n is a positive integer, then Γ (n + 1) = n !. ∴

L−1 {1/ pn+1} = t n / n !, p > 0 , n is a positive integer.

(iii) Inverse L.T. of 1/( p − a), p > a. ∵ L { e at} = 1/( p − a), p > a,

∴

L−1{1/( p − a)} = e at , p > a.

(iv) Inverse L.T. of 1 /( p2 + a2 ), p > 0. ∵

L {sin at} =

a 2

, p > 0, ∴

2

p +a

 p  1 = sin at L−1  2 2 p + a  a

2

(v) Inverse L.T. of p /( p + a2 ), p > 0. ∵

L {cos at} =

p 2

2

p +a

 p  = cos at. ∴ L−1  2 2 p + a 

,

(vi) Inverse L.T. of 1 /( p2 − a2 ), p > | a | . ∵

L {sinh at} =

a 2

2

p −a

,

 1  1 ∴ L−1  2 = sinh at. 2 p − a  a

T-45

(vii) Inverse L.T. of p /( p2 − a2 ), p > | a | . L {cosh at} =

∵

p 2

2

p −a

,

∴

 p  = cosh at . L−1  2 2 p − a 

2.6 Table of Some Inverse Laplace Transforms L−1 { f ( p)} = F(t )

f ( p) 1 p

1. 2.

1 p

n +1

1

, n is a positive integer

3.

1 p− a

4.

2

e at

1

1 sin at a

2

p +a p

5.

cos at

2

p + a2 1

6.

1 sinh at a

p2 − a2 p

7.

cosh at

2

p − a2 1

8.

p

n +1

tn n!

,n> −1

tn Γ (n + 1)

If we know the inverse L.T. given in the above table then nearly all the inverse Laplace transforms can be obtained by using the general theorems, which we shall give later on.

 1 Example 1: Find (i) L−1  4  p 

 1  (ii) L−1  2 .  p + 4



 1  t 4 − 1 t3 Solution: (i) L−1  4  = =  p 

(ii)

3!

6

 1  1  1 −1  = sin 2 t L−1  2 =L  2 2  p +2  2  p + 4

(Meerut 2013B)

T-46

Prove that

Example 2:

−1

L

2   7   5  √ p − 1   t  7 −2 t / 3 . −  2 +  = 1 + 6 t − 4   − e  3 p + 2 π 3 p    p 

2   7   5  √ p − 1  We have L−1  2 +  −   3 p + 2 p    p 

Solution:

p− 2 √ p+1 7 1  5  = L−1  2 + − .  2 3 p + (2 / 3) p  p   1  7  1  1   1 = 6 L−1  2  + L−1   − 2 L−1  3 /2  − L−1    p  p + (2 / 3) p  3  p  =6

t 2 −1 t 3 /2 −1 7 + 1− 2 . − e −2 t /3 1! Γ (3 / 2) 3

= 6 t + 1 − 4 √ (t / π) − (7 / 3) e −2 t /3 . Show that

Example 3:

1 1 t3 t5 t7 L−1  sin  = t − + − + ... p (3 !)2 (5 !)2 (7 !)2  p Solution:

1 L−1  sin  p

3

5

(Avadh 2014) 7

   (1/ p) (1/ p) 1 −1 1  1 (1/ p) + − + ...  =L   − p 3! 5! 7!    p p

 1  1 −1  1  1 −1  1  1 −1  1  = L−1  2  − L  4 + L  6 − L  8  + ...  p   p  3!  p  5!  p  7! =t−

t3 (3 !)2

+

t5 (5 !)2

−

t7 (7 !)2

+ ... .

Comprehensive Exercise 1 Find the inverse Laplace transforms of each of the following functions. p 6p 3 1. 4 /( p − 2). 2. 2 + 2 + ⋅ p + 2 p − 16 p − 3 2p + 1 3. ⋅ p ( p + 1) (Purvanchal 2014; Kashi 14, 14B) 4. 5.

2p − 5 p2 − 9 3p − 2 5 /2

p

⋅ (Meerut 2013)

−

7 ⋅ 3p + 2

6.

3 + 4p

8 − 6p 6 ⋅ + − 2 2 p − 3 9 p − 16 16 p2 + 9

T-47

7. 8. 9.

3 2

p −3

+

3p + 2 p

3 ( p2 − 1)2 5

2p

−

3

+

3 p − 27 2

p +9

4 p − 18 2

9− p

p4

( p + 1) (2 − √ p)

+

1 Show that L−1  cos  p

6 − 30 √ p

+

p5 /2

⋅

⋅

1 t2 t4 t6 + − + ...  = 1− 2 2 p (2 !) (4 !) (6 !)2

A nswers 1 1.

4 e 2 t.

2. cos √ 2 t + 6 cosh 4 t + 3 e 3 t .

3.

e − t + 1.

4. 2 cosh 3 t − (5 / 3) sinh 3 t.

5.

t 8 t 7 6   − t   − e −2 t /3 .  π 3  π 3

6.

3 e 3 t /2 −

7.

√ 3 sinh (√ 3 t) + 3 t + t 2 − 3 cos 3 t + 9 sin 3 t + t 3 − 16 t 2 √ (t / π).

8.

1 3 2 1 4 8 − t + t − 4 cosh 3 t + 6 sinh 3 t + 4 √ (t / π) + t √ (t / π) − t. 2 2 16 3

1 4t 4 4t 2 3t 3 3t sinh − cosh + sin − cos ⋅ 4 3 9 3 3 4 8 4

2.7 First Translation or Shifting Theorem If L−1 { f ( p)} = F (t), then L−1 { f ( p − a)} = e at F (t) = e at L−1 { f ( p)}. Proof: We have f ( p) = ∞

∞ − pt

∫0

e

F (t) dt.

e −(p − a)t F (t) dt =

∞ − pt

f ( p − a) =

∴

L−1 { f ( p − a)} = e at F (t) = e at L−1 { f ( p)}.

∫0

∫0

e

.{ e at ⋅ F (t)} dt = L { e at F (t)}.

∴

2.8 Second Translation or Shifting Theorem If

 F (t − a), L−1 { f ( p)} = F (t), then L−1 { e − ap f ( p)} = G (t), where G (t) =  0, 

Proof: We have f ( p) = ∴

∞

∫0

e − pt F (t) dt .

e − ap f ( p) =

∞ − p (t + a)

∫0

e

F (t) dt

t> a t < a.

T-48 ∞ − px

e

=

∫0

=

∫0 e

=

∫0 e

=

∫0

a − px a − pt

∞ − px

⋅ 0 dx +

∫a

⋅ 0 dt +

∫a

∞ − pt

e

F ( x − a) dx, putting t + a = x, so that dt = dx e

∞ − pt

e

F ( x − a) dx F (t − a) dt

G (t) dt = L { G (t)} ,

 F (t − a), t > a G (t) =  0, t < a. 

where

Note: In terms of Heaviside’s unit step function H (t − a) this theorem can be stated as :

If

L−1 { f ( p)} = F (t),

then

L−1 { e − ap f ( p)} = F (t − a). H (t − a).

2.9 Change of Scale Property Theorem: If L−1 { f ( p)} = F (t), then L−1 { f (ap)} = (1 / a) F (t / a). Proof:

We have, f ( p) =

∴

f (ap) =

∞ − pt

∫0

e

∞ − apt

∫0

e

F (t) dt .

F (t) dt =

1 ∞ − px  x  e F   dx .  a a ∫0 putting at = x, so that dt = (1 / a) dx

= Hence

1 ∞ − pt  t  1  t  t  1 e F   dt = L  F    = L F    . ∫      0 a a a  a  a  a

L−1 { f (ap)} = (1 / a) F (t / a).

2.10 Use of Partial Fractions If f ( p) is of the form

g ( p) h ( p)

, where g and h are polynomials in p, then break f ( p) into

partial fractions and manipulate term by term.

Example 4:

Solution:

 3p + 7  Evaluate L−1  2 .  p − 2 p − 3 3 ( p − 1) + 10   3p + 7  −1  L−1  2  =L  2  ( p − 1) − 4   p − 2 p − 3

(Bundelkhand 2013)

T-49

 3 ( p − 1)  10 = L−1  +  2 2  ( p − 1) − 4 ( p − 1) − 4  p−1   −1 = 3 L−1   + 10 L 2 ( p 1 ) 4 − −  

  1   2  ( p − 1) − 4 

p    1  + 10 e t L−1  2 = 3 e t L−1  2 2 2  p −2   p −2  = 3 e t cosh 2 t + 5 e t sinh 2 t = 4 e 3 t − e − t .  e −1 / p  cos 2 √ t  e−a / p  Example 5: If L−1  1 /2  = , find L−1  1 /2  , where a > 0.  p

Solution:

∴



√ (πt)

 p



 e −1 / p  cos 2 √ t Since L−1  1 /2  = , √ (πt)  p 

 e −1 / pk  1 cos 2 √ (t / k ) L−1  = 1 /2   ( pk )  k √ (πt / k )

 e −1 / pk  cos 2 √ (t / k ) L−1  1 /2  = . √ (πt)  p 

or

 e − a / p  cos 2 √ (at) Taking k = 1 / a, we have L−1  1 /2  = ⋅ √ (πt)  p  Example 6:

Find a function F(t) for which  3 4 e − p 4 e −3 p  F (t) = L−1  − 2 + ⋅ p p2  p

Solution:

 3 4 e − p 4 e −3 p  F (t) = L−1  − 2 +  p p2   p  e−p   e −3 p   1 = 3 L−1   − 4 L−1  2  + 4 L−1  2  ⋅  p p   p 

Now

L−1 {1 / p} = 1, L−1 {1 / p2 } = t.

∴

L−1 { e − p / p2 } = (t − 1) H (t − 1),

…(1)

L−1 { e −3 p / p2 } = (t − 3) H (t − 3). Putting in (1), we have F (t) = 3 − 4 (t − 1) H (t − 1) + 4 (t − 3) H (t − 3). Example 7: Find the inverse Laplace transform of the following function.

p+1 2

p ( p + 4 p + 8) 

p+1

 −1 =L 2  p ( p + 4 p + 8)

Solution: L−1 

.

p−4 1  −  , 2 8 p 8 ( p + 4 p + 8)  resolving into partial fractions

T-50

Example 8:

 1 1 −1  − L  p 8

 ( p + 2) − 6    2  ( p + 2) + 4 

=

1 −1 L 8

=

1 1 −2 t −1  p − 6  L  2 − e  8 8  p + 4

=

1 1 −2 t − e 8 8

=

1 [1 − e −2 t (cos 2 t − 3 sin 2 t)]. 8

 −1  p    1 −6 L−1  2 L  2 2 2    p +2   p + 2  

p   2 1 t Prove that L−1  4 sinh . sin . √ 3 t. = 2 2 2 3 √ 1 p + p +   (Kanpur 2010; Meerut 13B; Kashi 14; Rohilkhand 14)

Solution:

p   −1 We have L−1  4 =L 2 1 p + p +  

p    2 2 2  ( p + 1) − p 

p   = L−1  2  2  ( p − p + 1) ( p + p + 1)  1 ( p2 + p + 1) − ( p2 − p + 1) = L−1   2 2  2 ( p − p + 1) ( p + p + 1)    1 1 = L−1  −  2 2  2 ( p − p + 1) 2 ( p + p + 1)  1 −1  = L  2 ( p − 

Example 9:

(ii)

 1  1 −1 − L 1 2 3 2 ) +  2 4

   ( p + 

 1  1 2 3 ) +  2 4

=

  1 t / 2 −1  1  1 − t / 2 −1 − e e L  L 1 2 2 2 2 + √ 3 p ( )     2

=

1 t /2 2 t 1 2 t sin  √ 3  − e − t /2 . sin  √ 3  e   2 2 2 2 √3 √3

=

2 t t 1 t sinh sin  √ 3  ⋅ (e t /2 − e − t /2 ) sin  √ 3  =    2 2 2 √3 √3

  1 Evaluate (i) L−1  2 , 2  ( p + 4) ( p + 1) 

 3 p3 − 3 p2 − 40 p + 36 L−1  ( p2 − 4)2 

 ⋅ 

  1    2 1 2  p + ( √ 3)    2

(Meerut 2013)

T-51



Solution: (i) We have L−1 

1 2

 2

 ( p + 4) ( p + 1) 

2p + 3   2 1 = L−1  + −  2 25 ( p2 + 4)  25 ( p + 1) 5 ( p + 1) =

1  −1  1  1  −1  2 L   +5L  2  25   p + 1  ( p + 1)    p  1  −1  − 2 L−1  2  −3 L  2   p + 4  p + 4  

1 [2 e − t + 5 e − t L−1 {1/ p2 } − 2 cos 2 t − (3 / 2) sin 2 t] 25 1 −t = [e (2 + 5 t) − 2 cos 2 t − (3 / 2) sin 2 t ]. 25 =

(ii)

 3 p3 − 3 p2 − 40 p + 36 L−1  ( p2 − 4)2 

  

 3 p3 − 3 p2 − 40 p + 36 = L−1  2 2  ( p − 2) ( p + 2)

  

  2 3 5 = L−1  − + +  2 2 p + 2 ( p + 2)   ( p − 2)    1  1 −1 + 3 L−1  = − 2 L−1   + 5L 2 + p 2 ( p − 2 )    

  1  2  ( p + 2) 

= − 2 e2 t L−1 {1/ p2 } + 3 e −2 t + 5 e −2 t L−1 {1/ p2 } = − 2 e2 t . t + 3 e −2 t + 5 e −2 t . t = (5 t + 3) e −2 t − 2 t e2 t . Example 10:

  p2 1   Show that L−1  4 = (cosh at sin at + sinh at cos at). 4 2a  p + 4 a 

Solution: We have

   p2 p2 −1  L−1  4 L =   4 2 2 2 2 2  p + 4a   ( p + 2a ) − 4a p    p2 = L−1  2 2 2 2   ( p − 2 ap + 2 a ) ( p + 2 ap + 2 a )  1 p ( p2 + 2 ap + 2 a2 ) − p ( p2 − 2 ap + 2 a2 ) = L−1  ⋅ 4a ( p2 + 2 ap + 2 a2 ) ( p2 − 2 ap + 2 a2 ) 

  

T-52

p p   = L−1  − 2 2 2 2   4 a ( p − 2 ap + 2 a ) 4 a ( p + 2 ap + 2 a ) 1 −1  ( p − a) + a  1 −1  ( p + a) − a  L  L  = − 2 2 2 2 4a  ( p − a) + a  4 a  ( p + a) + a  1 at −1  p + a  1 − at −1  p − a  e L  2 e L  2 − = 2 2 4a 4 a p + a  p + a  p  1 at  −1  1  −1  = e L  2 +aL  2 2 2  4a   p + a    p +a  p  1 − at  −1  1  −1  e L  2 −aL  2 2 2  4a   p + a    p +a  1 at 1 − at = e (cos at + sin at) − e (cos at − sin at) 4a 4a −

  e at − e − at  1   e at + e − at   sin at +   cos at  2 a   2 2     1 = (cosh at sin at + sinh at cos at). 2a =

Example 11:

Solution:

∴

−1

L

 ( p + 1) e − πp  Find L−1  2 .  p + p+1 

 ( p + 1) + 1   p+1  −1  2 2 We have L  2 =L  1 3 2  p + p + 1 ( p + ) +   2 4 1  p+  − t / /2 −1  2 L  =e 3  p2 +  4  p   1 − t / 2 −1   1 = e − t /2 L−1  2 + e L  2  2 2  p + (√ 3 / 2)  2  p + (√ 3 / 2)  1 = e − t /2 cos (√ 3 t / 2) + e − t /2 . (2 / √ 3) sin (√ 3 t / 2) 2 e − t /2 = [√ 3 cos (√ 3 t / 2) + sin (√ 3 t / 2)]. √3 −1

 − (t − π ) / 2  ( p + 1 ) e − πp   e √3  2 =  p + p + 1   0  =

e − (t − π ) / 2 √3

  √3 √3 √ 3 cos 2 (t − π) + sin 2 (t − π) , t > π   ,t< π

  √3 √3 √ 3 cos 2 (t − π) + sin 2 (t − π) H (t − π).  

T-53

Comprehensive Exercise 2 Evaluate the following : 1.

  1 (i) L−1  2 ⋅  p − 6 p + 10 

p−1   (ii) L−1  ⋅ 2  ( p + 3) ( p + 2 p + 2)

2.

 1  (i) L−1  n  ( p + a) 

p   (ii) L−1  ⋅ 5 /2   ( p + 1) 

 p  (Lucknow 2011) (i) L−1  5  ( p + 1)   p2 − 2 p + 3  (i) L−1   2  ( p − 1) ( p + 1)

 3p + 2  (ii) L−1  2 ⋅  4 p + 12 p + 9 

3.

4.

  1 (ii) L−1  ⋅ 2  ( p + 2) ( p − 1)  (Gorakhpur 2005)

5. 6.

p   1  32 p  If L−1  2 = t sin t, find L−1  ⋅ 2 2 2 2  ( p + 1)   (16 p + 1)  − 4p   e −5 p  −1  e Find (i) L−1  (ii) L ⋅   4 4  ( p − 2)   ( p − 3) 

7.

 e4 − 3 p  Find L−1  ⋅ 5 /2   ( p + 4) 

8.

Find the inverse L.T. of e − 3 p / p3 .

9.

Prove the following : (i)

(ii) 10.

(Lucknow 2009)

 e − pπ  L−1  2  = − sin t. H (t − π)  p + 1

(Kanpur 2008)

 p e −2 pπ /3  L−1  2  = cos 3 (t − 2 π / 3) . H (t − 2 π / 3).  p +9 

Prove the following : (i)

 pe − ap  L−1  2 = cos h ω (t − a). H (t − a), a > 0 2  p −ω 

 3 (1 + e − pπ ) (ii) L−1   = − sin 3 t. H (t − π) + sin 3 t. 2  p +9  11.

Evaluate (i)

p+2   L−1  2   p − 2 p + 5

(Kanpur 2008; Lucknow 10; Avadh 13)

T-54

p+8   (ii) L−1  2 ⋅  p + 8 p + 5 12.

13.

4p + 5   1 t 1 −2 t t Prove that L−1   = 3t e + e − e . 2 3 3 ( 1 ) ( 2 ) p − p +   4p + 5   Evaluate (i) L−1   2  ( p − 4) ( p + 3) (ii)

14.

15.

(Purvanchal 2007)

 5 p2 − 15 p − 11 . L−1  3  ( p + 1) ( p − 2) 

(Gorakhpur 2009)

2p + 1   1 Prove that L−1  = t (e t − e −2 t ). 2 2 ( ) ( ) 2 1 p + p −   3

(Gorakhpur 2007)

p   1 Prove that L−1  2  = sin t sinh t. 2  ( p − 2 p + 2) ( p + 2 p + 2) 2

A nswers 2 1.

(i) e 3 t sin t (ii) −

2. 3. 4. 5. 6. 7. 9.

13.

4 − 3t 1 − t e + e (4 cos t − 3 sin t) 5 5

(i) e − at ⋅

t n−1 (n − 1) !

(ii)

1 −t (ii) e (4 t 3 − t 4 ) 24 1 3 (i) (t − ) e t + e − t (ii) 2 2 1 1 t sin ( t) 4 4 1 (i) (t − 5)3 e 2 (t − 5) . H (t − 5) (ii) 9 4 8. (1/ √ π) . (t − 3)3 /2 e − 4 (t − 4) . H (t − 3) 3 3 (i) e t [cos 2 t + sin 2 t] 2 (ii) e − 4 t [cosh (√ 11t) + (4 / √ 11) sinh (√ 11t)] (i)

(i) −

1 − 3t 1 4t e + e + 3 te 4 t 7 7

(ii)

2 −t e 3

 t  (3 − 2 t)    π

1 −3 t /2 e (6 − 5 t) 8 1 [(3 t − 1) e t + e − 2 t ] 9

1 (t − 4)3 e 3 (t − 4) . H (t − 4) 6 1 (t − 3)3 . H (t − 3) 2

1 7 (− e − t + e2 t ) + 4 te2 t − t2 e2 t 3 2

T-55

2.11 Inverse Laplace Transform of Derivatives Theorem: If L−1 { f ( p)} = F (t), then n

L−1 { f

dn  ( p)} = L−1  n f ( p)  dp  = (− 1)n t n

= (− 1) t

n

F (t)

n

L−1 { f ( p)}, n = 1, 2, 3,...

Proof: Since, we have L { t n F (t)} = (− 1)n

dn dpn

f ( p)

= (− 1)n f n ( p),  dn  L−1 { f n ( p)} = L−1  n f ( p) = (−1)n t  dp 

∴

n

F (t)

= (−1)n t n L−1 { f ( p)}.

2.12 Inverse Laplace Transform of Integrals Theorem:

∞ F (t) If L−1 { f ( p)} = F (t), then L−1 ∫ ⋅ f ( x) dx =  p  t

Proof: Since, we have 1 L  F (t) =  t  L−1  

∴

∞

∫p

∞

∫p

 F (t) exists. t→0  t 

f ( x) dx, provided lim

F (t) f ( x) dx = ⋅  t

2.13 Multiplication by Powers of p Theorem: Proof:

If L−1 { f ( p)} = F (t) and F (0 ) = 0 , then L−1 { p f ( p)} = F ′ (t).

We have L {F ′ (t)} = p L {F (t)} − F (0 ) = p f ( p).

∴ Note 1:

−1

L

{ p f ( p)} = F ′ (t).

If F (0 ) ≠ 0 , then L−1 { p f ( p) − F (0 )} = F ′ (t)

or

L−1 { p f ( p)} = F ′ (t) + F (0 ) δ(t)

where δ (t) is the Dirac delta function or unit impulse function.

[ ∵ F (0 ) = 0 ]

T-56

Generalizations to L−1 { p n f ( p)} are possible, for n = 2, 3, ... i. e.,

Note 2:

in general L−1 { p n f ( p)} = F n (t) if F (0 ) = 0 .

2.14 Division by Powers of p Theorem I: If F(t) is sectionally continuous and of exponential order a and such that t  f ( p) F (t) exists then for p > a, L−1  lim  = ∫ 0 F ( x) dx. t→0 t p   Proof: Then

t

Let G (t) =

∫0

F ( x) dx.

G ′ (t) = F (t) and G (0 ) = 0 .

∴

L { G ′ (t)} = p L { G (t)} − G (0 ) = p L { G (t)}

or

L { F (t)} = f ( p) = p L { G (t)}. f ( p)  f ( p) L { G (t)} = . Hence L−1   = G (t) = p  p 

∴

 f ( p) L−1  2  =  p 

Theorem II:

Proof: Let G (t) = Then G ′ (t) =

t

∫0 ∫0

y

t

∫0 ∫0

y

t

∫ 0 F ( x) dx.

F ( x) dx dy.

F ( x) dx dy .

t

∫ 0 F ( x) dx, G ′ ′ (t) = F (t).

Also G (0 ) = G ′ (0 ) = 0 .

Now L { G ′ ′ (t)} = p2 L { G (t)} − p G (0 ) − G ′ (0 ) = p2 L { G (t)} L { F (t)} = f ( p) = p2 L { G (t)}.

or or

 f ( p) L−1  2  = G (t) =  p 

t

∫0 ∫0

y

 f ( p) L−1  n  =  p 

t

t

t

∫ 0 ∫ 0 ... ∫ 0

t

t

∫0 ∫0

p

F (t) dt n.

 1 d  1  = L−1  − ⋅  2  2 2 dp p +a  ( p + a )   

Solution: L−1 

2



2 2

p2

F (t) dt2 .

p   Find L−1  2 ⋅ 2 2 ( p + a ) 

Example 12:

f ( p)

F ( x) dx dy .

 f ( p) This may also be written as L−1  2  =  p  In general

L { G (t)} =

∴

(Purvanchal 2010)

 1 −1 = − L 2 

d  1      2 2   dp  p + a  

T-57

=−

  t 1 1 sin at. t . (− 1)1 L−1  2 = 2 a 2 2 p a +  

2   p Find L−1  2 ⋅ 2  ( p + 4)  p Solution: Let f ( p) = . 2 ( p + 4)2

Example 13:

p   t = sin 2 t = F (t) L−1 { f ( p)} = L−1  2 2  ( p + 4)  4

∴

[See Ex. 12, here a = 2] and ∴

F (0 ) = 0 . L−1 { p f ( p)} = F ′ (t)

or

  d 1 p2 1 L−1  2 = ( t sin 2 t) = (sin 2 t + 2 t cos 2 t). 2 dt 4 4 p + ( 4 )  

Example 14:

Solution:

 p+2  Find L−1  2 ⋅  p ( p + 3)

( p + 3) − 1   p+2  −1  We have L−1  2  =L  2  p ( p + 3)   p ( p + 3)   1  1 1 −1  1  −1  = L−1  2 − 2   = L  2 − L  2 p ( p + 3)  p ( p + 3)  p  p 

Now

…(1)

L−1 {1/ p2 } = t /1 ! = t.

 1  −3 t L−1  = F (t), say. =e p + 3   t   1 ∴ we have L−1   = ∫ 0 F ( x) dx.  p ( p + 3) (by theorem I of article 2.14) t −3 x 1 −3 t =∫ e dx = (1 − e ) = F1 (t) say, 0 3   t 1 1 t −3 x ∴ L−1  2  = ∫ 0 F1 ( x) dx = ∫ 0 (1 − e ) dx 3 ( ) 3 p p +   1 1 = t + (e −3 t − 1). 3 9 Also

∴ From (1), we have  p+2  1 1 −3 t 2 1 1 − 1) = t − e −3 t + ⋅ L−1  2  = t − t − (e 3 9 3 9 9  p ( p + 3) Example 15:

  1  Find (i) L−1  log 1 + 2   p    

(Agra 2001)

T-58

1  1  (ii) L−1  log 1 + 2   ⋅ p     p Solution:

(i) Let  p2   1  f ( p) = log 1 + 2  = − log  2 p    p + 1 = − 2 log p + log ( p2 + 1). 2p 2 + 2 ⋅ p p +1

∴

f ′ ( p) = −

∴

L−1 { f ′ ( p)} = − 2 + 2 cos t

or

− t L−1 { f ( p)} = − 2 (1 − cos t)

or

L−1 { log (1 + 1/ p2 )} = 2 (1 − cos t) / t.

(ii)

From part (i), we have F (t) = L−1 { log (1 + 1/ p2 )} = 2 (1 − cos t) / t.

∴

1  1  1  L−1  log 1 + 2   = L−1  f ( p) p   p  p =

t

∫0

=∫

t 0

F ( x) dx 2 (1 − cos x) dx . x

Comprehensive Exercise 3 Evaluate the following : 1.

  1 (i) L−1  3  ( p − a) 

  1 (ii) L−1  ⋅ 3  ( p + a)  (Rohilkhand 2010)

p    2 2 2 ( p − a ) 

(ii) L−1

p+1   ⋅  2 2  ( p + 2 p + 2) 

2.

(i) L−1

3.

  1  L−1  log 1 − 2   p   

4.

  1 L−1  3 ⋅  p ( p + 1)

5.

  1 L−1  3 2 ⋅  p ( p + 1)

6.

  1 ⋅ L−1  3  p ( p + 1) 

7.

p+2   (i) L−1  log  p+1 

p+2  1 (ii) L−1  log ⋅ p+1  p

T-59

8.

p+3   (i) L−1  log  p+2 

(Kanpur 2007)

p+3  1 (ii) L−1  log ⋅ p+2  p 9.

10.

p   1 If L−1  2 = t sin t , find L−1 2 ( ) 1 p +   2  L−1  tan−1 

  1 ⋅  2 2  ( p + 1) 

(Kanpur 2011)

1 ⋅ p

A nswers 3 1. 2.

1 2 1 (i) 2 (i)

1 2 − at t e . 2 1 (ii) te − t sin t. 2

t2 e at

(ii)

 t  sinh at    a

1 2 t − e− t . 2 1 6. 1 − e − t (1 + t + t2 ). 2 t 1 −x (ii) ∫ (e − e −2 x ) dx . 0 x

3.

2 (1 − cosh t) / t.

4. 1 − t +

5.

1 2 t + cos t − 1. 2

7.

(i) (e − t − e − 2 t ) / t

8.

(i) (e − 2 t − e − 3 t ) / t

(ii) ∫

9.

1 (sin t − t cos t). 2

10.

t 0

1 −2 x (e − e −3 x ) dx . x

sin t ⋅ t

2.15 Convolution Let F (t) and G (t) be two functions of class A, then the convolution of the two functions F (t) and G (t) denoted by F * G is defined by the relation F*G=

t

∫0

F ( x) G (t − x) dx.

This relation F * G is also called the resultant or falting of F and G.

Properties of Convolution: (i)

F * G is commutative i. e., F * G = G * F .

Proof:

F*G=

t

∫0

F ( x) G (t − x) dx

T-60

= −∫ =∫

t 0

t 0

F (t − y) G ( y) dy, putting t − x = y so that dx = − dy

G ( y) F (t − y) dy = G * F .

(ii)

F * G is associative i. e., ( F * G) * H = F * (G * H).

(ii)

F * G is distributive w.r.t. addition, i.e., F * (G + H) = F * G + F * H .

