Fundamentals of differential equations [Eighth edition /,Pearson new international edition] 1292023821, 9781292023823

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9 781292 023823

Fundamentals: Differential Equations Nagle et al. 8e

ISBN 978-1-29202-382-3

Fundamentals of Differential Equations Nagle Saff Snider Eighth Edition

Fundamentals of Differential Equations Nagle Saff Snider Eighth Edition

Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners.

ISBN 10: 1-292-02382-1 ISBN 13: 978-1-292-02382-3

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

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Table of Contents 1. Introduction R. Kent Nagle/Edward B. Saff/Arthur David Snider

1

2. First-Order Differential Equations R. Kent Nagle/Edward B. Saff/Arthur David Snider

37

3. Mathematical Models and Numerical Methods Involving First-Order Equations R. Kent Nagle/Edward B. Saff/Arthur David Snider

93

4. Linear Second-Order Equations R. Kent Nagle/Edward B. Saff/Arthur David Snider

159

5. Introduction to Systems and Phase Plane Analysis R. Kent Nagle/Edward B. Saff/Arthur David Snider

253

6. Theory of Higher-Order Linear Differential Equations R. Kent Nagle/Edward B. Saff/Arthur David Snider

335

7. Laplace Transforms R. Kent Nagle/Edward B. Saff/Arthur David Snider

369

8. Series Solutions of Differential Equations R. Kent Nagle/Edward B. Saff/Arthur David Snider

445

9. Matrix Methods for Linear Systems R. Kent Nagle/Edward B. Saff/Arthur David Snider

525

10. Partial Differential Equations R. Kent Nagle/Edward B. Saff/Arthur David Snider

599

Tables and Equations R. Kent Nagle/Edward B. Saff/Arthur David Snider

683

Index

687

I

II

Introduction

1

BACKGROUND In the sciences and engineering, mathematical models are developed to aid in the understanding of physical phenomena. These models often yield an equation that contains some derivatives of an unknown function. Such an equation is called a differential equation. Two examples of models developed in calculus are the free fall of a body and the decay of a radioactive substance. In the case of free fall, an object is released from a certain height above the ground and falls under the force of gravity.† Newton’s second law, which states that an object’s mass times its acceleration equals the total force acting on it, can be applied to the falling object. This leads to the equation (see Figure 1) m

d 2h  mg , dt 2

where m is the mass of the object, h is the height above the ground, d 2h / dt 2 is its acceleration, g is the (constant) gravitational acceleration, and mg is the force due to gravity. This is a differential equation containing the second derivative of the unknown height h as a function of time. Fortunately, the above equation is easy to solve for h. All we have to do is divide by m and integrate twice with respect to t. That is, d 2h  g , dt 2 so dh  gt  c1 dt and h  h AtB 

gt 2  c1t  c2 . 2

−mg

h

Figure 1 Apple in free fall † We are assuming here that gravity is the only force acting on the object and that this force is constant. More general models would take into account other forces, such as air resistance.

From Chapter 1 of Fundamentals of Differential Equations, Eighth Edition. R. Kent Nagle, Edward B. Saff and Arthur David Snider. Copyright © 2012 by Pearson Education, Inc. Publishing as Pearson Addison-Wesley. All rights reserved.

1

Introduction

We will see that the constants of integration, c1 and c2, are determined if we know the initial height and the initial velocity of the object. We then have a formula for the height of the object at time t. In the case of radioactive decay (Figure 2), we begin from the premise that the rate of decay is proportional to the amount of radioactive substance present. This leads to the equation dA  kA , dt

k0 ,

where A A 0 B is the unknown amount of radioactive substance present at time t and k is the proportionality constant. To solve this differential equation, we rewrite it in the form 1 dA  k dt A and integrate to obtain

冮 A1 dA  冮 k dt ln A  C1  kt  C2 . Solving for A yields A  A A t B  eln A  ekte C2 C1  Cekt , where C is the combination of integration constants e C2 C1 . The value of C, as we will see later, is determined if the initial amount of radioactive substance is given. We then have a formula for the amount of radioactive substance at any future time t. Even though the above examples were easily solved by methods learned in calculus, they do give us some insight into the study of differential equations in general. First, notice that the solution of a differential equation is a function, like h A t B or A A t B , not merely a number. Second, integration is an important tool in solving differential equations (not surprisingly!). Third, we cannot expect to get a unique solution to a differential equation, since there will be arbitrary “constants of integration.” The second derivative d 2h / dt 2 in the free-fall equation gave rise to two constants, c1 and c2, and the first derivative in the decay equation gave rise, ultimately, to one constant, C. Whenever a mathematical model involves the rate of change of one variable with respect to another, a differential equation is apt to appear. Unfortunately, in contrast to the examples for free fall and radioactive decay, the differential equation may be very complicated and difficult to analyze.

