Ordinary Differential Equations (Instructor Solution Manual, Solutions) [1 ed.] 1461436176, 9781461436171

Citation preview

William A. Adkins, Mark G. Davidson

ORDINARY DIFFERENTIAL EQUATIONS Instructor Solution Manual January 23, 2012

Springer

Chapter 1

Solutions

Section 1.1 1. The rate of change in the population P (t) is the derivative P ! (t). The Malthusian Growth Law states that the rate of change in the population is proportional to P (t). Thus P ! (t) = kP (t), where k is the proportionality constant. Without reference to the t variable, the differential equation becomes P ! = kP 2. a. This statement mathematically is b(t) = b0 P (t) where we have used b0 to represent the proportionality constant. b. This statement translates as d(t) = d0 P 2 (t) where we have used d0 to represent the proportionality constant. c. The overall growth rate is P ! (t). Thus the Logistic Growth Law is P ! (t) = b(t) − d(t) = b0 P (t) − d0 P 2 (t)

= (b0 − d0 P (t))P (t).

3. Torricelli’s law states that the ! change in height, h! (t) ! is proportional to the square root of the height, h(t). Thus h! (t) = λ h(t), where λ is the proportionality constant. 4. The highest order derivative is y ! so the order is 1 and the standard form is y ! = t3 /y 2 . 5. The highest order derivative is y !! so the order is 2. The standard form is y !! = t3 /y ! . 6. The highest order derivative is y ! so the order is 1 and the standard form is y ! = (et − ty)/t2 . 3

4

1 Solutions

7. The highest order derivative is y !! so the order is 2. The standard form is y !! = −(3y + ty ! )/t2 . 8. The highest order derivative is y !! so the order is 2 and the standard form is y !! = t2 − 3y ! − 2y. 9. The highest order derivative is y (4) so the order is 4. Solving for y (4) gives ! 3 (4) the standard form: y = (1 − (y !!! )4 )/t.

10. The highest order derivative is y ! so the order is 1 and the standard form is y ! = ty 4 − t2 y. 11. The highest order derivative is y !!! so the order is 3. Solving for y !!! gives the standard form: y !!! = 2y !! − 3y ! + y. 12. The following table summarizes the needed calculations: Function

y ! (t)

2y(t)

y1 (t) = 0

y1! (t) = 0

2y1 (t) = 0

y2 (t) = t2

y2! (t) = 2t

2y2 (t) = 2t2

y3 (t) = 3e2t

y3! (t) = 6e2t

2y3 (t) = 6e2t

y4 (t) = 2e3t

y4! (t) = 6e3t

2y4 (t) = 4e3t

To be a solution, the entries in the second and third columns need to be the same. Thus y1 and y3 are the only solutions. 13. The following table summarizes the needed calculations: Function y1 (t) = 0 y2 (t) = 3t y3 (t) = −5t y4 (t) = t

3

ty ! (t) ty1! (t) ty2! (t) ty3! (t) ty4! (t)

y(t)

=0

y1 (t) = 0

= 3t

y2 (t) = 3t

= −5t = 3t

3

y3 (t) = −5t y4 (t) = t3

To be a solution, the entries in the second and third columns need to be the same. Thus y1 , y2 , and y3 are solutions. 14. We first write the differential equation in standard form: y !! = −4y. The following table summarizes the needed calculations:

1 Solutions

5

Function y1 (t) = e

y !! (t) y1!! (t) y2!! (t) y3!! (t) y4!! (t)

2t

y2 (t) = sin 2t y3 (t) = cos(2t − 1) y4 (t) = t

2

−4y(t)

2t

−4y1 (t) = −4e2t

= 4e

= −4 sin 2t

−4y2 (t) = −4 sin 2t

− 4 cos(2t − 1) −4y3 (t) = −4 cos(2t − 1) −4y4 (t) = −4t2

=2

To be a solution, the entries in the second and third columns need to be the same. Thus y2 and y3 are solutions. 15. The following table summarizes the needed calculations: Function y1 (t) = 0 y2 (t) = 1 y3 (t) = 2 y4 (t) =

1 1−e2t

y ! (t) y1! (t) y2! (t) y3! (t)

2y(t)(y(t) − 1)

2y1 (t)(y1 (t) − 1) = 2 · 0 · (−1) = 0

=0

2y2 (t)(y2 (t) − 1) = 2 · 1 · 0 = 0

=0 =0

y4! (t) =

2t

2e (1−e2t )2

2y3 (t)(y3 (t) − 1) = 2 · 2 · 1 = 4 " # 1 1 2y4 (t)(y4 (t) − 1) = 2 1−e − 1 2t 2t 1−e 2t

1 e = 2 1−e 2t 1−e2t =

2e2t (1−e2t )2

Thus y1 , y2 , and y4 are solutions. 16. The following table summarizes the needed calculations: Function

2y(t)y ! (t)

1

y1 (t) = 1

2y1 (t)y1! (t) = 0

1

y2 (t) = t

2y2 (t)y2! (t) = 2t

1

y3 (t) = ln t √ y4 (t) = t − 4

t 2y3 (t)y3! (t) = 2 1t ln t = 2 ln t √ 2y4 (t)y4! (t) = 2 t − 4 2√1t−4 = 1

1 1

Thus y4 is the only solution. 17. The following table summarizes the needed calculations: Function √ y1 (t) = −t

√ y2 (t) = − et − t √ y3 (t) = t √ y4 (t) = − −t

2y(t)y ! (t) √ −1 2 −t √ = −1 2 −t t − 1) √ −(e −2 et − t √ = et − 1 2 et − t √ 1 2 t √ =1 2 t √ 1 2(− −t) √ = −1 2 −t

Thus y1 , y2 , and y4 are solutions.

y2 + t − 1 √ ( −t)2 + t − 1 = −1

√ (− et − t)2 + t − 1 = et − 1 √ ( t)2 + t − 1 = 2t − 1 √ (− −t))2 + y − 1 = −1

6

1 Solutions

18. The following table summarizes the needed calculations for the first three functions: Function

y ! (t)

y1 (t) = t

1

y2 (t) = 2t

2

y3 (t) = 3t

3

y 2 (t) − 4y(t)t + 6t2 t2 2 2 t − 4t + 6t2 =3 t2 2 2 2 4t − 8t + 6t =2 t2 2 2 2 9t − 12t + 6t =3 t2

3t + 2t2 t(3 + 2t) = the quotient rule and simplifying gives 1+t 1+t 2 2t + 4t + 3 y4! (t) = . On the other hand, (1 + t)2 For y4 (t) =

y42 (t) − 4y4 (t)t + 6t2 t2

t2 (3 + 2t)2 4t2 (3 + 2t) − + 6t2 (1 + t)2 (1 + t) = t2 2 (3 + 2t) − 4(3 + 2t)(1 + t) + 6(1 + t)2 = (1 + t)2 2 2t + 4t + 3 = . (1 + t)2

It follows that y2 , y3 , and y4 are solutions. 19. y ! (t) = 3ce3t 3y + 12 = 3(ce3t − 4) + 12 = 3ce3t − 12 + 12 = 3ce3t . Note that y(t) is defined for all t ∈ R. 20. y ! (t) = −ce−t + 3

−y(t) + 3t = −ce−t − 3t + 3 + 3t = −ce−t + 3. Note that y(t) is defined for all t ∈ R. 21. cet (1 − cet )2 1 1 1 − (1 − cet ) cet y 2 (t) − y(t) = − = = . (1 − cet )2 1 − cet (1 − cet )2 (1 − cet )2 y ! (t) =

1 Solutions

7

If c ≤ 0 then the denominator 1 − cet > 0 and y(t) has domain R. If c > 0 then 1 − cet = 0 if t = ln 1c = − ln c. Thus y(t) is defined either on the interval (−∞, − ln c) or (− ln c, ∞). 22. 2

y ! (t) = cet 2t = 2ctet

2

2

2ty(t) = 2tcet . 23. y ! (t) =

−cet cet − 1

t

−ey − 1 = −e− ln(ce

−1)

−1=

−1 −cet −1= t . t ce − 1 ce − 1

Since c > 0 then y(t) is defined if and only if cet − 1 > 0. This occurs if et > 1t which is true if t > ln 1c = − ln c. Thus y(t) is defined on the interval (− ln c, ∞). 24. We first calculate y ! (t) = −c(t + 1)−2 so (t + 1)y ! (t) + y(t) = (t + 1)

−c c −c c + = + = 0. (t + 1)2 t+1 t+1 t+1

Observe that y(t) is not defined at t = −1 so the two intervals where y is defined are (−∞, −1) and (−1, ∞). 25. y ! (t) = −(c − t)−2 (−1) = y 2 (t) =

1 . (c − t)2

1 (c − t)2

The denominator of y(t) is 0 when t = c. Thus the two intervals where y(t) is defined are (−∞, c) and (c, ∞). 26. This is a differential equation we can solve by simple integration: We get 2 y(t) = t2 + 3t + c. 27. Integration gives y(t) =

e2t 2

− t + c.

28. Integration (by parts) gives y(t) = −te−t − e−t + c. 29. Observe that

t+1 t

= 1 + 1t . Integration gives y(t) = t + ln |t| + c.

8

1 Solutions

30. We integrate two times. First, y ! (t) = t2 + t + c1 . Second, y(t) = t2 2 + c1 t + c2 .

t3 3

+

31. We integrate two times. First, y ! (t) = −2 cos 3t + c1 . Second, y(t) = − 23 sin 3t + c1 t + c2 . 32. From Problem 19 the general solution is y(t) = ce3t − 4. At t = 0 we get −2 = y(0) = ce0 − 4 = c − 4. It follows that c = 2 and y(t) = 2e3t − 4. 33. From Problem 20 the general solution is y(t) = ce−t + 3t − 3. At t = 0 we get 0 = y(0) = ce0 + 3(0) − 3 = c − 3. It follows that c = 3 and y(t) = 3e−t + 3t − 3. 34. From Problem 21 the general solution is y(t) = 1/(1 − cet ). At t = 0 we 1 get 1/2 = y(0) = 1−c . It follows that c = −1 and y(t) = 1/(1 + et ). 35. From Problem 24 the general solution is y(t) = c(t + 1)−1 . At t = 1 we get −9 = y(1) = c(1 + 1)−1 = c/2. It follows that c = −18 and y(t) = −18(t + 1)−1 . 36. From Problem 27 the general solution is y(t) = e2t /2 − t + c. Evaluation at t = 0 gives 4 = e0 /2 − 0 + c = 1/2 + c. Hence c = 7/2 and 37. From Problem 28 the general solution is y(t) = −te−t − e−t + c. Evaluation at t = 0 gives −1 = y(0) = −1+c so c = 0. Hence y(t) = −te−t −e−t . 38. From Problem 31 the general solution is y(t) = − 23 sin 3t + c1 t + c2 and a y ! (t) = −2 cos 3t + c1 . Evaluation at t = 0 gives 1 = y(0) = c2 and 2 = y ! (0) = −2 + c1 . If follows that c1 = 4 and c2 = 1. Thus y(t) = − 23 sin 3t + 4t.

1 Solutions

9

Section 1.2 1.

y! = t 5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

−4

−3

−2

2.

−1

0 t

1

2

3

4

5

2

3

4

5

3

4

5

y! = y2 5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

−4

−3

3.

−2

−1

0 t

1

y ! = y(y + t) 5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

−4

−3

−2

−1

0 t

1

2

10

1 Solutions

4.

4 3 2 1 0 −1 −2 −3 −4

5.