Proof: F * (G + H) = =∫

t

∫0 t 0

F ( x) . [G (t − x) + H (t − x)] dx F ( x) G (t − x) dx +

t

∫0

F ( x) H (t − x) dx

= F*G+ F* H.

2.16 Convolution Theorem (Convolution Property) Let F (t) and G (t) be two functions of class A and let L−1 { f ( p) } = F (t) and L−1 { g ( p)} = G (t), then L−1 { f ( p) g ( p)} = ∫

t

F ( x) G (t − x) dx = F * G.

0

(Rohilkhand 2003, 07; Avadh 06, 09, 14; Lucknow 06, 10, 11; Kanpur 09)

Proof: Here we shall prove that the Laplace transform of the convolution of two functions is equal to the product of their Laplace transforms i. e.,

t L ∫ F ( x) G (t − x) dx = f ( p) g ( p)  0 

from which the required result follows immediately. We have,

t L  ∫ F ( x) G (t − x) dx = 0  

∞

∫ t =0 ∞

t e − pt ∫ F ( x) G (t − x) dx dt  0 



t

e − pt F ( x) G (t − x) dx dt ,  …(1) x P the integration being performed first with respect to =

∫ t =0 ∫ x =0

x and then with respect to t. In the double integral (1), the region of integration is the area in the tx-plane lying below the infinite line OP

t=0

x

=

t

t

whose equation is x = t and above the line Ot i.e., t-axis whose equation is x = 0. Here t is taken along the line Ot and x is taken along the perpendicular line Ox. The

O

integration is first performed with respect to x regarding t as constant and so the strip is taken parallel to Ox.

x x=0

t

T-61

If we change the order of integration, then the strip is taken parallel to Ot so that the limits of t are from x to ∞ and those of x are from 0 to ∞. ∴

changing the order of integration in the double integral (1), we have ∞ t  ∞ e − pt F ( x) G (t − x) dt dx L  ∫ F ( x) G (t − x) dx = ∫ x = 0 ∫ t = x   0  ∞

∞ e − px F ( x) ∫ e px e − pt G (t − x) dt dx  t = x 

∞

∞ e − px F ( x) ∫ e − p (t − x)G (t − x) dt dx  t = x 

∞

∞ e − px F ( x) ∫ e − pz G (z ) dz  dx,  z = 0 

=

∫ x =0

=

∫ x =0

=

∫ x =0

putting t − x = z , so that dt = dz ; when t = x, we have z = 0 and when t → ∞, z → ∞ ∞ ∞ = ∫ e − px F ( x) dx ∫ e − pz G (z ) dz   x = 0   z = 0  ∞ ∞ = ∫ e − pt F (t) dt ∫ e − pt G (t) dt  t = 0   t = 0 

∵ 

b

∫a

f ( x) dx =

b

∫a

f (t) dt =

b

∫a

f (z ) dz  

= L{ F (t)}. L { G (t)} = f ( p) g ( p). Thus

t L  ∫ F ( x) G (t − x) dx = f ( p) g ( p) .   0

Hence

L−1{ f ( p) g ( p)} =

t

∫0

F ( x) G (t − x) dx = F (t) * G (t) .

2.17 Heaviside’s Expansion Theorem or Formula Let F (p) and G (p) be two polynomials in p where F (p) has degree less than that of G (p). If G (p) has n distinct zeros α r , r = 1, 2, ..., n, i.e. G ( p) = ( p − α1) ( p − α 2 )...( p − α n), then n  F ( p)  F (α r ) α r t L−1  e . = Σ  G ( p)  r = 1 G ′ (α r )

Proof: Since F (p) is a polynomial of degree less than that of G ( p) and G ( p) has n distinct zeros α r , r = 1, 2, ..., n F ( p) F ( p) ∴ = G ( p) ( p − α1) ( p − α 2 )...( p − α n) =

A1 A2 Ar An . + + ... + + ... + p − α1 p − α2 p − αr p − αn

Multiplying both sides by ( p − α r ) and taking limits as p → α r , we have

T-62

Ar = lim

p→ α r

F ( p) ⋅ ( p − α r ) G ( p)

= F (α r ) ⋅ lim

(p − αr) G ( p)

= F (α r ) ⋅ lim

1 G ′ ( p)

p→ α r

p → αr

= F (α r ) ⋅

Form 0   0  [by L’ Hospital’s rule]

1 ⋅ G ′ (α r )

F ( p) F (α1) F (α2 ) 1 1 = ⋅ + . + ... G ( p) G ′ (α1) ( p − α1) G ′ (α2 ) ( p − α2 )

∴

+

F (α r ) F (α n) 1 1 ⋅ + ... + ⋅ + ... . G ′ (α r ) ( p − α r ) G ′ (α n) p − α n

 F ( p)   F (α1) −1  1  F (α2 ) −1  1 Hence L−1  ⋅L  L  = +  + ...  G ( p )  G ′ (α1)  p − α1  G ′ (α2 )  ( p − α2 )  F (α r ) −1  1  F (α n) −1  1  L  .L   + ... +  G ′ (α r ) p − α G ′ (α n) r    p − αn  F (α1) α 1 t F (α r ) α r t F (α n) α n t = ⋅e + ... + ⋅e + ... + ⋅e G ′ (α1) G ′ (α r ) G ′ (α n) +

n

= Σ

r =1

F (α r ) α r t e . G ′ (α r )

2.18 The Beta Function If m > 0, n > 0, the Beta function is defined as B (m, n) =

1

∫0

x m −1 (1 − x)n−1 dx.

p2



Example 16: Use the convolution theorem to find L−1 

2

 2 2

( p + a ) 

Solution:

p   We have L−1  2 = cos at. 2  p +a 

∴ By the convolution theorem, we have    p2 p p  = L−1  2 ⋅ 2 L−1  2 2 2 2 2 p + a p + a  ( p + a ) 

⋅

T-63 t

=

∫ 0 cos ax cos a (t − x) dx

=

∫0

t

cos ax (cos at cos ax + sin at sin ax) dx

= cos at

t

2

∫ 0 cos

t

ax dx + sin at ∫ cos ax sin ax dx 0

=

t t 1 1 cos at ∫ (1 + cos 2 ax) dx + sin at ∫ sin 2 ax dx 0 0 2 2

=

1 1 1 1 cos at  x + sin 2 ax  + sin at. − cos 2 ax   0 2  2 a  0 2 2a

t

t

1 1 1 cos at t + sin 2 at  + sin at (1 − cos 2 at) 2 2a   4 a 1 1 1 = t cos at + sin at + (sin 2 at cos at − sin at cos 2 at) 2 4a 4a 1 1 1 = t cos at + [sin at + sin (2 at − at)] = [at cos at + sin at]. 2 4a 2a =

 Example 17: Find L−1 

 1  , by the convolution integral and deduce the value of  √ p . ( p − a)

  1 L−1  ⋅  p √ ( p + a)  1 1 t1 /2 − 1 −1  1  Solution: We have L−1  = F1 (t), say =  = L  1 /2  = 1 Γ (2 ) √π √t  √ p  p   1 L−1   p−

and

 at  = e = F2 (t), say. a

∴ By the convolution theorem , we have   1 L−1   = F1 (t) * F2 (t) =  √ p ⋅ ( p − a) =

e at √π

√(at)

∫0

t

∫0

1 ⋅ e a(t − x)dx √π √ x

2 u du √ a − u2 2 u e ⋅ du, putting ax = u2 , so that dx = u a a

e at 2 √(at) − u2 e at ⋅ e du = erf (√ (at)). ∫ √a √π 0 √a    1 1 −1  Deduction. L−1  =L    p √ ( p + a)  ( p + a − a) √ ( p + a) at   1 1 − at e = e − at L−1  erf (√ (at)). erf (√ (at)) = =e ⋅ √a √a  ( p − a) √ p =

Example 18:

Apply convolution theorem to prove that 1 Γ (m) ⋅ Γ(n) B (m, n) = ∫ x m −1 (1 − x)n−1 dx = , m > 0, n > 0 . 0 Γ(m + n) (Gorakhpur 2008; Agra 02)

Hence deduce that

T-64 π /2

∫0 Solution:

sin2 m −1 θ ⋅ cos2 n−1 θ dθ =

Γ (m) ⋅ Γ (n) 1 ⋅ B (m, n) = 2 2 Γ (m + n)

Consider the function

We have,

t

x m −1 ⋅ (t − x)n−1 dx.

t

F1 ( x) ⋅ F2 (t − x) dx, where F1 (t) = t m −1 and F2 (t) = t n−1

F (t) =

∫0

F (t) =

∫0

= F1 * F2 . L {F (t)} = L {F1 * F2 }

∴

= L {F1 (t)}. L {F2 (t)}, by convolution theorem Γ (m) Γ (n) Γ (m) Γ (n) = L { t m − 1} . L {t n − 1} = ⋅ n = ⋅ pm p pm+n t m −1

F (t) =

∴

∫0 x

 Γ (m) Γ (n) ⋅ (t − x)n − 1 dx = L−1  m+n   p 

 1  Γ (m) Γ (n) m + n − 1 t . = Γ (m) ⋅ Γ (n) ⋅ L−1  m + n  = Γ (m + n) p  Taking t = 1, we have B (m, n) = =

1

∫0

x m − 1 (1 − x)n − 1 dx

Γ (m) Γ (n) ⋅ Γ (m + n)

…(1)

Deduction: Taking x = sin2 θ, so that dx = 2 sin θ cos θ dθ. From (1), we have 2

π /2

∫0

π /2

Hence

∫0

Example 19:

sin2 m − 1 θ cos2 n − 1 θ dθ =

sin2 m − 1 θ ⋅ cos2 n − 1 θ dθ =

Γ (m) Γ (n) ⋅ Γ (m + n) Γ (m) Γ (n) 1 = B (m, n) 2Γ (m + n) 2

3p + 1   Using Heaviside’s expansion formula find L−1   2  ( p − 1) ( p + 1) (Gorakhpur 2008)

Solution:

and ∴

Here F ( p) = 3 p + 1 G ( p) = ( p − 1) ( p2 + 1) = ( p − 1) ( p − i) ( p + i).

G ( p) has 3 distinct zeros α1 = 1, α2 = i and α3 = − i.

Also G ′ ( p) = 3 p2 − 2 p + 1. ∴ By the Heaviside’s expansion formula, we have 3p + 1   F (1) t F (i) i t F (− i) − it L−1  e + e + e = 2 G ′ ( 1 ) G ′ ( i ) G ′ (− i) ( 1 ) ( 1 ) p − p +  

T-65

=

(3 i + 1) it (− 3 i + 1) − it 4et + e + e − (2 + 2 i) (− 2 + 2 i) 2

= 2e t −

(3 i + 1) (1 − i) it (3 i − 1) (1 + i) − it e + e 2 (1 + i)(1 − i) 2 (1 − i) (1 + i)

1 1 (i + 2) e it + (i − 2) e − it 2 2 1 = 2 e t − i (e i t − e − i t ) − (e i t + e − i t ) 2 1 = 2 e t − i . 2 i sin t − 2 cos t = 2 e t + sin t − 2 cos t. 2 = 2et −

Example 20:

Using Heaviside’s expansion formula find  2 p2 + 5 p − 4  L−1  3 ⋅ 2  p + p − 2 p 

Solution:

(Avadh 2010)

Here F ( p) = 2 p2 + 5 p − 4 G ( p) = p3 + p2 − 2 p = p ( p − 1) ( p + 2)

and

G ′ ( p) = 3 p2 + 2 p − 2. G ( p) has 3 distinct zeros α1 = 0 , α2 = 1 and α3 = − 2. ∴ By the Heaviside’s expansion formula, we have  2 p2 + 5 p − 4  F (0 ) 0 t F (1) t F (− 2) −2 t e + L−1  3 e + e = 2 G ′ (1) G ′ (− 2)  p + p − 2 p  G ′ (0 ) = 2 + e t − e −2 t .

Comprehensive Exercise 4

1.

Use the convolution theorem to find:   1 (i) L−1  (ii)  p − p + ( 1 ) ( 2 )   (iii)

2.

(i)

  1 L−1  ⋅ p − p + ( 1 ) ( 3 )   p   L−1  2 2 2 ( p + a ) 

  1 L−1   p + p − ( 1 ) ( 2 )  

(Kanpur 2012)

(ii)

  1 ⋅ L−1  2 2  p ( p + 4)  (Lucknow 2009, 11)

 1   p 1 −1  ⋅ = L−1  2 ⋅ 2 Hint. (ii) L  2 2 2   p ( p + 4)   p ( p + 4)  Take f1 ( p) =

1 2

p

, f2 ( p) =

p 2

 2

( p + 4) 

⋅

T-66

  1 L−1   2  ( p − 2) ( p + 1)

3.

(i)

4.

  1 (i) L−1  2 2  p ( p + 1) 

  1 L−1  2 ⋅  ( p + 4) ( p + 2)

(ii)

(Gorakhpur 2006)

  1 (ii) L−1  ⋅ 2  ( p + 2) ( p − 2) 5.

(Kanpur 2011)

  p2 Find L−1  2 ⋅ 2  ( p + 4) 

(Kanpur 2010)

[Hint. Proceed as in Ex. 16. Here a = 2]. 6.

  1 1 Show that L−1   = erf (2 √ t).  p √ ( p + 4) 2 Using Heaviside’s expansion formula, find

7.

 2 p2 − 6 p + 5  L−1  3 ⋅ 2  p − 6 p + 11p − 6 

9.

(i)

19 p + 37   L−1    ( p + 1) ( p − 2) ( p + 3)

8.

  p2 − 6 L−1  3 ⋅ 2  p + 4 p + 3 p

(ii)

p+5   L−1  ⋅ 2  ( p + 1) ( p + 1)

A nswers 4 1.

(i)

1 t (e − e − 2 t ) 3

(ii)

1 2t (e − e − t) 3

(ii)

1 (1 − t sin 2 t − cos 2 t) 16

(ii)

1 − 2t (e + sin 2 t − cos 2 t) 8

(ii)

1 2t [e − (4 t + 1) e 2 t ] 16

1 t (e − e − 3 t ) 4

(iii) 2.

(i) (1/ 2 a) t sin at

3.

(i)

4.

(i) (t + 2) e − t + t − 2

5.

1 (2 t cos 2 t + sin 2 t). 4

8.

−2+

9.

(i) − 3 e − t + 5 e2 t − 2 e −3 t

1 2t (e − 2 sin t − cos t) 5

7.

1 t 5 e − e2t + e3t 2 2

5 − t 1 − 3t e + e 2 2 (ii) 2 e − t − 2 cos t + 3 sin t

T-67

Example 21:

Solution:

∴

 e −√ p   − x √ p   x  −1  e Find L−1  ⋅  and hence deduce that L   = erfc  p  p   2 √ t    

Let f ( p) = e − √ p. F (t) = L−1 {e − √ p}   p p3 /2 p2 p5 /2 = L−1 1 − √ ( p) + − + − + … 2 ! 3 ! 4 ! 5 !   1 −1 1 −1 3 / 2 = L−1 {1} − L−1 { p1 /2} + L { p} − L {p } 2! 3! +

Now

1 −1 2 1 −1 5 / 2 L {p }− L { p } + ... 4! 5!

…(1)

  1 L−1{ pn + (1 /2)} = L−1  − n−(1 /2)  p  =

=

t

− n − (3 /2)

1 Γ (− n − ) 2

, for n = 0 , 1, 2, ...

(− 1)n + 1  1  3   5   2 n + 1       ...   t √ π  2  2  2  2 

− n − (3 /2)

.

   2  1 n+1  2   2   2   √ π ∵ Γ (− n − ) = (− 1)       ...  2 1 3 5  2 n + 1   Also L−1 { p n} = 0 , if n is a +ve integer. ∴ from (1), we have F (t) = −

(− 1) t −3 /2 1 1 (− 1)2  1  3  −5 /2 ⋅ − ⋅ ⋅    t 2 3 ! √ π  2  2 √π −

=

=

Since

 f ( p) L−1  =  p 

1 2 √ (π) t3 /2 1 2 √ (π) t3 /2 t

∫0

1 (− 1)3  1  3   5  −7 /2 + ...     t 5 ! √ π  2  2  2

2 3    1  1         1  4t  1 − + − 4 t + ... 2! 3!  4t     

e −1 / 4 t ⋅

F ( x) dx , where F (t) = L−1 { f ( p)}, we have

T-68

 e −√ p  t 1 L−1  e −1 /(4 x)dx  = ∫0 2 √ (π) x3 /2  p  =−

=

2 π

2 √π

1 /(2 √ t)

2

e − y dy, putting x =

∫∞

∞

∫1 / (2 √t) e

− y2

1 4y

2

so that dx = −

dy 2 y3

1  dy = erfc  ⋅  2 √ t

∵ Complementary Error Function erfc (t) = 1 − erf (t) = 1 − 2  √π

∫0 e

2  ∞ − x2 e dx +  √ π  ∫0

e

=

0

∫t

2 2 e − x dx =  √π

t − x2

∞ − x2

∫t

dx

 dx 

 e −√ p   1  Deduction: We have L−1  ⋅  = erfc   2 √ t  p  ∴

 e −√( x2 p)    1 1   L−1  2  = 2 erfc   , by change of scale property 2  2 √ (t / x )  x p  x

or

− x √p   e   x ⋅ L−1    = erfc  2 √ t  p 

Example 22:

 1  t2 t5 t8 t11 (i) Prove that L−1  3 − + − + ... =  p + 1 2 ! 5 ! 8 ! 11!

(ii) Applying Heaviside’s Expansion formula , prove that  1  1 −t 1 1 t /2 L−1  3  = [e − e { cos ( √ 3 t) − √ 3 sin ( √ 3 t)}]. 3 2 2 p + 1   1 1 Solution: (i) We have, 3 = 3 (1 + 1 / p3 )−1 p +1 p =

=

∴

 1  1 1 1 1 1 − 3 + 6 − 9 + 12 − ... 3  p  p p p p  1 p3

−

1 p6

+

1 p9

−

1 p12

+ ...

 1  t2 t5 t8 t11 − + − +… L−1  3 =  p + 1 2 ! 5 ! 8 ! 11 !

(ii) Here F ( p) = 1 and G ( p) = p 3 + 1 = ( p + 1) ( p2 − p + 1), G ′ ( p) = 3 p2 . G ( p) has 3 distinct zeros − 1, ∴

1 1 (1 + √ 3 i) and (1 − √ 3 i). 2 2

By the Heaviside’s expansion formula , we have

T-69 1

 1  F { 12 (1 + √ 3 i )} { (1 + √3 i)} t F (− 1) − t e 2 L−1  3 e + = G ′ { 12 (1 + √ 3 i )}  p + 1 G ′ (− 1) + 1 −t e + 3

=

1 3 4

(1 + √ 3 i )2

1 { (1 + √3 i )}t 2

e

F { 12 (1 − √ 3 i )} G ′ { 12 (1 − √ 3 i )} +

1 3 4

(1 − √ 3 i )2

1 { (1 − √3 i)} t 2

e

1 { (1 − √3 i)}t 2

e

1

{ (1 + √3 i )}t 2 (√ 3 i + 1) 1 = e− t + e 2 3 3 (√ 3 i − 1) (√ 3 i + 1) 1

+

{ (1 − √3 i )}t 2 (√ 3 i − 1) e2 3 (√ 3 i + 1) (√ 3 i − 1)

1 − t 1 t /2 e − e [(√ 3 i + 1) e √3 it /2 − (√ 3 i − 1) e −√3 it /2 ] 3 6 1 1 = e − t − e t /2 [√ 3 i (e √3 it /2 − e − √3 it /2 ) + (e √3 it /2 + e −√3 it /2 )] 3 6 1 1 1 = [e − t − e t /2 {− √ 3 sin ( √ 3 t) + cos ( √ 3 t)}]. 3 2 2 ∞ 1 Example 23: Show that ∫ (Avadh 2008) cos x2 dx = √ (π / 2) 0 2 =

Solution:

∴

Let F (t) =

L { F (t)} =

∞ − pt

∫0

=∫ =

∞

∫0 e

∞ 0

cos tx2 dx. F (t) dt =

∞ − pt

 ∫ 0

e

∞ − pt

∫0

e

 ∞ cos tx2 dx dt  ∫ 0 

∞ cos tx2 dt dx = ∫ L {cos tx2 } dx =  0

p

∞

∫0

2

p + x4

dx

π /2 1 dθ , putting x = √ ( p tan θ) so that ∫ 2 √ p 0 √ (tan θ)

dx =

p sec2θ dθ 2 √ ( p tan θ)

=

π /2 1 sin−1 /2 θ cos 1 /2 θ dθ ∫ 2√ p 0

=

1 3 1 1 π 1 1 Γ (4 ) Γ(4 ) 1 Γ (4 ) Γ (1 − 4 ) π = ⋅ = = ⋅ 1 2 4 √ p sin 4 π 2 √ (2 p) 2 √ p 2 Γ (1) 2√ p

  π , 0 < p < 1 ∵ Γ ( p) Γ (1 − p) = sin pπ   ∴

F (t) =

or

∫0

∞

 1  π π t(1 /2) −1 1  π  = L−1  1 /2  =   1 2  2t  2√2  p  2 2 Γ (2 )

cos tx2 dx =

Now taking t = 1, we have

1  π   . 2  2t 

T-70 ∞

∫0 Example 24: Solution:

Prove that

Let F (t) =

1 √ (π / 2) 2 2 1 e − x dx = √ π. 2

cos x2 dx = ∞

∫0

(Meerut 2013B)

2

∞

e − tx dx.

∫0

∴ Proceeding as in Ex. 23, L { F (t)} = ∫ or

F (t) =

or

∫0

∞

∞ 0

π −1 L 2

∞

∫0

∞

 1 x  π = tan−1  = 2 √ p 0 2 √ p p+ x √ p dx

 1 π 1 1  π =     = ⋅  t 2 2 √ p √ ( t ) π  

2

e − t x dx =

Taking t = 1, we have

2

L { e − tx } dx =

∞

∫0

1  π   . 2  t 2

e − x dx =

1 √ π. 2

  8 Prove that L−1  2 = (3 − t2 ) sint − 3 t cos t. 3 ( p + 1 )     1 Solution: We have L−1  2  = sin t . p  + 1 ∴ by the convolution theorem, we have  1  t 1 L−1  2 ⋅ 2  = ∫ 0 sin x sin (t − x) dx ( p + 1) (( p + 1) Example 25:

=∫

t 0

sin x (sin t cos x − cos t sin x) dx t

t

0

0

= sin t ∫ sin x cos x dx − cos t ∫

sin2 x dx

t t 1 1 sin t ∫ sin 2 x dx − cos t ⋅ ∫ (1 − cos 2 x) dx 0 0 2 2 1 1 1 1 = sin t . (1 − cos 2 t) − cos t . (t − sin 2 t) 2 2 2 2 1 1 1 = sin t.sin2 t − t cos t + cos t sin t cos t 2 2 2   1 1 t L−1  2 = sin t − cos t. 2 2 2 ( ) p + 1  

=

or ∴

   1 1 1  L−1  2 = 8 L−1  2 ⋅ 2  3 2 ( p + 1)  ( p + 1) ( p + 1) t 1 x = 8 ∫  sin x − cos x sin (t − x) dx, by the convolution theorem  0 2 2 t

= 4 ∫ (sin x − x cos x) (sin t cos x − cos t sin x) dx 0

= 4 sin t ∫

t 0

(sin x cos x − x cos2 x) dx − 4 cos t

t

∫0

(sin2 x − x sin x cos x) dx

T-71

= 2 sin t ∫

t 0

t

{sin 2 x − x (1 + cos 2 x) } dx − 2 cos t ∫ {(1 − cos 2 x) − x sin 2 x} dx 0

 t2 1 − cos 2 t t 1 − cos 2 t  = 2 sin t − + − sin 2 t +  2 2 4  2  1 t cos 2 t sin 2 t  −2 cos t t − sin 2 t + −  2 2 4  = − t 2 sin t +

3 3 sin t − sin t cos 2 t − t sin t sin 2 t 2 2 − 2 t cos t +

= − t 2 sin t +

3 sin 2 t cos t − t cos t cos 2 t 2

3 3 sin t + (− sin t cos 2 t + sin 2 t cos t) 2 2 − t (cos 2 t cos t + sin t sin 2 t) − 2 t cos t

2

= (3 − t ) sin t − 3 t cos t.

Comprehensive Exercise 5 1.

  1 Find L−1  . 5  ( p − 1) ( p + 2)

(Kanpur 2009, 11)

2.

  1 Prove that L−1  = 2 2   p √ ( p + a )

3.

Apply Heaviside’s expansion formula, to prove that

4. 5.

t

∫0

J0 (ax) dx .

 1  1 t 1 1 −1 / 2 {cos ( √ 3 t) + √ 3 sin ( √ 3 t)}] L−1  3  = [e − e 3 2 2 p − 1   ∞ 1 Show that ∫ sin x2 dx = √ (π /2). 0 2 Apply the convolution theorem to show that t t ∫ 0 sin u cos (t − u) du = 2 sin t . ∞ sin2

(Gorakhpur 2007, 11)

x

π ⋅ 2

6.

Use Laplace transform to prove that

7.

 2 2 Prove that L−1  tan−1 2  = sin t sinh t. p  t 

8.

Show that 1 * 1 * 1 * … * 1 (n times ) = t n−1 / (n − 1)! , where n = 1, 2, 3,…

∫0

x2

dx =

(Agra 2002)

(Rohilkhand 2002)

T-72

9.

10.

  1 Find L−1  2 ⋅ 2 3 /2  ( p + a )  Prove that

t

t

∫0 ∫0

......

t

∫0

F (t) dt n =

− u)n−1 F (u) du. (n − 1) !

t (t

∫0

A nswers 5 1. 9.

1 t 4 4 3 4 2 8 8 1 −2 t e t − t + t − t + e  −   72 3 3 9 27 243 t J1 (at) a

2.19 Table of Inverse Laplace Transform Theorems Inverse Laplace Transform Theorems No. 1.

f (p)

L−1 { f ( p)} = F(t )

a1 f1 ( p ) + a2 f2 ( p )

a1 L−1 { f1 ( p )} +

Operation Linearity property

a2 L−1 { f2 ( p )} 2.

First translation or shifting theorem

3.

Second translation or shifting theorem

4.

f ( p − a)

e at L−1 { f ( p )}

e − ap f ( p )

 F (t − a), t > a G (t) =  t< a  0,

Change of scale property

f (ap )

1  t F  a  a

5.

Differentiation theorem

f

6.

Integral theorem

7.

Multiplication theorems

∞

∫p

n

( p)

f ( x) dx

(− 1)n t

n

F (t)

1 F (t) t

p f ( p)

F ′ (t)

pn f ( p )

F n (t)

T-73

f ( p) 8.

t

p

Division theorems

t

f ( p)

Convolution theorem

10.

Heaviside’s expansion theorem

f ( p). g ( p )

F ( p) G ( p)

t

t

∫0 ∫0 …∫0

pn 9.

F ( x) dx

∫0

F*G=

n

Σ

,

t

∫0

r =1

F (t) (dt)n

F ( x) ⋅ G (t − x) dx

F (α r ) α e G ′ (α r )

r t

where α r , r = 1, 2, ... n, are roots of G ( p ) = 0 and are all distinct.

degree F ( p ) < degree G ( p)

Objective Type Questions

Muliple Choice Questions

1.

2.

3.

Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d).  1 The value of L−1  4  is p  (a)

t3 3!

(b)

t4 4!

(c)

t5 5!

(d) None of these.

 1 The value of L−1   is  √ p (a)

 π    t

(c)

1 √ (πt)

 1  The value of L−1  5 /2  is p  4 3 /2 (a) t 3 4 t (c) t    3 π

(b)

 t    π

(d) √ (πt).

4 5 /2 [ t 3 4 π (d) t   ⋅  3 t (b)

T-74

4.

If L−1 { f ( p)} = F (t) then L−1 { f ( p − a)} is (a) e at F (t) (c) (−1)n t

5.

−1

If L

n

F (t) −1

{ f ( p)} = F (t) then L

t

∫0

(d)

F (u) du

Inverse Laplace transform of

1 p2 + a2

(Rohilkhand 2002)

F (t) ⋅ t

(Rohilkhand 2002)

(b) cos at (d) sinh at.

(Rohilkhand 2003)

 p  The value of L−1  2  is  2 p − 8

(c)

(b) sinh 2t

1 sinh 2 t 2

(d)

1 cosh 2 t. 2

If L−1 { f ( p)} = F (t) and F (0 ) = 0 , then L−1 { p f ( p )} is (a) (−1)n t (c)

n

(c)

(b) F ′ (t)

F (t)

F (t) t

The value of ∫ (a)

11.

F (u) du

is

(a) cosh 2t

10.

t

(b) F ′ (t)

(a) sin at 1 (c) sin at a

9.

∫0

 f ( p) If L−1 { f ( p)} = F (t) then L−1   is  p 

(c)

8.

(d)

(b) F (at) 1 t (d) F   . (Rohilkhand 2003, 10)  a a

(a) (−1)n t n F (t )

7.

1  t F  a  a

{ f (ap)} is

(a) aF (at) t (c) F    a 6.

(b)

(d) ∫ ∞ 0

t 0

F (u) du.