A

Figure 2 Radioactive decay

2

Introduction

R

L

+ emf

C

−

Figure 3 Schematic for a series RLC circuit

Differential equations arise in a variety of subject areas, including not only the physical sciences but also such diverse fields as economics, medicine, psychology, and operations research. We now list a few specific examples. 1. In banking practice, if P A t B is the number of dollars in a savings bank account that pays a yearly interest rate of r% compounded continuously, then P satisfies the differential equation (1)

dP r  P , t in years. dt 100 2. A classic application of differential equations is found in the study of an electric circuit consisting of a resistor, an inductor, and a capacitor driven by an electromotive force (see Figure 3). Here an application of Kirchhoff’s laws leads to the equation

(2)

L

d 2q dt

2

R

dq dt



1 q  E AtB , C

where L is the inductance, R is the resistance, C is the capacitance, E A t B is the electromotive force, q A t B is the charge on the capacitor, and t is the time. 3. In psychology, one model of the learning of a task involves the equation (3)

dy / dt

y 3/ 2 A 1  y B 3/2



2p 1n

.

Here the variable y represents the learner’s skill level as a function of time t. The constants p and n depend on the individual learner and the nature of the task. 4. In the study of vibrating strings and the propagation of waves, we find the partial differential equation (4)

2 0 2u 20 u  c  0 , †† 0t 2 0x 2

where t represents time, x the location along the string, c the wave speed, and u the displacement of the string, which is a function of time and location.

††

Historical Footnote: This partial differential equation was first discovered by Jean le Rond d’Alembert (1717–1783) in 1747.

3

Introduction

To begin our study of differential equations, we need some common terminology. If an equation involves the derivative of one variable with respect to another, then the former is called a dependent variable and the latter an independent variable. Thus, in the equation (5)

d 2x dx  a  kx  0 , 2 dt dt

t is the independent variable and x is the dependent variable. We refer to a and k as coefficients in equation (5). In the equation (6)

0u 0u   x  2y , 0x 0y

x and y are independent variables and u is the dependent variable. A differential equation involving only ordinary derivatives with respect to a single independent variable is called an ordinary differential equation. A differential equation involving partial derivatives with respect to more than one independent variable is a partial differential equation. Equation (5) is an ordinary differential equation, and equation (6) is a partial differential equation. The order of a differential equation is the order of the highest-order derivatives present in the equation. Equation (5) is a second-order equation because d 2x / dt 2 is the highest-order derivative present. Equation (6) is a first-order equation because only first-order partial derivatives occur. It will be useful to classify ordinary differential equations as being either linear or nonlinear. Remember that lines (in two dimensions) and planes (in three dimensions) are especially easy to visualize, when compared to nonlinear objects such as cubic curves or quadric surfaces. For example, all the points on a line can be found if we know just two of them. Correspondingly, linear differential equations are more amenable to solution than nonlinear ones. Now the equations for lines ax  by  c and planes ax  by  cz  d have the feature that the variables appear in additive combinations of their first powers only. By analogy a linear differential equation is one in which the dependent variable y and its derivatives appear in additive combinations of their first powers. More precisely, a differential equation is linear if it has the format (7)

an A x B

dn y dx

n

ⴙ an1 A x B

d nⴚ1y dx

nⴚ1

ⴙ

p

ⴙ a1 A x B

dy dx

ⴙ a0 A x B y ⴝ F A x B ,

where an A x B , an1 A x B , . . . , a0 A x B and F A x B depend only on the independent variable x. The additive combinations are permitted to have multipliers (coefficients) that depend on x; no restrictions are made on the nature of this x-dependence. If an ordinary differential equation is not linear, then we call it nonlinear. For example, d2y dx 2

 y3  0

is a nonlinear second-order ordinary differential equation because of the y3 term, whereas t3

dx  t3  x dt

is linear (despite the t 3 terms). The equation d 2y dx

2

y

dy dx

 cos x

is nonlinear because of the y dy / dx term.

4

Introduction

Although the majority of equations one is likely to encounter in practice fall into the nonlinear category, knowing how to deal with the simpler linear equations is an important first step (just as tangent lines help our understanding of complicated curves by providing local approximations).

1

EXERCISES

In Problems 1–12, a differential equation is given along with the field or problem area in which it arises. Classify each as an ordinary differential equation (ODE) or a partial differential equation (PDE), give the order, and indicate the independent and dependent variables. If the equation is an ordinary differential equation, indicate whether the equation is linear or nonlinear. 1.

d 2y dx 2

 2x

dy dx

 2y  0

(aerodynamics, stress analysis)

(Hermite’s equation, quantum-mechanical harmonic oscillator) 2. 5

d 2x dx  4  9x  2 cos 3t 2 dt dt

(mechanical vibrations, electrical circuits, seismology) 3.

0 2u 0 2u  0 0x 2 0y 2 (Laplace’s equation, potential theory, electricity, heat, aerodynamics)

4.

dy dx



y A 2  3x B

x A 1  3y B

(competition between two species, ecology) 5.

dx  k A 4  x B A 1  x B , where k is a constant dt (chemical reaction rates)

dy 2 6. y c 1  a b d  C , where C is a constant dx (brachistochrone problem,† calculus of variations) 7. 11  y

d2y dx

2

dp  kp A P  p B , where k and P are constants dt (logistic curve, epidemiology, economics) d4y 9. 8 4  x A 1  x B dx (deflection of beams) d2y dy 10. x 2   xy  0 dx dx 8.