−4

−3

−2

−1

0

1

2

3

4

−4

−3

−2

−1

0 t

1

2

3

4

5

−4

−3

−2

−1

0 t

1

2

3

4

5

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

6.

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

1 Solutions

7.

11

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

8.

−4

−3

−2

−1

0 t

1

2

3

4

5

−4

−3

−2

−1

0 t

1

2

3

4

5

−4

−3

−2

−1

0 t

1

2

3

4

5

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

9.

5 4 3 2

y

1 0 −1 −2 −3 −4 −5 −5

10. We set y 2 = 0 and see that y = 0 is the only constant (= equilibrium) solution. 11. We set y(y + t) = 0. We look for constant solutions to y(y + t) = 0, and we see that y = 0 is the only constant (= equilibrium) solution.

12

1 Solutions

12. The equation y − t = 0 has no constant solution. Thus, there are no equilibrium solutions. 13. The equation 1 − y 2 = 0 has two constant solutions: y = 1 and y = −1 14. We substitute y = at + b into y ! = y − t to get a = (a − 1)t + b. Equality for all t forces a − 1 = 0 and a = b. Thus a = 1 and b = 1 and the only linear solution is y = t + 1. 15. We substitute y = at + b into y ! = cos(t + y) to get a = cos((a + 1)t + b). Equality for all t means that cos((a+1)t+b) must be a constant function, which can occur only if the coefficient of t is 0. This forces a = −1 leaving us with the equation −1 = cos b. This implies b = (2n+1)π, where n is an integer. Hence y = −t + (2n + 1)π, n ∈ Z is a family of linear solutions. 16. Implicit differentiation with respect to t gives 6t + 8yy ! = 0. 17. Implicit differentiation with respect to t gives 2yy ! − 2t − 3t2 = 0. 18. Differentiation gives y ! = 2ce2t + 1. However, from the given function we have ce2t = y − t. Substitution gives y ! = 2(y − t) + 1 = 2y − 2t + 1. 19. Differentiation gives y ! = 3ct2 + 2t. However, from the given function 2 ! we have ct3 = y − t2 and hence ct2 = y−t t . Substitution gives y = 2

3 y−t + 2t = t

3y t

− t.

Section 1.3 1. separable; h(t) = 1 and g(y) = 2y(5 − y) 2. In standard form we get y ! = (1 − y)/y. This is separable; h(t) = 1 and g(y) = (1 − y)/y. 1−2ty 3. First write in standard form: y ! = 1−2ty as a t2 . We cannot write t2 product of a function of t and a function of y. It is not separable.

4. In standard form we get y ! = y(y − t). We cannot write y(y − t) as a product of a function of t and a function of y. It is not separable. 5. Write in standard form to get: y ! = y−2yt . Here we can write t 1−2t y 1−2t . It is separable; h(t) = and g(y) = y. t t

y−2ty t

=

6. We can factor to get y ! = y 2 (t − 1) + t − 1 = (y 2 +1)(t − 1). It is separable; h(t) = t − 1 and g(y) = y 2 + 1.

1 Solutions

13

−2ty ! 7. In standard form we get y ! = t2−2ty +3y 2 . We cannot write y = t2 +3y 2 as a product of a function of t and a function of y. It is not separable

8. It is not separable as t2 + y 2 cannot be written as a product of a function of t and a function of y. 9. In standard form we get: y ! = e−t (y 3 − y) It is separable; h(t) = e−t and g(y) = y 3 − y 10. The variables are already separated, so integrate both sides of the equation to get y 2 /2 = t2 /2 + c, which we can rewrite as y 2 − t2 = k where k = 2c ∈ R is a constant. Since y(2) = −1, substitute t = 2 and y = −1 to get that k = (−1)2 − 22 = −3. Thus the solution is given implicitly by √ the equation y 2 − t2 = −3 or we can solve explicitly to get y = − t2 − 3, where the negative square root is used since y(2) = −1 < 0. 2

11. In standard form we get y ! = 1−y ty . Clearly, y = ±1 are equilibrium solutions. Separating the variables gives y 1 dy = dt. 1 − y2 t Integrating both sides of this equation (using the substitution u = 1 − y 2 , du = −2y dy for the integral on the left) gives 1 − ln |1 − y 2 | = ln |t| + c. 2 Multiplying by −2, taking the exponential of both sides, and removing the absolute values gives 1 − y 2 = kt−2 where k is a nonzero constant. However, when k = 0 the equation becomes 1−y 2 = 0 and hence y = ±1. By considering an arbitrary constant (which we will call c), the implicit equation t2 (1 − y 2 ) = c includes the two equilibrium solutions for c = 0. 12. The variables are already separated, so integrate both sides to get y 4 /4 = t2 /2 + c, c a real constant. This can be simplified to y 4 = 2t2 + c. (where we replace 4c by c) We leave the answer in implicit form. 13. The variables are already separated, so integrate both sides to get y 5 /5 = t2 /2 + 2t + c, c a real constant. Simplifying gives y 5 = 52 t2 + 10t + c. We leave the answer in implicit form 14. There is an equilibrium solution y = 0. Separating variables give y −2 y ! = t and integrating gives −y −1 = t2 /2 + c. Thus y = −2/(t2 + 2c), c a real constant. This is equivalent to writing y = −2/(t2 + c), c a real constant, since twice an arbitrary constant is still an arbitrary constant.

14

1 Solutions

15. In standard form we get y ! = (1 − y) tan t so y = 1 is a solution. Sepady rating variables gives 1−y = tan t dt. The function tan t is continuous on the interval (−π/2, π/2) and so has an antiderivative. Integration gives − ln |1 − y| = − ln |cos t| + k1 . Multiplying by −1 and exponentiating gives |1 − y| = k2 |cos t| where k2 is a positive constant. Removing the absolute value signs gives 1 − y = k3 cos t, with k3 &= 0. If we allow k3 = 0 we get the equilibrium solution y = 1. Thus the solution can be written y = 1 − c cos t, c any real constant. 16. An equilibrium solution is y = 0. Separating variables gives y −n dy = 1−n m+1 tm dt and integrating gives y1−n = tm+1 +c, c a real constant. Simplifying 1−n m+1 gives y 1−n = m+1 t + c, and the equilibrium solution y = 0. 17. There are two equilibrium solutions; y = " 0 and y#= 4. Separating vari1 1 1 ables and using partial fractions gives 4 y + 4−y dy = dt. Integrating $ $ $ y $ y and simplifying gives ln $ 4−y $ = 4t+k1 which is equivalent to 4−y = ce4t , 4t

4ce c a nonzero constant. Solving for y gives y = 1+ce 4t . When c = 0 we get the equilibrium solution y = 0. However, there is no c which gives the other equilibrium solution y = 4.

18. There are no equilibrium solutions. Separating variables gives y2y+1 dy = dt and integrating gives 12 ln(y 2 + 1) = t + k. Solving for y 2 gives y 2 = ce2t − 1, where c > 0. 19. Separating variables gives y2dy+1 = dt and integrating gives tan−1 y = t+c. Thus y = tan(t + c), c a real constant. % & 2 20. Separating variables gives y dy = − 1t − t dt and integrating gives y2 = 2 − ln |t| − t2 + c. Simplifying gives y 2 + t2 + ln t2 = c, c a real constant. 1 21. In standard form we get y ! = −(y+1) y−1 1+t2 from which we see that y = −1 solution. Separating variables and simplifying gives " is an equilibrium # 2 dt 2 y+1 − 1 dy = t2 +1 . Integrating and simplifying gives ln(y + 1) − y =

tan−1 t + c.

22. Separating variables gives 2y dy = et dt and integrating gives y 2 = et + c, c a constant. 23. The equilibrium solution is y = 0. Separating variables gives y −2 dy = dt 1 1−t . Integrating and simplifying gives y = ln|1−t|+c , c real constant. 24. In standard form we get y ! = y(y+1) from which we see y = 0 and y = −1 are equilibrium solutions. The equilibrium solution y(t) = 0 satisfies the initial condition y(0) = 0 so y(t) = 0 is the required solution.

1 Solutions

15

25. y = 0 is the only equilibrium solution. The equilibrium solution y(t) = 0 satisfies the initial condition y(1) = 0 so y(t) = 0 is the required solution. dy 26. Rewriting we get y ! = dx = x+2 = 0 is x y from which we see% that y & dy 2 an equilibrium solution. Separating variables gives y = 1 + x dx and integrating gives ln |y| = x + ln x2 + k, k a constant. Solving for y by taking the exponential of both sides gives y = cx2 ex , and allowing c = 0 gives the equilibrium solution. The initial condition gives e = y(1) = ce so c = 1. Thus y = x2 ex .

27. In standard form we get y ! = −2ty so y = 0 is a solution. Separating variables and integrating gives ln |y| = −t2 + k. Solving for y gives 2 y = ce−t and allowing c = 0 gives the equilibrium solution. The initial 2 condition implies 4 = y(0) = ce0 = c. Thus y = 4e−t . 28. Since cot y = 0 at y = π2 + mπ for all integers m we have equilibrium lines at y = π2 + mπ, none of which satisfy the initial condition y(1) = π4 . Separating variables gives tan y dy = dt t and integrating gives − ln |cos y| = ln t + c. We can" solve for c here using the initial condition: # we get c = − ln cos π4 = − ln

√

2 2

. Solving for y gives y = cos−1

√1 2t

u 29. Separating variables gives dy y = u2 +1 du√and integrating gives ln |y| = √ 2 ln u2 + 1 + k. Solving for y gives y = √ c u + 1, for c &= 0. The initial 2 condition gives 2 = y(0) = c. So y = 2 u + 1.

30. In standard form we get y ! = Separating variables gives

dy y

so y = 0 is an equilibrium solution. # 2 = 1 − t+2 dt. Integrating we get ln |y| = t y t+2 "

t

e t − 2 ln |t + 2| + k. Solving for y we get y = c (t+2) 2 , for c &= 0. However, allowing c = 0 gives the equilibrium solution.

31. Since y 2 + 1 ≥ 1 there are no equilibrium solutions. Separating the variables gives dy dt = 2, 2 y +1 t and integration of both sides gives tan−1 y = − 1t + c. Solve for y by applying the tangent function to both sides of the equation. Since tan(tan−1 y) = y, we get % 1 & y(t) = tan − + c . t √ To find c observe that 3 = y(1) = tan(−1 + c), which implies that c − 1 = π/3, so c = 1 + π/3. Hence

16

1 Solutions

' ( 1 π y(t) = tan − + 1 + . t 3 To determine the maximum domain on which this solution is defined, note that the tangent function is defined on the interval (−π/2, π/2), so that y(t) is defined for all t satisfying −

π 1 π π 6/(6 + 5π). The second equality is solved to give t < 6/(6 − π). Thus the maximum domain for the solution y(t) is the% interval (a,& b) = (6/(6 + 5π), 6/(6 − π)). limt→b− y(t) = limt→b− tan − 1t + 1 + π3 = limx→π/2− tan x = ∞.