2

e − x dx is

1 √π 2 √π

1  π   2  2 (d) None of these. (b)

 1  is The value of L−1  3  ( p − 1)  (a) t2 e t

(b)

te t 2

t2 e t 2

(d)

None of these.

(c)

(Rohilkhand 2004)

T-75

Fill In The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1. 2.

If f ( p ) is the Laplace transform of a function F (t), then F (t) is called the …… Laplace transform of the function f ( p ). 1 Inverse Laplace transform of , p > a is …… . p− a

3.

If L−1 { f ( p )} = F (t) then L−1 {e − ap f ( p )} = G (t), where G (t) = …… .

4.

Let F (t) and G (t) be two functions of class A and let L−1 { f ( p )} = F (t) and

L−1 { g ( p )} = G (t), then L−1 { f ( p ) g ( p )} = …… .

5.

The value of ∫

∞ 0

cos x2 dx = ......

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

 4  − 4t The value of L−1   is 3 e . p + 4  

2.

 1  is t2 e − t . The value of L−1  2 ( 1 ) p +  

3.

  1 − 4t The value of L−1  2  is te .  p + 8 p + 16 

4.

The convolution of two functions F and G obeys the commutative law.

5.

  1 −t 1 2t The value of L−1   is (e − e ).  ( p − 1) ( p + 2) 3

A nswers Multiple Choice Questions 1.

(a)

2. (c)

3. (c)

4.

(a)

5. (d)

6. (c)

7.

(c)

8. (d)

9. (b)

10.

(a)

11. (b)

T-76

Fill in the Blank(s) 1.

inverse

4.

∫0

t

2. e a t

F ( x) G (t − x) dx

3.

F (t − a), t > a and 0 , t < a

5.

1  π   2  2

True or False 1.

T

2.

F

4

T

5.

F

3. T

¨

T-77

3 A pplications

of

L aplace

T ransform (To Solutions of Differential Equations and Integral Equations)

3.1 Solution of an Ordinary Differential Equation with Constant Coefficients

T

he Laplace transform is very useful in solving ordinary linear differential equations with constant coefficients.

Let us consider a linear differential equation with constant coefficients dn y

d n−1 y dy + C + ... + Cn−1 + Cn y = F (t) 1 n n − 1 dt dt dt

…(1)

where F (t) is a function of the independent variable t. Let

y (0 ) = A0 , y ′ (0 ) = A1,......, y n−1(0 ) = An−1

…(2)

be the given initial or boundary conditions where A0 , A1, A2 ,..., An−1 are constants. On taking the Laplace transform of both sides of equation (1) and using conditions (2) we obtain an algebraic equation known as “subsidiary equation” from which y ( p ) = L { y(t)} is determined. The required solution is then obtained by finding the inverse Laplace transform of y ( p ).

T-78

Example 1: Solve y ′ ′ (t) + y(t) = t with y ′ (0 ) = 1, y (π) = 0 . (Kanpur 08; Lucknow 10)

Solution: Taking Laplace transform of both sides of the given equation, we have

L{ y ′ ′ } + L{ y} = L{ t} or

p2 L{ y } − p y(0 ) − y ′ (0 ) + L{ y } = 1 / p2

or

( p2 + 1) L{ y} = Ap + 1 + 1 / p2 = Ap + ( p2 + 1) / p2 , where y (0 ) = A

or

L{ y } = A ⋅

∴

 p   1 y = y (t) = AL−1  + L−1   = A cos t + t.  2 2  p + 1 p 

p 2

p +1

+

1 p2

⋅

But y (π) = 0, therefore 0 = A cos π + π or A = π. ∴

y = π cos t + t, which is the required solution.

Example 2: Solve ( D2 + m2 ) x = a cos nt, t > 0 , x, Dx equal to x0 and x1, when t = 0, n ≠ m. Solution: (a) Taking Laplace transform of both sides of the given equation, we have

L { x ′ ′ } + m2 L { x } = a L {cos nt} or

p2 L { x} − px (0 ) − x ′ (0 ) + m2 L { x} =

or

ap ( p2 + m2 ) L {x} = px0 + x1 + p2 + n2

or

L { x} = x0 ⋅

= x0 .

= x0 .

p 2

2

p +m

p p2 + m2 p p2 + m2

+ x1.

+ x1 .

+ x1 ⋅

1 2

2

p +m

1 p2 + m2 1 p2 + m2

+

ap 2

p + n2

a. p 2

( p + m2 ) ( p2 + n2 )

+

( p2 + m2 ) − ( p2 + n2 )  m2 − n2  ( p2 + m2 ) ( p2 + n2 ) 

+

  p p ⋅ ⋅  − (m2 − n2 )  p2 + n2 p2 + m2 

ap

   

a

Taking the inverse transform, we have

or

    p 1 x = x0 . L−1  + x1 . L−1  2 2 2 2  p +m   p +m       p p a + ⋅  L−1  − L−1    2 2 2 2 (m2 − n2 )   p +n   p + m   x a x = x0 cos mt + 1 ⋅ sin mt + (cos nt − cos mt), 2 m (m − n2 )

which is the required solution.

T-79 Example 3:

Solve ( D2 + 9) y = cos 2 t if

y (0 ) = 1, y (π / 2) = − 1.

(Rohilkhand 2000, 02, 04; Agra 01; Gorakhpur 07, 10, 11; Kashi 14)

Solution: Taking the Laplace transform of both sides of the given equation, we have

L { y ′ ′ } + 9 L { y } = L {cos 2 t} or

p2 L { y } − p y(0 ) − y ′ (0 ) + 9 L { y } = p /( p2 + 4)

or

( p2 + 9) L { y } − p − A = p / ( p2 + 4), where y ′ (0 ) = A

or

L{ y} = =

p+ A p2 + 9 p p2 + 9

p

+ +

( p2 + 9) ( p2 + 4) A p2 + 9

+

p 5 ( p2 + 4)

−

p 5 ( p2 + 9)

⋅

 p   1  1 −1  p  1 −1  p  −1 y = L−1   + AL  2 + L  2 − L  2  2  p + 9  p + 9 5  p + 4 5  p + 9 1 1 1 = cos 3 t + A sin 3 t + cos 2 t − cos 3 t 3 5 5 4 1 1 = cos 3 t + A sin 3 t + cos 2 t. 5 3 5 But y (π / 2) = − 1. 4 3 1 3 1 ∴ − 1 = cos π + A sin π + cos π 5 2 3 2 5 1 1 or or A = 12 / 5. − 1= − A − 3 5 4 4 1 Hence the required solution is y = cos 3 t + sin 3 t + cos 2 t. 5 5 5 ∴

Example 4:

Solve ( D + 1)2 y = t given that y = − 3, when t = 0 and y = − 1, when t = 1. (Rohilkhand 2006)

Solution:

2

The given equation can be written as ( D + 2 D + 1) y = t.

∴

L { y ′ ′ } + 2 L { y ′ } + L { y} = L { t}

or

p2 L { y} − py (0 ) − y ′ (0 ) + 2 [ pL { y} − y (0 )] + L { y} = 1/ p2

or

( p2 + 2 p + 1) L { y } − p (− 3) − A − 2 (− 3) = 1/ p2 where y ′ (0 ) = A

or

( p + 1)2 L { y} = (1 / p2 ) − 3 p − 6 + A

or

L{ y} = =

1 2

2

p ( p + 1) 1 2

2

p ( p + 1)

− −

3p + 6 2

( p + 1)

+

A ( p + 1)2

3 ( p + 1) + 3 2

( p + 1)

+

A ( p + 1)2

=−

2 1 2 1 3 3 A + + + − − + p p2 p + 1 ( p + 1)2 ( p + 1) ( p + 1)2 ( p + 1)2

=−

A−2 2 1 1 + − + ⋅ 2 p p ( p + 1) ( p + 1)2

T-80

 1   1  1  1  y = − 2 L−1   + L−1   − L−1  + ( A − 2) L−1   2 2  p  p + 1  ( p + 1)  p 

∴

= − 21 . + t − e − t + ( A − 2) e − t L−1 {1/ p2} = − 2 + t − e − t + ( A − 2) te − t . Now, since y = − 1, when t = 1, − 1 = − 2 + 1 − e − 1 + ( A − 2) e − 1 or A = 3.

∴

Hence the complete solution is y = − 2 + t − e − t + te − t . Example 5:

dy dt

= 0 when t = 0.

(Avadh 2006, 07; Meerut 13B)

Taking the Laplace transform of both the sides of the given equation, we L { y ′ ′ } + L { y } = L { t cos 2 t} d p2 L { y } − p y (0 ) − y ′ (0 ) + L { y } = − ( L {cos 2 t}) dp

Solution:

have or

2 p2 1 d  p  =−  ( p2 + 1) L { y} = − + dp  p2 + 4  p2 + 4 ( p2 + 4)2

or

or

Solve ( D2 + 1) y = t cos 2 t, y = 0 ,

L{ y} =

p2 − 4 ( p2 + 1) ( p2 + 4)2

=−

5 5 8 + + 9 ( p2 + 1) 9 ( p2 + 4) 3 ( p2 + 4)2

 5 −1  1  5 −1  1  8 −1  1 L  + L  2 + L  2  2 2 9  p + 1 9  p + 4 3  ( p + 4)  5 5 8 t1 1 = − sin t + sin 2 t + ∫ sin 2 x ⋅ sin 2 (t − x) dx , 0 9 18 3 2 2

y=−

∴

   1  1 −1 By the convolution theorem since L  2  = sin 2t    p + 4 2 5 5 1 t = − sin t + sin 2 t + {cos (2 t − 4 x) − cos 2 t} dx 9 18 3 ∫0 =−

5 5 1 sin t + sin 2 t + 9 18 3

t − 1 sin (2 t − 4 x) − x cos 2 t   4  0

5 5 1 1 1 sin t + sin 2 t + sin 2 t − t cos 2 t + sin 2 t 9 18 12 3 12 5 4 1 y = − sin t + sin 2 t − t cos 2 t , which is the required solution. 9 9 3 =−

or Example 6:

Solve ( D3 − D2 − D + 1) y = 8 t e − t if y = D2 y = 0 , Dy = 1 when t = 0.

Solution: Taking the Laplace transform of both sides of the given equation, we have

L { y ′ ′ ′ } − L { y ′ ′ } − L { y ′ } + L { y } = 8 L { te − t} or

p3 L { y } − p2 y (0 ) − py ′ (0 ) − y ′ ′ (0 ) − [ p2 L { y } − p y (0 ) − y ′ (0 )] d − [ pL { y } − y (0 )] + L { y } = − 8 [ L {e − t}] dp

T-81

or

d  1  ( p3 − p2 − p + 1) L { y } − p + 1 = − 8   dp  p + 1

or

( p − 1)2 ( p + 1) L { y } = p − 1 +

or

L{ y} =

=

8 ( p + 1)2

1 8 + ( p − 1) ( p + 1) ( p − 1)2 ( p + 1)3 1 1 1  3 1 3 − + +   − 2  p − 1 p + 1 2 ( p − 1) ( p − 1)2 2 ( p + 1) +

=−

2 ( p + 1)2

+

2 ( p + 1)3

1 1 1 2 2 + + + + ⋅ 2 2 p − 1 p + 1 ( p − 1) ( p + 1) ( p + 1)3

  1  1  −1  1  −1 y = − L−1  +L  +L  2  p − 1  p + 1  ( p − 1) 

∴

  1  1  + 2 L−1  + 2 L−1   2 3  ( p + 1)   ( p + 1)   1  1  1 = − e t + e − t + e t L−1   + 2 e − t L−1   + 2 e − t L−1   2 2 3 p  p  p  = − e t + e − t + e t t + 2 e − t t + 2 e − t (t 2 / 2 !) = (1 + 2 t + t 2 ) e − t − (1 − t) e t , which is the required solution. Example 7:

Solve ( D 4 + 2 D2 + 1) y = 0 , where y (0 ) = 0 , y ′ (0 ) = 1, y ′ ′ (0 ) = 2 and

y ′ ′ ′ (0 ) = − 3. Solution: Taking the Laplace transform of both sides of the given equation, we have

L { y iv} + 2 L { y ′ ′ } + L { y } = 0 p4 L { y } − p3 y (0 ) − p2 y ′ (0 ) − p y ′ ′ (0 ) − y ′ ′ ′ (0 )

or

+ 2 [ p2 L { y } − p y (0 ) − y ′ (0 )] + L { y } = 0 or

( p4 + 2 p2 + 1) L { y } − p2 − 2 p + 3 + 2 (− 1) = 0

or

( p2 + 1)2 L { y } = p2 + 2 p − 1

or

∴

L{ y} =

p2 + 2 p − 1 2

2

( p + 1)

=

1 2

p +1

+

2p − 2 2

2

( p + 1)

=

1 2

p +1

+

2p 2

2

( p + 1)

 1   2p    1 −1 y = L−1  − 2 L−1  ⋅ +L  2 2 2 2 2  p + 1  ( p + 1)   ( p + 1) 

−

2 2

( p + 1)2

⋅ …(1)

T-82

Now

and

 d  1    2p   1    = − (− t) L−1  L−1  = − L−1    = t sin t  2 2 2 2  dp  p + 1   ( p + 1)   p + 1    1 1 1  L−1  = L−1  ⋅  = F (t) * F (t), 2 2 2 2  ( p + 1)   p + 1 p + 1  1  where F (t) = L−1   = sin t 2  p + 1 t 1 t = ∫ sin x ⋅ sin (t − x) dx = ∫ [cos (2 x − t) − cos t] dx 0 2 0 =

t 1 1 1 sin (2 x − t) − x cos t  = (sin t − t cos t). 2 2  0 2

Putting in (1), the required solution is y = sin t + t sin t − (sin t − t cos t) or

y = t (sin t + cos t).

Example 8: Find the general solution of the differential equation

x ′ ′ (t) + k 2 x (t) = F (t); x(0 ) = A, x ′ (0 ) = B. Solution: Taking the Laplace transform of both sides of the given equation, we have

L { x ′ ′ } + k 2 L { x } = L {F (t)} or

p2 L { x } − px(0 ) − x ′ (0 ) + k 2 L { x} = L {F (t)}

or

( p2 + k 2 ) L { x } = pA + B + f ( p), where L {F (t)} = f ( p)

or

L { x } = A.

∴

      p 1 1 x = AL−1  + BL−1  + L−1  ⋅ f ( p) ⋅ 2 2 2 2 2 2  p +k   p +k   p +k 

p 2

p + k2

+ B.

1 1 + f ( p). p2 + k 2 p2 + k 2

= A cos kt + ( B / k ) sin kt + {(1/ k ) sin kt} * F (t) x (t) = A cos kt + ( B / k ) sin kt + (1/ k )

t

∫0

sin k (t − x) ⋅ F ( x) dx. [By the convolution theorem]

This is the required general solution. Example 9: Solve

d2 x dt2

+ x = F (t), if x = x ′ = 0 for t = 0.

Solution: Taking the Laplace transform of both sides of the given equation, we have

L{ x ′ ′ } + L{ x} = L {F (t)} or or

p2 L{ x} − px(0 ) − x ′ (0 ) + L{ x} = f ( p), where L{F (t)} = f ( p) L { x} =

1 p2 + 1

f ( p).

T-83

 1  x = L−1  f ( p) = (sin t) * F (t) 2  p +1 

∴ or

x=

t

∫0

sin (t − x) ⋅ F ( x) dx ,

(by the convolution theorem)

which is the required solution.

Comprehensive Exercise 1

1.

Solve the following problems by means of the Laplace transform : dy (i) + y = 1, given that y = 2 when t = 0. dt (ii)

2.

d2 y 2

dt

+ y = 0 , under the conditions that y = 1,

dy dt

= 0 when t = 0. (Rohilkhand 2002)

(i) ( D2 + 1) y = 6 cos 2 t, if y = 3, D y = 1, when t = 0. (Gorakhpur 2006, 09; Rohilkhand 07; Purvanchal 07) 2

(ii) ( D + D) x = 2, when x (0 ) = 3, x ′ (0 ) = 1. 3.

2

(Rohilkhand 2011; Kanpur 07)

2

(i) ( D + m ) x = a sin nt, t > 0 , where x, Dx equal to x0 and x1 , when t = 0, n ≠ m. (ii) ( D + 2)2 y = 4 e −2 t , y (0 ) = − 1 and y ′ (0 ) = 4.

4.

(i) ( D2 − D − 6) y = 2, t > 0 , if y = 1, Dy = 0 , when t = 0. (ii) ( D2 + 6 D + 9) y = sin x, where y (0 ) = 1, y ′ (0 ) = 0.

5.

( D2 + 9) y = 18 t, if y (0 ) = 0 , y (π / 2) = 0 .

6.

( D2 + 6 D + 25) y = 208 e3 t , t > 0 , if y = 1, D y = 0 when t = 0.

7.

( D2 − 4 D + 5) y = 125 t2 , if y = 0 = D y , when t = 0.

8.

( D2 + 2 D + 1) y = 3 t e − t , t > 0 , subject to the conditions, y = 4, D y = 2 when t = 0.

9.

( D2 − 3 D + 2) y = 1 − e 2 t , y = 1, D y = 0 when t = 0. (Avadh 08; Gorakhpur 05) 2

10.

( D + 1) y = sin t sin 2 t, t > 0 , if y = 1, Dy = 0 when t = 0.

11.

( D3 − D) y = 2 cos t, y = 3, D y = 2, D2 y = 1, when t = 0.

12.

( D3 + D) y = e2 t , y (0 ) = y ′ (0 ) = y ′ ′ (0 ) = 0 .

13.

( D3 − 2 D2 + 5 D) y = 0 , if y (0 ) = 0 , y ′ (0 ) = 1, y (π / 8) = 1.

14.

( D2 + D) y = t2 + 2 t, where y (0 ) = 4, y′ (0 ) = − 2.

T-84

A nswers 1 1.

(i) y = e − t + 1

(ii) y = cos t

2.

(i) y = 5 cos t + sin t − 2 cos 2 t

(ii) y = 2 + 2 t + e − t

3.

x a sin nt − n sin mt (i) x = x0 cos mt + 1 sin mt +   2 2   m m m −n (ii) y = e −2 t (2 t2 + 2 t − 1)

4.

5. 6.

1 4 −2 t 8 3 t + e + e 3 5 15 1 (ii) y = [(53 + 155 x) e −3 x − (3 cos x − 4 sin x)] 50 (i) y = −

y = π sin 3 t + 2 t 3 y = 4 e3 t − e −3 t (4 cos 4 t + 7 sin 4 t) 4

7.

y = 25 t2 + 40 t + 22 + 2 e2 t (2 sin t − 11 cos t)

8.

y=

9.

y = 1 + e t (sin 2 t − cos 2 t)

14. y =

10. 11. 12. 13.

1 −t 3 e . t + 4 e − t + 6 te − t 2 15 1 1 y= cos t + t sin t + cos 3 t 16 4 16 y = 3 sinh t − sin t + cosh t + 2 1 1 2t 2 1 y=− + e + cos t − sin t 2 10 5 5

y=

1 1 2t + e − te2 t 2 2

1 3 t + 2 (1 + e − t ) 3

3.2 Solution of Ordinary Differential Equations with Variable Coefficients The Laplace transform can be used in solving some ordinary differential equations with variable coefficients. The method is found useful in case of the equations having the dm terms of the form t m y n (t) whose Laplace transform is (− 1)m L { y n (t)}. dp m

Example 10: Solution:

Solve ty ′ ′ + y ′ + 4 t y = 0 if y (0 ) = 3, y ′ (0 ) = 0 .

(Purvanchal 2010)

Taking the Laplace transform of both sides of the given equation, we have L { t y ′ ′ } + L { y ′ } + 4 L { t y} = 0

T-85

d d L{ y ′ ′ } + L{ y ′ } + 4. (− 1) L{ y } = 0 dp dp d 2 d − [ p y − py (0 ) − y ′ (0 )] + [ p y − y(0 )] − 4 y = 0, dp dp

or

−

or

where y = L{ y } d y

d ( p2 y − 3 p) + ( p y − 3) − 4 =0 dp dp d y d y p − ( p2 + 4) − p y = 0 or + dp = 0. 2 dp y p +4

or

−

or

Integrating, log y +

1 log ( p2 + 4) = log C1 or 2

y=

C1 √ ( p2 + 4)

⋅

  1 y = L−1 { y } = C1 L−1   2  √ ( p + 4) (See Ex. 25, after article 1.23 of Chapter 1) y = C1 J0 (2 t)

∴ or

Since y(0 ) = 3, therefore 3 = C1 J0 (0 ) = C1.

[ ∵ J0 (0 ) = 1]

Hence y = 3 J0 (2 t), which is the required solution. Example 11: Solve [t D2 + (1 − 2 t) D − 2] y = 0 if y(0 ) = 1, y ′ (0 ) = 2. Solution:

∴ or or

The given equation can be written as ty ′ ′ + y ′ − 2 t y ′ − 2 y = 0. L {t y ′ ′} + L { y ′} − 2 L {t y ′} − 2 L { y } = 0 d d − L { y ′ ′} + L { y ′} + 2 L { y ′} − 2 L { y } = 0 dp dp d − [ p2 y − p y(0 ) − y ′ (0 )] + [ p y − y (0 )] dp d +2 [ p y − y (0 )] − 2 y = 0 , where y = L { y } dp d d ( p2 y − p − 2) + ( p y − 1) + 2 ( p y − 1) − 2 y = 0 dp dp

or

−

or

d y − ( p2 − 2 p) − p y =0 dp

or

d y 1 + dp = 0. y p−2

Integrating, log y + log ( p − 2) = log C1 or y = C1 / ( p − 2). ∴ y = L−1 { y } = C1 L−1 {1/( p − 2)} = C1 e 2 t . But ∴

y (0 ) = 1. 1 = C1.

Hence y = e2 t , which is the required solution. Example 12:

Solve y ′ ′ − ty ′ + y = 1,

if y (0 ) = 1, y ′ (0 ) = 2.

Solution: Taking the Laplace transform of both sides of the given equation, we have

L{ y ′ ′ } − L { t y ′ } + L { y} = L {1}

T-86

d 1 [ L { y ′ }] + y = dp p

or

p2 y − py (0 ) − y ′ (0 ) +

or

p2 y − p − 2 +

d 1 [ p y − y(0 )] + y = dp p

or

p2 y − p − 2 +

d 1 ( p y − 1) + y = dp p

or

p

or

d y  + p+ dp 

d y 1 + ( p2 + 2) y = p + 2 + dp p 2 2 1  y = 1+ + 2 p p p

…(1)

which is a linear differential equation in y. ∴

2 2 I. F. = e ∫ ( p + 2 / p) dp = e p /2 + 2 log p = p2 e p /2.

∴

solution of (1) is given by

∫

 2 2  2 p2 /2  p e 1 + + dp = c1 +  p p2  

= c1 +

∫

2 2 ( p2 + 1) e p /2 dp + 2 ∫ pe p /2 dp

= c1 +

∫

2 p2 e p /2 y = c1 +

(2 v + 1) e v ⋅

dv +2 √ (2 v)

putting = c1 +

or

∫

=

∫ √ (2v) e

v

2 ( p2 + 2 p + 1) e p /2 dp

dv , √ (2 v)

p2 = v so that p dp = dv or dp = dv / √ (2 v) 2

ev dv + 2 √ (2 v)

= c1 + √ (2 v) e v + 2 e v = c1 + 2 c 1 y = L { y } = 1 e − p /2 + + 2 p p c = 1 p2

∴

v ∫ √ (2v) e dv +

∫

∫e

v

dv

2 2 pe p /2 + 2 e p /2

2 p2

  1 2 p2 p4 1 − + −... + +   p p2 2 4 ⋅2 !  

(2 + c1) 2

p

c c 1 − 1 + 1 p2 − ... + ⋅ 2 8 p

1 1 y = (2 + c1) L−1 {1/ p2} − c1 L−1 {1} + c1 L−1 { p2} + ... + L−1 {1/ p} 2 8 = (2 + c1) t + 1,

since

L−1 { pn} = 0 , n = 0 , 1, 2, ...

But given y ′ (0 ) = 2. Therefore, 2 = 2 + c1

or

c1 = 0 .

Hence y = 2 t + 1, which is the required solution.

T-87

Comprehensive Exercise 2 Solve the following problems by means of the Laplace transform: 1. y ′ ′ + t y ′ − y = 0, if y (0 ) = 0 , y ′ (0 ) = 1. (Agra 2002; Purvanchal 11) 2. [t D2 + (t − 1) D − 1] y = 0 , if y (0 ) = 5, y (∞) = 0 . 3. 4. 5.

y ′ ′ (t) + at y ′ (t) − 2 a y (t) = 1, y (0 ) = y ′ (0 ) = 0 , a > 0 . ty ′ ′ + 2 y ′ + t y = 0 , y (0 +) = 1, y (π) = 0 . Solve ty ′ ′ (t) + (2 t + 3) y ′ (t) + (t + 3) y (t) = ae − t if y (t) and its derivatives have transforms.

A nswers 2 1. 3. 5.

y = 5e− t sin t 4. y = t 2.

y=t 1 y = t2 2 at y =  A +  e − t  3

3.3 Solution of Simultaneous Ordinary Differential Equations The Laplace transform can also be used in solving two or more simultaneous ordinary differential equations.

Example 13: Solve Dx + D y = t ; D 2 x − y = e − t if x (0 ) = 3, x ′ (0 ) = − 2, y (0 ) = 0 . (Rohilkhand 2010)

Solution: Taking the Laplace transform of both sides of the two equations, we have

L { x ′ ′ } − L { y} = L { e − t}

L { x ′ } + L { y ′ } = L { t} and or

p x − x (0 ) + p y − y (0 ) = 1 / p2

and

p2 x − p x (0 ) − x ′ (0 ) − y = 1 / ( p + 1)

or

p x + p y = 3 + 1/ p2

and

p2 x − y = 3 p − 2 + 1/( p + 1).

Solving for x and y, we have x=

3 p2 + 1 3

2

p (1 + p )

+

3p 2

1+ p

−

2 2

1+ p

−

1 ( p + 1) ( p2 + 1)

T-88

and

=

3p p 2 p4  1  2 1 1 1 + 2 p2 −  + − + − + 3  2 2 2 2 + 2 1 ( p ) 1+ p  1+ p 1+ p 2 ( p + 1) 2 ( p2 + 1) p 

=

p 2 1 1 3 + + + − 3 2 p p 2 ( p + 1) 2 ( 1 + p ) 2 ( p2 + 1)

y=

1 2

p ( p + 1) ( p + 1)

+

2 2

p +1

=

1 p 1 2 1 − − − + 2 2 2 p 2 ( p + 1) 2 ( p + 1) 2 ( p + 1) p + 1

=

1 p 3 1 − − + ⋅ p 2 ( p + 1) 2 ( p2 + 1) 2 ( p2 + 1)

 1  1 −1  1   1 x = 2 L−1   + L−1  + L   3  p  p + 1  p  2

∴

+ =2+ and

1 2 1 −t 1 3 t + e + cos t − sin t 2 2 2 2

 1 1  1  1 −1  p  3 −1  1  y = L−1   − L−1  − L  2 + L  2   p 2  p + 1 2  p + 1 2  p + 1 1 1 3 = 1 − e − t − cos t + sin t. 2 2 2

Example 14: Solution:

1 −1  p  3 −1  1  L  − L  2  2 2  p + 1 2  p + 1

Solve Dx + 2 D 2 y = e − t , ( D + 2) x − y = 1 if x (0 ) = y (0 ) = y ′ (0 ) = 0 .

Taking the Laplace transform of both sides of the given equations, we have L { x ′ } + 2 L { y ′ ′ } = L { e− t }

and

L { x ′ } + 2 L { x} − L { y} = L {1}

or

p x − x (0 ) + 2 [ p2 y − py (0 ) − y ′ (0 )] =

1 , p+1

where x = L { x } and y = L { y} 1 p x − x (0 ) + 2 x − y = p 1 1 and ( p + 2) x − y = ⋅ p x + 2 p2 y = p+1 p

and or

Solving for x and y, we have 1 2 x= + p ( p + 1) (2 p2 + 4 p + 1) (2 p2 + 4 p + 1) =

1  2 − √ 2 2 p ( p + 1)  p +  2  

 2 + √ 2 p+  2  

+

1  2 − √ 2  2 + √ 2 p+  p+  2 2    

T-89

1 1 + − p p+1

=

1 1 −    2 + √ 2 2 − √2 (2 − √ 2)  p +  (2 + √ 2)  p +  2  2    +

=

and

y=

 2 − √ 2 √2 p+  2  

−

1  2 + √ 2 √2 p+  2  

1 1 1 1 + − − p p+1  2 − √ 2  2 + √ 2 p+  p+  2   2   1

p ( p + 1) (2 p2 + 4 p + 1) 1 1 1 1 = + − − p p+1   2 + √ 2 2 − √ 2 (2 − √ 2)  p +  (2 + √ 2)  p +  2  2    =

∴

1

 2 − √ 2  2 + √ 2 1 1 1 1 − + −   p p+1  2   2   2 + √ 2 2 − √2   p+  p+  2  2   

 1  1  −1  1  −1  1  x = L−1   + L−1  −L  −L    p  p + 1  p + a  p + b = 1 + e − t − e − at − e − b t

and

 1  1  −1  1  −1  1  y = L−1   + L−1  −bL   − aL    p  p + 1  p + a  p + b = 1 + e − t − be − a t − ae − b t , where

Example 15:

if

a=

2 − √2 2 + √2 and b = ⋅ 2 2

Solve ( D2 − 3) x − 4 y = 0 , x + ( D2 + 1) y = 0, t > 0 x = y = D y = 0 , D x = 2 when t = 0.