 2x

dy dx

0

(Kidder’s equation, flow of gases through a porous medium)

0N 0 2N 1 0N  2   kN, where k is a constant 0t r 0r 0r (nuclear fission) dy d 2y 12.  0.1 A 1  y 2 B  9y  0 2 dx dx (van der Pol’s equation, triode vacuum tube) 11.

In Problems 13–16, write a differential equation that fits the physical description. 13. The rate of change of the population p of bacteria at time t is proportional to the population at time t. 14. The velocity at time t of a particle moving along a straight line is proportional to the fourth power of its position x. 15. The rate of change in the temperature T of coffee at time t is proportional to the difference between the temperature M of the air at time t and the temperature of the coffee at time t. 16. The rate of change of the mass A of salt at time t is proportional to the square of the mass of salt present at time t. 17. Drag Race. Two drivers, Alison and Kevin, are participating in a drag race. Beginning from a standing start, they each proceed with a constant acceleration. Alison covers the last 1 / 4 of the distance in 3 seconds, whereas Kevin covers the last 1 / 3 of the distance in 4 seconds. Who wins and by how much time?

Historical Footnote: In 1630 Galileo formulated the brachistochrone problem A bra´ xi´sto  shortest, xro´ no  time B , that is, to determine a path down which a particle will fall from one given point to another in the shortest time. It was reproposed by John Bernoulli in 1696 and solved by him the following year. †

5

5

Introduction

2

SOLUTIONS AND INITIAL VALUE PROBLEMS An nth-order ordinary differential equation is an equality relating the independent variable to the nth derivative (and usually lower-order derivatives as well) of the dependent variable. Examples are d 2y

(second-order, x independent, y dependent)

d2 y 1  a 2b  y  0 dt B

(second-order, t independent, y dependent)

d 4x  xt dt 4

(fourth-order, t independent, x dependent).

dx

2

x

dy

 y  x3

x2

dx

Thus, a general form for an nth-order equation with x independent, y dependent, can be expressed as (1)

F ax, y,

dy dny , . . . , nb  0 , dx dx

where F is a function that depends on x, y, and the derivatives of y up to order n; that is, on x, y, . . . , d n y / dx n . We assume that the equation holds for all x in an open interval I (a 6 x 6 b, where a or b could be infinite). In many cases we can isolate the highest-order term d n y / dx n and write equation (1) as (2)

d ny dx n

 f ax, y,

dy dx

,...,

b ,

d n1y dx n1

which is often preferable to (1) for theoretical and computational purposes.

Explicit Solution Definition 1. A function f A x B that when substituted for y in equation (1) [or (2)] satisfies the equation for all x in the interval I is called an explicit solution to the equation on I.

Example 1

Show that f A x B  x 2  x 1 is an explicit solution to the linear equation (3)

d 2y dx

2



2 y0, x2

but c A x B  x3 is not. Solution

The functions f A x B  x 2  x 1, f¿ A x B  2x  x 2, and f– A x B  2  2x 3 are defined for all x  0. Substitution of f A x B for y in equation (3) gives A 2  2x 3 B 

6

6

2 2 A x  x 1 B  A 2  2x 3 B  A 2  2x 3 B  0 . x2

Introduction

Since this is valid for any x  0, the function f A x B  x 2  x 1 is an explicit solution to (3) on A q, 0 B and also on A 0, q B . For c A x B  x 3 we have c¿ A x B  3x 2, c– A x B  6x, and substitution into (3) gives 6x 

2 3 x  4x  0 , x2

which is valid only at the point x  0 and not on an interval. Hence c(x) is not a solution. ◆ Example 2

Show that for any choice of the constants c1 and c2, the function f A x B  c1e x  c2e 2x is an explicit solution to the linear equation (4)

Solution

y–  y¿  2y  0 .

We compute f¿ A x B  c1e x  2c2 e 2x and f– A x B  c1e x  4c2 e 2x . Substitution of f, f¿ , and f– for y, y ¿ , and y – in equation (4) yields A c1e x  4c2e 2x B  A c1e x  2c2e 2x B  2 A c1e x  c2e 2x B

 A c1  c1  2c1 B e x  A 4c2  2c2  2c2 B e 2x  0 .

Since equality holds for all x in A q, q B , then f A x B  c1e x  c2e 2x is an explicit solution to (4) on the interval A q, q B for any choice of the constants c1 and c2. ◆ The methods for solving differential equations do not always yield an explicit solution for the equation. We may have to settle for a solution that is defined implicitly. Consider the following example. Example 3

Show that the relation (5)

y2  x 3  8  0

implicitly defines a solution to the nonlinear equation (6)

dy 3x 2  dx 2y

on the interval A 2, q B . Solution

When we solve (5) for y, we obtain y   2x 3  8 . Let’s try f(x)  2x 3  8 to see if it is an explicit solution. Since df / dx  3x 2 / A22x 3  8 B , both f and df / dx are defined on A 2, q B . Substituting them into (6) yields 3x2 22x3  8



3x2

2 A 2x3  8 B

,

which is indeed valid for all x in A 2, q B . [You can check that c A x B  2x 3  8 is also an explicit solution to (6).] ◆

7

Introduction

Implicit Solution Definition 2. A relation G A x, y B  0 is said to be an implicit solution to equation (1) on the interval I if it defines one or more explicit solutions on I.