32. We assume the decay model N (t) = N (0)e−λt . If t is the age of the bone then N (t) = 13 N (0). Thus 13 = e−λt . Solving for t gives t = lnλ3 = 5730 ln 3 ≈ 9082 years ln 2 33. Let m denote the number of Argon-40 atoms in the sample. Then 8m is the number of Potassium-40 atoms. Let t be the age of the rock. Then t years ago there were m + 8m = 9m atoms of Potassium-40. Hence N (0) = 9m. On the other hand, 8m = N (t) = N (0)e−λt = 9me−λt . This − ln 8 −τ 8 implies that 89 = e−λt and hence t = λ 9 = ln 2 ln 9 ≈ 212 million years old. 34. We need only solve .3N (0) = N (0)e−λt for t. We get t = − lnλ.3 = ln .3 − 5.27 = 9.15 years. ln 2 35. The ambient temperature is 32◦ F, the temperature of the ice water. From Equation (13) we get T (t) = 32 + kert . At t = 0 we get 70 = 32 + k, so k = 38 and T (t) = 32 + 38ert . After 30 minutes we have 1 55 = T (30) = 32 + 38e30r and solving for r gives r = 30 ln 23 38 . To find the rt time t when T (t) = 45 we solve 45 = 32 + 38e , with r as above. We get 13−ln 38 t = 30 ln ln 23−ln 38 ≈ 64 minutes. 36. The ambient temperature is Ta = 70◦ . Equation (13) gives T (t) = 70 + kert for the temperature of the coffee at time t. Since the initial temperature of the coffee is T (0) = 180 we get 180 = T (0) = 70 + k. Thus k = 110. The constant r is determined from the temperature at a 7 second time: 140 = T (3) = 70 + 110e3r so r = 13 ln 11 ≈ −.1507. Thus rt T (t) = 70 + 110e , with r as calculated. The temperature requested is % 7 & 53 T (5) = 70 + 110 11 ≈ 121.8◦ .

37. The ambient temperature is Ta = 65◦ . Equation (13) gives T (t) = 65 + kert for the temperature at time t. Since the initial temperature of the

1 Solutions

17

thermometer is T (0) = 90 we get 90 = T (0) = 65 + k. Thus k = 25. The constant r is determined from the temperature at a second time: 85 = T (2) = 65 + 25e2r so r = 12 ln 45 . Thus T (t) = 65 + 25ert , with r = 12 ln 45 . To answer the first question we solve the equation 75 = T (t) = 65 + 25ert 2−ln 5 for t. We get t = 2 ln ≈ 8.2 minutes. The temperature at t = 20 is % 4 &ln104−ln 5 ◦ T (20) = 65 + 25 5 ≈ 67.7 .

38. The ambient temperature is Ta = 70◦ . Equation (13) gives T (t) = 70 + kert for the temperature of the soda at time t. Since the initial temperature of the soda is T (0) = 40 we get 40 = T (0) = 70 + k. Thus k = −30. The constant r is determined from the temperature at a second time: 60 = T (2) = 70 − 30e2r so r = 12 ln 13 . Thus T (t) = 70 − 30ert , with 1 1 30 r = 12 ln 13 . The temperature at t = 1 is T (1) = 70−30e 2 ln 3 = 70− √ ≈ 3 ◦ 52.7 . 39. The ambient temperature is Ta = 70◦ . Equation (13) gives T (t) = 70 + kert for the temperature of the coffee at time t. We are asked to determine the initial temperature of the coffee so T (0) is unknown. However, we have the equations 150 = T (5) = 70 + ke5r 142 = T (6) = 70 + ke6r or 80 = ke5r 72 = ke4r . r Dividing the second equation by the first gives 72 80 = e so r = ln 0.9. −5r From the first equation we get k = 80e ≈ 135.5. We now calculate T (0) = 70 + k ≈ 205.5◦

40. The ambient temperature is Ta = 40◦ . Equation (13) gives T (t) = 40 + kert for the temperature of the beer at time t. Since the initial temperature of the beer is T (0) = 80 we get 80 = T (0) = 40 + k. Thus k = 40. The constant r is determined from the temperature at a second time: 60 = T (1) = 40 + 40er so r = − ln 2. Thus T (t) = 40 + 40ert , with r = − ln 2. We now solve the equation 50 = T (t) = 40 + 40ert for t and ln 4 get t = − − ln 2 = 2. She should therefore put the beer in the refrigerator at 2 p.m. 41. Let us start time t = 0 at 1980. Then P (0) = 290. The Malthusian growth model gives P (t) = 290ert . At t = 10 (1990) we have 370 = 290e10r 1 30r and hence r = 10 ln 37 = 29 . At t = 30 (2010) we have P (30) = 290e % 37 &3 290 29 ≈ 602.

18

1 Solutions

42. The initial population is 40 = P (0). Since the population doubles in 3 hours we have P (3) = 80 or 80 = 40e3r . Hence r = ln32 . Now we can compute the population after 30 hours: P (30) = 40e30r = 40(210 ) = 40, 960. 43. We have 3P (0) = P (5) = P (0)e3r . So r = 2P (0) = P (t) = P (0)ert for t. We get t =

ln 3 5 . Now we solve the equation ln 2 5 ln 2 r = ln 3 ≈ 3.15 years.

44. In the logistic growth equation m = 800 and P (0) = 290. Thus P (t) = 800·290 800·290 290+510e−rt . To determine r we use P (10) = 370 to get 370 = 290+510e−10r . 1 1887 A simple calculation give r = 10 ln 1247 . Now the population in 2010 is 800·290 P (30) = 3 ≈ 530 290+510( 1247 1887 ) 45. In the logistics equation m = 5000 and P0 = 2000. Thus P (t) = 10,000,000 10,000 10,000 2,000+3,000e−rt = 2+3e−rt . Since P (2) = 3000 we get 3000 = 2+3e−rt . Solv10,000 10,000 ing this equation for r gives r = ln 32 . Now P (4) = 2+3e 4 ≈ −4r = 2+3( 23 ) 3857 46. Let x = e−rt0 . Then x2 = e−2rt . The equation P (t0 ) = P1 implies that P0 (m−P2 ) 0 (m−P1 ) 2 x= P P1 (m−P0 ) . The equation P (2t0 ) = P2 implies x = P2 (m−P0 ) . These P02 (m−P1 )2 0 (m−P2 ) = P P2 (m−P0 ) . Cross multiplying and P12 (m−P0 )2 (P0 P2 −P12 )m+(P12 P0 +P12 P2 −2P0 P1 P2 ) = 0. Solving −rt0

equation together imply

simplifying leads to for m gives the result. Now replace the formula for m into e =x= P0 (m−P1 ) P0 P2 −P1 −rt0 . Simplifying gives e = . The formula for r follows P1 (m−P0 ) P2 P1 −P0 after taking the natural log of both sides.

47. We have P (0) = P0 = 400, P (3) = P1 = 700, and P (6) = P2 = 1000. Using the result of the previous problem we get m = 700(700(400+1000)−2·400·1000) = (700)2 −400·1000 1, 400

Section 1.4 1. This equation is already in standard form with p(t) = 3. An antiderivative ) of p(t) is P (t) = 3 dt = 3t so the integrating factor is µ(t) = e3t . If we multiply the differential equation y ! +3y = et by µ(t), we get the equation e3t y ! + 3e3t y = e4t , and the left hand side of this equation is a perfect derivative, namely, (e3t y)! . Thus, (e3t y)! = e4t . Now take antiderivatives of both sides and multiply by e−3t . This gives

1 Solutions

19

y=

1 t e + ce−3t 4

for the general solution of the equation. To find the constant c to satisfy the initial condition y(0) = −2, substitute t = 0 into the general solution to get −2 = y(0) = 14 + c. Hence c = − 94 , and the solution of the initial value problem is 1 9 y = et − e−3t . 4 4 2. Divide by cos t to put the equation in the standard form y ! + (tan t)y = sec t. In this case p(t) = tan t, an antiderivative is P (t) = ln(sec t), and the integrating factor is µ(t) = sec t. (We do not need | sec t| since we are working near t = 0 where sec t > 0.) Now multiply by the integrating factor to get (sec t)y ! + (sec t tan t)y = sec2 t, the left hand side of which is a perfect derivative. Thus ((sec t)y)! = sec2 t and taking antiderivatives of both sides gives (sec t)y = tan t + c where c ∈ R is a constant. Now multiply by 1/ sec t = cos t to get y = sin t+c cos t for the general solution. Letting t = 0 gives 5 = y(0) = sin 0 + c cos 0 = c so c = 5 and y = sin t + 5 cos t. 3. This equation is already in standard form. In this case p(t) = −2, an antiderivative is P (t) = −2t, and the integrating factor is µ(t) = e−2t . Now multiply by the integrating factor to get e−2t y ! − 2e−2t y = 1, the left hand side of which is a perfect derivative ((e−2t )y)! . Thus ((e−2t )y)! = 1 and taking antiderivatives of both sides gives (e−2t )y = t + c, where c ∈ R is a constant. Now multiply by e2t to get y = te2t + ce2t for the general solution. Letting t = 0 gives 4 = y(0) = c so y = te2t + 4e2t . 4. Divide by t to put the equation in the standard form 1 et y! + y = t t

20

1 Solutions

In this case p(t) = 1/t, an antiderivative is P (t) = ln t, and the integrating factor is µ(t) = t. Now multiply the standard form equation by the integrating factor to get ty ! + y = et , the left hand side of which is a perfect derivative (ty)! . (Note that this is just the original left hand side of the equation. Thus if we had recognized that the left hand side was already a perfect derivative, the preliminary steps could have been skipped for this problem, and we could have proceeded directly to the next step.) Thus the equation can be written as (ty)! = et and taking antiderivatives of both sides gives ty = et + c where c ∈ R is a constant. Now divide by t to get et c y= + t t for the general solution. 5. The general solution from Problem 4 is y = t 0 = e + c. So c = −e and y = et − et .

et t

+ ct . Now let t = 1 to get

6. Divide by t to put the equation in the standard form y! +

m y = ln t. t

m In this case p(t) = m t , an antiderivative is P (t) = m ln t = ln t , and the m integrating factor is µ(t) = t . Now multiply the standard form equation by the integrating factor to get tm y ! + mtm−1 y = tm ln t, the left hand side of which is a perfect derivative ((tm )y)! . Thus ((tm )y)! = tm ln t. To integrate tm ln t we consider the cases m = −1 and m &= −1 separately. ) 2 Case m = −1: A simple substitution gives t−1 ln t dt = (ln2t) + c.

Hence, t−1 y =

Case m &= −1: tm+1 (m+1)2

(ln t)2 2

+ c and so y =

t(ln t)2 2

+ ct

Use integration by parts to get

+ c. Then y =

t ln t m+1

−

t (m+1)2

+

tm . c

)

tm ln t dt =

tm+1 ln t m+1

−

7. We first put the equation in standard form and get 1 y ! + y = cos(t2 ). t In this case p(t) = 1t , an antiderivative is P (t) = ln t, and the integrating factor is µ(t) = t. Now multiply by the integrating factor to get ty ! + y = t cos(t2 ), the left hand side of which is a perfect derivative (ty)! . Thus (ty)! = t cos(t2 ) and taking antiderivatives of both sides gives ty = 12 sin(t2 ) + c

1 Solutions

21

where c ∈ R is a constant. Now divide by t to get y = general solution.

sin(t2 ) 2t

+ ct . for the

!

8. In this case p(t) = 2 and the integrating factor is e 2 dt = e2t . Now multiply to get e2t y ! + 2e2t y = e2t sin t, which simplifies to (e2t y)! = 1 e2t sin t. Now integrate both sides to get e2t y = (− cos t + 2 sin t)e2t + c, 5 ) 2t where we computed e sin t by parts two times. Dividing by e2t gives 1 y = (2 sin t − cos t) + ce−2t . 5 !