Solution: Taking the Laplace transform of both sides of the two equations, we have

L { x ′ ′ } − 3 L { x } − 4 L { y } = 0 and L { x } + L { y ′ ′ } + L { y } = 0 or

p2 x − px (0 ) − x ′ (0 ) − 3 x − 4 y = 0 , where x = L { x } and 2

and

x + p y − py (0 ) − y ′ (0 ) + y = 0

or

( p2 − 3) x − 4 y = 2 and

x + ( p2 + 1) y = 0 .

Solving for x and y, we have x=

2 ( p2 + 1) 2

2

( p − 1)

=

1 2

( p − 1)

+

1 ( p + 1)2

y = L{ y}

T-90

−2

1 2

 1 1 1 1  + − − − ⋅ 2 ( p − 1)2   p + 1 p − 1 ( p + 1)

and

y=

∴

  1  1  x = L−1  + L−1   2 2  ( p − 1)   ( p + 1) 

( p + 1)2 ( p − 1)2

=

= e t L−1 {1/ p2} + e − t L−1 {1/ p2} = (e t + e − t ) t and

1  −1  1  − L  + 2 p + 1   1 = [− e − t + e t − t e − 2

y=

     1  1 1 −1 −1 L−1  −L  −L   2 2  p − 1  ( p + 1)   ( p − 1)   t

1 1 − te t ] = (1 − t) e t − (1 + t) e − t . 2 2

Comprehensive Exercise 3 Solve the following problems by means of the Laplace transform: 1.

( D − 2) x + 3 y = 0 , 2 x + ( D − 1) y = 0 , if x (0 ) = 8 and y (0 ) = 3.

2.

( D2 + 2) x − Dy = 1, Dx + ( D2 + 2) y = 0 , if x = 0 = Dx = y = Dy, when t = 0.

3.

( D2 − 1) x + 5 Dy = t, − 2 Dx + ( D2 − 4) y = − 2 , if x = 0 = Dx = y = D y when t = 0.

4.

5.

Applying Laplace transform solve the equations : dx dy d2 x + = t and − y = e− t, dt dt dt2 dx given that x (0 ) = 0 = y (0 ) and = 0 when t = 0. dt

(Rohilkhand 2007)

Solve ( D − 2) x − ( D + 1) y = 6 e3 t , (2 D − 3) x + ( D − 3) y = 6 e3 t , if x = 3, y = 0 when t = 0.

6.

Solve ( D − 2) x − ( D − 2) y = sin t, ( D2 + 1) x + 2 D y = 0 , if x = 0 = x ′ (0 ) = y (0 ).

A nswers 3 1.

x = 5 e − t + 3 e4 t , y = 5 e − t − 2 e4 t .

2.

x=

3.

x = − t + 5 sin t − 2 sin 2 t, y = 1 − 2 cos t + cos 2 t. 1 1 1 1 1 1 1 x = t 2 − 1 + e − t + cos t + sin t, y = 1 − e − t − cos t − sin t. 2 2 2 2 2 2 2

4.

1 1 1 − (2 cos t + cos 2 t), y = (sin 2 t − 2 sin t). 2 6 6

T-91

5.

x = e t + 2 t e t + 2 e 3 t, y = e t − t e t − 2 e 3 t.

6.

x=

1 4 2t 1 1 (1 + 3 t) + e − t + e − (cos t + 2 sin t), y = [(1 + 3 t) e − t − e2 t ]. 9 45 5 9

3.4 Solution of Partial Differential Equations Laplace transform is also useful in solving partial differential equations when the boundary conditions are given. Laplace transforms of some partial derivatives: Theorem: If y ( x, t) is a function of x and t, then  ∂y  (a) L   = p y ( x, p) − y ( x, 0 )  ∂t   ∂2 y  (b) L  2  ∂t

  2  = p y ( x, p) − p y ( x, 0 ) − yt ( x, 0 )   ∂2 y  d 2 y and (d) L  where L { y ( x, t)} = y ( x, p). = 2 2 ∂ x dx  

 ∂y  d y (c) L  = dx  ∂x  Proof:

(a)

 ∂y  L  =  ∂t 

∞

∫0

e − pt

∂y dt = lim ∂t s→ ∞

s

∫0

e − pt

∂y dt ∂t

s s   = lim  e − pt y ( x, t) + p ∫ e − pt y ( x, t) dt 0 0 s→ ∞ 

{

=p (b) ∴

Let V =

∂y ∂t

∞

∫0

}

e − pt y ( x, t) dt − y ( x, 0 ) = py ( x, p) − y ( x, 0 )

⋅

 ∂2 y  L 2  ∂t

   ∂V   = p . L {V } − V ( x, 0 )  = L  ∂t   = p [ pL { y} − y ( x, 0 )] − yt ( x, 0 ) = p2 y ( x, p) − py ( x, 0 ) − yt ( x, 0 ).

(c)

(d)

 ∂y  L  =  ∂x   ∂2 y  L 2  ∂x

∞

∫0

e − pt

∂y d y d ∞ − pt dt = e y dt = ⋅ ∫ 0 ∂x dx dx

 ∂y   ∂U   , where U =  = L ∂x  ∂x  

∂y   = yt  ∵ V = ∂ t  

T-92 2  ∂y  d d d d y  d y = L {U } = ⋅ L   = = dx dx  ∂x  dx  dx  dx2

Example 16:

Solve

∂y ∂2 y where y (0 , t) = 0 = y (5, t) and y ( x, 0 ) = 10 sin 4 πx. =2 ∂t ∂x 2 (Meerut 2008)

Solution:

Taking the Laplace transform of both sides of given equation, we have  ∂2 y   ∂y  L   = 2L  2  ∂t   ∂x  d2 y

or

2

dx

−

or

p y − y ( x, 0 ) = 2

dx2

p y = − 5 sin 4 πx whose general solution is 2

y = C1e √( p /2 ) x + C2 e − √( p /2 ) x − y = C1e √( p /2 ) x + C2 e − √( p /2 ) x +

or

d2 y

5 sin 4 πx − (4 π)2 − p / 2 10 32 π2 + p

sin 4 πx

…(1)

But y (0 , t) = 0 = y (5, t). Therefore, y (0 , p) = 0 , y (5, p) = 0 . ∴ from (1), we have 0 = C1 + C2 0 = C1e5 √( p /2 ) + C2 e −5 √( p /2 ) +

and

10 32 π2 + p

sin 20 π

= C1e5 √( p /2 ) + C2 e −5 √( p /2 ) + 0 . Solving C1 = 0 = C2 . ∴ from (1), we have

y=

10 32 π2 + p

 10  y = L−1  sin 4 πx  2  32 π + p  which is the required solution.

∴

Example 17:

sin 4 πx.

or

Find the bounded solution of

where y (0 , t) = 1, y ( x, 0 ) = 0 . Solution:

have

or

2 y = 10 e −32 π t sin 4 πx,

∂y ∂2 y = , x > 0, t > 0, ∂t ∂x2 (Meerut 2007)

Taking the Laplace transform of both the sides of the given equation, we  ∂2 y   ∂y  L  = L 2  ∂t   ∂x  d2 y dx2

− p y =0

or

p y ( x, p) − y ( x, 0 ) =

d2 y dx2

T-93

whose general solution is y = c1 e √ px + c2 e −√ px . But y ( x, t) must be bounded as x → ∞. ∴

y ( x, p) = L { y ( x, t)} must also be bounded as x → ∞

∴

c1 = 0

so that y = c2 e −√ px , if √ p > 0.

…(1)

But y (0 , t) = 1. Therefore, L { y (0 , t)} = L {1} or

…(2)

y (0 , p) = 1/ p

Thus, from (1) and (2) we have 1/ p = c2 . ∴

y = (1/ p) e − √ px

or

 1 y = L−1   pe √ px

Example 18: Solution:

Solve

∂2 y

−

2

∂x

∂2 y 2

∂t

  x   , which is the required solution.  = erfc   2 √ t  [See Ex. 21, after article 2.18 of Chapter 2] = xt where y = 0 =

∂y at t = 0 and y (0 , t) = 0 . ∂t

Taking the Laplace transform of both the sides of the given equation, we

have  ∂2 y L 2  ∂x or or

Since ∴ Again

d2 y dx2 d2 y

 ∂2 y  − L  2   ∂t 

  = L {xt} 

− [ p2 y ( x, p) − p y ( x, 0 ) − yt ( x, 0 )] = x L { t}

− p2 y = x / p2 whose general solution is dx2 y = c1 e px + c2 e − px − x / p4 .

y = 0 for all values of x, therefore, c1 = 0 , otherwise y = ∞, as x → ∞. y = c2 e − px − x / p4 . y = 0 when x = 0; 4

y=−x/ p

or

∴

c2 −1

y=−L

= 0.

{ x / p4} = −

1 3 xt , 6

which is the required solution.

Comprehensive Exercise 4 Solve the following problems by means of the Laplace transform: ∂y ∂y 1. =2 + y, y ( x, 0 ) = 6 e −3 x which is bounded for x > 0 , t > 0 . ∂x ∂t  ∂y  ∂y ∂2 y π  2. = 3 2 , where y  , t  = 0 ,  = 0 and y ( x, 0 ) = 30 cos 5 x. 2  ∂t ∂x  ∂x  x =0

(Meerut 2006)

T-94

3.

∂y ∂2 y = 2 , y ( x, 0 ) = 3 sin 2 π x , y (0 , t) = 0 = y (1, t), 0 < x < 1, t > 0 . ∂t ∂x (Meerut 2006) 2

4.

∂y ∂ y = 2 2 , y (0 , t) = 0 , y (5, t) = 0 , y ( x, 0 ) = 10 sin 4 πx − 5 sin 6 πx. ∂t ∂x (Meerut 2008)

5.

∂u ∂u − = 1 − e − t , 0 < x < 1, t > 0 , u ( x, 0 ) = x. ∂x ∂t

(Meerut 2007)

A nswers 4 1.

y ( x, t) = 6 e − 2 t − 3 x .

3.

y ( x, t) = 3 e − 4 π

4.

y ( x, t) = − 5 e − 72 π

5.

u ( x, t) = ( x + 1) − e − t .

2

t

2. y = 30 e − 75 t cos 5 x.

sin 2 πx. 2

t

sin 6 πx + 10 e − 32 π

2

t

sin 4 πx.

3.5 Integral Equations 1. Integral equation:

An equation of the form

F (t) = y (t) +

b

∫a

K (u, t) F (u) du

is called an “Integral equation” where y (t) and K (u, t) are known, a and b are either constants or functions of t. Here the function F (t) which appears under the sign of integration is to be determined. 2. Abel’s integral equation: An equation of the form t F (u) du G (t) = ∫ 0 (t − u)n is called Abel’s integral equation where F (u) is unknown and G (t) is known and n is a constant between 0 and 1, i. e., 0 < n < 1. 3. Integral equation of Convolution type: An integral equation of the form F (t) = y (t) +

t

∫ 0 K (t − u) F (u) du

which may also be expressed as F (t) = y (t) + K (t) * F (t) is called an integral equation of convolution type. 4. Integro-differential equation: An integral equation in which various derivatives of the unknown function F (t) can also be present is called an Integro-differential equation. For example,

F ′ (t) = F (t) + y (t) +

t

∫ 0 sin (t − u) F (u) du.

T-95

3.6 Application of Laplace Transform to Integral Equations Laplace transform may be used to solve various integral equations. Example 19:

Solve the integral equation F (t) = e − t − 2 ∫

Solution:

t 0

cos (t − u) F (u) du.

The given integral equation may be expressed as F (t) = e − t − 2 F (t) * cos t.

Taking the Laplace transform of both the sides, we have L { F (t)} = L { e − t} − 2 L { F (t) * cos t} =

1 − 2 . L { F (t)} . L {cos t} p+1

=

2p 1 − ⋅ L { F (t)} p +1 p2 +1

or

 2p  1 1 + 2  L { F (t)} = p +1 p + 1 

or

L { F (t)} =

∴

 p 2 + 1  −1 F (t) = L−1  =L  ( p + 1)3 

( p 2 + 1) ( p + 1)3

⋅ 2  ( p + 1 − 1) + 1     ( p + 1)3

 ( p − 1)2 + 1 − t −1 = e − t L− 1  =e L 3 p  

 1 2 2   − 2 + 3 p   p p

 t t3  = e − t 1 − 2 ⋅ +2⋅  Γ (2) Γ (3)  = e − t (1 − 2 t + t 2 ) = e − t (1 − t)2 . Example 20:

Solve the integral equation F (t) = 1 +

t

∫0

F (u) ⋅ sin (t − u) du

and verify your solution . Solution:

The given integral equation may be expressed as F (t) = 1 + F (t) * sin t.

Taking the Laplace transform of both the sides, we have L { F (t)} = L {1} + L { F (t) * sin t} = L {1} + L { F (t)} . L {sin t} 1 1 = + L { F (t)} ⋅ 2 p p +1

(Rohilkhand 2011)

T-96

or

1  1  or  L { F (t)} = 1 − 2 p p + 1 

∴

 p 2 + 1  1 F (t) = L−1  3  = L−1   + L−1  p   p = 1+

L { F (t)} =

p2 +1 p3

⋅

 1   3   p 

t2 t2 = 1+ ⋅ Γ (3) 2

Verification: We have F (t) = 1 +

t2 ⋅ 2

Putting in the R.H.S. of the given equation, we have t  u2  R.H.S. = 1 + ∫ 1 +  ⋅ sin (t − u) du 0  2 t

  u2   cos (t − u) − = 1 + 1 + 2 0  = 1+ 1+

Example 21:

t

∫0

u ⋅ cos (t − u) du

t2 − cos t − [− u ⋅ sin (t − u)]0t − 2

t

∫0

sin (t − u) du

=2+

t2 t2 t − cos t − [cos (t − u)]0 = 2 + − cos t − (1 − cos t) 2 2

= 1+

t2 = F (t) = L.H. S. 2

Solve the following equation for F (t) with the condition that F (0 ) = 0 , F ′ (t) = sin t +

t

∫0

F (t − u) cos u du .

The given equation may be expressed as F ′ (t) = sin t + F (t) * cos (t). Taking the Laplace transform of both the sides, we have L { F ′ (t)} = L {sin t} + L { F (t) * cos t} 1 or pL { F (t)} − F (0 ) = 2 + L { F (t)} . L {cos t} p +1 p 1 or pL { F (t)} = 2 + L { F (t)} . 2 p +1 p +1 Solution:

or

 p  1  p − 2  L { F (t)} = 2 or p + 1 p +1 

∴

 1  t2 t2 = ⋅ F (t) = L−1  3  =  p  Γ (3) 2

Example 22: Solution:

Solve the integral equation

t

∫0

L { F (t)} =

p3

⋅

F (u) F (t − u) du = 16 sin 4 t.

The given integral equation may be expressed as F (t) * F (t) = 16 sin 4 t.

1

T-97

Taking the Laplace transform of both the sides, we have L { F (t) * F (t)} = 16 L {sin 4 t} 4 L { F (t)} . L { F (t)} = 16 ⋅ 2 p + 42 8 L { F (t)} = ± ⋅ 2 √ ( p + 42 )

or or

 1  = ± 8 J0 (4 t). F (t) = ± 8 L−1  2 2   √ ( p + 4 )

∴

(See Q. 2 of Ex ercise 5 Chapter 2) Example 23: Solution:

Solve the integral equation

∫0

F (u) du (t − u)1 /3

= t (1 + t).

The given integral equation may be expressed as t

∫0 or

t

F (u) ⋅ (t − u)−1 /3 du = t + t2

F (t) * t

− 1 /3

= t + t 2.

Taking the Laplace transform of both the sides, we have L { F (t) * t − 1 /3} = L { t + t 2} or

L { F (t)} . L { t

− 1 /3

} = L { t} + L { t 2}

1 + 1) Γ (2) Γ (3) 3 L { F (t)} ⋅ − 1 /3 + 1 = 2 + 3 p p p Γ (2 / 3) 1 2 L { F (t)} ⋅ = 2 + 3 p 2 /3 p p Γ (−

or or or

∴

L { F (t)} =

 1 1 2  ⋅  4 /3 + 7 /3  ⋅ Γ (2 / 3)  p p 

 1    7 /3      p  t (4 /3) − 1 1 t (7 /3) − 1  = ⋅ +2⋅  Γ (2 / 3)  Γ (4 / 3) Γ (7 / 3)    t 1 /3  t 4 /3 1 = ⋅ +2⋅  4 1 Γ (2 / 3)  1 ⋅ Γ (1/ 3) ⋅ ⋅ Γ (1/ 3)  3  3 3 1 /3 3t 1 + 3 t = Γ (1 − 1 ) Γ (1 )  2 

F (t) =

 1 ⋅  L−1 Γ (2 / 3) 

3t

3 1 /3

 1  4 /3  p

3

1 + 3 t   1 (π / sin π)  2  3 3 √ 3 1 /3 = ⋅ t (2 + 3 t). 4π =

 −1  +2 L 

 π   ∵ Γ (n) Γ (1 − n) =  sin nπ  

T-98 Example 24: Solution:

Solve F ′ (t) = t +

t

F (t − u) cos u du, F (0 ) = 4.

∫0

The given equation may be expressed as F ′ (t) = t + F (t) * cos t.

Taking Laplace transform of both the sides, we have L { F ′ (t)} = L { t} + L { F (t) * cos t} 1 or pL { F (t)} − F (0 ) = 2 + L { F (t)} . L {cos t} p pL { F (t)} − 4 =

or

1 p

3

p

or

L { F (t)} ⋅

or

L { F (t)} =

2

p +1

= ∴

=

p2 + 1 p5 1 3

p

+

+ L { F (t)} ⋅

2

1

5

p

p +1

+4

p2

4 ( p2 + 1)

+

1

p 2

p3 +

4 4 4 5 1 + 3 = + 3 + 5 ⋅ p p p p p

 4 5 1  5 t2 t4 + F (t) = L−1  + 3 + 5  = 4 + Γ (3) Γ (5) p   p p =4+

Example 25:

5 t2 t4 5 t4 + = 4 + t2 + ⋅ 2 4! 2 24

Express 2 F ′ ′ (t) − 3 F ′ (t) − 2 F (t) = 4 e − t + 2 cos t, F (0 ) = 4, F ′ (0 ) = − 1

into an integral equation. Solution: We have 2 F ′ ′ (t) − 3 F ′ (t) − 2 F (t) = 4 e − t + 2 cos t,

…(1)

F (0 ) = 4, F ′ (0 ) = − 1

…(2)

Method 1: Let F ′ ′ (t) = G (t). Integrating (3), we get t

F ′ (t) =

∫0

G (u) du + C1.

For t = 0, (4) gives − 1 = F ′ (0 ) = 0 + C1 ∴

t

F ′ (t) =

∫0

…(3)

or C1 = − 1.

G (u) du − 1.

…(5)

Again integrating, we get F (t) =

t

∫0

(t − u) G (u) du − t + C2 .

For t = 0, this gives 4 = F (0 ) = 0 − 0 + C2

…(4)

or C2 = 4.

T-99

∴

t

F (t) =

∫0

…(6)

(t − u) G (u) du − t + 4

Putting the values of F ′ ′ (t), F ′ (t) and F (t) from (3), (5) and (6) in (1), we get 2 G (t) − 3

t

∫0

t

G (u) du + 3 − 2 ∫

(t − u) G (u) du + 2 t − 8

0

= 4 e − t + 2 cos t or

2 G (t) +

t

∫0

(− 2 t + 2 u − 3) G (u) du = 4 e − t + 2 cos t − 2 t + 5 ,

which is the required integral equation. Method 2: Integrating (1) between the limits 0 to t, we get t

t

2 [ F ′ (u)] 0 − 3 [ F ′ (u)] 0 − 2 or

t

F (u) du = − 4 [e − u] 0t + 2 [sin u] 0t

∫0

t

2 F ′ (t) − 2 F ′ (0 ) − 3 F (t) + 3 F (0 ) − 2

∫0

F (u) du = 4 (1 − e − t ) + 2 sin t.

or

t

2 F ′ (t) − 3 F (t) − 2

∫0

F (u) du = − 4 e − t + 2 sin t − 10 . [Using (2)]

Again integrating, between the limits 0 to t, we get t

2 [ F (u)] 0 − 3

t

∫0

F (u) du − 2

t

∫0

(t − u) F (u) du t

[ ]0 − 2 [cos u] 0t − 10 t.

= 4 e−u or

2 [ F (t) − F (0 )] +

t

∫0

(− 2 t + 2 u − 3) F (u) du = 4 (e − t − 1) − 2 (cos t − 1) − 10 t.

or

2 F (t) +

t

∫0

(− 2 t + 2 u − 3) F (u) du = 4 e − t − 2 cos t − 10 t + 6

which is the required integral equation. Example 26:

Express the given differential equation F ′ ′ (t) + a F (t) = 0 , F (0 ) = 0 = F (1)

as an integral equation. Solution: We have

Method 1:

F ′ ′ (t) + aF (t) = 0

…(1)

F (0 ) = 0 = F (1).

…(2)

Let F ′ ′ (t) = G (t).

…(3)

Integrating (3) between the limits 0 to t, we get F ′ (t) =

t

∫ 0 G (u) du + C1

…(4)

Again integrating between the limits 0 to t, we get F (t) =

t

∫0

(t − u) G (u) du + C1 t + C2 .

…(5)

T-100

For t = 0, (5) gives 0 = F (0 ) = 0 + C2

or C2 = 0 .

Putting this value in (5), we get F (t) =

t

…(6)

∫ 0 (t − u) G (u) du + C1 t

For t = 1, (6) gives 0 = F (1) = or

C1 = −

1

∫0

1

∫0

(1 − u) G (u) du + C1

(1 − u) G (u) du.

Putting this value of C1 in (6), we get F (t) =

or

1

t

∫ 0 (t − u) G (u) du − ∫ 0 t

=

∫0

=

∫0

=

∫0

F (t) =

∫0

t

t

1

t (1 − u) G (u) du

1

(t − u) G (u) du +

∫0

(t − u) G (u) du +

∫ 0 (ut − t) G (u) du + ∫ t (ut − t) G (u) du

(t − 1) u G (u) du +

(ut − t) G (u) du

t

1

1

t (u − 1) G (u) du

∫t

(t − 1) u , K (t, u) G (u) du, where K (t, u) =   t (u − 1) ,

if u < t if u > t.

Hence from (1), we get G (t) + a

t

∫0

K (t, u) G (u) du = 0 , where

(t − 1) u , u < t K (t, u) =  (u − 1) t , u > t.

This is the required integral equation of the given differential equation. Method 2:

Integrating (1) between the limits 0 to t, we get t

∫0

F ′ ′ (u) du + a

t

F (u) du = 0

∫0

t

or

[ F ′ (u)] 0t a + ∫ 0 F (u) du = 0

or

F ′ (t) − F ′ (0 ) + a

t

∫ 0 F (u) du = 0.

Again integrating between the same limits, we get t

[ F (u)] 0t − F ′ (0) [u]0t + a ∫ 0 (t − u) F (u) du = 0 or

F (t) − F (0 ) − tF ′ (0 ) + a

or

F (t) − t F ′ (0 ) + a

t

∫ 0 (t − u) F (u) du = 0

…(1)

t

∫ 0 (t − u) F (u) du = 0

For t = 1, this gives F (1) − F ′ (0 ) + a

1

∫ 0 (1 − u) F (u) du = 0

[Using (2)]

T-101

or

0 − F ′ (0 ) + a

or

F ′ (0 ) = a

1

∫0

(1 − u) F (u) du = 0

1

∫ 0 (1 − u) F (u) du.

Putting this value in (I), we get 1

F (t) − ta

(1 − u) F (u) du + a

∫0

t

∫0

1

t

t

1

(t − u) F (u) du = 0

or

F (t) + a

∫ 0 t (u − 1) F (u) du + a ∫ 0 (t − u) F (u) du = 0

or

F (t) + a

∫ 0 t (u − 1) F (u) du + a ∫ t t (u − 1) F (u) du +a t

t

∫0

(t − u) F (u) du = 0

1

or

F (t) + a

∫ 0 u (t − 1) F (u) du + a ∫ t t (u − 1) F (u) du = 0

or

F (t) + a

∫0

Example 27:

1

 u (t − 1) , K (u, t) F (u) du = 0 , where K (u, t) =   t (u − 1) ,

u< t u > t.

Convert the given integral equation F (t) −

t

∫0

(t − u) sec t ⋅ F (u) du = t

into differential equation and associated conditions. Solution: The given integral equation is F (t) −

t

…(1)

∫ 0 (t − u) sec t ⋅ F (u) du = t

We know that, d dt

b (t)

∫ a (t)

K (u, t) du =

b (t)

∫ a (t)

∂K db da − K (a, t) ⋅ du + K (b, t) ∂t dt dt

…(2)

Differentiating (1) using the fact (2), we get t

F ′ (t) −

∫ 0 [sec t + (t − u) sec t tan t] F (u) du = 1

or

F ′ (t) −

∫ 0 sec t ⋅ F (u) du − tan t ∫ 0 (t − u) sec t F (u) du = 1

or

F ′ (t) −

∫ 0 sec t ⋅ F (u) du + tan t ⋅ [t − F (t)] = 1, [Using (1)]

t

t

t

…(3)

Again differentiating (3), we get F ′ ′ (t) + sec2 t ⋅ [t − F (t)] + tan t ⋅ [1 − F ′ (t)] −sec t ⋅ tan t ∫ or

t 0

F (u) du − sec t ⋅ F (t) = 0 .

F ′ ′ (t) + sec2 t . [t − F (t)] + tan t . [1 − F ′ (t)] − sec t . F (t) + tan t . [1 − F ′ (t) − t tan t + tan tF (t)] = 0 , [Using (2)]

or

F ′ ′ − F ′ . [tan t + tan t] + [− sec 2 t + tan2 t − sec t] F + [t sec2 t + tan t − t . tan2 t + tan t] = 0

T-102

or

F ′ ′ (t) − 2 tan t . F ′ (t) − (1 + sec t) F (t) + (t + 2 tan t) = 0 .

…(4)

Putting t = 0 in (1) and (3) respectively, we get …(5)

F (0 ) = 0 , F ′ (0 ) = 1.

Hence, (4) is the required differential equation and its associated conditions are given by (5).

Comprehensive Exercise 5 1.

Solve the integral equation t

∫ 0 F (u) ⋅ cos (t − u) du.

F (t) = a sin t − 2 t

2.

Solve y (t) = t − 1 +

3.

Show that the solution of the integral equation

∫0

y (Γ) ⋅ sin (t − Γ) dΓ. t

F (t) = 4 t − 3 is

F (u) sin (t − u) du

∫0

3 sin 2 t. 2

F (t) = t +

4. Solve the integral equation F (t) = 1 + 2 1 6

t

Solve y (t) = t +

6.

Solve the integral equation F (t) =

7.

Solve 2 F (t) = 2 − t +

8.

Solve

9.

Convert the given differential equation

t

F (u) du

F (t − u) ⋅ e −2 u du.

(t − u)3 ⋅ y (u) du.

5.

∫0

t

∫0

t

∫0

1 2 t − 2

t

∫0

(t − u) F (u) du.

(Agra 2003)

F (t − u) F (u) du. 2

∫ 0 √ (t − u) = 1 + t + t

.

F ′ ′ (t) + 2 F ′ (t) − 8 F (t) = 5 t2 − 3 t, F (0 ) = − 2, F ′ (0 ) = 3 10.

into an integral equation. Convert the given integral equation F (t) = t2 − 3 t + 4 − 3

t

∫0

(t − u)2 F (u) du

into differential equation and associated conditions.

A nswers 5 1.

F (t) = at e − t

2. y (t) = − 1 + t −

4.

F (t) = 1 + 2 t

5. y (t) =

1 2 1 3 t + t 2 6

1 (sinh t + sin t) 2

T-103

6. 8.

F (t) = 1 − cos t 1 8 F (t) = t − 1 /2 + 2 t 1 /2 + t 3 /2    π 3

9.

G (t) +

t

2

∫ 0 [2 − 8t + 8u] G (u) du − 5t

or F (t) + 10.

7. F (t) = 1, − 1

− 21t + 22 = 0 5 t4

t

∫ 0 (− 8t + 8u + 2) F (u) du = 12

t3 − t−2 2

−

F ′ ′ ′ (t) + 6 F (t) = 0 with F (0 ) = 4, F ′ (0 ) = − 3, F ′ ′ (0 ) = 2

Objective Type Questions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 1.

2.

The solution of ( D + 1) y = 0 , t > 0 , given that y = y0 when t = 0 is (a) y0 e − t

(b) y0 e t

(c) 2 y0 e − t

(d) 2 y0 e t

∂ If y ( x, t) is a function of x and t, then L  ∂ (a) (c)

∂ y ∂x

y  is x

(b)

∂2 y

(d)

∂x2

dy dx d2 y dx 2

where L { y ( x, t)} = y ( x, p). 3.