Example 4

Show that (7)

x  y  e xy  0

is an implicit solution to the nonlinear equation (8) Solution

A 1  xe xy B

dy  1  ye xy  0 . dx

First, we observe that we are unable to solve (7) directly for y in terms of x alone. However, for (7) to hold, we realize that any change in x requires a change in y, so we expect the relation (7) to define implicitly at least one function y A x B . This is difficult to show directly but can be rigorously verified using the implicit function theorem† of advanced calculus, which guarantees that such a function y A x B exists that is also differentiable (see Problem 30). Once we know that y is a differentiable function of x, we can use the technique of implicit differentiation. Indeed, from (7) we obtain on differentiating with respect to x and applying the product and chain rules, dy dy d A x  y  e xy B  1   e xy ay  x b  0 dx dx dx or A 1  xe xy B

dy  1  ye xy  0 , dx

which is identical to the differential equation (8). Thus, relation (7) is an implicit solution on some interval guaranteed by the implicit function theorem. ◆ Example 5

Verify that for every constant C the relation 4x 2  y 2  C is an implicit solution to (9)

y

dy  4x  0 . dx

Graph the solution curves for C  0, 1, 4. (We call the collection of all such solutions a one-parameter family of solutions.) Solution

When we implicitly differentiate the equation 4x 2  y 2  C with respect to x, we find 8x  2y

†

dy 0 , dx

See Vector Calculus, 5th ed, by J. E. Marsden and A. J. Tromba (Freeman, San Francisco, 2004).

8

Introduction

y C = −4

C=0

C = −1

C=0

2 C=1

C=1 −1

x

1 C=4

C=4 −2

Figure 4 Implicit solutions 4x 2  y 2  C

which is equivalent to (9). In Figure 4 we have sketched the implicit solutions for C  0, 1, 4. The curves are hyperbolas with common asymptotes y  2x. Notice that the implicit solution curves (with C arbitrary) fill the entire plane and are nonintersecting for C  0. For C  0, the implicit solution gives rise to the two explicit solutions y  2x and y  2x, both of which pass through the origin. ◆ For brevity we hereafter use the term solution to mean either an explicit or an implicit solution. In the beginning of Section 1, we saw that the solution of the second-order free-fall equation invoked two arbitrary constants of integration c1, c2: h AtB 

gt 2  c1t  c2 , 2

whereas the solution of the first-order radioactive decay equation contained a single constant C: A A t B  Ce kt . It is clear that integration of the simple fourth-order equation d 4y dx 4

0

brings in four undetermined constants: y A x B  c1x 3  c2 x 2  c3 x  c4 . It will be shown later in the text that in general the methods for solving nth-order differential equations evoke n arbitrary constants. In most cases, we will be able to evaluate these constants if we know n initial values y A x0 B , y A x0 B , . . . , y An1B A x0 B .

9

Introduction

Initial Value Problem Definition 3.

By an initial value problem for an nth-order differential equation

F ax, y,

dy d ny , . . . , nb  0 , dx dx

we mean: Find a solution to the differential equation on an interval I that satisfies at x0 the n initial conditions y A x0 B  y0 , dy A x B  y1 , dx 0 · · · d n1y dx n1

A x 0 B  yn1 ,

where x0 僆 I and y0, y1, . . . , yn – 1 are given constants.

In the case of a first-order equation, the initial conditions reduce to the single requirement y A x0 B  y0 , and in the case of a second-order equation, the initial conditions have the form y A x0 B  y0 ,

dy A x B  y1 . dx 0

The terminology initial conditions comes from mechanics, where the independent variable x represents time and is customarily symbolized as t. Then if t0 is the starting time, y A t0 B  y0 represents the initial location of an object and y¿ A t0 B gives its initial velocity. Example 6

Show that f A x B  sin x  cos x is a solution to the initial value problem (10)

Solution

d 2y dx

2

y0 ;

y A 0 B  1 ,

dy dx

A0B  1 .

Observe that f A x B  sin x  cos x, df / dx  cos x  sin x, and d 2f / dx 2  sin x  cos x are all defined on A q, q B . Substituting into the differential equation gives A sin x  cos x B  A sin x  cos x B  0 ,

which holds for all x 僆 A q, q B . Hence, f A x B is a solution to the differential equation in (10) on A q, q B . When we check the initial conditions, we find f A 0 B  sin 0  cos 0  1 , df A 0 B  cos 0  sin 0  1 , dx which meets the requirements of (10). Therefore, f A x B is a solution to the given initial value problem. ◆

10

Introduction

Example 7

As shown in Example 2, the function f A x B  c1e x  c2e 2x is a solution to d 2y dx

2



dy dx

 2y  0

for any choice of the constants c1 and c2. Determine c1 and c2 so that the initial conditions y A0B  2

and

dy A 0 B  3 dx

are satisfied. Solution

To determine the constants c1 and c2, we first compute df / dx to get df / dx  c1e x  2c2e 2x. Substituting in our initial conditions gives the following system of equations:

冦

f A0B

 c1e 0  c2e 0  2 ,

df A 0 B  c1e 0  2c2e 0  3 , dx

or

冦

c1  c2  2 , c1  2c2  3 .