9. In this case p(t) = −3 and the integrating factor is e −3 dt = e−3t . Now multiply to get e−3t y ! − 3e−3t y = 25e−3t cos 4t, which simplifies to (e−3t y)! = 25e−3t cos 4t. Now integrate both) sides to get e−3t y = (4 sin 4t − 3 cos 4t)e−3t + c, where we computed 25e−3t cos 4t by parts twice. Dividing by e−3t gives y = 4 sin 4t − 3 cos 4t + ce3t . 10. In standard form this equation becomes y! −

1 2 y= . t(t + 1) t(t + 1)

−1 1 Using partial fractions we get p(t) = t(t+1) = t+1 − 1t , an antiderivative is % t+1 & P (t) = ln(t+1)−ln t = ln t , and the integrating factor is µ(t) = t+1 t . Now multiply by the integrating factor to get

t+1 ! 1 2 y − 2y = 2, t t t ! the left hand side of which is a perfect derivative ( t+1 t y) . Thus

(

t+1 ! 2 y) = 2 t t

−2 and taking antiderivatives of both sides gives t+1 t y = t + c where c ∈ R t −2 ct is a constant. Now multiply by t+1 to get y = t+1 + t+1 = ct−2 t+1 for the general solution.

11. In we get z ! − 2tz = −2t3 . An integrating factor is ! standard form 2 2 −2t dt −t2 e = e . Thus (e−t z)! = −2t3 e−t . Integrating both sides gives 2 2 e−t z = (t2 + 1)e−t + c, where the integral of the right hand side is done 2 2 by parts. Now divide by the integrating factor e−t to get z = t2 +1+cet . 12. The given differential equation is in standard form, p(t) = a, an antiderivative is P (t) = at, and the integrating factor is µ(t) = eat . Now multiply by the integrating factor to get eat y ! + aeat y = beat ,

22

1 Solutions

the left hand side of which is a perfect derivative ((eat )y)! . Thus ((eat )y)! = beat . If a &= 0 then taking antiderivatives of both sides gives eat y = ab eat + c where c ∈ R is a constant. Now multiply by e−at to get y = ab + ce−at for the general solution. In the case a = 0 then y ! = b and y = bt + c. 13. The given equation is in standard form, p(t) = cos t, an antiderivative is P (t) = − sin t, and the integrating factor is µ(t) = e− sin t . Now multiply by the integrating factor to get e− sin t y ! + (cos t)e− sin t y = (cos t)e− sin t , the left hand side of which is a perfect derivative ((e− sin t )y)! . Thus ((e− sin t )y)! = (cos t)e− sin t and taking antiderivatives of both sides gives (e− sin t )y = e− sin t + c where c ∈ R is a constant. Now multiply by esin t to get y = 1 + cesin t for the general solution. To satisfy the initial condition, 0 = y(0) = 1 + cesin 0 = 1 + c, so c = −1. Thus, the solution of the initial value problem is y = 1 − esin t −2 14. The given equation is in standard form, p(t) = t+1 , an antiderivative −2 is P (t) = −2 ln(t + 1) = ln((t + 1) ), and the integrating factor is µ(t) = (t + 1)−2 . Now multiply by the integrating factor to get

(t + 1)−2 y ! −

2 y = 1, (t + 1)3

the left hand side of which is a perfect derivative (((t + 1)−2 )y)! . Thus (((t + 1)−2 )y)! = 1 and taking antiderivatives of both sides gives ((t + 1)−2 )y = t + c where c ∈ R is a constant. Now multiply by (t + 1)2 to get y = (t + c)(t + 1)2 for the general solution. 15. The given linear differential equation is in standard form, p(t) = −2 t , an antiderivative is P (t) = −2 ln t = ln t−2 , and the integrating factor is µ(t) = t−2 . Now multiply by the integrating factor to get t−2 y ! −

2 t+1 y = 3 = t−2 + t−3 , 3 t t

the left hand side of which is a perfect derivative (t−2 y)! . Thus

1 Solutions

23

(t−2 y)! = t−2 + t−3 −2

and taking antiderivatives of both sides gives (t−2 )y = −t−1 − t 2 + c where c ∈ R is a constant. Now multiply by t2 to and we get y = −t − 1 −2 for the general solution. Letting t = 1 gives −3 = y(1) = −3 2 + ct 2 +c −3 so c = 2 and 1 3 y(t) = −t − − t−2 . 2 2 16. The given equation is in standard form, p(t) = a, an antiderivative is P (t) = at, and the integrating factor is µ(t) = eat . Now multiply by the integrating factor to get eat y ! + aeat y = 1, the left hand side of which is a perfect derivative (eat y)! . Thus (eat y)! = 1 and taking antiderivatives of both sides gives eat y = t + c where c ∈ R is a constant. Now multiply by e−at to get y = te−at + ce−at for the general solution. 17. The given equation is in standard form, p(t) = a, p(t) = a, an antiderivative is P (t) = at, and the integrating factor is µ(t) = eat . Now multiply by the integrating factor to get eat y ! + aeat y = e(a+b)t , the left hand side of which is a perfect derivative (eat y)! . Thus (eat y)! = e(a+b)t and taking antiderivatives of both sides gives (eat )y =

1 (a+b)t e +c a+b

where c ∈ R is a constant. Now multiply by e−at to get y=

1 bt e + ce−at a+b

for the general solution. 18. The given differential equation is in standard form, p(t) = a, an antiderivative is P (t) = at, and the integrating factor is µ(t) = eat . Now multiply by the integrating factor to get eat y ! + aeat y = tn , the left hand side of which is a perfect derivative (eat y)! . Thus (eat y)! = tn . Now assume n &= −1. Then taking antiderivatives of both sides gives n+1 (eat )y = tn+1 + c where c ∈ R is a constant. Now multiply by e−at to n+1

get y = tn+1 e−at + ce−at for the general solution. If n = 1 then taking antiderivatives leads to eat y = ln t + c and hence y = (ln t)e−at + ce−at is the general solution in this case.

19. In standard form we get y ! − (tan t)y = sec t. In this case p(t) = − tan t, an antiderivative is P (t) = ln cos t, and the integrating factor is µ(t) = eP (t) = cos t. Now multiply by the integrating factor to get (cos t)y ! − (sin t)y = 1, the left hand side of which is a perfect derivative ((cos t)y)! . Thus ((cos t)y)! = 1 and taking antiderivatives of both

24

1 Solutions

sides gives (cos t)y = t + c where c ∈ R is a constant. Now multiply by 1/ cos t = sec t and we get y = (t + c) sec t for the general solution. 20. Divide by t to put the equation in the standard form y! +

2 ln t 4 ln t y= . t t

t 2 In this case p(t) = 2 ln t , an antiderivative is P (t) = (ln t) , and the 2 integrating factor is µ(t) = e(ln t) . Now multiply by the integrating factor to get 2 2 ln t (ln t)2 4 ln t (ln t)2 e(ln t) y ! + e y= e , t t 2

the left hand side of which is a perfect derivative (e(ln t) y)! . Thus 2 t (ln t)2 (e(ln t) y)! = 4 ln and taking antiderivatives of both sides gives t e 2

2

e(ln t) y = 2e(ln t) + c 2

where c ∈ R is a constant. Now multiply by e−(ln t) and we get y = 2 2 + ce−(ln t) for the general solution. 21. The given differential equation is in standard form, p(t) = −n/t, an antiderivative is P (t) = −n ln t = ln(t−n ), and the integrating factor is µ(t) = t−n . Now multiply by the integrating factor to get t−n y ! − nt−n−1 y = et , the left hand side of which is a perfect derivative (t−n y)! . Thus (t−n y)! = et and taking antiderivatives of both sides gives (t−n )y = et + c where c ∈ R is a constant. Now multiply by tn to and we get y = tn et + ctn for the general solution. 22. The given differential equation is in standard form, p(t) = −1, an antiderivative is P (t) = −t, and the integrating factor is µ(t) = e−t . Now multiply by the integrating factor to get e−t y ! − e−t y = tet , the left hand side of which is a perfect derivative (e−t y)! . Thus (e−t y)! = tet . Taking antiderivatives of both sides and using integration by parts gives e−t y = tet − et + c = (t − 1)et + c where c ∈ R is a constant. Now multiply by et to get y = (t − 1)e2t + cet for the general solution. Letting t = 0 gives a = y(0) = −1 + c so c = a + 1 and y = (t − 1)e2t + (a + 1)et . 23. Divide by t to put the equation in the standard form 3 y ! + y = t. t

1 Solutions

25

In this case p(t) = 3/t, an antiderivative is P (t) = 3 ln t = ln(t3 ), and the integrating factor is µ(t) = t3 . Now multiply the standard form equation by the integrating factor to get t3 y ! +3t2 y = t4 , the left hand side of which is a perfect derivative (t3 y)! . Thus (t3 y)! = t4 and taking antiderivatives of both sides gives t3 y = 15 t5 + c where c ∈ R is a constant. Now multiply by t−3 and we get y = 15 t2 + ct−3 for the general solution. Letting t = −1 gives 2 = y(−1) = 15 − c so c = −9 5 and y=

1 2 9 −3 t − t . 5 5

24. In this case the given differential equation is in standard form. We have p(t) = 2t, an antiderivative is P (t) = t2 , and the integrating 2 factor is µ(t) = et . Now multiply by the integrating factor to get 2 2 2 et y ! + 2tet y = et , the left hand side of which is a perfect derivative 2 2 2 (et y)! . Thus (et y)! = et and taking antiderivatives of both sides gives ) 2 2 et y = et dt + c where c ∈ R is a constant. However, the right hand side has no closed form antiderivative. Using Corollary 8 to can write * t * t 2 2 2 2 2 2 y = e−t eu du + y(0)e−t = e−t eu du + e−t . 0

0

25. Divide by t2 to put the equation in the standard form 2 y ! + y = t−2 . t In this case p(t) = 2/t, an antiderivative is P (t) = 2 ln t = ln t2 , and the integrating factor is µ(t) = t2 . Now multiply by the integrating factor to get t2 y ! +2ty = 1, the left hand side of which is a perfect derivative (t2 y)! . Thus (t2 y)! = 1 and taking antiderivatives of both sides gives t2 y = t + c where c ∈ R is a constant. Now multiply by t−2 to get y = 1t + ct−2 for the general solution. Letting t = 2 gives a = y(2) = 12 + 4c so c = 4a − 2 and 1 y = + (4a − 2)t−2 . t gal lbs lbs ×0 =0 . min gal min gal y(t) lbs 4 lbs output rate = 4 × = y(t) . min 100 gal 100 min

26. input rate: input rate = 4 output rate: Since

y ! = input rate − output rate

we get the initial value problem

26

1 Solutions

y! = 0 −

4 y, 100

y(0) = 80.

1 Simplifying and putting in standard form gives y ! + y = 0. The coeffi25 ) cient function is p(t) = 1/25, P (t) = p(t) dt = t/25, and the integrating factor is µ(t) = et/25 . Thus (et/25 y)! = 0. Integrating and simplifying gives y = ce−t/25 . The initial condition implies c = 80 so y = 80e−t/25 . The concentration of the brine solution is now obtained by dividing by 80 −t/25 the volume which is 100 gallons: 100 e = 0.8e−t/25 . 27. Let V (t) denote the volume of fluid in the tank at time t. Initially, there are 10 gal of brine. For each minute that passes there is a net decrease of 4 − 3 = 1 gal of brine. Thus V (t) = 10 − t gal. gal lbs lbs input rate: input rate = 3 ×1 =3 . min gal min gal y(t) lbs 4y(t) lbs output rate: output rate = 4 × = . min V (t) gal 10 − t min ! Since y = input rate−output rate, it follows that y(t) satisfies the initial value problem 4 y! = 3 − y(t) , y(0) = 2. 10 − t Put in standard form, this equation becomes y! +

4 y = 3. 10 − t

) 4 The coefficient function is p(t) = 10−t , P (t) = p(t) dt = −4 ln(10 − t) = ln(10 − t)−4 , and the integrating factor is µ(t) = (10 − t)−4 . Multiplying the standard form equation by the integrating factor gives ((10 − t)−4 y)! = 3(10 − t)−4 . Integrating and simplifying gives y = (10 − t) + c(10 − t)4 . The initial condition y(0) = 2 implies 2 = y(0) = 10 + c104 and hence c = −8/104 so 8 y = (10 − t) − 4 (10 − t)4 . 10 Of course, this formula is valid for 0 ≤ t ≤ 10. After 10 minutes there is no fluid and hence no salt in the tank. L g g × 10 = 10 . min L min L y(t) g y(t) g output rate: output rate = 1 × = . min 10 L 10 min ! Since y = input rate−output rate, it follows that y(t) satisfies the initial value problem

28. input rate: input rate = 1

1 Solutions

27

y ! = 10 −

1 y, 10

y(0) = 0.