An integral equation of the form F (t) = y (t) +

t

∫0

K (t − u) F (u) du is called

(a) an integral equation of convolution type (b) Abel’s integral equation (c) integro-differential equation (d) Cauchy’s integral equation

Fill In The Blank(s) Fill in the Blanks ‘……’ so that the following statements are complete and correct. 1.

On taking the Laplace transform of both sides of the given differential equation and using the given conditions, we obtain an algebraic equation known as …… from which y ( p) = L { y (t)} is determined.

2.

The solution of the given differential equation is obtained by finding the …… transform of y ( p), where y ( p) = L { y ( x, t)}.

T-104

3.

 ∂ y If y ( x, t) is a function of x and t, then L   = …… .  ∂t 

4.

2  ∂ y  If y ( x, t) is a function of x and t, then L  2  = ……  ∂x 

5.

An equation of the form F (t) = y (t) +

b

∫a

K (u, t) F (u) du is called …… .

equation where y (t) and K (u, t) are known, a and b are either constants or functions of t.

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

If y ( x, t) is a function of x and t then  ∂2 y  L  2  = y ( x, p) − py ( x, 0 ) − yt ( x, 0 ) where L { y ( x, t)} = y ( x, p).  ∂t 

2.

The differential equation takes the form

3.

d2 y dx2

−

∂y ∂2 y = 2 2 with the condition y ( x, 0 ) = 10 sin 4 π x is ∂t ∂x

p y = − 5 sin 4 π x, where L { y ( x, t)} = y ( x, p). 2

An equation of the form G (t) =

t

∫0

F (u) (t − u)n

du is called Abel’s integral equation

where F (u) is unknown and G (t) is known and n is a constant between 0 and 1, i. e., 0 < n < 1. 4.

An integral equation in which various derivatives of the unknown function F (t) can also be present is called an Integro transform equation.

A nswers Multiple Choice Questions 1.

(a)

2. (b)

3. (a)

Fill in the Blank(s) 1.

subsidiary equation

3.

p y ( x, p) − y ( x, 0 ), where L { y ( x, t)} = y ( x, p)

4.

d2 y dx2

, where L { y ( x, t)} = y ( x, p)

2. inverse Laplace

5. an integral

True or False 1.

F

2. T

3. T

4.

F

¨

T-105

4 F ourier T ransforms

4.1 Dirichlet’s Conditions

A (ii)

function f ( x) is said to satisfy Dirichlet conditions in the interval (a, b), if (i) f ( x) is defined and is single-valued except possibly at a finite number of points in the interval (a, b), and f ( x) and f ′ ( x) are piecewise continuous in the interval (a, b).

These conditions play an important role in the study of Fourier series and Fourier Transforms.

4.2 Fourier Series (Meerut 2013, 13B; Rohilkhand 14)

If f ( x) is a periodic function with period 2 l i. e. f ( x + 2 l ) = f ( x) and satisfies Dirichlet conditions in the interval (− l, l ), then at every point of continuity we have f ( x) =

∞ 1 nπx nπx  a0 + ∑ an cos + bn sin  2 l l  n =1 

…(1)

T-106

1 l 1 bn = l

where

an =

and

nπx dx l l nπx ∫ − l f ( x) sin l dx. l

∫ −l

f ( x) cos

…(2) …(3)

The series (1) with coefficients an and bn given by (2) and (3) respectively is called the Fourier series of f (x), and the coefficients an and bn are called the Fourier coefficients corresponding to f ( x). At a point of discontinuity 1 f ( x) = [ f ( x + 0 ) + f ( x − 0 )]. 2 If the function f ( x) defined in the interval (− l, l ) be an even function of x i.e. if f (− x) = f ( x), then 1 l 2 l nπx nπx an = ∫ f ( x) cos dx = f ( x) cos dx l −l l l ∫0 l 1 l nπx and bn = ∫ f ( x) sin dx = 0 . l −l l Therefore in this case we get Fourier cosine series. Again if f ( x) be an odd function of x i.e. if f (− x) = − f ( x), then 1 l nπx an = ∫ f ( x) cos dx = 0 − l l l 1 l nπx 2 l nπx and bn = ∫ f ( x) sin dx = f ( x) sin dx ∫ 0 − l l l l l and thus in this case we get Fourier sine series. Note: If f ( x) is a function of period 2l but is defined only in (0 , l ), we can extend it to (− l, 0 ) so as to be an even or an odd function of x in the interval (− l , l )

4.3 Fourier’s Integral Formula Let f ( x) be a function satisfying Dirichlet conditions in every finite interval − l ≤ x ≤ l 1 and defined as [ f ( x + 0 ) + f ( x − 0 )] at every point of discontinuity. 2 Further let

∞

∫ − ∞ | f ( x)| dx converge i.e.

f ( x) is absolutely integrable in − ∞ < x < ∞,

then by Fourier’s integral formula ∞ 1  ∞  f ( x) = f (v) ∫ cos w ( x − v) dw dv ∫  − ∞  2 π −∞ ∞ ∞ 1 or f ( x) = dw ∫ f (v) cos w ( x − v) dv. ∫ −∞ 2 π −∞ The representation (1) of f ( x) is known as Fourier’s integral formula. The proof is out of scope of this book. Another form: We have

…(1)

T-107

0=

1 2π

∞

∫ −∞

f (v)

 ∞  sin w ( x − v) dw dv  ∫ − ∞ 

…(2)

Multiplying both sides of (2) by i and then adding to (1), we get ∞ 1  ∞ iw ( x − v )  f ( x) = f (v) ∫ e dw dv  − ∞  2 π ∫ −∞ ∞ ∞ 1 or f ( x) = e iw x dw ∫ f (v) e − iwv dv. ∫ −∞ 2 π −∞

4.4 Fourier Transform or Complex Fourier Transform (Purvanchal 2014)

Let f ( x) be a function defined on (− ∞, ∞) and be piecewise continuous in each finite partial interval and absolutely integrable in (− ∞, ∞), then ∞ 1 F { f ( x)} = e ipx f ( x) dx ∫ √ (2π) − ∞ ~

is called the Fourier Transform of f ( x) and is denoted by F { f ( x)} or f ( p). ~

The function f ( x) is called the inverse Fourier transform of f ( p) i.e., ~ f ( x) = F −1 { f ( p)}.

4.5 Inversion Theorem for Complex Fourier Transform ~

If f ( p) is the Fourier transform of f ( x) and if f ( x) satisfies the Dirichlet conditions in every ∞

finite interval (− l, l ) and further if

∫ − ∞ | f ( x)| dx is convergent, then at every point of

continuity of f ( x), f ( x) = Proof:

1 √ (2π)

∞

∫ −∞

~

f ( p) e − ipx dp.

From Fourier integral formula, we have 1 ∞  ∞ iw ( x − v )  f ( x) = f (v) ∫ e dw dv  − ∞  2π ∫ − ∞ ∞ ∞ 1 = e iwx dw ∫ f (v) e − iwv dv −∞ 2 π ∫ −∞ ∞ ∞ 1 1 = e − ipx dp f ( x) e ipx dx, ∫ ∫ − ∞ − ∞ √ (2 π) √ (2 π) putting w = − p so that dw = − dp 1 = √ (2 π)

∞

∫ −∞

e

− ipx

~

f ( p) dp.

Note: Some authors also define the Fourier transform in the following forms :

(1)

~

f ( p) = ∫

∞ −∞

e − ipx f ( x) dx

T-108

and f ( x) =

1 2π

~

(2)

f ( p) = ∫

~

(3)

−∞

e ipx f ( x) dx

and f ( x) =

~

∞

e − ipx f ( p) dp.

∫ −∞

1 √ (2π)

f ( p) =

f ( p) ⋅ e ipx dp.

∫ −∞

∞

1 2π

and f ( x) =

~

∞

∞

e − ipx f ( x) dx

∫ −∞

1 √ (2π)

~

∞

f ( p) ⋅ e ipx dp.

∫ −∞

4.6 Fourier Sine Transform ~

The infinite Fourier sine transform of f ( x), 0 < x < ∞, is denoted by Fs { f ( x)} or f s ( p) , and is defined as ~ 2 Fs { f ( x)} = f s ( p) =    π

∞

∫0

f ( x) sin px dx . ~

The function f ( x) is called the inverse Fourier sine transform of f s ( p) i. e.,

~

f ( x) = Fs −1 { f s ( p)}.

Note: Some authors also define ~

f s ( p) = ∫

∞ 0

f ( x) sin px dx.

4.7 Inversion Formula for Fourier Sine Transform ~

If f s ( p) is the Fourier sine transform of the function f ( x) which satisfies the Dirichlet conditions in every finite interval (0 , l ) and is such that 2 f ( x) =    π

∞

∫0

| f ( x)| dx exists, then

∞ ~ f s ( p) sin px dp

∫0

at every point of continuity of f ( x). This is an inversion formula for infinite Fourier sine transform. Proof: From Fourier integral formula, we have ∞ 1 ∞ f ( x) = dw ∫ f (v) cos w ( x − v) dv ∫ 0 − ∞ π

T-109

1 π 1 = π =

or

∞

∫0

∞

∫0

f ( x) =

1 π

dp

∞

∫ −∞

{ f (v) cos px cos pv + f (v) sin px sin pv} dv, where w = p

cos px dp ∞

∞

f (v) cos pv dv +

∫ −∞

cos px dp

∫0

∞

1 π

∞

sin px dp

∫0

∞

∫ −∞

f (v) sin pv dv

f ( x) sin px dx

…(1)

f ( x) cos px dx

∫ −∞ +

1 π

∞

sin px dp

∫0

∞

∫ −∞

Now define f ( x) in (− ∞, 0 ) such that f ( x) is an odd function of x in (− ∞, ∞). Then obviously f ( x) cos px is an odd function of x and f ( x) sin px is an even function of x in (− ∞, ∞). ∴ From (1), we have

or

2 f ( x) =    π

∫0

2 f ( x) =    π

∫0

∞

2 sin px dp ×    π

∞

∫0

f ( x) sin px dx

∞ ~ f s ( p) sin px dp .

Note: According to the authors who define ~

f s ( p) =

we have

f ( x) =

2 π

∞

∫0

f ( x) sin px dx,

∞

~ f s ( p) sin px dp.

∫0

4.8 Fourier Cosine Transform The infinite Fourier cosine transform of f ( x), 0 < x < ∞, is denoted by Fc { f ( x)} or ~ f c ( p), and is defined as

~ 2 Fc { f ( x)} = f c ( p) =    π

∞

∫0

f ( x) cos px dx. ~

The function f ( x) is called the inverse Fourier Cosine transform of f c ( p) i.e.,

~

f ( x) = Fc −1 { f c ( p)} ~

∞

Note: Some authors also define f c ( p) = 0

∫

f ( x) cos px dx .

4. 9 Inversion Formula for Fourier Cosine Transform ~

If f c ( p) is the Fourier cosine transform of the function f ( x) which satisfies the Dirichlet conditions in every finite interval (0 , l ) and is such that

∞

∫0

| f ( x)| dx exists, then

T-110 ∞ ~ f c ( p) cos px dp

2 f ( x) =    π

∫0

at every point of continuity of f ( x). This is an inversion formula for infinite Fourier cosine transform. Proof: Proceeding as in article 4.7, we have ∞ 1 ∞ f ( x) = cos px dp ∫ f ( x) cos px dx −∞ π ∫0 ∞ 1 ∞ + sin px dp ∫ f ( x) sin px dx. ∫ −∞ π 0

…(1)

Now define f ( x) in (−∞, 0 ) such that f ( x) is an even function of x in (− ∞, ∞). Then obviously f ( x) cos px is an even function of x and f ( x) sin px is an odd function of x in (− ∞, ∞). ∴ from (1), we have

or

2 f ( x) =    π

∫0

2 f ( x) =    π

∫0

∞

2 cos px dp ×    π

∞

∫0

f ( x) cos px dx

∞ ~ f c ( p) cos px dp.

Note: According to the authors who define ~

f c ( p) = ∫

we have

f ( x) =

2 π

∞

f ( x) cos px dx,

0

∞

∫0

~ f c ( p) cos px dx.

4.10 Linearity Property of Fourier Transform ~

~

If f ( p) and g ( p) are Fourier transforms of f ( x) and g ( x) respectively, then ~

~

F { a f ( x) + bg ( x)} = a f ( p) + b g ( p), where a and b are constants. Proof:

We have ~

1 √ (2π)

∫−∞

∞

e ipx f ( x) dx

~

1 √ (2π)

∫−∞

∞

e ipx g( x) dx.

F { f ( x)} = f ( p) = and

F { g ( x)} = g ( p) =

∴

F { a f ( x) + b g ( x)} ∞ 1 e ipx { af ( x) + bg ( x)} dx = ∫ √ (2 π) − ∞ ∞ ∞ a b = e ipx f ( x) dx + e ipx g ( x) dx ∫ ∫ √ (2 π) − ∞ √ (2 π) − ∞ ~

~

= a f ( p) + b g ( p).

T-111

4.11 Change of Scale Property ~

Theorem 1: (For Complex Fourier Transform).

If f ( p) is the complex Fourier

transform of f ( x), the complex Fourier transform of f (ax) is

1 ~  p f  . a  a

Proof. We have ~

f ( p) = F { f ( x)} =

Now

F { f (ax)} = =

1 √ (2π)

1 √ (2π) ∞

e ipx f ( x) dx.

…(1)

e ipx f (ax) dx

∫−∞

1 1 a √ (2 π)

∞

∫−∞

∞

∫−∞

e ip( t / a ) f (t) dt, putting ax = t so that dx =

=

1 1 ⋅ a √ (2 π)

=

1 ~  p f   , from (1). a  a

∞

∫−∞

1 dt a

e i( p / a ) t f (t) dt

~

Theorem 2: (For Fourier Sine Transform). If f s ( p) is the Fourier sine transform of f ( x), 1 ~  p then the Fourier sine transform of f (ax) is f s   ⋅ a  a Proof:

~

We have f s ( p) = Fs { f ( x)} 2 =    π

∫0

2 Now Fs { f (ax)} =    π

∫0

∞

∞

…(1)

f ( x) sin px dx f (ax) sin px dx

=

1  2 ∞ p  ⋅   ∫ f (t) . sin  t dt, putting ax = t so that dx = (1 / a) dt   0 a  a π

=

1 ~  p f s   , from (1).  a a ~

Theorem 3: (For Fourier Cosine Transform). If f c ( p) is the Fourier Cosine 1 ~  p Transform of f ( x) , then the Fourier Cosine transform of f (ax) is f c   . a  a Proof:

We have ~ f c ( p) = Fc { f ( x)}

2 =    π

∞

∫0

f ( x) cos px dx

…(1)

T-112

2 Fc { f (ax)} =    π

Now

=

1  2 ⋅   a  π

∞

∫0

f (ax) cos px dx

p  f (t) cos  t dt , a 

∞

∫0

putting ax = t so that dx = (1 / a) dt 1 ~  p = f c   , from (1).  a a

4.12 Shifting Property ~

If f ( p) is the complex Fourier transform of f ( x), then the complex Fourier transform of f ( x − a) ~ is e ipa f ( p).

Proof:

(Purvanchal 2014)

We have ~

f ( p) = F { f ( x)} =

Now

F { f ( x − a)} =

1 √ (2π)

∞

∫ −∞

1 √ (2π)

∞

∫−∞

e ipx f ( x) dx

e ipx f ( x − a) dx =

…(1)

∞ 1 e ip( a + t ) f (t) dt, ∫ − √ (2 π) ∞

putting x − a = t so that dx = dt =e

ipa

∞ ipt 1 ⋅ e f (t) dt √ (2 π) ∫ − ∞

~ = e ipa f ( p) , from (1).

4.13 Modulation Theorem ~

If f ( p) is the Complex Fourier transform of f ( x), then the Fourier transform of f ( x) cos ax is ~ 1 ~ [ f ( p − a) + f ( p + a)]. 2

Proof. We have ~

f ( p) = F { f ( x)} =

Now

1 √ (2π)

∞

∫ −∞

e ipx f ( x) dx

…(1)

F { f ( x) cos ax} e iax + e − iax e ipx . f ( x) ⋅ dx 2

=

1 √ (2 π)

=

1 1  2  √ (2 π)

=

~ 1 ~ [ f ( p + a) + f ( p − a)]. 2

∞

∫ −∞ ∞

∫ −∞

e i( p + a ) x f ( x) dx +

 ∞ 1 e i ( p − a ) x f ( x) dx ∫ − ∞ √ (2 π) 

T-113

4.14 Important Results Theorem: (i) (ii) (iii)

~

~

If f s ( p) and f c ( p) are Fourier sine and cosine transforms of f ( x) respectively, then

1 [ 2 1 Fc { f ( x) sin ax} = [ 2 1 Fs { f ( x) sin ax} = [ 2 Fs { f ( x) cos ax} =

Proof:

~ ~ f s ( p + a) + f s ( p − a)] ~ ~ f s ( p + a) − f s ( p − a)] ~ ~ f c ( p − a) − f c ( p + a)].

(i) We have 2 Fs { f ( x) cos ax} =    π

∞

∫0

2 1 =   ⋅  π 2 =

f ( x) cos ax sin px dx ∞

∫0

f ( x) ⋅ [sin ( p + a) x + sin ( p − a) x] dx

1   2 ∞ f ( x) ⋅ sin ( p + a) x dx    2   π ∫ 0 2 +    π

=

∞

∫0

 f ( x) ⋅ sin ( p − a) x dx 

~ 1 ~ [ f s ( p + a) + f s ( p − a)]. 2

(ii) We have 2 Fc { f ( x) sin ax} =    π

∞

∫0

f ( x) sin ax cos px dx

2 1 ∞ =   ⋅ ∫ f ( x) [sin ( p + a) x − sin ( p − a) x] dx  π 2 0 =

1   2    2   π

∞

∫0

f ( x) sin ( p + a) x dx 2 −    π

=

∞

∫0

 f ( x) ⋅ sin ( p − a) x dx 

~ 1 ~ [ f s ( p + a) − f s ( p − a)]. 2

(iii) We have ∞ 2 Fs f { ( x) sin ax} =   ∫ f ( x) sin ax sin px dx  π 0

2 1 ∞ =   ⋅ ∫ f ( x) [cos ( p − a) x − cos ( p + a) x] dx  π 2 0

T-114

=

1   2 ∞ f ( x) cos ( p − a) x dx    2   π  ∫0 −

=

 2  ∞ f ( x) cos ( p + a) x dx   ∫   π 0 

~ 1 ~ [ f c ( p − a) − f c ( p + a)]. 2

4.15 Theorem If φ ( p) is the Fourier sine transform of f ( x) for p > 0, then Fs { f ( x)} = − φ (− p) Proof:

for p < 0 .

We have ∞ 2 Fs { f ( x)} =   ∫ F ( x) sin px dx  π 0

= φ ( p), for p > 0.

…(1)

For p < 0, let p = − s where s > 0. ∴

2 Fs { f ( x)} =    π

∞

∫0

2 = −    π

f ( x) sin (− sx) dx ∞

∫0

f ( x) sin sx dx = − φ (s)

= − φ (− p ), for p < 0. Hence in general  φ (| p |), p > 0 Fs { f ( x)} =   − φ (| p |), p < 0 or

Fs { f ( x)} = φ (| p |). Sgn p,  + 1, p > 0 where the symbol Sgn p =   − 1, p < 0 .

4.16 Multiple Fourier Transforms Let f ( x, y) be a function of two variables x and y. Regarding f ( x, y), temporarily, as a function of x, its Fourier transform is ~ ∞ 1 f ( p, y) = f ( x, y) e ipx dx . ∫ √ (2π) − ∞ ~

Now regarding f ( p, y) as a function of y, its Fourier transform is ~

F ( p, q) =

1 √ (2π)

∞

∫ −∞

~

f ( p, y) e ipy dy

T-115 ~

or

1 ∞ 2 π ∫ −∞

F ( p, q) =

∞

f ( x, y) e i ( px + qy )dx dy

∫ −∞

which is Fourier transform of f ( x, y). Inversion formula: Using inversion formula for Fourier transforms, we have ~ ∞ 1 f ( x, y) = f ( p, y) e − ipx dp ∫ − ∞ √ (2π) ~

and Hence

f ( p, y) =

1 √ (2π)

f ( x, y) =

1 2π

∞

∫ −∞

∞

∫ −∞

~

F ( p, q) e − iqy dq.

~

F ( p, q) e − i( px + qy )dp dq,

which is the inversion formula for the Fourier transform of f ( x, y).

4.17 Convolution The function H ( x) = F * G =

1 √ (2π)

∞

∫ −∞

F (u). G ( x − u) du

is called the convolution or Falting of two integrable functions F and G over the interval (− ∞, ∞). Note: Some authors also define

F*G=

∞

∫ −∞

F (u). G ( x − u) du.

4.18 The Convolution or Falting Theorem for Fourier Transforms If F { f ( x)} and F { g ( x)} are the Fourier transforms of the functions f ( x) and g ( x) respectively, then the Fourier transform of the convolution of f ( x) and g ( x) is the product of their Fourier transforms i.e. Proof:

F { f ( x) * g ( x)} = F { f ( x)} . F { g ( x)}. We have F { f ( x) * g ( x)}  1 = F  √ (2 π)

∞

∫ −∞

 f (u) ⋅ g ( x − u) du 

=

 ∞  1 ∞ 1 f (u) g ( x − u) du e ipx dx  ∫ ∫ √ (2 π) − ∞  √ (2 π) − ∞ 

=

1 ∞  ∞  f (u) ∫ g ( x − u) e ipx dx du  − ∞  2π ∫ − ∞

T-116

=

1 ∞  ∞  f (u) ∫ g ( y) e ip ( u + y )dy du,  − ∞  2π ∫ − ∞ putting x − u = y so that dx = dy,

1 = 2π

∫ −∞

1 2π

∫ −∞

=

∞

f (u)

∞

∞

  e ipu g ( y) e ipy dy du  ∫ − ∞ 

∞   f (u) e ipu ∫ g ( y) e ipy dy du   −∞

  ∞ ∞ 1 1 f (u) e ipu g ( x) e ipx dx du ∫ ∫ − ∞ − ∞ √ (2 π) √ (2 π)   ∞ 1 ipu = f (u) [e F { g ( x)}] du √ (2 π) ∫ − ∞ =

 1  ∞ = f (u) e ipu du F { g ( x)} ∫ − ∞  √ (2 π)   1  ∞ = f ( x) e ipx dx F { g( x)} ∫ − ∞  √ (2 π)  = F { f ( x)} ⋅ F { g ( x)} .

4.19 Relationship Between Fourier and Laplace Transforms Let us consider the function  e − x t g(t), t > 0 f (t) =  , t < 0. 0 ∴

…(1)

The Fourier transform of f (t) is given by ∞

F { f (t)} =

∫ −∞

=∫ =

0 −∞ ∞

∫0

e ipt . f (t) dt, (Taking non-symmetrical form of F. T.)

0 ⋅ e ipt dt +

∞

∫0

e − xt g(t) ⋅ e ipt dt

e( ip− x ) t g (t) dt =

∞

∫0

e − st g (t) dt , putting x − ip = s

= L { g (t)}. Hence the Fourier transform of the function f (t) defined by (1) is the Laplace transform of the function g (t).

4.20 Fourier Transform of The Derivatives of A Function ~

~

(a) The Fourier transform of f ′ ( x) , the derivative of f ( x) is − ip f ( p), where f ( p) is the Fourier transform of f ( x).

T-117

Proof: By definition F { f ′ ( x)} = =

1 √ (2π)

∞

∫ −∞

f ′ ( x) ⋅ e ipx dx

∞ f ( x + h) − f ( x) ipx 1 lim ⋅ e dx ∫ h √ (2 π) − ∞ h→ 0

∞ f ( x + h) ipx 1 1 ⋅ e dx − lim ∫ − ∞ h h→ 0 √ (2 π) h→ 0 √ (2 π)

= lim

∞

∫ −∞

f ( x) ipx ⋅ e dx h

∞ f ( x + h) ip( x + h ) − iph 1 = lim ⋅e e d ( x + h) − lim ∫ − ∞ h h→ 0 √ (2π) h→ 0

~

f ( p) h

~

f ( p) e − iph ∞ 1 = lim ⋅ f ( y) e ipy dy − lim ∫ h→ 0 √ (2 π) − ∞ h h→ 0 h ~

e − iph f ( p) = lim − lim h h→ 0 h→ 0 ~

~

f ( p) h

~ e − iph − 1 = (− ip) f ( p). h h→ 0

= f ( p) ⋅ lim

(b) The Fourier transform of f n( x), the nth derivative of f ( x) is (− ip)n times the Fourier transform of f ( x) provided that the first (n − 1) derivatives of f ( x) vanish as x → ± ∞ . Proof:

By definition F { f n ( x)} =

1 √ (2π)

∞

∫ −∞

f n ( x) . e ipx dx.

Integrating by parts, we have F { f n ( x)} = =

(− ip) √ (2 π)

∞

∫ −∞

∞ 1 1 [ f n−1( x) ⋅ e ipx ] ∞ f n−1( x) ipe ipx dx −∞ − √ (2 π) √ (2 π) ∫ − ∞

f n−1 ( x) e ipx dx, since

lim f n−1 ( x) = 0 . x→ ± ∞

Repeating the same process of integration by parts (n − 1) times more, we have ∞ 1 F { f n ( x)} = (− ip)n f ( x) e ipx dx ∫ √ (2π) − ∞ ~

or

~

f n ( p) = (− ip)n f ( p).

(c) The Fourier cosine and sine transforms of the derivatives of f ( x) are given by n −1 ~ ~ 2 f c2 n ( p) = −   (− 1)r α2 n−2 r −1 p2 r + (−1)n p2 n f c ( p);  π r=0

∑

T-118 n

~

2 f c2 n+1 ( p) = −    π ~ 2 f s2 n ( p) = −    π

r =1 n

∑ (− 1)r α2 n−2 r p2 r −1 + (−1)n+1 p2 n

~ f s ( p);

r =1

~ 2 f c2 n+1 ( p) = −    π

and

~

∑ (− 1)r α2 n−2 r p2 r + (−1)n p2 n+1 f s ( p);

n

~

∑ (− 1)r α2 n−2 r −1 p2 r −1 + (−1)n+1 p2 n+1 f c ( p);

r =1

provided that first (n − 1) derivatives of f ( x) vanish as x → ∞ and d n−1 f dx n−1 Proof:

→ α n−1 etc . as x → 0 .

By definition, we have

and

~ f cn ( p) =

 2    π

∫0

~ f cn ( p) =

 2    π

∫0

∞

f n ( x) cos px dx

…(1)

∞

f n ( x) sin px dx .

…(2)

Integrating R.H.S. of (1) by parts, we have ~ f cn ( p) =

 2  [ f n−1( x)cos px] ∞ + p  2      0  π  π

∞

∫0

f n−1( x)sin px dx

~ ~ 2 f cn ( p) = −   α n−1 + p f sn−1( p).  π

or

…(3)

Similarly integrating R.H.S. of (2), we have ~ ~ f sn ( p) = − p f cn−1( p).

…(4)

From (3) and (4), we have ~ ~ 2 f cn ( p) = −   α n−1 − p2 f nn−2 ( p).  π

…(5)

~

~

By repeated application of these rules f c n ( p) is obtained as a sum of α’s and f c ′ ( p) or ~ f c ( p). ~

It is clear that f c ′ ( p) will occur when n is odd and in that case it may be replaced by ~ 2 −   α0 + p f s ( p).  π

Thus, we have ~ 2n

fc

2 ( p) = −    π

n −1

∑

r=0

~

(− 1)r α2 n−2 r −1 p2 r + (−1)n p2 n f c ( p)

T-119 ~ 2n + 1

and

n

2 ( p) = −    π

fc

~

∑ (− 1)r α2 n−2 r p2 r + (−1)n p2 n+1 f s ( p)

r=0

Also from (3) and (4), we have ~   2 f sn( p) = − p −   α n−2 + p f sn−2 ( p)   π  

~

~

~ 2 f sn( p) = p   α n−2 − p2 f sn−2 ( p).  π

or

~

~

By repeated application of these results f s n ( p) is obtained as a sum of α’s and f s ′ ( p) or ~ f s ( p). ~

~

It is clear that f s ′ ( p) will occur when n is odd and it may be replaced by − p f c ( p). Thus, we have ~ 2 f s2 n( p) = −    π

and

n

∑ (− 1)r α2 n−2 r p2 r −1 + (−1)n+1 p2 n

~ f s ( p)

r =1

2 n f c2 n + 1( p) = −   ∑ (− 1)r α2 n−2 r +1 p2 r −1  π r =1

~

~ + (− 1)n+1 p2 n+1 f c ( p).

Note. The infinite sine and cosine transforms can be applied when the range of the

variable selected for exclusion is 0 to ∞.

Example 1:

Find the Fourier complex transform of f ( x), if  e iωx f ( x) =  0

Solution:

a< x< b x < a, x > b.