Adding the last two equations yields 3c2  1, so c2  1 / 3 . Since c1  c2  2, we find c1  7 / 3. Hence, the solution to the initial value problem is f A x B  A 7 / 3 B e x  A 1 / 3 B e 2x. ◆ We now state an existence and uniqueness theorem for first-order initial value problems. We presume the differential equation has been cast into the format dy  f A x, y B . dx Of course, the right-hand side, f A x, y B , must be well defined at the starting value x0 for x and at the stipulated initial value y0  y A x0 B for y. The hypotheses of the theorem, moreover, require continuity of both f and 0f / 0y for x in some interval a x b containing x0, and for y in some interval c y d containing y0. Notice that the set of points in the xy-plane that satisfy a x b and c y d constitutes a rectangle. Figure 5 depicts this “rectangle of continuity” with the initial point A x0, y0 B in its interior and a sketch of a portion of the solution curve contained therein.

Existence and Uniqueness of Solution Theorem 1. Consider the initial value problem dy  f A x, y B , dx

y A x0 B  y0 .

If f and 0f / 0y are continuous functions in some rectangle R  E A x, y B : a 6 x 6 b, c 6 y 6 dF

that contains the point A x0, y0 B , then the initial value problem has a unique solution f A x B in some interval x0  d x x0  d, where d is a positive number. 11

11

Introduction

y

d

y = (x)

y0

c

a

x0 − 

x0

x0 + 

b

x

Figure 5 Layout for the existence–uniqueness theorem

The preceding theorem tells us two things. First, when an equation satisfies the hypotheses of Theorem 1, we are assured that a solution to the initial value problem exists. Naturally, it is desirable to know whether the equation we are trying to solve actually has a solution before we spend too much time trying to solve it. Second, when the hypotheses are satisfied, there is a unique solution to the initial value problem. This uniqueness tells us that if we can find a solution, then it is the only solution for the initial value problem. Graphically, the theorem says that there is only one solution curve that passes through the point A x0, y0 B . In other words, for this first-order equation, two solutions cannot cross anywhere in the rectangle. Notice that the existence and uniqueness of the solution holds only in some neighborhood A x 0  d, x 0  d B . Unfortunately, the theorem does not tell us the span A 2d B of this neighborhood (merely that it is not zero). Problem 18 elaborates on this feature. Problem 19 gives an example of an equation with no solution. Problem 29 displays an initial value problem for which the solution is not unique. Of course, the hypotheses of Theorem 1 are not met for these cases. When initial value problems are used to model physical phenomena, many practitioners tacitly presume the conclusions of Theorem 1 to be valid. Indeed, for the initial value problem to be a reasonable model, we certainly expect it to have a solution, since physically “something does happen.” Moreover, the solution should be unique in those cases when repetition of the experiment under identical conditions yields the same results.† The proof of Theorem 1 involves converting the initial value problem into an integral equation and then using Picard’s method to generate a sequence of successive approximations that converge to the solution. The conversion to an integral equation and Picard’s method are discussed in Group Project B at the end of this chapter.

†

At least this is the case when we are considering a deterministic model, as opposed to a probabilistic model.

12

Introduction

Example 8

For the initial value problem (11)

3

dy  x2  xy3 , dx

y(1)  6 ,

does Theorem 1 imply the existence of a unique solution? Solution

Example 9

Dividing by 3 to conform to the statement of the theorem, we identify f A x, y B as A x2  xy3 B /3 and 0f / 0y as xy2. Both of these functions are continuous in any rectangle containing the point (1, 6), so the hypotheses of Theorem 1 are satisfied. It then follows from the theorem that the initial value problem (11) has a unique solution in an interval about x  1 of the form A 1  d, 1  d B , where d is some positive number. ◆ For the initial value problem (12)

dy  3y 2/3 , dx

y A2B  0 ,

does Theorem 1 imply the existence of a unique solution? Solution

Here f A x, y B  3y 2/3 and 0f / 0y  2y 1/3. Unfortunately 0f / 0y is not continuous or even defined when y  0. Consequently, there is no rectangle containing A 2, 0 B in which both f and 0f / 0y are continuous. Because the hypotheses of Theorem 1 do not hold, we cannot use Theorem 1 to determine whether the initial value problem does or does not have a unique solution. It turns out that this initial value problem has more than one solution. We refer you to Problem 29 for the details. ◆ In Example 9 suppose the initial condition is changed to y A 2 B  1 . Then, since f and 0f / 0y are continuous in any rectangle that contains the point A 2, 1 B but does not intersect the x-axis— say, R  E A x, y B : 0 6 x 6 10, 0 6 y 6 5F—it follows from Theorem 1 that this new initial value problem has a unique solution in some interval about x  2.