1 Put in standard form, this equation becomes y ! + 10 y = 10. The coefficient ) 1 function is p(t) = 10 , P (t) = p(t) dt = t/10, and the integrating factor is µ(t) = et/10 . Multiplying the standard form equation by the integrating factor gives (et/10 y)! = 10et/10 . Integrating and simplifying gives y = 100 + ce−t/10 . The initial condition y(0) = 0 implies c = −100 so y = 100 − 100e−t/10 . After 10 minutes we have y(10) = 100 − 100e−1 g of salt. The concentration is thus (100 − 100e−1 )/10 = 10 − 10e−1 g/L

29. Let V (t) denote the volume of fluid in the container at time t. Initially, there are 10 L. For each minute that passes there is a net gain of 4−2 = 2 L of fluid. So V (t) = 10 + 2t. The container overflows when V (t) = 10 + 2t = 30 or t = 10 minutes. L g g input rate: input rate = 4 × 20 = 80 . min L min L y(t) g 2y(t) g output rate: output rate = 2 × = . min 10 + 2t L 10 + 2t min Since y ! = input rate−output rate, it follows that y(t) satisfies the initial value problem 2y y ! = 80 − , y(0) = 0. 10 + 2t Simplifying and putting in standard form gives the equation y! +

1 y = 80. 5+t

) 1 The coefficient function is p(t) = 5+t , P (t) = p(t) dt = ln(5+t), and the integrating factor is µ(t) = 5 + t. Multiplying the standard form equation by the integrating factor gives ((5 + t)y)! = 80(5 + t). Integrating and simplifying gives y = 40(5+t)+c(5+t)−1 , where c is a constant. The initial condition y(0) = 0 implies c = −1000 so y = 40(5 + t) − 1000(5 + t)−1 . At the time the container overflows t = 10 we have y(10) = 600 − 1000 15 ≈ 533.33 g of salt. 30. Let V (t) denote the volume of fluid in the tank at time t. Initially, there are 10 gallons of fluid. For each minute that goes by there is a net increase of 4 − 2 = 2 gallons. It follows that V (t) = 10 + 2t. The tank will overflow when 100 = V (t). Solving 100 = 10 + 2t gives t = 45. Thus T = 45 minutes. Next we find y(t): gal lbs lbs input rate: input rate = 4 × 0.5 =2 . min gal min gal y(t) lbs 2y(t) lbs output rate: output rate = 2 × = . min 10 + 2t gal 10 + 2t min

28

1 Solutions

Since y ! = input rate−output rate, it follows that y(t) satisfies the initial value problem y y! = 2 − , y(0) = 0. 5+t Putting this equation in standard form gives y! +

1 y = 2. 5+t

) 1 The coefficient function is p(t) = 5+t , P (t) = p(t) dt = ln(5+t), and the integrating factor is µ(t) = 5 + t. Thus ((5 + t)y)! = 2(5 + t). Integrating and simplifying gives y = (5+t)+c(5+t)−1 . The initial condition y(0) = 0 implies c = −25 so y = (5+t)−25(5+t)−1 , for 0 ≤ t ≤ 45. At t = T = 45 we get y(45) = 50 − 25 50 = 49.5 lbs salt. Once the tank is full, the inflow and outflow rates will be equal and the brine in the tank will (in the limit as t → ∞) stabilize to the concentration of the incoming brine, i.e., 0.5 lb/gal. Since the tank holds 100 gal, the total amount present will approach 0.5 × 100 = 50 lbs. Thus limt→∞ y(t) = 50. 31. input rate: input rate = rc output rate:

output rate = r PV(t)

Let P0 denote the amount of pollutant at time t = 0. Since P ! = input rate − output rate it follows that P (t) is a solution of the initial value problem rP (t) P ! = rc − , P (0) = P0 . V Rewriting this equation in standard form gives the differential equation P ! + Vr P = rc. The coefficient function is p(t) = r/V and the integrating rt rt factor is µ(t) = ert/V . Thus (e V P )! = rce V . Integrating and simplifying −rt gives P (t) = cV +ke V , where k is the constant of integration. The initial −rt condition P (0) = P0 implies c = P0 − cV so P (t) = cV + (P0 − cV )e V . (a) limt→∞ P (t) = cV.

(b) When the river is cleaned up at t = 0 we assume the input concentration is c = 0. The amount of pollutant is therefore given by −rt P (t) = P0 e V . This will reduce by 1/2 when P (t) = 12 P0 . We solve −rt the equation 12 P0 = P0 e V for t and get t1/2 = V lnr2 . Similarly, the pollutant will reduce by 1/10 when t1/10 = V lnr10 . (c) Letting V and r be given as stated for each lake gives: Lake Erie: t1/2 = 1.82 years, t1/10 = 6.05 years. Lake Ontario: t1/2 = 5.43 years, t1/10 = 18.06 years 32. Let y1 (t) and y2 (t) denote the amount of salt in Tank 1 and Tank 2, respectively, at time t.

1 Solutions

29

L g g × 100 = 400 . min L min L y1 (t) g 4y1 (t) g output rate for Tank 1: output rate = 4 × = . min 10 L 10 m The initial value problem for Tank 1 is thus: input rate for Tank 1: input rate = 4

y1! = 400 −

4 y1 , 10

y1 (0) = 0.

4 Simplifying and putting this equation in standard form gives y1! + 10 y1 = 4t/10 4t/10 ! 4t/10 400. The integrating factor is µ(t) = e . Thus (e y1 ) = 400e . Integrating and simplifying gives y1 = 1000 + ce−4t/10 . The initial condition y1 (0) = 0 implies c = −1000 so y1 = 1000 − 1000e−4t/10 . Now the brine solution in Tank 1 has concentration y1 (t)/10 = 100 − 100e−4t/10 and flows into Tank 2 at a rate of 4 liters per minute. Thus L g input rate for Tank 2: input rate = 4 × (100 − 100e−4t/10 ) = min L g 400 − 400e−4t/10 . min L y2 (t) g output rate for Tank 2: output rate = 4 × = min 10 L 4y2 (t) g . 10 min The initial value problem for Tank 2 is thus:

y2! = 400 − 400e−4t/10 −

4 y2 , 10

y2 (0) = 0.

Simplifying and putting this equation in standard form gives y2! +

4 y2 = 400 − 400e−4t/10 . 10

The integrating factor is again (as for the Tank 1 equation) µ(t) = e4t/10 . Thus multiplying by the integrating factor gives (e4t/10 y2 )! = 400e4t/10 − 400. Integrating and simplifying gives y2 (t) = 1000 − 400te−4t/10 + ce−4t/10 . The initial condition y2 (0) = 0 implies c = −1000 so y2 (t) = 1000 − 400te−4t/10 − 1000e−4t/10 .

30

1 Solutions

33. Let y1 (t) and y2 (t) denote the amount of salt in Tank 1 and Tank 2, respectively, at time t. The volume of fluid at time t in Tank 1 is V1 (t) = 10 + 2t and Tank 2 is V2 (t) = 5 + t. L g g input rate for Tank 1: input rate = 4 × 10 = 40 . min L min L y1 (t) g output rate for Tank 1: output rate = 2 × = min 10 + 2t L 2y(t) g . The initial value problem for Tank 1 is thus 10 + 2t min y1! = 40 −

2 y1 , 10 + 2t

y1 (0) = 0.

Simplifying this equation and putting it in standard form gives y1! +

1 y1 = 40. 5+t

The integrating factor is µ(t) = 5 + t. Thus ((5 + t)y1 )! = 40(5 + t). Integrating and simplifying gives y1 (t) = 20(5 + t) + c/(5 + t). The initial condition y(0) = 0 implies c = −500 so y1 = 20(5 + t) − 500/(5 + t). L y1 (t) g input rate for Tank 2: input rate = 2 × = 20 − min 10 + 2t L 500 g . (5 + t)2 min output rate for Tank 2:

output rate = 1

The initial value problem for Tank 2 is thus y2! = 20 − 500/(5 + t)2 −

L y2 (t) g y2 (t) g × = . min 5 + t L 5 + t min

1 y2 , (5 + t)

y2 (0) = 0.

When this equation is put in standard form we get y2! +

1 500 y2 = 20 − . (5 + t) (5 + t)2

The integrating factor is µ(t) = 5 + t. Thus ((5 + t)y2 )! = 20(5 + t) −

500 . 5+t

Integrating and simplifying gives y2 (t) = 10(5 + t) −

500 ln(5 + t) c + . 5+t 5+t

The initial condition y2 (0) = 0 implies c = 500 ln 5 − 250 so

1 Solutions

31

y2 (t) = 10(5 + t) −

500 ln(5 + t) 500 ln 5 − 250 + . 5+t 5+t

Section 1.5 y 2 + yt + t2 1. In standard form we get y ! = which is homogeneous since t2 the degrees of the numerator and denominator are each two. Let y = tv. Then v + tv ! = v 2 + v + 1 and so tv ! = v 2 + 1. Separating variables gives dv dt = . Integrating gives tan−1 v = ln |t| + c. So v = tan(ln |t| + c). v2 + 1 t Substituting v = y/t gives y = t tan(ln |t| + c). The initial condition implies 1 = y(1) = 1 · tan c = tan c and hence c = π/4. Therefore y(t) = t tan(ln |t| + π/4). 2. Since the numerator and denominator are homogeneous of degree 1 the 4 − 3v quotient is homogeneous. Let y = tv. Then v + tv ! = and so 1−v (v − 2)2 tv ! = . Clearly, v = 2 is an equilibrium solution. Separating 1−v −dv dv dt the variables and using partial fractions gives − = . 2 (v − 2) v−2 t Integrating gives (v − 2)−1 − ln |v − 2| = ln |t| + c. Simplifying and ex−1 ponentiating gives e(v−2) = kt(v − 2), k &= 0. Now let v = y/t then t e y−2t = k(y − 2t) for k &= 0. The equilibrium solution v = 2 gives y = 2t as another solution. 3. Since the numerator and denominator are homogeneous of degree 2 the quotient is homogeneous. Let y = tv. Then v + tv ! = v 2 − 4v + 6. So tv ! = v 2 − 5v + 6 = (v − 2)(v − 3). There are two equilibrium solutions v '= 2, 3. Separating the variables and using partial frac( 1 1 dt tions gives − dv = . Integrating and simplifying gives v − 3 v − 2 t $ $ $v − 3$ $ = ln |t| + c. Solving for v gives v = 3 − 2kt , for k &= 0, and so ln $$ v − 2$ 1 − kt 3t − 2kt2 y = , for k &= 0. When k = 0 we get v = 3 or y = 3t, which 1 − kt is the same as the equilibrium solution v = 3. The equilibrium solution 3t − 2kt2 v = 2 gives y = 2t. Thus we can write the solutions as y = , 1 − kt k ∈ R and y = 2t. The initial condition y(2) = 4 is satisfied for the linear

32

1 Solutions

equation y = 2t but has no solution for the family y = y = 2t is the only solution.