(Meerut 2013B; Avadh 14)

We have F { f ( x)} = =

1 √ (2π)

∞

∫ −∞

e ipx f ( x) dx

1  a 0 ⋅ e ipx dx + √ (2 π)  ∫ − ∞

b ipx

∫a

e

⋅ e iωx dx + ∫

 e i( p + ω ) x  b i ( p + ω)x 1 1 = = ⋅ e dx   √ (2 π) ∫ a √ (2 π)  i ( p + ω)  =

 ei ( p + ω ) a − ei ( p + ω ) b 

1  √ (2 π) 

p+ω

⋅ 

∞ b b a

 0 ⋅ e ipx dx 

T-120

Find the Fourier transform of F ( x) defined by

Example 2:

 1, | x | < a F ( x) =   0 ,| x | > a. (Kanpur 2008; Meerut 13; Bundelkhand 13; Avadh 13)

and hence evaluate ∞ sin pa cos px (a) ∫ dp, −∞ p

and

∞ sin p

(b)

∫0

p

dp.

We have

Solution:

~

F ( p) =

1 √ 2π)

1 √ (2 π)

=

a

∞

∫ −∞

e ipx F ( x) dx

e ipx dx =

∫ −a

1  e ipx    √ (2 π)  ip 

a −a

1  e ipa e − ipa  = −   ip  √ (2 π)  ip =

2 i sin pa ip √ (2 π)

=

2 sin pa p √ (2 π)

, p ≠ 0.

~

For p = 0 , F ( p) = 2 a / √ (2 π). (a)

We know that if ~

F ( p) =

1 √ (2π)

∫ − ∞ F ( x) e

then

F ( x) =

1 √ (2π)

∫ −∞

∴

1 √ (2 π)

∫ −∞

But L.H.S. = =

1 π

1 π

∞

ipx

dx

~

∞

F ( p) e − ipx dp .

2 sin pa

 1, | x | < a ⋅ e − ipx dp =  p √ (2 π) 0 , | x | > a.

∞

sin pa cos px

∞

∫ −∞

∫ −∞

∞

p sin pa cos px p

dp −

i π

∞

∫ −∞

sin pa sin px p

dp

dp,

since the integrand in the other integral is an odd function of p. ∞ sin pa cos px  π, | x | < a ∴ dp =  ∫ −∞ p 0 , | x | > a. (b) If x = 0 and a = 1 in (a), then ∞ sin p ∫ − ∞ p dp = π or

2

∞

∫0

sin p p

dp = π

or

∞ sin p

∫0

p

dp =

π ⋅ 2

…(1)

T-121 Note: Putting x = 0 in (a) and on simplification, we get ∞

∫0

sin ap

dp =

p

π ⋅ 2

…(2)

The results (1) and (2) can be used as standard formulae. Find Fourier sine and cosine transforms of e − x and using the inversion formulae recover the original functions, in both the cases. (Kanpur 2009) Example 3:

Solution:

Let f ( x) = e − x .

Then

~ f s ( p) =

 2    π

∞

∫0

2 f ( x) sin px dx =    π

∞ −x

e

∫0

sin px dx

∞

 2  e− x =    (− sin px − p cos px)  π  1 + p2 0  = and

p 2

1+ p

~ f c ( p) =

 2    π  2    π

∞

∫0

2 f ( x) cos px dx =    π

∞

∫0

e − x cos px dx

∞

 2  e− x =    (− cos px + p sin px)  π  1 + p2 0  =

1 2

1+ p

 2 ⋅    π

Applying inversion to the sine transform, we have 2 f ( x) =    π =

2 π

∞

∫0

∞ ~ f s ( p) ⋅ sin px dp

∫0

p sin px 1 + p2

dp

…(1)

and applying inversion to the cosine transform, we have 2 f ( x) =    π =

2 π

∞

∫0

∞ ~ f c ( p) cos px dp

∫0

cos px 1 + p2

dp.

Now from Fourier integral theorem, we have ∞ 1 ∞ f ( x) = dp ∫ f (v) cos p ( x − v) dv ∫ 0 − ∞ π ∞ 1 ∞ or f ( x) = cos px dp ∫ f (v) cos pv dv −∞ π ∫0 ∞ 1 ∞ + sin px dp ∫ f (v) sin pv dv. ∫ 0 − ∞ π

…(2)

…(3)

T-122

Case I:

Defining f ( x) in (− ∞, 0 ) such that f ( x) is an even function of x, from (3), we

have f ( x) =

2 π

∞

∞

cos px dp

∫0

f (v) cos pv dv.

∫0

Taking f ( x) = e − x , we have e− x =

2 π

∞

cos px dp

∫0

∞

∫0

e − v cos pv dv ∞

 e− v  2 ∞ = ∫ cos px  (− cos pv + p sin pv) dp 2 π 0 1 + p 0 2 ∞ cos px = dp. π ∫ 0 1 + p2 ∞

cos px

dp =

∴

∫0

∴

from (2) we have f ( x) =

Case II:

2

1+ p

π −x e . 2

2 π −x ⋅ e = e− x . π 2

Again defining f ( x) in (− ∞, 0 ) such that f ( x) is an odd function of x, from

(2), we have f ( x) =

2 π

∞

∫0

sin px dp

∞

∫0

f (v) sin pv dv.

Taking f ( x) = e − x and simplifying, we have p sin px

∞

∫0 ∴

1 + p2

from (1), f ( x) =

Example 4:

π −x e . 2

2 π −x ⋅ e = e− x . π 2

Find Fourier cosine transform of f ( x) =

transform of F ( x) = Solution:

dp =

x 1 + x2

⋅

(Kanpur 2007, 10; Meerut 13; Rohilkhand 14)

We have ~ f c ( p) =

1 and hence find Fourier sine 1 + x2

 2    π

∫0

2 =    π

∫0

∞

f ( x) cos px dx

∞ cos px dx . 2

1+ x

Differentiating both sides w.r.t. p, we have d ~ 2 f c ( p) = −    π dp

∞ x sin px dx 2

∫0

1+ x

T-123 ∞ ( x2 + 1 − 1) sin px dx 2

2 = −    π

∫0

2 = −    π

∫0

x (1 + x )

∞ sin px

x

2 dx +    π

2 π 2 = −   ⋅ +    π 2  π

sin px

∞

∫0

∞

∫0

2

x (1 + x )

sin px x (1 + x2 )

dx  ∵ 

dx .

∞ sin px

∫0

x

dx =

π 2 

Differentiating again w.r.t. p, we have 2 d2 ~ f ( p) =   2 c  π dp

∫

~ ∞ cos px dx = f c ( p) 2 0

1+ x

~

( D2 − 1) f c ( p) = 0

or

whose general solution is ~ f c ( p) = Ae p + Be − p .

…(1)

Now when p = 0 , ~ f c ( p) =

=

 2    π π 2

∞

∫0

∞ 2 =   tan−1 x   0 π 1+ x

dx

2

[

]

 2 =  π      π  2

d ~ π f c ( p) = −   .  2 dp

and

π ∴ from (1), we have   = A + B  2 π −   = A − B .  2

and

π Solving, A = 0 , B =   .  2 ∴

~ π from (1), we have f c ( p) =   e − p.  2

Second part:

We have ~ f c ( p) =

 2    π

∞ cos px dx = 2

∫0

1+ x

 π  ⋅ e − p.    2

Now differentiating both sides w.r.t. p, we have ∞ x sin px π − ∫ dx = − ⋅ e − p . 2 0 2 1+ x ∴

~ 2 Fs ( p) =    π

∞

∫0

x π sin px dx =   e − p .  2 1 + x2

T-124

Find the sine and cosine transforms of x ne − ax .

Example 5: Solution:

Let f ( x) = x ne − ax . ~ f s ( p) =

∴

We have

∞

∫0

 2    π

∫0

2 =    π

∫0

∞

f ( x) sin px dx

∞ n − ax

x e

…(1)

sin px dx ∞

e

− ax

 e − ax  sin px dx =  (− a sin px − p cos px) 2 2  a + p 0 =

p 2

2

a +p

=

1 1 1  − ⋅  2 i  a − ip a + ip

Differentiating both sides w.r.t. a, n times, we have (−1)n =

∞

x n e − ax sin px dx

∫0

 1  dn dn −1 a ip − − ( ) (a + ip)−1  n n 2 i  da da 

1 (− 1)n (n !) [(a − ip)−( n+1) − (a + ip)−( n+1)] 2i 1 = (− 1)n (n !) [2 i r −( n+1) sin (n + 1) θ], putting a = r cos θ, p = r sin θ 2i =

= (− 1)n n !(1 / r)n+1 sin (n + 1) θ. ∴

∞

∫0

x n e − ax sin px dx = (n !) . [1 / (a2 + p2 )( n+1)/2 ] sin {(n + 1) tan−1 ( p / a)}. [ ∵ r = (a2 + p2 )1 /2 and θ = tan−1 ( p / a)]

Hence, from (1),

Also

We have,

~ f s ( p) =

−1  2  ⋅ n !sin {(n + 1) tan ( p / a)}    π (a2 + p2 )( n + 1) /2

~ f c ( p) =

 2    π

∫0

2 =    π

∫0

∞

f ( x) cos px dx

∞ n − ax

x e

…(2)

cos px dx . ∞

 e − ax  − ax ∫ 0 e cos px dx =  a2 + p2 (− a cos px + p sin px)  0 a = a2 + p2 ∞

=

1 2

 1 1  + ⋅  a ip a ip − + 

T-125

Differentiating both sides w.r.t. a, n times, we have (− 1)n ∫

∞ 0

x n e − ax cos px dx =

1 (− 1)n (n !) [(a − ip)−( n+1) + (a + ip)−( n+1)] 2

= (− 1)n (n !) (1 / r)n+1 cos (n + 1) θ, putting a = r cos θ, p = r sin θ and on simplification. ∴

cos { (n + 1) tan−1 ( p / a)} x n e − ax cos px dx = (n !) . ⋅ (a2 + p2 )( n + 1)/2

∞

∫0

Hence from (2), we have −1  2  ⋅ n !cos {(n + 1) tan ( p / a)} .    π (a2 + p2 )( n + 1) /2

~ f c ( p) =

Alternative method to evaluate the integrals ∞ − ax

We have

∞

e − ax x n cos px dx + i

∫0 =∫ = = =

= =

∞

∞

x n sin px dx and

e

∫0

∫0 ∞

e − ax x n cos px dx.

e − ax x n sin px dx

∫0

e − ax x n (cos px + i sin px) dx

0 ∞

∫0

e − ax x n e ipx dx =

∞

∫0

e −( a − ip ) x x ( n + 1) −1 dx

Γ (n + 1)

 ∵

n+1

(a − ip )

Γ (n + 1) n+1

(cos θ − i sin θ) n + 1

n!

(cos θ + i sin θ) n + 1

r

n +1

r

n! r n +1

∞

∫0

Γ (n) e − az z n − 1 dz = a n 

, putting a = r cos θ and p = r sin θ   1 = cos θ + i sin θ ∵  cos θ − i sin θ 

[cos (n + 1) θ + i sin (n + 1) θ].

Equating real and imaginary parts on both sides of (1), we have ∞

∫0

∞

e − ax x n cos px dx = e − ax x n sin px dx =

n! r n +1 n!

cos (n + 1) θ

and

∫0

where

r2 = a2 + p2 i. e., r = (a2 + p2 )1 /2 and θ = tan−1 ( p / a).

r n +1

sin (n + 1) θ ,

...(1)

T-126

Hence

∞

= and

n! 2

2 ( n+1)/2

(a + p ) ∞

cos { (n + 1) tan−1 ( p / a)}

e − ax x n sin px dx

∫0 =

Example 6:

e − ax x n cos px dx

∫0

n! 2

2 ( n+1)/2

(a + p )

sin {(n + 1) tan−1 ( p / a)}.

Find the sine transform of e ax + e − ax

⋅ e πx − e − πx

Solution:

e ax + If f ( x) = πx e −

e − ax e − πx

(Meerut 2013B; Purvanchal 14; Kanpur 14)

, then we have

~ f s ( p) =

 2    π

2 =    π

∫

∞ e ax + e − ax sin px dx − πx 0 πx

2 =    π

∫0

e

∞

∫0

f ( x) sin px dx

−e

∞ e ax + e − ax e ipx − e − ipx ⋅ dx 2i e πx − e − πx

2  1 ∞ e( a + ip) x − e −( a + ip) x =    ∫ dx  π  2 i 0 e π x − e − πx  −

1 ∞ e( a − ip) x − e −( a − ip) x  dx 2i ∫ 0 e π x − e− π x 

a + ip 1 1 a − ip 2 1 1 =   ⋅ tan − ⋅ tan  π  2 i 2 2 2i 2 2   ∞ e az − e − az 1 a ∵ From definite integrals,  ∫ 0 e π z − e− π z dz = 2 tan 2   a + ip a − ip   sin sin 2  1 1  2 2  =    − a + ip 4 i a − ip   π  4i cos  cos   2 2  2 =    π

sin

a + ip a − ip a − ip a + ip cos − sin cos 2 2 2 2 a + ip a − ip 4 i cos cos 2 2

T-127

2 sin a + sin ip − (sin a − sin ip) =    π 2 ⋅ 2 i [cos ip + cos a] sin ip sinh p = = √ (2π) i [cos ip + cos a] √ (2π) (cosh p + cos a) =

⋅ √ (2 π) (e p + e − p + 2 cos a)

Find the Fourier sine transform of 1 f ( x) = ⋅ 2 x (a + x2 )

Example 7:

Solution:

e p − e− p

We have ~ f s ( p) =

Let

Then

 2    π

∫0

dI d = dp dp =∫ =

∴

d2 I dp2

2

x (a + x2 ) 1

∞

I =

1

∞

∫0

2

x (a + x2 )

…(2)

sin px dx.

sin px

∞

∫0

…(1)

sin px dx .

x (a2 + x2 )

dx

∞∂ 

sin px      dx 0 ∂p x (a2 + x2 )     cos px

∞

∫0

=−

a2 + x2 ∞

∫0

∞

x sin px a2 + x2

dx

x2 sin px

=−

∫0

=−

∫0

=−

π + a2 I. 2

∞

…(3)

dx.

x (a2 + x2 ) sin px x

∞

dx = −

dx + a2

∫0 ∞

∫0

( x2 + a2 ) − a2 x (a2 + x2 ) sin px

x (a2 + x2 )

sin px dx

dx

 ∵ 

∞ sin px

∫0

x

dx =

π ⋅ 2 

2

∴ or

d I

π − a2 I = − 2 dp 2

π d ( D2 − a2 ) I = − , where D ≡ ⋅ 2 dp

The solution of the above differential equation is π I = Ae − ap + Be ap + ⋅ 2 a2

…(4)

T-128

∴

dI = − Aae − ap + Bae ap. dp

…(5)

Now from (2), when p = 0, we have I = 0 and from (3), when p = 0, we have dI = dp

1

∞

∫0

dx =

a2 + x2

∞ tan−1 x  = π ⋅  a 0 2 a

1 a

So putting p = 0 in (4) and (5), we get A+ B= − and

π

…(6)

2 a2

a (− A + B) =

π π i. e., − A + B = ⋅ 2a 2 a2

Solving (6) and (7), we get B = 0 , A = −

π 2 a2

…(7)

⋅

Putting the values of A and B in (4), we get I =

∞

∫0

sin px

dx = −

x (a2 + x2 )

π 2 a2

e − ap +

π 2 a2

=

π 2 a2

(1 − e − ap).

Now putting the value of I in (1), we get ~ f s ( p) =

Example 8: Solution:

 2  ⋅ π (1 − e − ap) = 1    π  2 a2 a2

 π  ⋅ (1 − e − ap).    2

2 Find the Fourier cosine transform of e − x .

We have ~ 2 2 Fc { e − x } =    π

∞

∫0

2

e − x cos px dx = I

…(1)

Differentiating w.r.t.‘p’ we have 2 dI = −    π dp

∞

∫0

2

xe − x sin px dx

=

1  2   2  π

=

∞ − x2 1  2   − x2  sin px)0∞ − p ∫ e cos px dx   (e    0 2 π 

∞

∫0

2

(− 2 xe − x ) ⋅ sin px dx

(Integrating by parts taking sin px as first function) =−

p 2

I.

∴

p dI =− dp. I 2

Integrating, we have log I = −

p2 4

2 + log A or I = Ae − p /4 .

…(2)

T-129

2 But when p = 0, from (1) I =    π

∞ − x2

∫0

e

dx =

1 ⋅ √2

∴ from (2), A = 1 / √ 2. 2 2 Hence I = Fc { e − x } = (1 / √ 2) e − p /4 .

Example 9: Solution:

~

Use the sine inversion formula to obtain f ( x) if f s ( p) =

1 + p2

p

∞

∫0

1 + p2

⋅ sin px dp

p2

2 =    π

 2 ∫ 0 p (1 + p2 ) ⋅ sin px dp =  π 

2 =    π

∫0

∞

∞ sin px

p

∫

∞ ( p2 + 1) − 1 ⋅ sin px dp 2 0

p (1 + p )

∞ sin px 2 dp −   ∫ dp  π  0 p (1 + p2 )

π 2 f ( x) =   −    2  π

sin px

∞

∫0

p (1 + p2 )

…(1)

dp  ∵ 

∴

and

⋅

Using Fourier sine inversion formula, we have 2 f ( x) =    π

or

p

df 2 = −    dx π d2 f 2

dx

2 =    π

∞ cos px dp 2

∫0

1+ p

∞ sin px

∫0

p

dp =

π  2

…(2)

∞ p sin px dp 2

∫0

1+ p

2

or

d f dx2

− f =0

whose solution is f = A e x + B e − x .

…(3)

df = A e x − B e− x . dx

…(4)

∴

π Now when x = 0 , f =   , from (1)  2 and

df 2 = −    π dx

∞

∫0

dp π = −   , from (2).  2 1 + p2

π π ∴ from (3) and (4),   = A + B and −   = A − B .  2  2 π π Solving, A = 0 , B =   . Hence f ( x) =   e − x .  2  2

T-130

Comprehensive Exercise 1 1. (i) Find the Fourier transform of f ( x), if  √ (2 π) , | x| ≤ ε  f ( x) =  2 ε  0 , | x| > ε.  (ii) Find the Fourier transform of  x, | x| ≤ a f ( x) =   0 , | x| > a.

(Kanpur 2008)

2 2 2. (i) Show that the Fourier transform of f ( x) = e − x /2 is e − p /2 . (Kanpur 2011)

3.

(ii) Find the Fourier transform of the function 1 + x , for − a < x < 0  a  x f ( x) = 1 − , for 0 < x < a a   0 , other wise.  1 − x2 , | x | ≤ 1 (i) Find the Fourier transform of F ( x) =  0, | x | > 1  (Kanpur 2010; Purvanchal 14)

and hence evaluate

∫

∞  x cos x − sin  0  x3

x x  cos dx.  2

(ii) Find the cosine transform of the function f ( x), if cos x, 0 < x < a f ( x) =  x > a.  0, 4.

(i) Find the Fourier sine and cosine transform of f ( x), if 0 < x 2 .  (ii) Find the Fourier sine and cosine transform of the function f ( x) = x m − 1.  Hint. 

∞

x m − 1 cos px dx − i

∫0

=

∞

∫0

∞

∫0

x m − 1 sin px dx

Γ (m) Γ (m)  mπ mπ   e ipx x m − 1 dx = = − i sin cos  m m  2 2   (ip) p

T-131

5.

(i) Find the Fourier sine transform of x /(1 + x2 ). (ii) Find the cosine transform of

6.

Find the sine transform of

e ax + e − ax

⋅ e πx + e − πx

and deduce that e πx − e − πx

Fs (cosech πx) =

1 tanh ( p / 2). √ (2 π)

7.

Find the Fourier sine transform of f ( x) , if 0 , 0 < x < a  f ( x) =  x, a ≤ x ≤ b 0 , x > b. 

8.

Find f ( x) if its cosine transform is 1/(1 + p2 ).

9.

~ 1 Find f ( x) if f c ( p) = πe − p. 2

10.

Find f ( x) if its sine transform is π / 2.

11.

Find f ( x) if (i) its sine transform is e − ap, (ii) its cosine transform is e

12.

(Purvanchal 2014; Kanpur 14)

1

(Kanpur 2012)

− ap

.

~ Find the inverse Fourier transform of f ( p) = e −| p | y .

~ 13. Find f ( x) if f s ( p) = pn e − ap.

14.

Find f ( x) if its cosine transform is p  1   ~  a −  , if p < 2 a  f c ( p) =  √ (2 π)  2   0, if p ≥ 2 a

15.

Find f ( x) if f s ( p) =

~

e − ap . Hence deduce Fs −1 {1/ p}. p

A nswers 1 sin pε

1.

(ii)

2.

(ii)

3.

2  p cos p − sin (i) −2   ⋅   π  p3

(ii) −

pε 1 2

ap

⋅

i

2 ⋅   (ap cos ap − sin ap)  π p 2

2 [1 − cos pa] π p 3π ; −  16 

(ii)

1 sin (1 + p) a sin (1 − p) a  +   1− p  √ (2 π)  1 + p

T-132

4.

2 sin p 2 cos p (i) 2   ⋅ (1 − cos p); 2   ⋅ (1 − cos p).  π  p2  π p2 (ii)

Γ(m)  2  mπ mπ Γ(m)  2  ; m   cos   sin m     π 2 π 2 p p

5.

(i) √ (π / 2) . e − p

6.

1 2 √ (2 π)

7. 8. 9.

(ii)

ep − 1 ep + 1

 2   − b cos pb + a cos pa + sin pb − sin pa       π  p p2   √ (π / 2) e − x  π ⋅ 1    2  1 + x2

10. √ (π / 2).(1 / x)

11.

x  2 ⋅ ;   2  π  a + x2

13.

−1  2  ⋅ n !sin {(n + 1) tan ( x / a)}    π (a2 + x2 )( n + 1)/2

15.

p /2 + e − p /2 )  2  cos (a / 2) ⋅ (e   p  π 2 cos a + e + e − p

a  2 ⋅   2  π  a + x2

12.

14.

y √2 √ π( y2 + x2 ) π −1 x −2 sin2 ax

√ (2 / π) tan−1 ( x / a) ; √ (π / 2)

Objective Type Questions

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). ~

1.

If f ( p) is the complex Fourier transform of f ( x), then the complex Fourier transform of f (ax) is ~  p 1 ~  p (b) f   (a) f    a a  a 1~ (c) (d) None of these. f ( p) a

2.

If f s ( p) and f c ( p) are infinite Fourier sine and cosine transforms of f ( x) respectively, then Fc { f ( x) sin ax} is ~ ~ ~ 1 ~ (a) [ f s ( p + a) − f s ( p − a)] (b) [ f s ( p + a) − f s ( p − a)] 2 ~ ~ ~ 1 ~ (c) 2 [ f s ( p + a) − f s ( p − a)] (d) [ f s ( p − a) − f s ( p + a)] 2

~

~

T-133

3.

The infinite Fourier sine transform of e − x is p 1  2  2 (a) (b)     1 + p2  π  1 + p2  π  1  2 1  2 (d) (c)     p  π p2  π 

4.

What is f ( x) if its infinite Fourier sine transform is

e − ap ? p

(a)

 2  tan−1 x    π a

(b)

 2  tan−1 a    π x

(c)

 2  tan−1 2 x    π a

(d)

 2  tan−1 2 a ⋅    x x

Fill In The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. 1.

If the function f ( x) is defined on (− ∞ , ∞) and is piecewise continuous in each finite partial interval and absolutely integrable in (− ∞ , ∞), then the integral ∞ 1 e ipx f ( x) dx is called the ........... transform of f ( x). √ (2 π) ∫ − ∞

2.

The infinite Fourier sine transform f s ( p) of f ( x) in 0 < x < ∞ , is ............ .

3.

If f ( p) is the complex Fourier transform of f ( x), then the complex Fourier

~

~

transform of f ( x − a) is ........... . 4.

~

~

If f s ( p) and f c ( p) are infinite Fourier sine and cosine transforms of f ( x) respectively, then Fs { f ( x) cos ax} = ........... .

5.

 1, | x|< a The Fourier transform of F ( x) =  is ......... . 0 , | x|> a

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

The Fourier transform is a linear transformation.

2.

If

~ f c ( p) is the infinite Fourier cosine transform of f ( x), then the infinite ~  p

Fourier cosine transform of f (ax) is f c   ⋅  a 3. 4.

x  2 ⋅ is the inverse Fourier sine transform of e − ap.   2  π  a + x2 ~

~

If f s ( p) and f c ( p) are infinite Fourier sine and cosine transforms of f ( x) ~ 1 ~ respectively, then Fs { f ( x) sin ax} = [ f c ( p − a) + f c ( p + a). 2

T-134

5.

~

~

The Fourier transform of f ′ ( x), the derivative of f ( x) is (− ip) f ( p), where f ( p) is the Fourier transform of f ( x).

6.

The Fourier transform of the function  e − xt g(t), t > 0 is the Laplace transform of the function g (t). f (t) =  t 0 pπ  pπ / 4 0 p2 π2

and if p = 0, then f c ( p) =

4

∫0

2 x ⋅ 1 dx = 16.

Example 2: Find the finite Fourier cosine transform of f ( x) if

(i ) f ( x) =

π x2 −x+ ⋅ 3 2π

(ii) f ( x) = sin nx.

Solution: (i) We have ~ f c ( p) =

π

∫0

~

f ( x) cos px dx . ∴ f c ( p) = π

x2   −x+  cos px dx 2π  3

π π

∫0

 π  x2  1 1 π x  sin px − ∫  −1 +  sin px dx =  − x +  0 2π  p π p  3 0

T-142 π

 x 1 1 1 π1 = − −  −1 +  cos px − ∫ cos px dx   p π p p2 0 π 0 1 1 1 π = − sin px]0 = , if p > 0 2 3 [ p p π p2 and when p = 0, x2    dx = 0 − x + ∫0 3 2π 

~ f c ( p) = ~ f c ( p) =

(ii)

π π

π

∫0

sin nx cos px dx

=

1 2

=

1  cos (n + p) x cos (n − p) x  − −  , if p ≠ n 2  n+ p n− p 0

π

∫0

[sin (n + p) x + sin (n − p) x] dx π

~

∴

if p ≠ n, f c ( p) =

1 2

 cos (n + p) π cos (n − p) π 1 1 − + + − n + p n − p n + p n − 

 ⋅ p

If n − p is even, then n + p is also even and so ~ 1 f c ( p) = −

1 1 1 1 − + + 2  n+ p n− p n+ p n−

 =0. p

If n − p is odd, then n + p is also odd and so ~ 1 2 f c ( p) = 

2 n + p

+

2  2n = ⋅  2 n − p n − p2

If p = n, then

~ π 1 π f c ( p) = sin nx cos nx dx = sin 2 nx dx 0 2 0

∫

= ∴ Example 3:

∫

π 1  cos 2 nx  − =0. 2  2 n 0

~ f c ( p) = 0 or

2n 2

n − p2

according as n − p is even or odd.

Find the finite cosine transform of f ( x) if f ( x) = −

cos k (π − x) ⋅ k sin kπ

Solution: We have ~ π cos { k (π − x)} f c ( p) = − cos px dx 0 k sin k π

∫

=−

1 2 k sin kπ

π

∫0

[cos { k (π − x) + px} + cos { k (π − x) − px}] dx

T-143 π

sin (kπ − kx + px) sin (kπ − kx − px) 1 −  2 k sin kπ  p− k p+ k 0 sin pπ sin (− pπ) sin kπ sin kπ  1 =− − − + 2 k sin kπ  p − k p+ k p− k p + k  =−

=

1 2k

 1 1  1 − , k ≠ 0 , 1, 2, 3, ......   = 2 − + p k p k   p − k2

Example 4: Find finite Fourier sine transform of f ( x) if

f ( x) =

π sin kx 2

2 k sin kπ

−

x cos k (π − x) ⋅ 2 k sin kπ

Solution: We have ~ f s ( p) =

=∫ =

π

∫0

f ( x) sin px dx]

π  π sin kx  2 0

−

 2 k sin kπ π π

2 k sin2 kπ

x cos k (π − x)  sin px dx 2 k sin kπ 

sin kx sin px dx

∫0

−

=

π

π

2

4 k sin kπ −

=

∫0

1 2 k sin kπ

π

∫0

x cos k (π − x) sin px dx

[cos ( p − k ) x − cos ( p + k ) x] dx

1 4 k sin kπ

π

∫0

x [sin (kπ − kx + px) + sin ( px − kπ + kx)] dx

sin ( p − k ) x sin ( p + k ) −  2 p− k p+ k 4 k sin kπ  π

π

x  0

π

  cos (kπ − kx + px) cos ( px − kπ + kx)  1 − ⋅  x− −  4k sin kπ   ( p − k) ( p + k)  0 + =

1 4 k sin kπ

π

 cos (kπ − kx + px) cos ( px − kπ + kx) −  dx ( p − k) ( p + k) 

∫ 0 1⋅  −

sin ( p − k ) π sin ( p + k ) π  −   p− k p+ k 4 k sin kπ   π

2

+

π 4 k sin kπ

1 − 4 k sin kπ

  1 1  +  cos pπ ⋅   p − k p + k   π

sin (kπ − kx + px) sin ( px − kπ + kx) +   ( p − k )2 ( p + k )2 0 

T-144

=

  1 1  ⋅ sin pπ cos kπ ⋅  −   p − k p + k 4 k sin kπ  π

2

 1 2p 1  π − cos pπ sin kπ ⋅  + ⋅ cos pπ  + 2 p − k p + k k sin k 4 π   p − k2 +

  1 1 1   ⋅ sin kπ ⋅  −  2 4 k sin kπ  ( p + k )2    ( p − k ) 

2 2 1 ( p + k) − ( p − k) ⋅ 4k ( p2 − k 2 )2 p = , (| k | ≠ 0 , 1, 2,...). 2 ( p − k 2 )2

=

Example 5:

Find f ( x) if pπ





− cos pπ 6 sin ~   2 2 for p = 1, 2, 3,...... and for p = 0, f c ( p) = (2 p + 1) π

π

where 0 < x < 4. Solution: We have f ( x) =

pπx 1~ 2 ∞ ~ f c (0 ) + ∑ f c ( p) cos l l l p=1

pπ − cos pπ 6 sin pπx   1 2 2 2 = ⋅ + ⋅ ∑ cos 4 π 4 p=1 (2 p + 1) π 4 ∞

sin pπ − cos pπ   pπx   1 3 ∞  2 = + cos  ⋅ ∑  4  2π π p = 1 2p + 1 Example 6: Find f ( x) if its finite sine transform is given by ~ 1 − cos pπ , where 0 < x < π. f s ( p) = 2 2

p π

Solution: We have

f ( x) =

2 ∞ ~ ∑ f s ( p) sin px π p=1

=

2 ∞ 1 − cos pπ   sin px ∑ π p = 1  p2 π2 

=

∞  1 − cos pπ    sin px.  π3 p = 1  p2 

2

∑

(Kanpur 2008)

T-145

Comprehensive Exercise 1 1. 2.