2

EXERCISES

1. (a) Show that y 2  x  3  0 is an implicit solution to dy / dx  1 / A 2y B on the interval A q, 3 B . (b) Show that xy 3  xy 3 sin x = 1 is an implicit solution to A x cos x  sin x  1 B y dy  3 A x  x sin x B dx on the interval A 0, p / 2 B . 2. (a) Show that f A x B  x 2 is an explicit solution to x

dy  2y dx

on the interval A q, q B . (b) Show that f A x B  e x  x is an explicit solution to dy  y 2  e 2x  A 1  2x B e x  x 2  1 dx

on the interval A q, q B .

(c) Show that f A x B  x 2  x 1 is an explicit solution to x 2d 2y / dx 2  2y on the interval A 0, q B . In Problems 3–8, determine whether the given function is a solution to the given differential equation. 3. x  2 cos t  3 sin t , x –  x  0 d 2y  y  x2  2 4. y  sin x  x 2 , dx 2 dx  tx  sin 2t 5. x  cos 2t , dt 6. u  2e 3t  e 2t , 7. y  3 sin 2x  e x , 8. y  e 2x  3e x ,

du d 2u u  3u  2e 2t dt 2 dt y–  4y  5e x dy d 2y  2y  0  2 dx dx

13

Introduction

In Problems 9–13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship does define y implicitly as a function of x and use implicit differentiation. dy 2xy  9. y  ln y  x 2  1 , dx y1 dy x  dx y dy e xy  y  xy dx e x

10. x 2  y 2  4 , 11. e xy  y  x  1 , 12. x  sin A x  y B  1 , 2

interval can be quite small (if c is small) or quite large (if c is large). Notice also that there is no clue from the equation dy / dx  2xy 2 itself, or from the initial value, that the solution will “blow up” at x  c.

19. Show that the equation A dy / dx B 2  y2  4  0 has no (real-valued) solution.

20. Determine for which values of m the function f A x B  e mx is a solution to the given equation. (a)

dy  2x sec A x  y B  1 dx (b)

13. sin y  xy  x  2 , 3

y– 

6xy¿  A y¿ B 3sin y  2 A y¿ B 2 3x 2  y

14. Show that f A x B  c1 sin x  c2 cos x is a solution to d 2y / dx 2  y  0 for any choice of the constants c1 and c2. Thus, c1 sin x  c2 cos x is a two-parameter family of solutions to the differential equation. 15. Verify that f A x B  2 / A 1  ce x B , where c is an arbitrary constant, is a one-parameter family of solutions to y A y  2B dy  . dx 2 Graph the solution curves corresponding to c  0, 1, 2 using the same coordinate axes. 16. Verify that x 2  cy 2  1, where c is an arbitrary nonzero constant, is a one-parameter family of implicit solutions to dy xy  2 x 1 dx and graph several of the solution curves using the same coordinate axes. 17. Show that f A x B  Ce 3x  1 is a solution to dy / dx  3y  3 for any choice of the constant C. Thus, Ce3x  1 is a one-parameter family of solutions to the differential equation. Graph several of the solution curves using the same coordinate axes. 18. Let c 7 0. Show that the function f A x B  (c2  x 2)1 is a solution to the initial value problem dy / dx  2xy 2 , y(0)  1 / c2 , on the interval c 6 x 6 c. Note that this solution becomes unbounded as x approaches c. Thus, the solution exists on the interval (d, d) with d  c, but not for larger d. This illustrates that in Theorem 1 the existence

14

d 2y dx

6

2

d 3y dx

3

3

dy dx

 5y  0

d 2y dx

2

2

dy dx

0

21. Determine for which values of m the function f A x B  x m is a solution to the given equation. (a) 3x 2

d 2y dx

d 2y

2

 11x

dy dx

 3y  0

dy

 5y  0 dx dx 22. Verify that the function f A x B  c1e x  c2e 2x is a solution to the linear equation (b) x 2

2

d 2y dx

2

x



dy dx

 2y  0

for any choice of the constants c1 and c2. Determine c1 and c2 so that each of the following initial conditions is satisfied. (a) y A 0 B  2 , y¿ A 0 B  1 (b) y A 1 B  1 , y¿ A 1 B  0 In Problems 23–28, determine whether Theorem 1 implies that the given initial value problem has a unique solution. dy y A0B  7  y4  x4 , 23. dx 24.

dy  ty  sin2t , dt

25. 3x 26.

dx  4t  0 , dt

dx  cos x  sin t , dt

dy x , dx dy 3  3x  2 y1 , 28. dx 27. y

y A pB  5 x A 2 B  p x A pB  0 y A1B  0 y A2B  1

Introduction

29. (a) For the initial value problem (12) of Example 9, show that f1 A x B ⬅ 0 and f2 A x B  A x  2 B 3 are solutions. Hence, this initial value problem has multiple solutions. (b) Does the initial value problem y¿  3y 2/ 3, y(0)  107, have a unique solution in a neighborhood of x  0? 30. Implicit Function Theorem. Let G A x, y B have continuous first partial derivatives in the rectangle R  E A x, y B : a 6 x 6 b, c 6 y 6 dF containing the point A x 0, y0 B . If G A x 0, y0 B  0 and the partial derivative Gy A x 0, y0 B  0, then there exists a differentiable function y  f A x B , defined in some interval I  A x 0  d, x 0  d B , that satisfies G Ax, f A x B B  0 for all x 僆 I.