3t − 2kt2 . Thus 1 − kt

4. Since the numerator and denominator are homogeneous of degree 2 the t2 v 2 + 2t2 v quotient is homogeneous. Let y = tv. Then v + tv ! = = t2 + t2 v 2 v + 2v v . Subtract v from both sides to get tv ! = . Separate the 1+v v+1 variables to get ' ( 1 1 1+ dv = dt. v t

Integrating gives v + ln |v| = ln |t| + c. Now exponentiate, substitute v = y/t, and simplify to get yey/t = kt2 , k ∈ R.

5. Since the numerator and denominator are homogeneous of degree 2 the 3v 2 − 1 quotient is homogeneous. Let y = tv. Then v + tv ! = . Subtract 2v 2 v −1 v from both sides to get tv ! = . The equilibrium solutions are 2v 2v dv dt v = ±1. Separating variables gives 2 = and integrating gives v − 1 t $ 2 $ ln $v − 1$ = ln |t|+c. Exponentiating gives v 2 −1 = kt and by simplifying √ √ we get v = ± 1 + kt. Now v = y/t so y = ±t 1 + kt. The equilibrium solutions v = ±1 become y = ±t. These occur when k = 0, so are already included in the general formula. 6. Since t2 +y 2 and ty are homogeneous of degree 2 their quotient (t2 +y 2 )/ty is a homogeneous function. Let y = tv. Then y ! = v + tv ! and the given differential equation becomes v + tv ! =

1 + v2 1 = + v. v v

Simplifying and separating variables √ gives v dv = dt/t. Integrating we get√v 2 /2 = ln t + c and so v = ± 2 ln t + 2c. Since v = y/t we √ get y = ±t 2 ln t + 2c. The initial condition implies that 2e = y(e) = +e 2 + 2c √ and therefore c = 1. Thus y = t 2 ln t + 2. ! ! y + t2 − y 2 ! 7. In standard form we get y = . Since (αt)2 − (αy)2 = t ! ! α2! (t2 − y 2 ) = α t2 − y 2 for α > 0 it is easy to see that y ! = √ y + t2 − y 2 is homogeneous. Let y = tv. Then v + tv ! = v + 1 − v 2 . t √ Simplifying gives tv ! = 1 − v 2 . Clearly v = ±1 are equilibrium solution. dv dt Separating variables gives √ = . Integrating gives sin−1 v = 2 t 1−v

1 Solutions

33

ln |t| + c and so v = sin(ln |t| + c). Now substitute v = y/t to get y = t sin(ln |t| + c). The equilibrium solutions imply y = ±t are also solutions. ! y y t2 + y 2 8. In standard form we get y ! = + . It is straightforward to t t2 ! ! y y t2 + y 2 see that t2 + y 2 is homogeneous of degree one. So y ! = + t t2 ! is√a homogeneous differential equation. Let y = tv then v + tv = v+ √ v 1 + v 2 or tv ! = v 1 + v 2 . It follows that v = 0 is an equilibrium dv dt solution. Separating variables gives √ = . Integrating gives 2 t v 1 + v $ $ $ $ v $ = ln |t| + c. (To integrate the left hand side use the trig √ ln $$ 1 + 1 + v2 $ v √ substitution v = tan θ.) Exponentiating gives = kt, k &= 0. 1 + 1 + v2 y ! Now let v = y/t. Then = kt. The case where k = 0 gives t + t2 + y 2 the equilibrium solution. 9. Note that although y = 0 is part of the general solution it does not satisfy the initial value. Divide both sides by y 2 to get y −2 y ! − y −1 = t. Let z = y −1 . Then z ! = −y −2 y ! . Substituting gives −z ! − z = t or z ! + z = −t. An integrating factor is et . So (et z) = −tet . Integrating both sides gives) et z = −tet + et + c, where we have used integration by parts to compute −tet dt. Solving for z gives z = −t + 1 + ce−t . Now substitute 1 z = y −1 and solve for y to get y = . The initial condition −t + 1 + ce−t 1 1 implies 1 = and so c = 0. The solution is thus y = . 1+c 1−t

10. Note that y = 0 is a solution. Divide by y 2 to get y −2 y ! + y −1 = 1. Let z = y −1 . Then z ! = −y −2 y ! and substituting gives −z ! + z = 1. In the standard form for linear equations this becomes z ! − z = −1. Multiplying by the integrating factor e−t gives (e−t z)! = −e−t so that e−t z = e−t + c. Hence z = 1 + cet . Now go back to the original function y by solving 1 z = y −1 for y. Thus y = z −1 = (1 + cet )−1 = 1+ce t . The general solution 1 1 is y = 1+cet and y = 0. The initial condition implies 1 = y(0) = 1+c and therefore c = 0. So y = 1 is the solution. 11. Note that y = 0 is a solution. First divide both sides by y 3 to get z! y −3 y ! + ty −2 = t. Let z = y −2 . Then z ! = −2y −3 y ! , so = y −3 y ! . −2 z! Substituting gives + tz = t, which in standard form is z ! − 2tz = −2t. −2 ! 2 2 2 An integrating factor is e −2t dt = e−t , so that (e−t z)! = −2te−t . 2 2 Integrating both sides gives e−t z = e−t + c, where the integral of the

34

1 Solutions

right hand side is done by the substitution u = −t2 . Solving for z gives 1 2 z = 1 + cet . Since z = y −2 we find y = ± √ . 1 + cet2 12. Note that y = 0 is a solution. First divide both sides by y 3 to get z! y −3 y ! + ty −2 = t3 . Let z = y −2 . Then z ! = −2y −3 y ! , so = y −3 y ! . −2 z! Substituting gives + tz = t3 , which in standard form is z ! − 2tz = −2 ! 2 2 2 −2t3 . An integrating factor is e −2t dt = e−t . Thus (e−t z)! = −2t3 e−t . 2 2 Integrating both sides gives e−t z = (t2 + 1)e−t + c, where the integral of the right hand side is computed using integration by parts with u = t2 2 2 and dv = −2te−t dt. Solving for z gives z = t2 + 1 + cet . Since z = y −2 1 we find y = ± √ . 2 t + 1 + cet2 13. Note that y = 0 is a solution. Divide by y 2 and (1 − t2 ) to get y −2 y ! − t 5t y −1 = . Let z = y −1 . Then z ! = −y −2 y ! and substituting 1 − t2 1 − t2 t 5t t gives −z ! − z= . In standard form we get z ! + z= 2 2 1−t (1 − t ) 1 − t2 −5t . Multiplying by the integrating factor 1 − t2 µ(t) = e

!

t 1−t2

dt

1

= e− 2 ln(1−t

2

)

= (1 − t2 )−1/2

gives (z(1 − t2 )−1/2 )! = −5t(1 − t2 )−3/2 . Integrating gives z(1 − t2 )−1/2 = √ 2 −1/2 2 −5(1 − t ) + c and hence z = −5 + c 1 − t . Since z = y −1 we have 1 √ y= . −5 + c 1 − t2 14. Note that y = 0 is a solution. Divide both sides by y 2/3 to get y −2/3 y ! + y 1/3 1 z = 1. Let z = y 1/3 . Then z ! = y −2/3 y ! and hence 3z ! + = 1. In t 3 t! z 1 1 ! standard form we get z + = . The integrating factor is e 3t dt = 3t 3 1 ln t e 3 = t1/3 . Multiplying by t1/3 gives (t1/3 z)! = t1/3 and integrating 3 1 t gives t1/3 z = t4/3 + c. Solving for z we get z = + ct−1/3 . Since 4 4 z = y 1/3 we can solve for y to get y = (t/4 + ct−1/3 )3 . 15. If we divide by y we get y ! +ty = ty −1 which is a Bernoulli equation with n = −1. Note that since n < 0, y = 0 is not a solution. Dividing by y −1 z! gets us back to yy ! + ty 2 = t. Let z = y 2 . Then z ! = 2yy ! so + tz = t 2 2 ! and in standard form we get z + 2tz = 2t. An integrating factor is et t2 ! t2 t2 t2 −t2 so (e z) = 2te . Integration gives e z = e + c so z = 1 + ce .

1 Solutions

35

√ Since z = y 2 √ we get y = ± 1 + ce−t2 . The initial condition implies √ −2 = y(0) = − 1 + c so c = 3. Therefore y = − 1 + 3e−t2 . 1 t − 1 −1 16. First divide by 2y to get y ! − y = y , which is a Bernoulli equation 2 2 with n = −1. Since n < 0, y = 0 is not a solution. Now divide by y −1 to 1 t−1 get yy ! − y 2 = . Let z = y 2 . Then z ! = 2yy ! and substituting gives 2 2 1 ! 1 t−1 z − z= . In standard form we get z ! −z = (t−1). An integrating 2 2 ! 2 factor is e −1 dt = e−t . Multiplying by e−t gives (e−t z)! = (t − 1)e−t . −t −t t Integration by parts gives √ e z = −te + c and thus z = −t + ce . Since 2 t z = y we have y = ± ce − t, c ∈ R. 17. Note that y = 0 is a solution. First divide both sides by y 3 to get z! y −3 y ! + y −2 = t. Let z = y −2 . Then z ! = −2y −3 y ! . So + z = t. −2 ! In standard form we get z ! − 2z = −2t. An integrating factor is e −2 dt = e−2t and hence (e−2t z)! = −2te−2t . Integration by parts gives e−2t z = 1 1 (t + )e−2t + c and hence z = t + + ce2t . Since z = y −2 we get 2 2 1 y = ±+ . t + 12 + ce2t

P 18. The logistic differential equation is P ! = r(1− )P which can be written m r P ! − rP = − P 2 . Note that P = 0 is a solution. Divide by P 2 to get m −r P −2 P ! − rP −1 = . Let z = P −1 . Then z ! = −P −2 P ! and −z ! − rz = m −r r r rt or z ! + rz = . An integrating factor is ert so (ert z)! = e . m m m rt e 1 Integrating gives ert z = + c. Solving for z we get z = + ce−rt = m m 1 + mce−rt m . Since z = P −1 we get P = . Now P0 = P (0) = m 1 + mce−rt m m − P0 and solving for c we get c = . Substituting and simplifying 1 + mc mP0 mP0 gives P (t) = . P0 + (m − P0 )e−rt 19. Let z = 2t−2y +1. Then z ! = 2−2y ! and so y ! =

2 − z! . Substituting we 2

2 − z! get = z −1 and in standard form we get z ! = 2 − 2z −1 , a separable 2 differential equation. Clearly, z = 1 is an equilibrium solution. Assume for now that z &= 1. Then separating and simplifying using " variables # z 1 1 −1 1/(1 − z ) = z−1 = 1 + z−1 gives 1 + z−1 dz = 2 dt. Integrating we