(i) Find the finite Fourier sine and cosine transforms of f ( x) = 1. (ii) Find the finite Fourier sine and cosine transforms of f ( x) = x. (i) Find the finite Fourier sine transforms of 1 − x  and x ⋅    π 4π (ii) Find the finite Fourier cosine transform of 1 − x  and x ⋅    π 4π

3.

Find the finite Fourier sine transform of f ( x) if 0 ≤ x≤ π/2  x, (i) f ( x) =   π − x, π / 2 ≤ x ≤ π . x< c  − x, (ii) f ( x) =   π − x, x > c , where 0 ≤ c ≤ π .

4. (i) Find the finite Fourier cosine transform of f ( x) if  1, 0 < x < π / 2 f ( x) =   −1, π / 2 < x < π.

(Kanpur 2012)

2

(ii) Find the finite cosine transform of (1 − x / π) . 5.

Find the finite Fourier sine transforms of (ii) x (π2 − x2 ).

(i) x (π − x) 6.

Find the finite Fourier sine transform of f ( x) if π x2 −x+ ⋅ 3 2π Find the finite sine transform of f ( x), if

(i) f ( x) = sin nx 7.

(Avadh 2013)

(i)

(ii)

f ( x) =

(ii) f ( x) = x3

f ( x) = cos kx

(iii) f ( x) = e cx .

8.

Find finite Fourier cosine transform of f ( x) if f ( x) =

9.

Find finite Fourier sine transform of f ( x) , if sin k (π − x) f ( x) = ⋅ sin (kπ)

cosh { c (π − x)} ⋅ sinh (πc )

10. Find the finite Fourier sine and cosine transforms of f ( x) = x2 , 0 < x < 4. ~

11. Find f ( x) if f c ( p) =

cos(2 pπ / 3) (2 p + 1)2

, if 0 < x < 1.

12. When f ( x) = sin mx, where m is a positive integer, show that ~ ~ f s ( p) = 0 if p ≠ m and f s ( p) = π / 2 if p = m.

T-146

A nswers 1 1.

(i) (ii)

1 [1 − (− 1) p]; 0 p π (−1) p + 1 (−1) p − 1 π2 ; , if p = 1, 2, 3,... and , if p = 0 ⋅ 2 p 2 p

1 (−1) p + 1 ; p 4p 1 1 (ii) [1 − (−1) p] ; [(−1) p − 1] 2 πp 4 πp2

2.

(i)

3.

(i) (2 / p2 ) sin ( pπ / 2) ,

4.

(i) (2 / p) sin ( pπ / 2), p > 0 and 0, if p = 0. 2 π (ii) , if p > 0 and , if p = 0 2 3 πp

5.

(i)

6.

(i) 0, if p ≠ n , and π / 2, if p = n π 1 (ii) {(− 1) p + 2} + {(− 1) p − 1} 6p πp3 p (i) [1 − (−1) p cos kπ], 2 2 p −k p (iii) [1 − (−1) p e cπ ] 2 c + p2 c

7.

8. 10.

2 3

p

(ii) (π / p) cos pc .

[1 − (− 1) p],

(ii)

6π p3

(−1) p + 1

6 π2   , (ii) π (− 1) p  − 3 p  p

9.

p

, k ≠ 0 , 1, 2,... . p − k2 c +p 64 128 128 64 − cos pπ + (cos pπ − 1) ; cos p π, if p > 0; , if p = 0 3 3 2 2 pπ 3 p π p π 2

2

∞

11. 1 + 2

∑

p=1

cos (2 pπ / 3) (2 p + 1)2

2

cos pπx

Objective Type Questions

Multiple Choice Questions

1.

Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). If f ( x) is continuous and f ′ ( x) is sectionally continuous then Fs { f ′ ( x)} is (a) − Fc { f ( x)}

(b) Fc { f ( x)}

(c) − p Fc { f ( x)}

(d) − p2 Fc { f ( x)}

T-147

2.

The finite Fourier sine transform of f ( x) = x is π π (a) (b) (− 1) p (− 1) p + 1 p p (c)

3.

π p2

(− 1) p + 1

p2

2 ∞ (− 1) p − 1 sin px ∑ π p=1 p3 ∞

(c) 4

∑

(− 1) p − 1 p3

p=1

(c)

p3 ∞

(b) 4

∑

p3

sin px

4 ∞ (− 1) p − 1 sin px ∑ π p=1 p3

(d)

sin px

, where 0 < x < π ?

(− 1) p

p =1

The finite Fourier sine transform of (a)

(− 1) p

2 π (− 1) p − 1

~

What is f ( x) if its finite sine transform f s ( p) is (a)

4.

π

(d)

x is π

(− 1) p

(− 1) p + 1

(b)

2

p

(− 1) p − 1 p

p2 (− 1) p + 1 . p

(d)

Fill In The Blank(s) Fill in the blanks “……” so that the following statements are complete and correct. ~

1.

The finite Fourier sine transform f s ( p) of f ( x) on the interval (0 , π) is ........

2.

On the interval (0 , π) , f c ( p) =

~

π

∫0

f ( x) cos px dx, then the function f ( x) is ~

called ................. Fourier cosine transform of f c ( p). 3.

If F ( x) and G ( x) are two functions defined on the interval − 2 π < x < 2 π ,then the function F ( x) * G ( x) =

π

∫−π

F ( x − y) G ( y) dy is called the ......... of F ( x) and

G ( x) on the interval − π < x < π. 4.

The finite Fourier sine transform of the function f ( x) = 2 x , 0 < x < 4 is .............

True or False Write ‘T’ for true and ‘F’ for false statement. 1.

~

If f s ( p) is the finite Fourier sine transform of f ( x) on the interval (0 , π) then the inversion formula for sine transform is f ( x) =

2.

2 ∞ ~ ∑ f s ( p)sin px π p=1

If f ( x) is continuous and f ′ ( x) is sectionally continuous then Fc { f ′ ( x)} = p Fs { f ( x)} − f (0 ) + f (π).

T-148

3.

The finite Fourier cosine transform of the function f ( x) = 1 is 0.

4.

If the finite sine transform is given by f s ( p) =

~

the value of f ( x) is

2 π

3

1 − cos pπ p2 π2

, where 0 < x < π , then

∞

1 − cos pπ    sin px.   p2  p =1 

∑

A nswers Multiple Choice Questions 1.

(c)

2. (a)

3.

(c)

4.

(d)

2.

inverse finite

3. Convolution

Fill in the Blank(s) π

f ( x ) sin px dx

1.

∫0

4.

− 32 cos pπ pπ

True or False 1. T

2.

F

3.

T

4. T

¨

T-149

6 A pplications

of F ourier T ransforms in I nitial and B oundary V alue P roblems

6.1 Application of Infinite Fourier Transforms irst we shall consider the problems in which one of the variables ranges from − ∞ to ∞ or 0 to ∞ . Such problems are solved by taking the infinite Fourier transform of both the sides and then taking the corresponding inverse Fourier transform of the solution of the differential equation thus obtained.

F

6.2 Choice of Infinite Sine or Cosine Transform The choice of sine or cosine transform is decided by the form of the boundary conditions at the lower limit of the variable selected for exclusion. ∂2V

If we want to remove a term

∂x2

in a differential equation, applying sine transform,

we have ∞

∫0

sin px ⋅

=−p

∞

∂2V 2

∂x

∞

 ∂V  dx =  sin px − p ∂ x  0 ∂V

∫0 cos px ⋅ ∂x

dx, if

∞

∫0

cos px ⋅

∂V → 0 as x → ∞ ∂x

∂V dx ∂x

T-150

 ∞ = − p (V ⋅ cos px )0 + p 

∞



∫0 V sin px dx

~

~

= p (V ) x = 0 − p2 V s where V s is the sine transform of V and assuming that V → 0 as x → ∞ . Again applying the cosine transform, we have ∞

∫0

cos px ⋅

∂2V 2

∂x

∞

  ∂V dx =  cos px + p 0  ∂x

 ∂V  ∞ =− + p [V sin px]0 − p2   ∂x  x = 0

∞

∫0 sin px

∂V dx ∂x

∞

∫0 V cos px ⋅ dx if

∂V → 0 as x → ∞ ∂x

~  ∂V  =− − p2 V c if V → 0 as x → ∞,   ∂x  x = 0 ~

where V c is the cosine transform of V. Thus we see that for the exclusion of

∂2V ∂x2

from a differential equation we require

(V ) x = 0 in sine transform and

 ∂V  in cosine transform.    ∂x  x = 0

Note 1: By the Fourier sine or cosine transform we cannot exclude a derivative of odd

order from the given differential equation. Note 2: When one of the variables in a differential equation ranges from − ∞ to ∞ then

that variable can be excluded with the help of complex Fourier transform.

Example 1: The temperature U in the semi-infinite rod 0 ≤ x < ∞ is determined by the

differential equation ∂U ∂2U =k ∂t ∂x2 subject to the conditions (i)

U = 0 when t = 0 , x ≥ 0 ∂U (ii) = − µ (a constant) when x = 0 and t > 0 . ∂x Making use of cosine transform, show that

T-151

U ( x, t) =

2µ π

∞

∫0

2  cos px  1 − e − kp t  dp . 2   p

 ∂U  is given, so taking the Fourier cosine transform of both the   ∂x  x = 0

Solution: Since 

sides, we have  2    π

∞

∫0

 2 d    π  dt

or

∂U  2 cos px dx = k    π ∂t ∞

∫0

∞ ∂2U

∫0

∂x2

cos px dx ∞

U cos px dx = k

  2   ∂U cos px     π   ∂x 0

 2 + kp    π

∞

∫0

∂U sin px dx ∂x

d ~ ∞  2   ∂U   2 (U c ) = − k   ⋅  + kp   (U sin px )0        π ∂x x = 0 π dt

or

− kp2 if ~

~  2 2   ⋅ kµ − kp U c  π

or

dUc = dt

or

~ d ~ U c + kp2 U c = dt

 2    π

∞

∫0 U cos px dx

∂U → 0 as x → ∞ ∂x

if U → 0 as x → ∞

 2   kµ ,  π

which is a linear differential equation of first order.

∴

I.F. = e ∫ its solution is 2

e kp

t

kp2 dt

2

= e kp t .

~ 2  2 ⋅ U c = A +   k µ e kp t dt  π

∫

2  2 µ = A +   2 e kp t .  π p

~

But when

t = 0 ,Uc = 0 .

∴

 2 µ 0 = A+   2  π p

or

A= −

 2 ⋅√  .  π p µ

2

∴

2 ~ 2   2 µ  e kp t U c =   ⋅ 2  − 1 + e kp t   π p  

or

~ 2  2 µ U c =   ⋅ 2 (1 − e − kp t ).  π p

Applying the inverse Fourier cosine transform, we have

[∵ U = 0 when t = 0 ]

T-152

or

∞ ~

 2    π

Uc =

U c cos px dp

∞ cos

2µ π

Uc =

∫0

2  px  ⋅ 1 − e − kp t  dp .   p

∫0

2

∂U ∂2U =2 2 ∂t ∂x

Example 2: Solve

if U (0 , t) = 0 , U ( x, 0 ) = e − x , x > 0 , U ( x, t) is bounded where x > 0 , t > 0 . (Meerut 2013B) Solution: Since U (0 , t) is given, so taking the Fourier sine transform of both the sides,

we have  2    π or

d dt

∞

∫0

 2    π

∂U  2 sin px dx = 2    π ∂t ∞

∫0

∞ ∂2U

∫0

∂x2

sin px dx ∞

U sin px dx = 2

  2   ∂U sin px   ⋅  π   ∂x 0

 2 −2 p    π if or

∞

∫0

∂U cos px dx ∂x

∂U → 0 as x → ∞ . ∂x

d ~ ∞  2  2 U s = − 2 p   (U cos px )0 − 2 p2      π dt π

∞

∫0 U sin px dx

if U → 0 as x → ∞ ~  2 = 2 p   U (0 , t) − 2 p2 U s  π ~

or

~ dUs + 2 p2 U s = 2 p dt ~

2

whose solution is U s = Ae −2 p

 2   U (0 , t) = 0  π

t

…(1)

−x

But U ( x, 0 ) = e . ~

∴

 2    π

when t = 0 , U s =

∞

∫0

e − x sin px dx ∞

−x   2  e (− sin px − p cos px)    2  π 1 + p  0

=

= ∴

p 1 + p2

 2   .  π

from (1) when t = 0 ~

Us =

p 1 + p2

 2   = A.  π

T-153

Hence

~

2 p  2 e −2 p t .    π  1 + p2

Us =

Now applying the inverse Fourier sine transform, we have ∞

2 U ( x, t) = π

∫0

2

pe −2 p t sin px 1 + p2

dp .

Example 3: Using the Fourier sine transform, solve the partial differential equation

∂V ∂2V = k 2 for x > 0 , t > 0 , under the boundary conditions V = V0 when x = 0 , t > 0 , and the ∂t ∂x initial condition V = 0, when t = 0 , x > 0 . Solution: Taking the Fourier sine transform of both the sides, we have ∞

 2    π

∫0

∂2V ∂x2

~

∞

d  2   dt  π 

or

∞

∂V  2 sin px dx = k    π ∂t

∫0

V sin px dx = − kp2 V s + kp

∫0

sin px dx  2   V (0 , t)  π [See Ex. 2.]

~

~ dV s + kp2 V s = kp dt

or

 2   ⋅ V0 ,  π

[∵ V (0 , t) = V0 ]

which is linear differential equation of first order. ∴

I.F. = e ∫

kp2 dt

2

= e kp t .

Hence its solution is ~

2

t

=c+

~

2

t

2  2 V = c +   0 e kp t  π p

V s e kp or

V s e kp

 2    π

∫ kp V0 ⋅ e

kp2 t

dt …(1)

~

But when t = 0 , V s = 0 . ∴

from (1), we have 0=c+ ~

∴ or

[∵ V = 0 when t = 0]

2

from (1), V s ⋅ e kp ~

Vs =

 2    π

V0 ⋅ p t

=

V0 p

or c = −

V0 ⋅ p

 2   .  π

 2  kp2 t − 1)   (e  π

 2  V0  − kp2 t    1 − e .  π p  

Now applying the inverse Fourier transform, we have 2 V = ⋅ V0 π

2 ∞ (1 − e − kp t )

∫0

p

sin px dp

T-154

or

 2V0  π  

V =

∫0

p

 2V0 π  − π 2 

=

or

∞ sin

px

dp −

2 ∞ e − kp t

∫0

p

2 ∞ e − kp t

 2 V ( x, t) = V0 1 − π  

∫0

p

2 ∞ e − kp t

∫0

 ⋅ sin px dp  

p

 sin px dp    sin px dp ⋅  

Note: In terms of Error function or Complementary error function the solution may be

written as   x  x   as shown below. V ( x, t) = V0 1 − erf  = V0 erf c  2 √ (kt)  2 √ (kt)  We know that ∞

∫0

e−α

2 2

x

cos 2βx dx =

√ π −β2 / α 2 e . 2α

Integrating both sides w.r.t. β from 0 to β, we have ∞

∫0

∞

∫0

or

e−α

2 2

e−α

x

  

β



√π

1 √π sin 2βx dx = α 2x 2α

2 2

x

β

∫0 cos 2βx dβ dx = 2α ∫0 e β/α

∫0

−β2 /α 2

2

e − u du, putting

∞

∫0

or

e−α

2 2

x

⋅

sin 2βx π 2 dx = ⋅ x 2 √π =

∴

∞

β/ α

∫0

e − kp

2 π

∞

=

=

 x   x  2 π  = erf   . ⋅ erf  π 2  2 √ (kt)  2 √ (kt)

∫0

⋅

2

e − u du

sin px dp p

∫0

t

β = u so that dβ = α du α

π erf (β / α) . 2

2 π

2

dβ

2

e −4 ktu ⋅

sin 2 ux du , putting p = 2 u so that dp = 2 du u

Example 4: Use the method of Fourier transform to determine the displacement y ( x, t) of an

infinite string, given that the string is initially at rest and that the initial displacement is f ( x), − ∞ < x < ∞ . Show that the solution can also be put in the form 1 (Meerut 2006) y ( x, t) = [ f ( x + ct) + f ( x − ct)] . 2

T-155 Solution: Displacement of a string is governed by one dimensional wave equation

∂2 y 2

∂t

∂2 y

= c2

…(1)

∂x2

where y ( x, t) is the displacement at any time t. − ∞ < x < ∞, t > 0 T c2 = ⋅ ρ

and

Taking the Fourier transform of both the sides of equation (1), we have 1 √ (2 π)

∫− ∞

d2

1 dt √ (2 π)

or

2

d2 ~ y ( p, t)

or

dt2 d2 ~ y ( p, t)

or

∂2 y

∞

dt2

e ipx dx = c 2

2

∂t ∞

∫− ∞

1 √ (2 π)

∞

∫− ∞

∂2 y 2

∂x

e ipx dx

ye ipx dx = c 2 (− ip)2 ~ y ( p, t)

= − c 2 p2 ~ y ( p, t) + c 2 p2 ~ y ( p, t) = 0

whose solution is ~ y ( p, t) = A cos cpt + B sin cpt.

…(2)

Initially the string is at rest. ∂y ∴ = 0, at t = 0 . ∂t ∞ ∂y ∞ 1 1 d ⋅ e ipx dx = ∴ ye ipx dx dt √ (2 π) − ∞ √ (2 π) − ∞ ∂t d ~ y ( p, t) = = 0 , at t = 0 . dt

∫

∴

∫

from (2), we have 0 = Bcp or B = 0 .

Also at t = 0, y = f ( x) . 1 √ (2 π)

~

∞

∴

at t = 0 , ~ y ( p, 0 ) =

∴

from (2), we have f ( p) = A .

f (u) e ipu du = f ( p) .

∫− ∞

~

~

~

Hence y ( p, t) = f ( p) cos cpt. Taking the inverse Fourier transform, we have ∞ ~ 1 y ( x, t) = f ( p) cos cpt e − ipx dp √ (2π) − ∞

∫

=

1 2π

∞



∞

∫ − ∞ ∫ − ∞

 f (u) e ipu du cos cpt e − ipx dp 

T-156 ∞

=

1 4π

=

1 1  2 2π

  

∞

∞

  

 f (u) e ipu du (e icpt + e − icpt ) e − ipx dp 

∫− ∞ ∫− ∞

  f (u) e − iαudu (e − icαt + e icαt ) e iαx dα ,  

∞

∫− ∞ ∫− ∞

putting p = − α so that dp = − dα 1 1 =  2 2π

∞

 f (u) e − iαu  

∫− ∞

+ =

∞

∫− ∞

1 2π

 e iα ( x + ct)dα du 

∞

 f (u) e − iαu  

∫− ∞

∞

∫− ∞

  e iα ( x − ct) dα du  

1 [ f ( x + ct) + f ( x − ct)] . 2 (From Fourier integral formula, see article 4.3)

Example 5: If the flow of heat is linear so that the variation of θ (temperature) with z and y may

be neglected and if it is assumed that no heat is generated in the medium, then solve the differential ∂θ ∂2θ = k 2 (one dimensional heat equation) where − ∞ < x < ∞ and θ = f ( x) where ∂t ∂x t = 0, f ( x) being a given function of x. equation

Solution: Taking the Fourier transform of both the sides of the given equation, we have

1 √ (2 π)

∞

d 1 dt √ (2 π)

or

∂2θ

∞

∂θ ipx 1 e dx = k ∂t √ (2 π)

∫− ∞

∫ − ∞ ∂x2 e ~

∞

θ ⋅ e ipx dx = k (− ip)2 θ

∫− ∞

~

ipx

dx [From article 4.20 (b)]

~

where θ = θ ( p, t) is the Fourier transform of θ ( x, t) ~ d ~ θ = − kp2 θ dt

or

whose solution is ~

2

θ = A e − kp

t

…(1)

Now at t = 0 , θ = f ( x) . ~

1 √ (2 π)

~

∞

f ( x) e ipx dx = f ( p) .

∫− ∞

∴

at

∴

from (1), at t = 0, θ = f ( p) = A ;

t = 0, θ =

~

~

~

~

2

hence from (1), θ = f ( p) e − kp t . Taking the inverse Fourier transform, we have ∞ ~ 2 1 θ ( x, t) = f ( p) e − kp t e − ipx dp √ (2 π) − ∞

∫

or

θ ( x, t) =

1 √ (2 π)

∞

∫− ∞

~

2

f ( p) e − kp

t − ipx

dp.

T-157 Example 6: Solve

∂4V

+

4

∂x

∂2V

= 0 , − ∞ < x < ∞, y ≥ 0

∂ y2

satisfying the conditions (i) V and its partial derivatives tend to zero as x → ± ∞ and (ii) V = f ( x),

∂V = 0 on y = 0 . ∂y

Solution: Taking the Fourier transform of both the sides, we have

1 √ (2 π)

∞

∂4V

∫−∞ ∂x4

e ipx dx +

1 √ (2 π)

∞

∫−∞

∞

or

ip 1  ∂3V ipx   ⋅ e  − 2 π) √ (2 π)  ∂x3 √ (  −∞

∞

or

∞  2   ∂ V e ipx  − − ip  √ (2 π)  ∂x2 −∞ 

∞

e ipx dx = 0

∂ y2 ∂3V

∫ −∞ ∂x3 +

ip

∂2V

e ipx dx

d2

1 dy √ (2 π) 2

∂2V

∫ −∞ ∂x2

e

ipx

or

∞

or

or

(ip)4 √ (2 π)

or

(

4

~

∂x3

→ 0 as x → ± ∞

∂2V ∂x2

→ 0 as x → ± ∞

~

∞

)− ∞ − ip∫ −∞ V e

ipx

 d2 V dx + =0 dy2 

~

∞

∫ −∞ V e

p V +

∂3V

∂V ipx  d2 V e dx + = 0, ∂x dy2  since

∞

dx = 0 .

~

∫ −∞

(ip)3  Ve ipx − √ (2 π) 

ipx

~  2 d V =0 , dx +  dy2 

since ∞ (ip)2  ∂V ipx  e  − ip  −∞ √ (2 π)  ∂x 

∞

∫ −∞ V ⋅ e

ipx

dx +

d2 V dy2

=0

~

d2 V dy2

=0

whose solution is ~

V = A cos p2 y + B sin p2 y. Since on

y = 0, V = f ( x) and

∂V = 0, ∂y

…(1)

T-158

∴ on

taking Fourier transform, we have ∞ ~ ~ 1 y = 0, V = f ( x) e ipx dx = f ( p) √ (2 π) −∞

∫

1 √ (2 π)

and ∴

~

∞

∫ −∞

∂V ipx dV e dx = 0 i. e., =0. ∂y dy

from (1), we have ~

f ( p) = A ~

dV = 0 = Bp2 dy

and ∴

or

B=0.

the solution (1) reduces to ~

~

V = f ( p) cos p2 y. ∴

applying inversion theorem for Fourier transform, we have ∞ ~ 1 V = f ( p) cos ( p2 y) ⋅ e − ipx dp. √ (2 π) −∞

∫

Example 7: Use a cosine transform to show that the steady temperature in the semi-infinite solid

y > 0 when the temperature on the surface y = 0 is kept at unity over the strip| x | < a and at zero outside the strip, is 1  −1 tan π The result

∞

∫0

a +   y

a − x  + tan−1    y

e − sx x −1 sin rx dx = tan−1

x  ⋅ 

r , r > 0 , s > 0 , may be assumed. s

Solution: Here the steady temperature U ( x, y) in the semi-infinite solid is governed by

two-dimensional Laplace equation ∂2U

+

2

∂x

∂2U ∂ y2

…(1)

=0

0 < y < ∞, − ∞ < x < ∞ subject to the conditions and

U = 1, y = 0

− a< x< a

U = 0, y = 0

when | x | < | a.

Taking Fourier cosine transform of both sides of (1), we have  2    π

∞ ∂2U

∫0

2

∂x

 2 cos px dx +    π ∞

or

  2   ∂U  2 ⋅ cos px + p       π   ∂x  π 0

∞

∫0 ∞

∫0

∂2U ∂ y2

cos px dx = 0 ~

sin px dx +

d2 U c dy2

=0

∂U  ∂U  → 0 as x → ∞ and → 0 as x → 0 . By symmetry   ∂x  ∂x

T-159 ∞  2  2 p   [U sin px)] 0 − p2    π  π

or

~

d2 U c

or

dy

∞

∫0

~

U cos px dx +

~

− p2 U c = 0

2

d2 U c dy2

=0

(if U → 0 as x → ∞)

~

whose solution is U c = Ae py + Be − py.

…(2)

~

Now as y → ∞, U c → 0 . ∴ A = 0 . ~

U c = Be − py.

∴

…(3)

Also when y = 0 , a

~  2 Uc =    π

∫0 1⋅ cos px dx +

 2    π

∞

∫0

0 ⋅ cos px dx

a

∴

=

 2   sin px     π  p  0

=

 2  sin pa ⋅    π p

 2  sin pa from (3),   = B.  π p

~  2  sin pa − py Hence U c =   ⋅e .  π p

Taking the inverse Fourier cosine transform, we have ∞

 2 U ( x, y) =    π

∫0

∞ e − py

 2  sin pa − py e cos px dp    π p

=

1 π

=

1 π

=

 a − x 1 −1  a + x   + tan−1   tan  π  y   y 

∫0

p

∞

∫0

assuming the result

e − py ⋅ p−1 sin (a + x) p dp +

∞

∫0

⋅ [sin (a + x) p + sin (a − x) p] dp

e − sx x −1 sin rx dx = tan−1

1 π

∞

∫0

e − py ⋅ p−1 sin (a − x) p dp

r ,r>0 ,s>0 . s

6.3 Application of Finite Fourier Transforms In the problems in which the range of one of the variables is finite, finite Fourier transforms are applied.

T-160

6.4 Finite Fourier Transforms of Partial Derivatives (a) The finite Fourier sine and cosine transforms of

∂U where U is a function of x and t for ∂x

0 < x < l, t > 0 . We have  ∂U  Fs  =  ∂x 

l

∂U

∫0 ∂x sin

pπx dx l

l

pπx  pπ  = U ( x, t) sin −  l  0 l or

pπ  ∂U  Fs  Fc {U } =− l  ∂x 

and

 ∂U  Fc  =−  ∂x 

l

∂u

∫0 ∂x cos

l

∫0 U cos

pπx dx l l

pπx  pπ  + = U ( x, t) cos   l 0 l or

(Integrating by parts)

pπx dx l

l

∫0 U sin

pπx dx l

 ∂U  pπ Fc  Fs {U } − {U (0 , t) − U (l, t) cos pπ} . = l  ∂x 

(b) The finite Fourier sine and cosine transforms of

∂2U ∂x2

where U is a function of x and t, for

0 < x < l, t > 0 .  ∂2U  Fs  2  =  ∂x 

l ∂2U

∫0 ∂x2

l

sin

pπx  pπ pπx  ∂U sin − dx =  l  0 l l  ∂x

l  pπx  pπ  + U cos   l l  0

l

l

pπ l

=−

pπ  pπ  Fs {U } − {U (0 , t) − U (l, t)cos pπ}   l  l

∫0

or

 ∂2U  p2 π2 pπ Fs  2  = − Fs (U ) + {U (0 , t) − U (l, t) cos pπ} l l  ∂x 

and

 ∂2U  Fc  2  =  ∂x  =

l ∂2U

∫0 ∂x2

l

cos

pπx dx l

pπx  dx l 

=−

U ⋅ sin

∂U

∫0 ∂x cos

pπx pπx  pπ  ∂U dx =  + cos  l l 0 l  ∂x

pπ  ∂U  Fs   − {U x (0 , t) − U x (l, t) cos pπ} l  ∂x 

l

∂U

∫0 ∂x sin

pπx dx l

T-161

or

 ∂2U  p2 π2 Fc  2  = − 2 Fc {U } − {U x (0 , t) − U x (l, t) cos pπ} l  ∂x 

where U x denotes the partial derivative w.r.t. ‘x’.