3

The implicit function theorem gives conditions under which the relationship G A x, y B  0 defines y implicitly as a function of x. Use the implicit function theorem to show that the relationship x  y  e xy  0, given in Example 4, defines y implicitly as a function of x near the point A 0, 1 B . 31. Consider the equation of Example 5, dy (13) y  4x  0 . dx (a) Does Theorem 1 imply the existence of a unique solution to (13) that satisfies y A x 0 B  0? (b) Show that when x 0  0, equation (13) can’t possibly have a solution in a neighborhood of x  x0 that satisfies y A x 0 B  0. (c) Show that there are two distinct solutions to (13) satisfying y A 0 B  0 (see Figure 4).

DIRECTION FIELDS The existence and uniqueness theorem discussed in Section 2 certainly has great value, but it stops short of telling us anything about the nature of the solution to a differential equation. For practical reasons we may need to know the value of the solution at a certain point, or the intervals where the solution is increasing, or the points where the solution attains a maximum value. Certainly, knowing an explicit representation (a formula) for the solution would be a considerable help in answering these questions. However, for many of the differential equations that we are likely to encounter in real-world applications, it will be impossible to find such a formula. Moreover, even if we are lucky enough to obtain an implicit solution, using this relationship to determine an explicit form may be difficult. Thus, we must rely on other methods to analyze or approximate the solution. One technique that is useful in visualizing (graphing) the solutions to a first-order differential equation is to sketch the direction field for the equation. To describe this method, we need to make a general observation. Namely, a first-order equation dy  f A x, y B dx specifies a slope at each point in the xy-plane where f is defined. In other words, it gives the direction that a graph of a solution to the equation must have at each point. Consider, for example, the equation (1)

dy  x2  y . dx

The graph of a solution to (1) that passes through the point A 2, 1 B must have slope A 2 B 2  1  3 at that point, and a solution through A 1, 1 B has zero slope at that point. A plot of short line segments drawn at various points in the xy-plane showing the slope of the solution curve there is called a direction field for the differential equation. Because the

15

Introduction

y

y

1

1

0

x

x

0

1

(a)

1

(b)

Figure 6 (a) Direction field for dy / dx  x 2  y

(b) Solutions to dy / dx  x 2  y

y

y

1

0

(a)

1

1

x

dy = −2y dx

0

(b)

Figure 7 (a) Direction field for dy / dx  2y

1

x

dy y =− x dx

(b) Direction field for dy / dx  y / x

direction field gives the “flow of solutions,” it facilitates the drawing of any particular solution (such as the solution to an initial value problem). In Figure 6(a) we have sketched the direction field for equation (1) and in Figure 6(b) have drawn several solution curves in color. Some other interesting direction field patterns are displayed in Figure 7. Depicted in Figure 7(a) is the pattern for the radioactive decay equation dy / dx  2y (recall that in Section 1 we analyzed this equation in the form dA / dt  kA ). From the flow patterns, we can see that all solutions tend asymptotically to the positive x-axis as x gets larger. In other words, any material decaying according to this law eventually dwindles to practically nothing. This is consistent with the solution formula we derived earlier, A  Ce kt ,

16

or y  Ce 2x .

Introduction

y

y

y0 y0 0

x

x0

0

(a)

x0

x

(b)

Figure 8 (a) A solution for dy / dx  2y (b) A solution for dy / dx  y / x

From the direction field in Figure 7(b), we can anticipate that all solutions to dy / dx  y / x also approach the x-axis as x approaches infinity (plus or minus infinity, in fact). But more interesting is the observation that no solution can make it across the y-axis; 0 y A x B 0 “blows up” as x goes to zero from either direction. Exception: On close examination, it appears the function y A x B ⬅ 0 might just make it through this barrier. As a matter of fact, in Problem 19 you are invited to show that the solutions to this differential equation are given by y  C / x, with C an arbitrary constant. So they do diverge at x  0, unless C  0. Let’s interpret the existence–uniqueness theorem of Section 2 for these direction fields. For Figure 7(a), where dy / dx  f A x, y B  2y, we select a starting point x0 and an initial value y A x0 B  y0, as in Figure 8(a). Because the right-hand side f A x, y B  2y is continuously differentiable for all x and y, we can enclose any initial point A x0, y0 B in a “rectangle of continuity.” We conclude that the equation has one and only one solution curve passing through A x 0, y0 B , as depicted in the figure. For the equation dy y  f A x, y B   , dx x the right-hand side f A x, y B  y / x does not meet the continuity conditions when x  0 (i.e., for points on the y-axis). However, for any nonzero starting value x0 and any initial value y A x0 B  y0, we can enclose A x0, y0 B in a rectangle of continuity that excludes the y-axis, as in Figure 8(b). Thus, we can be assured of one and only one solution curve passing through such a point. The direction field for the equation dy  3y 2/3 dx is intriguing because Example 9 of Section 2 showed that the hypotheses of Theorem 1 do not hold in any rectangle enclosing the initial point x0  2, y0  0. Indeed, Problem 29 of that section demonstrated the violation of uniqueness by exhibiting two solutions, y A x B ⬅ 0