36

1 Solutions

get z + ln |z − 1| = 2t + c. Now substitute z = 2t − 2y + 1 and simplify to get −2y + ln |2t − 2y| = c, c ∈ R. (We absorb the constant 1 in c.) The equilibrium solution z = 1 becomes y = t. 20. Let z = t − y. Then z ! = 1 − y ! and so y ! = 1 − z ! . Substituting we get 1−z ! = z 2 and in standard form we get z ! = 1−z 2 . We see that z = ±1 are dz equilibrium solutions. Separating variables we get = dt. Partial 1 − z2 ' ( 1 1 fractions gives + dz = 2 dt. Integrating and simplifying 1+z $ $ 1−z 2t $1 + z $ $ = 2t + c, c ∈ R. Solving for z we get z = ke − 1 , k &= 0. gives ln $$ $ 1−z ke2t + 1 However, the case k = 0 gives the equilibrium solution z = −1. Now ke2t − 1 substitute z = t − y and simplify to get y = t − 2t , k ∈ R. The ke + 1 equilibrium solution z = 1 becomes y = t − 1. 21. Let z = t + y. Then z ! = 1 + y ! and substituting we get z ! − 1 = z −2 . In 1 + z2 standard form we get z ! = . Separating variables and simplifying z2 ' ( 1 we get 1 − dz = dt. Integrating we get z − tan−1 z = t + c. 1 + z2 Now let z = t + y and simplify to get y − tan−1 (t + y) = c, c ∈ R. 22. Let z = t − y. Then z ! = 1 − y ! and substituting we get 1 − z ! = sin z. π In standard form we get z ! = 1 − sin z. Notice that z = + 2πn, n ∈ Z, 2 dz are equilibrium solutions. Separating variable gives = dt. Now 1 − sin z 1 1 + sin z 1 + sin z = = = sec2 z + sec z tan z. 2 1 − sin z cos2 z 1 − sin z

So integrating gives tan z + sec z = t + c. Substituting, we get the implicit solution tan(t − y) + sec(t − y) = t + c. For the equilibrium solution π π z = + 2πn we get y = t − − 2πn. 2 2 23. This is the same as Exercise 16 where the Bernoulli equation technique there used the substitution z = y 2 . Here use the given substitution to get z ! = 2yy ! + 1. Substituting we get z ! − 1 = z and in standard form z ! = 1+z. Clearly, z = −1 is an equilibrium solution. Separating variables dz gives = dt and integrating gives ln |1 + z| = t + c, c ∈ R. Solving 1+z for z we get z = ket − 1, where k &= 0. Since z =√y 2 + t − 1 we get t y 2 + t − 1 = ket − 1 and solving for y gives √ y = ± ke − t. The case k = 0 gives the equilibrium solutions y = ± −t.

1 Solutions

37

24. If z = sin y then z ! = (cos y)y ! . Multiply the given differential equation by cos y to get (cos y)y ! = sin y + 2 cos t. Substituting we get z ! = z + 2 cos t and in standard form z ! − z = 2 cos t. An integrating factor is e−t so (ze−t )! = 2(cos t)e−t . Integrating by parts twice leads to ze−t = (sin t − cos t)e−t + c and hence z = sin t − cos t + cet . Solving for y gives y = sin−1 (sin t − cos t + cet ), c ∈ R. 25. If z = ln y then z ! =

y! . Divide the given differential equation by y. y

y! + ln y = t and substitution gives z ! + z = t. An integrating factor y is et so (et z)! = tet . Integration (by parts) gives et z = (t − 1)et + c and −t so z = t − 1 + ce−t . Finally, solving for y we get y = et−1+ce , c ∈ R. Then

−z ! 26. Let z = e−y . Then z ! = −e−y y ! so −z ! /z = y ! . Substituting gives = z −1 − 1. Multiply both sides by z and put in standard form to get z ! − z z = 1. An integrating factor is e−t so (e−t z)! = e−t . Integrating we get e−t z = −e−t +c and so z = cet −1. Since z = e−y , z > 0 and this requires c > 0. We thus get y = − ln(cet − 1), c > 0.

Section 1.6 1. This can be written in the form M (t, y) + N (t, y)y ! = 0 where M (t, y) = y 2 + 2t and N (t, y) = 2ty. Since ∂M/∂y = 2y = ∂N/∂t, the equation is exact, and the general solution is given implicitly by V (t, y) = c where the function V (t, y) )is determined by the solution method for exact equations. Thus V (t, y) = (y 2 + 2t) dt + φ(y) = y 2 t + t2 + φ(y). The function φ(y) satisfies ∂V ∂ 2 dφ dφ = (y t + t2 ) + = 2ty + = N (t, y) = 2ty, ∂y ∂y dy dy so that dφ/dy = 0. Thus, V (t, y) = y 2 t + t2 and the solutions to the differential equation are given implicitly by t2 + ty 2 = c. 2. This can be written in the form M (t, y) + N (t, y)y ! = 0 where M (t, y) = y − t and N (t, y) = t + 2y. Since ∂M/∂y = 1 = ∂N/∂t, the equation is exact, and the general solution is given implicitly by V (t, y) = c where the function V (t, y) )is determined by the solution method for exact equations. Thus V (t, y) = (y − t) dt + φ(y) = yt − t2 /2 + φ(y). The function φ(y) satisfies

38

1 Solutions

∂V ∂ dφ dφ = (yt − t2 /2) + =t+ = N (t, y) = t + 2y, ∂y ∂y dy dy so that dφ/dy = 2y. Hence φ(y) = y 2 and thus, V (t, y) = yt − t2 /2 + y 2 and the solutions to the differential equation are given implicitly by ty + y 2 − t2 /2 = c. 3. In this equation M = 2t2 − y and N = t + y 2 . Since ∂M/∂y = −1, while ∂N/∂t = 1, the equation is not exact. 4. This can be written in the form M (t, y) + N (t, y)y ! = 0 where M (t, y) = y 2 + 3t2 and N (t, y) = 2ty. Since ∂M/∂y = 2y = ∂N/∂t, the equation is exact, and the general solution is given implicitly by V (t, y) = c where the function V (t, y))is determined by the solution method for exact equations. Thus V (t, y) = (y 2 + 3t2 ) dt + φ(y) = y 2 t + t3 + φ(y). The function φ(y) satisfies ∂V ∂ 2 dφ dφ = (y t + t3 ) + = 2yt + = N (t, y) = 2yt, ∂y ∂y dy dy so that dφ/dy = 0. Hence V (t, y) = ty 2 + t3 and the solutions to the differential equation are given implicitly by ty 2 + t3 = c. 5. In this equation M = 3y − 5t and N = 2y − t. Since ∂M/∂y = 3, while ∂N/∂t = −1, the equation is not exact. 6. This can be written in the form M (t, y) + N (t, y)y ! = 0 where M (t, y) = 2ty and N (t, y) = t2 + 3y 2 . Since ∂M/∂y = 2t = ∂N/∂t, the equation is exact, and the general solution is given implicitly by V (t, y) = c where the function V (t, y) )is determined by the solution method for exact equations. Thus V (t, y) = 2ty dt + φ(y) = t2 y + φ(y). The function φ(y) satisfies ∂V ∂ 2 dφ dφ = t y+ = t2 + = N (t, y) = t2 + 3y 2 , ∂y ∂y dy dy

so that dφ/dy = 3y 2 . Hence φ(y) = y 3 and thus V (t, y) = t2 y + y 3 so that the solutions to the differential equation are given implicitly by ty 2 + t3 = c. To satisfy the initial condition y(1) = 1, substitute t = 1, y = 1 into this equation to get 2 = c. Thus the solution of the initial value problem is t2 y + y 3 = 2 7. This can be written in the form M (t, y) + N (t, y)y ! = 0 where M (t, y) = 2ty+2t3 and N (t, y) = t2 −y. Since ∂M/∂y = 2t = ∂N/∂t, the equation is exact, and the general solution is given implicitly by V (t, y) = c where the function V (t, y) )is determined by the solution method for exact equations. Thus V (t, y) = (2ty + 2t3 ) dt + φ(y) = t2 y + t4 /2 + φ(y). The function φ(y) satisfies

1 Solutions

39

∂V ∂ 2 dφ dφ = (t y + t4 /2) + = t2 + = N (t, y) = t2 − y, ∂y ∂y dy dy so that dφ/dy = −y. Hence φ(y) = −y 2 /2 so that V (t, y) = t2 y + t4 /2 − y 2 /2 and the solutions to the differential equation are given implicitly by t2 y + t4 /2 − y 2 /2 = c. Multiplying by 2 and completing the square (and replacing the constant 2c by c) gives (y − t2 )2 − 2t4 = c. 8. This can be written in the form M (t, y) + N (t, y)y ! = 0 where M (t, y) = t2 − y and N (t, y) = −t. Since ∂M/∂y = −1 = ∂N/∂t, the equation is exact, and the general solution is given implicitly by V (t, y) = c where the function V (t, y) )is determined by the solution method for exact equations. Thus V (t, y) = (t2 − y) dt + φ(y) = t3 /3 − yt + φ(y). The function φ(y) satisfies ∂V ∂ 3 dφ dφ = (t /3 − yt) + = −t + = N (t, y) = −t, ∂y ∂y dy dy so that dφ/dy = 0. Hence V (t, y) = t3 /3 − yt and the solutions to the differential equation are given implicitly by t3 /3 − yt = c, which can be 1 c solved for y to give y = t2 − . 3 t 9. This can be written in the form M (t, y) + N (t, y)y ! = 0 where M (t, y) = −y and N (t, y) = y 3 − t. Since ∂M/∂y = −1 = ∂N/∂t, the equation is exact, and the general solution is given implicitly by V (t, y) = c where the function V (t, y))is determined by the solution method for exact equations. Thus V (t, y) = (−y) dt + φ(y) = −yt + φ(y). The function φ(y) satisfies ∂V ∂ dφ dφ = (−yt) + = −t + = N (t, y) = y 3 − t, ∂y ∂y dy dy

so that dφ/dy = y 3 . Hence, φ(y) = y 4 /4 so that V (t, y) = y 4 /4 − yt and the solutions to the differential equation are given implicitly by y 4 /4 − yt = c. 10. This can be written in the form M (t, y)+N (t, y)y ! = 0 where M (t, y) = at + by and N (t, y) = −(ct + dy). Since ∂M/∂y = b, while ∂N/∂t = −c, the equation is exact if and only if b = −c.

Section 1.7 1. We first change the variable t to u and write y ! (u) = uy(u). Now integrate )t )t both sides from 1 to t to get 1 y ! (u) du = 1 uy(u) du. Now the left side )t )t is 1 y ! (u) du = y(t) − y(1) = y(t) − 1. Thus y(t) = 1 + 1 uy(u) du.