6.5 Choice of Finite Sine or Cosine Transform The choice of finite sine or cosine transform is decided by the form of the boundary conditions. From the last article we see that for exclusion of

∂2U ∂x2

from a differential equation, we

require U (0 , t) and U (l, t) in finite sine transform and

U x (0 , t) and U x (l, t) in finite cosine transform.

Example 8: Use finite Fourier transforms to solve

∂U ∂2U = 2, ∂t ∂x where

U (0 , t) = 0 , U (4, t) = 0 , U ( x, 0 ) = 2 x, interpretation of the problem.

Give

0 < x < 4, t > 0 .

the

physical

(Meerut 2013; Rohilkhand 14)

Solution: Taking the finite Fourier sine transform (with l = 4) of both sides of the given

partial differential equation, we have 4

pπx ∂U sin dx = 4 ∂t

∫0

4

∫0

∂2U 2

∂x

sin

pπx dx 4

~

p2 π2 ~ pπ dUs = − 2 Us+ [U (0 , t) − U (4, t) cos pπ] 4 dt 4

or

[See article 6.4 (b)

~

where U s is finite Fourier sine transform of U] ~

p2 π2 ~ dUs =− Us 16 dt

or

~

2 2

whose solution is U s = Ae − p

π t /16

…(1)

.

Since U ( x, 0 ) = 2 x, where 0 < x < 4 , taking finite Fourier sine transforms, we have ~

at t = 0 , U s =

4

∫0 2 x ⋅ sin

pπx dx 4 4

 pπx pπx  4 4 4 = − 2 x ⋅ +2⋅ ⋅ cos sin 4 4 0 p π p π p π 

T-162

32 (− cos pπ) 32 { − (−1)p } = ⋅ pπ pπ

= ∴

from (1),

32 (− 1)p +1 = A. pπ 32 (− 1)p +1 − tp2 π 2 /16 ⋅e . pπ

~

Hence

Us =

Taking the inverse finite Fourier sine transform, we have U ( x, t) =

=

2 4

∞

∑ p =1 ∞

16 π

∑ p =1

32 (− 1)p + 1 − p2 π 2 t /16 pπx ⋅e ⋅ sin pπ 4 (− 1)p + 1 − p2 π 2 t /16 pπx ⋅e ⋅ sin , p 4

which is the required solution. Physical Interpretation: Physically U ( x, t) represents the temperature at any point x at any time t in solid bounded by the planes x = 0 and x = 4 (or a bar on the x-axis with the ends x = 0 and x = 4, whose surface is insulated laterally). The conditions U (0 , t) = 0 and U (4, t) = 0 imply that the ends are kept at zero temperature while U ( x, 0 ) = 2 x implies that the initial temperature is a function of x. Example 9: Use the finite cosine transform to solve

∂V ∂2V = k 2 (one dimensional heat equation) ∂t ∂x ∂V = 0, when x = 0 and x = π, t > 0 and the initial condition ∂x (Kanpur 2014) V = f ( x) , when t = 0 , 0 < x < π . with the boundary conditions

Solution: Taking the finite Fourier cosine transform (with l = π) of both the sides of the

given equation, we have π

∫0

∂V cos px dx = k ∂t

π

∫0

∂2V ∂x2

cos px dx

~

~ dV c = k [− p2 V c − {V x (0 , t) − V x (π, t) cos pπ}], dt

or ~

where V c is finite Fourier cosine transform of V

[See article 6.4 (b)]

~

or

~ dV c = − k p2 V c dt

whose solution is ~

2

V c = A e − kp t . Now taking finite Fourier cosine transform of initial condition, we have

…(1)

T-163 ~

at ∴

π

∫0

t = 0, V c = from (1), A = ~

2

Hence V c = e − kp

t

π

∫0 π

∫0

f ( y) cos py dy . f ( x) cos px dx .

f ( y) cos py dy .

Taking the inverse finite Fourier cosine transform, we have V ( x, t) =

=

1 ~ 2 V c (0 ) + π π 1 π

π

∫0

∞

~

∑ V c ( p) cos px p =1

f ( y) dy +

2 π

∞



π

2

∑= e−kp t cos px. ∫0

p 1

 f ( y) cos py dy ⋅ 

∂U ∂2U = 2 , 0 < x < 6, t > 0 ∂t ∂x

Example 10: Solve

subject to the conditions 1, 0 < x < 3 U (0 , t) = 0 , U (6, t) = 0 , U ( x, 0 ) =  0 , 3 < x < 6 and interpret physically.

(Kanpur 2011)

Solution: Taking the finite Fourier sine transform (l = 6) of both the sides of the given

partial differential equation, we have 6

∫0

pπx ∂U dx = ⋅ sin 6 ∂t

6

∫0

∂2U 2

∂x

sin

pπx dx 6

~

or

p2 π2 ~ pπ dUs = − 2 Us + [U (0 , t) − U (6, t) cos pπ] 6 dt 6

or

p2 π2 ~ dUs =− U s, 36 dt

~

whose solution is ~

2 2

U s = Ae − p Now at

∴

at

π t /36

.

…(1)

 1, 0 < x < 3 t = 0, U =  0 , 3 < x < 6. ~

t = 0, U s = =

6

pπx dx 6

3

pπx dx + 6

∫0 U ⋅ sin ∫0 1⋅ sin

6

∫3 0 ⋅ sin

pπx pπ  6  dx = 1 − cos .  6 2 pπ

T-164

∴

pπ  6  1 − cos  = A. pπ  2

from (1), when t = 0 ,

Hence

~

pπ  − p2 π 2 t /36 6  . 1 − cos  e pπ  2

Us =

Taking the inverse Fourier sine transform, we have U ( x, t) =

=

2 ⋅ 6 2 π

∞

∑ p =1 ∞

∑ p =1

pπ  − p2 π 2 t /36 pπx 6  sin ⋅ 1 − cos  e pπ  2 6 pπ  − p2 π 2 t /36 pπx 1 ⋅ sin ⋅ 1 − cos  e   p 2 6

Physical Interpretation: Physically U ( x, t) represents the temperature at any point x at any time t in a bar with the ends x = 0 and x = 6 kept at zero temperature which is insulated laterally. Initially the temperature in the half bar from x = 0 to x = 3 is constant equal to 1 unit while the half bar from x = 3 to x = 6 is at zero temperature. Example 11: Solve the boundary value problem

∂U ∂2U = 2 , U (0 , t) = 1, U (π, t) = 3, U ( x, 0 ) = 1 , ∂t ∂x where 0 < x < π , t > 0 . Give the physical interpretation of the problem. Solution: Taking the finite Fourier sine transform (l = π) of both the sides of the given

differential equation, we have π

∫0

∂U sin px dx = ∂t

π

∫0

∂2U ∂x2

⋅ sin px dx

~

or

~ dUs = − p2 U s + p [U (0 , t) − U (π, t) cos pπ] dt

or

~ dUs = − p2 U s + p [1 − 3 cos pπ] dt

or

~ dUs + p2 U s = p (1 − 3 cos pπ) dt

~

~

…(1)

which is a linear diff. equation of first order. I. F. = e ∫ ∴

p2 dt

2

= ep t .

solution of (1) is ~

2

U s⋅ ep

t

= A+

∫ p (1 − 3 cos pπ) e

= A+

(1 − 3 cos pπ) p2 t e p

p2 t

dt

T-165 ~

2

U s = Ae − p

or

t

+

(1 − 3 cos pπ) ⋅ p

…(2)

Now when t = 0 , U = 1 . Taking finite Fourier sine transform ~

Us = ∴

At

π

∫0

1 ⋅ sin px dx =

t = 0, from (2),

or

A=

1 ⋅ (1 − cos pπ) ⋅ p

1 ⋅ (1 − cos pπ) 1 − 3 cos pπ = A+ p p

2 cos pπ ⋅ p

Hence from (2), ~

Us =

2 cos pπ − p2 t (1 − 3 cos pπ) e + ⋅ p p

Taking the inverse infinite Fourier sine transform, we have U ( x, t) =

2 π

∞

2 cos pπ − p2 t ⋅e sin px p

∑ p =1

+

Now

= and

π

∫0

Fs {1} =

2 π

π

∫0

=− ∴

1=

2 π

and

x=

2 π

∑ p =1

1 − cos pπ p

x ⋅ sin px dx π

∞

(1 − 3 cos pπ) ⋅ sin px p

1 − (− 1)p ,( p = 1, 2,...) p

Fs { x} =

2 π

∑ p =1

1 ⋅ sin px dx =

 x  =  − cos px + p  0

∴

∞

π

∫0

cos px dx p

π π cos pπ = (− 1)p + 1. p p ∞

∑ p =1 ∞

∑ p =1

1 − (− 1)p sin px p π (− 1)p +1 sin px = 2 p

(1 − 3 cos pπ) 2 sin px = p π

∞

∑

p =1

∞

∑

p =1

(− 1)p +1 sin px. p

1 − 3 (− 1)p sin px p

…(3)

T-166 ∞

2 π

=

∑ p =1

= 1+

1 − (− 1)p 4 sin px + p π

∞

∑ p =1

(− 1)p +1 sin px p

2 x. π

Hence from (3), U ( x, t) =

4 π

∞

∑ p=

1

cos pπ − p2 t 2x e sin px + 1 + ⋅ p π

Example 12: Using finite Fourier transform, find the solution of the wave equation

∂2U ∂t2

=4

∂2U ∂x2

,

subject to the conditions U (0 , t) = 0 , U (π, t) = 0 , U ( x, 0 ) = (0 ⋅1) sin x + (0 ⋅ 01) sin 4 x and

U t ( x, 0 ) = 0 for 0 < x < π, t > 0 .

(Meerut 2007)

Solution: Taking the finite Fourier sine transform of both the sides of the given

equation, we have ∂2U

π

∫0

2

∂t

sin px dx = 4

~

d2 U s

or

dt2

or

dt2

∂2U ∂x2

sin px dx

~

= − 4 p2 U s + 4 p [U (0 , t) − U (π, t) cos pπ]

~

d2 U s

π

∫0

~

+ 4 p2 U s = 0

whose solution is ~

…(1)

U s = A cos 2 pt + B sin 2 pt. Now at

t = 0 , U = (0 ⋅ 1) sin x + (0 ⋅ 01) sin 4 x

and

Ut =

∂U =0. ∂t

Taking finite sine transforms, we have At

~

t = 0, U s =

π

∫0 {(0 ⋅ 1) sin x + (0 ⋅ 01) sin 4 x} sin px dx

= (0 ⋅ 1)

π

π

∫0 sin x ⋅ sin px dx + (0 ⋅ 01) ∫0 sin 4 x sin px dx

~

and ∴

dUs =0. dt from (1), we have (0 ⋅1)

π

π

∫0 sin x sin px dx + (0 ⋅ 01)∫0 sin 4 x sin px dx = A

T-167 ~

dUs = 0 = 2 Bp dt

or

or

B=0.

Hence from (1), π π ~   U s = (0 ⋅ 1) sin x sin px dx + (0 ⋅ 01) sin 4 x sin px dx cos 2 pt. 0 0   Taking the inverse finite Fourier sine transform, we have

∫

U ( x, t) =

=

2 π

∞

∫

~

U s ⋅ sin px ∑ p =1

2 (0 ⋅ 1) π

∞



+

=

π

∫ ∑ 0 p =1 

 2 (0 ⋅ 1)  π 

π

∫0

 sin π sin px dx cos 2 pt sin px 

2 (0 ⋅ 01) π

∞



π

∑=1  ∫0

p

 sin 4 x sin px dx cos 2 pt sin px 

 sin x ⋅ sin x dx cos 2 t sin x  +

 2 (0 ⋅ 01)  π 

π

∫0

 sin 4 x ⋅ sin 4 x dx cos 8 t sin 4 x 

(Since the integrals in first summation are all zero if p ≠ 1 and the integrals in second summation are all zero if p ≠ 4) or

U ( x, t) = (0 ⋅ 1) cos 2 t sin x + (0 ⋅ 01) cos 8 t sin 4 x.

Comprehensive Exercise 1 1.

Solve (i) (iii)

2.

∂U ∂2U = 2 , x > 0 , t > 0 subject to the conditions ∂t ∂x  1, 0 < x < 1 when t = 0 (ii) U =  U = 0 , when x = 0 , t > 0 0 , x ≥ 1 U ( x, t) is bounded.

If the function U ( x, y) is determined by the differential equation ∂U ∂2U for x ≥ 0, − ∞ < y < ∞ and U = f ( y) when x = 0, = ∂x ∂ y2 2 1 ∞ ~ show that U ( x, y) = f ( p) e − p x − ipydp 2π − ∞ ~

∫

where f ( p) is the Fourier transform of f ( y). [Hint: Replace k → 1, θ → U , t → x and x → y in Ex. 5]

T-168

3. Use finite Fourier transform to solve ∂V ∂2V = 2 , 0 < x < 6, t > 0 , V x (0 , t) = 0 , V x (6, t) = 0 , V ( x, 0 ) = 2 x . ∂t ∂x (Kanpur 2009) 4. Solve the boundary value problem

∂2U ∂t2

=9

∂2U ∂x2

subject to the conditions

U (0 , t) = 0 , U (2, t) = 0 , U ( x, 0 ) = (0 ⋅ 05) x (2 − x) and U t ( x, 0 ) = 0 , where 0 < x < 2, t > 0 . Find the steady temperature V ( x, y) in a long square bar of side π when one face is kept at constant and the other faces at zero temperature. Also V ( x, y) is bounded. Or Determine a function V ( x, y) which is harmonic in the open square, 0 < x < π, 0 < y < π, takes a constant value V0 on the edge y = π and vanishes on the other edges of the square. 6. Determine the displacement U ( x, t) in a horizontal string stretched from origin to the point (π, 0 ) when the motion is due to the weight of the string alone. The string may be taken to be initially at rest in the position U = 0 .

5.

[Hint: Here displacement is governed by ∂2U

= c2

∂2U

+ g, 0 < x < π, t > 0 ∂x2 ∂U Conditions U = 0 = , at t = 0 and U = 0, when x = 0 or π.] ∂t ∂t2

A nswers 1 1. U ( x, t) =

∞1−

2 π

∫0

3. V ( x, t) = 6 +

4. U ( x, t) =

π

U ( x, t) =

3

4V0 π

2g πc

2

∞

 cos pπ − 1 − p2 π 2 t /36 pπx  e cos 2  6 p  p =1

 π2 ∑ 

(1 ⋅ 6)

5. V ( x, y) =

6.

24

cos p − p2 t e sin px dp p

∞

1

cos ∑ (2 n − 1)3 n =1 ∞

∑ n=0 ∞

(2 n − 1) πx 3 (2 n − 1) πt sin 2 2

sinh (2 n + 1) y sin (2 n + 1) x (2 n + 1) sinh (2 n + 1) π

1

[1 − (− 1)p ] (1 − cos pat) sin px . ∑ 3 p p =1

¨

T-169

7 F ourier S eries

7.1 Some Important Results

T (i)

(ii)

he following results are useful in Fourier Series : sin nπ = 0 , cos nπ = (−1)n, 1 sin  n +  π = (− 1)n, cos  n +   2

1  π = 0 , where n ∈ I . 2

∫ uv = uv1 − u′ v3 + u′ ′ v2 − u′ ′ ′ v4 + … , where u′ = 2π

(iii)

∫0

(v)

∫0

(vii)

∫0

(ix)

∫0

2π

2π 2π

d2 u du , u′ ′ = 2 , … and v1 = dx dx

∫ v dx ,

2π

sin nx dx = 0 .

(iv)

∫0

sin2 nx dx = π.

(vi)

∫0

sin nx . sin mx dx = 0 .

(viii) ∫

sin nx . cos mx dx = 0 .

(x) ∫

2π

2π 0

cos nx dx = 0 . cos2 nx dx = π.

2π 0

v2 = ∫ v1 dx, … .

cos nx . cos mx dx = 0 .

sin nx . cos nx dx = 0 .

T-170

7.2 Determination of Fourier Coefficients (Euler’s Formulae) The Fourier series is f ( x) =

a0 + a1 cos x + a2 cos 2 x + … + an cos nx + … 2 + b1 sin x + b2 sin 2 x + … + bn sin nx + …

…(1)

To find a0 . Integrating both sides of (1) from x = 0 to x = 2π, we have 2π 2π 2π 2π a ∫ 0 f ( x) dx = 20 ∫ 0 dx + a1 ∫ 0 cos x dx + a2 ∫ 0 cos 2 x dx + … + an ∫

2π 0

cos nx dx + … + b1

2π

∫0

sin x dx + b2

2π

∫0

+ … + bn ∫ a0 2 π dx, 2 ∫0 2π a ∫ 0 f ( x) dx = 20 2π. 1 2π a0 = ∫ f ( x) dx. π 0 =

or ∴

2π 0

sin 2 x dx sin nx dx + …

[Other integrals vanish]

.…(2)

To find an. Multiplying each side of (1) by cos nx and integrating from x = 0 to x = 2π, we have 2π 2π 2π a ∫ 0 f ( x) cos nx dx = 20 ∫ 0 cos nx dx + a1 ∫ 0 cos x cos nx dx + … + an ∫

2π 0

= an ∫ ∴

cos2 nx dx … + b1

2π 0

2π

∫0

sin xnx dx + b2

2π

∫0

cos2 nx dx

sin 2 x cos nx dx + …

[Other integrals vanish]

= an π. 1 2π an = ∫ f ( x) cos nx dx. π 0

…(3)

By taking n = 1, 2, … we get the values of a1, a2 , … . To find bn. Multiplying each side of (1) by sin nx and integrating from x = 0 to x = 2π, we have 2π 2π 2π a ∫ 0 f ( x) sin nx dx = 20 ∫ 0 sin nx dx + a1 ∫ 0 cos x sin nx dx + … + an ∫ + b1 = bn ∫

2π 0

= bn π.

sin2 nx dx

2π

∫0

2π 0

cos nx sin nx dx + …

sin x sin nx dx + … + bn ∫

2π 0

sin2 nx dx + ...

[Other integrals vanish]

T-171

bn =

∴

1 π

2π

∫0

…(4)

f ( x) sin nx dx.

By taking n = 1, 2 , … we get the values of b1, b2 , …… . 1 Note. To get similar formula of a0 , has been written with a0 in Fourier series. 2 ∞

If we write the Fourier series as f ( x) = a0 + Σ [an cos nx + bn sin nx], n =1

then the values of constants is given as : 1 1 2π 1 2π a0 = f ( x) dx ; an = ∫ f ( x) cos nx dx and bn = ∫ 0 0 π 2π π

2π

∫0

f ( x) sin nx dx .

7.3 Fourier Series Expansion In The Interval α < x < α + 2 π The Fourier series is ∞ ∞ a f ( x) = 0 + Σ an cos nx + Σ bn sin nx. n =1 2 n =1

…(1)

To find a0 . Integrating both sides of (1) from x = α to x = α + 2 π, we have 2π + α  ∞ 2π + α 2π + α  a f ( x) dx = 0 ∫ dx + ∫  Σ an cos nx dx ∫α α α 2  n =1  +

2π + α

∫α

 ∞   Σ bn sin nx dx n = 1  

a0 ⋅ 2π + 0 + 0. 2 1 2π + α a0 = ∫ f ( x) dx . π α =

∴

To find an. Multiplying both sides of (1) by cos nx and then integrating from x = α to x = α + 2 π , we have 2π + α 2π + α  ∞ 2π + α  α cos nx dx + ∫ f ( x) cos nx dx = 0 ∫  Σ an cos nx cos nx dx ∫α α   n =1 2 α +

∴ Similarly

2π + α

∫α

  ∞  Σ bn sin nx cos nx dx   n =1

= 0 + π an + 0 . 1 2π + α an = ∫ f ( x) cos nx dx . π α 1 2π + α bn = ∫ f ( x) sin nx dx. π α

Note 1.

Putting α = 0, the interval becomes 0 < x < 2 π and 1 1 2π 1 2π a0 = ∫ f ( x) dx, an = ∫ f ( x) cos nx dx and bn = 0 0 π π π Putting α = − π, the interval becomes − π < x < π and 1 1 π 1 π a0 = ∫ f ( x) dx, an = ∫ f ( x) cos nx dx and bn = − π − π π π π

2π

∫0

f ( x) sin nx dx.

Note 2.

π

∫−π

f ( x) sin nx dx.

T-172

7.4 Fourier Series For Discontinuous Functions  f1 ( x), α < x < x0 Let the function f ( x) be defined by f ( x) =   f2 ( x), x0 < x < α + 2 π, in the interval (α, α + 2 π). Here x0 is the point of discontinuity. The Fourier series for f ( x) in such cases is obtained in the usual way. The values of a0 , an , bn are given by α + 2π 1  x0  a0 = f1 ( x) dx + ∫ f2 ( x) dx ; ∫   α x π 0 α + 2π 1  x0  an = f1 ( x) cos nx dx + ∫ f2 ( x) cos nx dx ;  x0 π ∫ α bn =

1  x0 f1 ( x) sin nx dx + π ∫ α

α + 2π

∫ x0

 f2 ( x) sin n x dx . 

At the point of discontinuity, x = α, the Fourier series gives the value of f ( x) as the arithmetic mean of left and right limits. 1 At x = α, f ( x) = [ f (α − 0 ) + f (α + 0 ) ]. ∴ 2

Find a Fourier series to represent x − x2 from x = − π to x = π .

Example 1:

Deduce that Solution:

1 12

−

1 22

+

1 32

−

∞

π2 ⋅ 12 ∞

n =1

a0 =

1 2π

1 an = π

π

∫− π π

∫− π

an cos nx + Σ

n =1

( x − x2 ) dx =

1 2π

bn sin nx.  x2 x3  −   3 2

π

=− −π

π2 ; 3

( x − x2 ) cos nx dx

π sin nx 1  cos nx   sin nx   − (1 − 2 x)  − + ( − 2 ) − ( x − x2 )      π n n2  n3   − π  − 4 (−1)n = ; [ ∵ cos nπ = (−1)n ] n2 1 π ( x − x2 ) sin nx dx bn = ∫ π −π π 1  cos nx    cos n x   sin nx  = ( x − x2 )  − + ( − 2 )    − (1 − 2 x)  −   n3   − π   π n  n2   = − 2 (−1)n / n.

=

and

42

+…=

The Fourier series for f ( x) in (− π, π) is f ( x) = a0 + Σ

Here

1

T-173

∴

The required Fourier series is π2 cos x cos 2 x cos 3 x cos 4 x  x − x2 = − +4 − + − +… 2 2 2  12  3 2 3 4 sin x sin 2 x sin 3 x sin 4 x  +2 − + − + … ⋅ …(1)  1  2 3 4

Deduction:

Putting x = 0 in (1), we get π2 1 1 1 1 0=− + 4  2 − 2 + 2 − 2 + … 1  3 2 3 4

or

1 2

1

−

1 2

2

+

1 2

3

−

1 2

4

+…=

Example 2:

Find the Fourier series expansion for f ( x), if − π , − π < x < 0 1 1 1 π2 Deduce that 2 + 2 + 2 + … = f ( x) =  ⋅ 8 1 3 5  x, 0< x< π.

Solution:

The Fourier series of f ( x) in (− π, π) is ∞

f ( x) = a0 + Then

and

∴

n =1

an cos nx +

∞

Σ

n =1

bn sin nx.

π 1  0 (− π) dx + ∫ x dx  0 2 π ∫ − π  π  x2   1   1 π2   π 0 = − π ( x )− π +    = −  π2 +  = − ; π 2   2  2  0  π     π 1 π 1 0 (− π) cos nx dx + ∫ x cos nx dx an = ∫ f ( x) cos nx dx = ∫  0 π −π π  − π 0 π 1  sin nx   x sin nx cos nx   =  −π + +     n −π  π n n2  0   1 1 1 1 = 0 + 2 cos nπ − 2  = (cos nπ − 1) ;  π  n n  π n2

a0 =

1 2π

Σ

π2 ⋅ 12

1 π

π

∫−π

f ( x) dx =

1 0 (− π) sin nx dx + π ∫ − π 0 π 1   π cos nx   cos nx sin nx   =   + − +    −π  π  n n n2  0   1 π π 1 =  (0 − cos nπ) − cos nπ = (1 − 2 cos nπ).  n π  n n The required Fourier series is cos 3 x cos 5 x π 2  f ( x) = − − cos x + + + …  4 π 32 52 bn =

π

∫−π

f ( x) sin nx dx =

+ 3 sin x − 

π

∫0

x sin nx dx 

2 sin 2 x 3 sin 3 x sin 4 x + − + ...  2 3 4

Putting x = 0 in (1), we get π 2 1 1 f (0 ) = − − 1 + 2 + 2 + … ∞ ⋅  4 π 3 5 But f ( x) is discontinuous at x = 0, and we have f (0 − 0 ) = − π 1 ∴ f (0 ) = [ f (0 − 0 ) + f (0 + 0 )] = − (π / 2) 2

…(1)

Deduction:

…(2) and

f (0 + 0 ) = 0 . …(3)

T-174

Hence from (2) and (3), we have π π 2 1 1 1 1 1 1 π2 − = − −  2 + 2 + 2 + … or + + + … = ⋅  2 4 π 1 8 3 5 12 32 52  π − x Example 3: Obtain the Fourier series of f ( x) =   in the interval (0 , 2π) and hence deduce  2  π 1 1 1 = 1− + − + … 4 3 5 7 Solution:

The Fourier series for f ( x) in (0 , 2π) is ∞

∞

f ( x) = a0 + Σ an cos nx + Σ bn sin nx n =1

1 a0 = π

Here

2π

∫0

n =1

1 f ( x) dx = π

2π

∫0

π −   2

x  dx 

2π

1  x2  1 = = [2 π2 − 2 π2 ] = 0 ; πx −  2π  2 2π 0 1 2π 1 2π an = ∫ f ( x) cos nx dx = (π − x) cos nx dx π 0 2 π ∫0 2π

sin nx   − cos nx  = 0; (π − x) n − (− 1)  n2   0 1 2π 1 2π bn = ∫ f ( x) sin nx dx = (π − x) sin nx dx 2 π ∫0 π 0 =

1 2π

=

1 2π

2π

cos nx sin nx  1  − (π − x) − = ⋅ 2   n n n 0

∞ 1 π−x 1 1 =0 +0 + Σ sin nx = sin x + sin 2 x + sin 3 x + … n =1 n 2 2 3 π π 1 1 1 Deduction: Putting x = in (1), we get = 1 − + − + … 2 4 3 5 7

∴

f ( x) =

Comprehensive Exercise 1 1.

Find a series of sines and cosines of multiples of x which will represent π e x in the interval − π < x < π. 2 sinh π

2.

Find the Fourier series of the function defined as 0≤ x≤ π  x+π ; and f ( x + 2π) = f ( x). f ( x) =  − x − π ; − π≤ x 1. 1 − n2 =

When n = 1, we have π

1 2

u sin u cos u du =

∫0

x sin x = 1 +

∴

π

∫0

u sin 2 u du =

2 π  − cos x + π 4

∞

Σ

1 2

π

π  u cos 2 u sin 2 u − + =− ⋅  4 2 4  0

π cos nπ

n=2

2

1− n

 cos nx 

 1  1 1 1 = 1 + 2 − cos x − cos 2 x + cos 3 x − cos 4 x + … 4 1 3 2 4 . . 3 . 5   1 Deduction: Putting x = π, we get 2  1  π π 1 1 1 1 1 1 = + − + −… = 1+ 2  − + − … or 2 4 2 1. 3 3 . 5 5 . 7 1 . 3 3 ⋅ 5 5 . 7  π −2 1 1 1 or = − + − …. 4 1. 3 3 . 5 5 . 7 Example 7: Obtain Fourier Series of the function

 x, f ( x) =   − x,

− π< x0 0 > x > − π.

Thus f ( x) is an even fucntion in (− π , π) and so bn = 0. The Fourier series in this case is f ( x) = a0 + 1 Here a0 = π an =

2 π

π

∫0

π

∫0

1 f ( x) dx = π

∫0

2 π

∫0

f ( x) cos nx dx =

π

π

∞

Σ

n =1

an cos n x. π

1  x2  1  π2  π − x dx = −   =−  =− ; π 2  π 2  2 0 π

− x cos nx dx = −

2  sin nx cos nx  x + n π  n2  0

T-181

n is even 0 , 2  (− 1)n 1  2 [ 1 − (− 1)n ] =   2 − 2 = 2 2 π n n  πn 4 / πn , n is odd ⋅ π 4 cos x cos 3 x cos 5 x  The required series is f ( x) = − + + + +...  2 π  12 32 52 =−

∴

Example 8: A periodic function of period 4 is defined as

f ( x) = | x |, − 2 < x < 2. Obtain its Fourier series expansion. Solution: We have

| x |, − 2 < x < 2  0< x