17

Introduction

y

y

1

y(x) = (x − 2)3

1

y(x) = 0 0

x 1

2

0

(a) Figure 9 (a) Direction field for dy / dx  3y 2/3

x 1

2

(b) (b) Solutions for dy / dx  3y 2/3, y A 2 B  0

and y A x B  A x  2 B 3, passing through A 2, 0 B . Figure 9(a) displays this direction field, and Figure 9(b) demonstrates how both solution curves can successfully “negotiate” this flow pattern. Clearly, a sketch of the direction field of a first-order differential equation can be helpful in visualizing the solutions. However, such a sketch is not sufficient to enable us to trace, unambiguously, the solution curve passing through a given initial point A x0, y0 B . If we tried to trace one of the solution curves in Figure 6(b), for example, we could easily “slip” over to an adjacent curve. For nonunique situations like that in Figure 9(b), as one negotiates the flow along the curve y  A x  2 B 3 and reaches the inflection point, one cannot decide whether to turn or to (literally) go off on the tangent A y  0 B . Example 1

The logistic equation for the population p (in thousands) at time t of a certain species is given by (2)

dp  p A2  pB . dt

(Of course, p is nonnegative.) From the direction field sketched in Figure 10, answer the following: (a) If the initial population is 3000 3 that is, p A 0 B  3 4 , what can you say about the limiting population limtS q p A t B ?

(b) Can a population of 1000 ever decline to 500? (c) Can a population of 1000 ever increase to 3000? Solution

18

(a) The direction field indicates that all solution curves 3 other than p A t B ⬅ 0 4 will approach the horizontal line p  2 as t S q; that is, this line is an asymptote for all positive solutions. Thus, limtSq p A t B  2.

Introduction

p (in thousands) 4 3 2 1

0

t 1

2

3

4

Figure 10 Direction field for logistic equation

(b) The direction field further shows that populations greater than 2000 will steadily decrease, whereas those less than 2000 will increase. In particular, a population of 1000 can never decline to 500. (c) As mentioned in part (b), a population of 1000 will increase with time. But the direction field indicates it can never reach 2000 or any larger value; i.e., the solution curve cannot cross the line p  2. Indeed, the constant function p A t B ⬅ 2 is a solution to equation (2), and the uniqueness part of Theorem 1, precludes intersections of solution curves. ◆ Notice that the direction field in Figure 10 has the nice feature that the slopes do not depend on t; that is, the slopes are the same along each horizontal line. The same is true for Figures 8(a) and 9. This is the key property of so-called autonomous equations y¿  f A y B , where the right-hand side is a function of the dependent variable only. Group Project C, investigates such equations in more detail. Hand sketching the direction field for a differential equation is often tedious. Fortunately, several software programs have been developed to obviate this task†. When hand sketching is necessary, however, the method of isoclines can be helpful in reducing the work.

The Method of Isoclines An isocline for the differential equation y¿  f A x, y B is a set of points in the xy-plane where all the solutions have the same slope dy / dx ; thus, it is a level curve for the function f A x, y B . For example, if (3)

y¿  f A x, y B  x  y ,

the isoclines are simply the curves (straight lines) x  y  c or y  x  c. Here c is an arbitrary constant. But c can be interpreted as the numerical value of the slope dy / dx of every solution curve as it crosses the isocline. (Note that c is not the slope of the isocline itself; the latter is, obviously, 1.) Figure 11(a) depicts the isoclines for equation (3). †

An applet, maintained on the web at http://alamos.math.arizona.edu/~rychlik/JOde/index.html, sketches direction fields and automates most of the differential equation algorithms discussed in this text.

19

Introduction

y

y

c=5 c=4

1

0

1

1

x c=3

0

x 1

c=2 c=1 c=0 c = −5 c = −4 c = −3 c = −2 c = −1 (b)

(a) y

1

0

x 1

(c) Figure 11 (a) Isoclines for y¿  x  y (b) Direction field for y¿  x  y

(c) Solutions to y¿  x  y

To implement the method of isoclines for sketching direction fields, we draw hash marks with slope c along the isocline f A x, y B  c for a few selected values of c. If we then erase the underlying isocline curves, the hash marks constitute a part of the direction field for the differential equation. Figure 11(b) depicts this process for the isoclines shown in Figure 11(a), and Figure 11(c) displays some solution curves. Remark. The isoclines themselves are not always straight lines. For equation (1) at the beginning of this section, they are parabolas x 2  y  c. When the isocline curves are complicated, this method is not practical.

20

Introduction

3

EXERCISES

1. The direction field for dy / dx = 2x + y is shown in Figure 12. (a) Sketch the solution curve that passes through A 0, 2 B . From this sketch, write the equation for the solution. (b) Sketch the solution curve that passes through A 1, 3 B . (c) What can you say