40

1 Solutions

2. Change the variable t to u and write y ! (u) = y 2 (u). Now integrate both )t )t sides from 0 to t to get 0 y ! (u) du = 0 y 2 (u) du. The left side is y(t) + 1 )t so y(t) = −1 + 0 y 2 (u) du. 3. Change the variable t to u and write y ! (u) = both sides from 0 to t to get

)t 0

y ! (u) du =

) t u − y(u) is y(t) − 1 so y(t) = 1 + 0 du. u + y(u)

)t 0

u − y(u) . Now integrate u + y(u) u − y(u) du. The left side u + y(u)

4. Change the variable t to u and write y ! (u) = 1 + u2 . Now integrate both )t )t sides from 0 to t to get 0 y ! (u) du = 0 (1 + u2 ) du. The left side is y(t) )t so y(t) = 0 (1 + u2 ) du. )t 5. The corresponding integral equation is y(t) = 1 + 1 uy(u) du. We then have y0 (t) = 1

($t u2 $$ t2 1 1 + t2 y1 (t) = 1 + u · 1 du = 1 + = 1 + − = 2 $1 2 2 2 1 ( ' ($ * t ' t 1 + u2 u2 u4 $$ 5 t2 t4 y2 (t) = 1 + u du = 1 + + = + + $ 2 4 8 1 8 4 8 1 ( ' 2 ($ * t' 3 5 4 6 $t 5u u u 5u u u $ y3 (t) = 1 + + + du = 1 + + + 8 4 8 16 16 48 $1 1 =

*

t

'

29 5t2 t4 t6 + + + . 48 16 16 48

6. The corresponding integral equation is * t y(t) = 1 + (u − y(u)) du. 0

We then have

1 Solutions

41

y0 (t) = 1

($t $ u2 t2 y1 (t) = 1 + (u − 1) du = 1 + − u $$ = 1 − t + 2 2 0 0 ' (( ' ($t * t' u2 u3 $$ y2 (t) = 1 + u− 1−u+ du = 1 + −u + u2 − 2 6 $0 0 *

'

t

t3 = 1 − t + t2 − 6 ' (( * t' u3 t3 t4 y3 (t) = 1 + u − 1 − u + u2 − du = 1 − t + t2 − + 6 3 4! 0 ' (( * t' 3 4 u u t3 t4 t5 y4 (t) = 1 + u − 1 − u + u2 − + du = 1 − t + t2 − + − . 3 4! 3 12 5! 0 7. The corresponding integral equation is y(t) = have

)t 0

(u + y 2 (u)) du. We then

y0 (t) = 0 * t t2 y1 (t) = (u + 0) du = 2 0 ' 2 (2 ( * t, * t' u u4 t2 t5 y2 (t) = u+ du = u+ du = + 2 4 2 20 0 0 , ' 2 (2 ( * t * t' u u5 u4 u7 u10 y3 (t) = u+ + du = u+ + + du 2 20 4 20 400 0 0 =

t2 t5 t8 t11 + + + . 2 20 160 4400

8. The corresponding integral equation is * t y(t) = 1 + ((y(u))3 − y(u)) du. 0

We now have y0 (t) = 1 y1 (t) = 1 + y2 (t) = 1 y3 (t) = 1

*

t 0

(13 − 1) du = 1 +

9. The corresponding integral equation is

*

t

0 du = 1 0

42

1 Solutions

y(t) =

*

t 0

(1 + (u − y(u))2 ) du.

We then have y0 (t) = 0 ' ($t * t % & u3 $$ t3 y1 (t) = 1 + (u − 0)2 du = u + =t+ $ 3 0 3 0 ' ' (( ' ( * t, * 2 t u3 u6 y2 (t) = 1+ u− u+ du = 1+ du 3 9 0 0 ' ($t u7 $$ t7 = u+ =t+ $ 63 0 7 · 32 ( * t' u14 t15 y3 (t) = 1 + 2 4 du = t + 7 ·3 15 · 72 · 34 0 ( * t' 30 u t31 y4 (t) = 1 + 2 4 8 du = t + 15 · 7 · 3 31 · 152 · 74 · 38 0 ( * t' u62 t63 y5 (t) = 1+ 2 du = t + 4 8 1 2 31 · 15 · 7 · 3 6 63 · 31 · 154 · 78 · 316 0 10. The right hand side is F (t, y) = 1 + y 2 . Then Fy (t, y) = 2y. Both F and Fy are continuous in the whole (t, y)-plane and thus are continuous on any rectangle containing the origin (0, 0). Picard’s theorem applies and we can conclude there is a unique solution on an interval about 0. √ 11. The right hand side is F (t, y) = y. If R is any rectangle about (1, 0) then there are y-coordinates that are negative. Hence F is not defined on R and Picards’ theorem does not apply. 1 y. Then Fy (t, y) = √ . Choose a 2 y rectangle R about (0, 1) that lies above the t-axis. Then both F and Fy are continuous on R. Picard’s theorem applies and we can conclude there is a unique solution on an interval about 0.

12. The right hand side is F (t, y) =

√

t−y −2t . Then Fy (t, y) = . Choose t+y (t + y)2 a rectangle R about (0, −1) that contains no points on the line t + y = 0. Then both F and Fy are continuous on R. Picard’s theorem applies and we can conclude there is a unique solution on an interval about 0.

13. The right hand side is F (t, y) =

t−y , which is not defined at the initial t+y condition (t0 , y0 ) = (1, −1). Thus Picard’s theorem does not apply.

14. The right hand side is F (t, y) =

1 Solutions

43

15. The corresponding integral equation is y(t) = 1 + have y0 (t) = 1 y1 (t) = 1 +

*

0

ay(u) du. We thus

t

a du = 1 + at 0

* t a2 t2 a(1 + au) du = 1 + (a + a2 u) du = 1 + at + 2 0 0 ( * t ' a2 u2 a2 t2 a3 t3 y3 (t) = 1 + a 1 + au + du = 1 + at + + 2 2 3! 0 .. . a 2 t2 a n tn yn (t) = 1 + at + + ··· + . 2 n! y2 (t) = 1 +

*

)t

t

a k tk . We recognize this sum as the first n k! terms of the Taylor series expansion for eat . Thus the limiting function is y(t) = limn→∞ yn (t) = eat . It is straightforward to verify that it is a solution. If F (t, y) = ay then Fy (t, y) = a. Both F and Fy are continuous on the whole (t, y)-plane. By Picard’s theorem, Theorem 5, y(t) = eat is the only solution to the given initial value problem. We can write yn (t) =

.n

k=0

16. 1. The equation is separable so separate the variables to get y −2 dy = dt. Integrating gives −y −1 = t + c and the initial condition y(0) = 1 implies that the integration constant c = −1, so that the exact solution is y(t) =

1 = 1 + t + t2 + t3 + t4 + · · · ; 1−t

|t| < 1.

2. To apply Picard’s method, let y0 = 1 and define * t * t 2 y1 (t) = 1 + (y0 (u)) du = 1 + 1 du = 1 + t; 0 0 * t * t t3 y2 (t) = 1 + (y1 (u))2 du = 1 + (1 + u)2 du = 1 + t + t2 + ; 3 0 0 ( * t * t' 2 u3 y3 (t) = 1 + (y2 (u))2 du = 1 + 1 + u + u2 + du 3 0 0 ( * t' 8 5 2 1 = 1+ 1 + 2u + 3u2 + u3 + u4 + u5 + u6 du 3 3 3 9 0 2 1 1 1 = 1 + t + t2 + t3 + t4 + t5 + t6 + t7 . 3 3 9 63

44

1 Solutions

Comparing y3 (t) to the exact solution, we see that the series agree up to order 3. 17. Let F (t, y) = cos(t + y). Then Fy (t, y) = − sin(t + y). Let y1 and y2 be arbitrary real numbers. Then by the mean value theorem there is a number y0 in between y1 and y2 such that |F (t, y1 ) − F (t, y2 )| = |sin(t + y0 )| |y1 − y2 | ≤ |y1 − y2 | . It follows that F (t, y) is Lipschitz on any strip. Theorem 10 implies there is a unique solution on all of R. 18. Let F (t, y) = −p(t)y + f (t). Since p(t) and f (t) are continuous on [a, b] it follows that F (t, y) is continuous on the strip {(t, y) : t ∈ [a, b], y ∈ R}. Further more p(t) is bounded: i.e. there is a number A such that |p(t)| ≤ A, for all t ∈ [a, b]. Let t ∈ [a, b] and y1 and y2 be arbitrary real numbers. Then |F (t, y1 ) − F (t, y2 )| ≤ |−p(t)y1 + f (t) − (−p(t)y2 + f (t))| = |p(t)| |y1 − y2 | ≤ A |y1 − y2 | .

It follows that F (t, y) is Lipschitz with Lipschitz constant A. By Theorem 10, y ! + p(t)y = f (t), y(t0 ) = y0 has a unique solution on the entire interval [a, b]. 19. 1. First assume that t &= 0. Then ty ! = 2y − t is linear and in standard !form becomes y ! − 2y/t = −1. An integrating factor is µ(t) = e (−2/t) dt = t−2 and multiplying both sides by µ gives t−2 y ! − 2t−3 y = −t−2 . This simplifies to (t−2 y)! = −t−2 . Now integrate to get t−2 y = t−1 + c or y(t) = t + ct2 . We observe that this solution is also valid for t = 0. Graphs are given below for various values of c. 4 3 2 1 0 −1 −2 −3 −4 −6

−4

−2

0 t

2

4

6

Graph of y(t) = t + ct3 for various c

1 Solutions

45

2. Every solution satisfies y(0) = 0. There is no contradiction to Theo2 rem 5 since, in standard form, the equation is y ! = y − 1 = F (t, y) t and F (t, y) is not continuous for t = 0. 20. 1. If F (t, y) = y 2 then Fy (t, y) = 2y. Both are continuous on any rectangle that contains (t0 , y0 ). Hence Theorem 5 applies and implies there is a unique solution on an interval that contains t0 . 1 2. y(t) = 0 is a solution defined for all t; y(t) = is a solution 1−t defined on (−∞, 1) . 21. No. Both y1 (t) and y2 (t) would be solutions to the initial value problem y ! = F (t, y), y(0) = 0. If F (t, y) and Fy (t, y) are both continuous near (0, 0), then the initial value problem would have a unique solution by Theorem 5. 22. No, Both y1 (t) and y2 (t) would be solutions to the initial value problem y ! = F (t, y), y(0) = 1. If F (t, y) and Fy (t, y) are both continuous near (0, 1), then the initial value problem would have a unique solution by Theorem 5. 23. For t < 0 we have y1! (t) = 0 and for t > 0 we have y1! (t) = 3t2 . For t = 0 y1 (h) − y1 (0) y1 (h) we calculate y1! (0) = limh→0 = limh→0 . To compute h−0 h this limit we show the left hand and right hand limits agree. We get y1 (h) h3 = lim+ = lim+ h2 = 0 h h h→0 h→0 h→0 y1 (h) 0 lim = lim+ = 0 h h→0− h→0 h / 0, for t < 0 ! It follows that y1 (t) = and so 3t2 for t ≥ 0 lim+

ty1! (t) =

/

3y1 (t) =

/

0, 3t3

for t < 0 for t ≥ 0

On the other hand, 0, 3t3

for t < 0 for t ≥ 0

It follows that y1 is a solution. It is trivial to see that y2 (t) is a solution. 3 There is no contraction to Theorem 5 since, in standard form y ! = y = t

46

1 Solutions

F (t, y) has a discontinuous F (t, y) near (0, 0). So Picard’s theorem does not even apply.

Section 2.1 1. Apply the Laplace transform to both sides of the equation. For the left hand side we get sY (s) − 2 − 4Y (s), while the right hand side is 0. Solve 2 for Y (s) to get Y (s) = . From this we see that y(t) = 2e4t . s−4 2. Apply the Laplace transform to both sides of the equation. For the left hand side we get sY (s) − 4Y (s), while the right hand side is 1/s. Solve 1 for Y (s) to get Y (s) = , which can be written in partial fractions s(s − 4) as ' ( 1 1 1 Y (s) = − . 4 s−4 s Therefore y(t) = 14 (e4t − 1).

3. Apply the Laplace transform to both sides of the equation. For the left hand side we get sY (s) − 4Y (s), while the right hand side is 1/(s − 4). 1 Solve for Y (s) to get Y (s) = . Therefore, y(t) = te4t . (s − 4)2 4. Apply the Laplace transform to both sides of the equation. For the left hand side we get sY (s) − 1 + aY (s), while the right hand side is 1/(s + a). 1 1 Solve for Y (s) to get Y (s) = + . Therefore, y(t) = e−at + s + a (s + a)2 te−at . 5. Apply the Laplace transform to both sides of the equation. For the left hand side we get sY (s) − 2 + 2Y (s), while the right hand side is 3/(s − 1). Solve for Y (s) to get Y (s) =

2 3 1 1 + = + . s + 2 (s − 1)(s + 2) s+2 s−1

Thus y(t) = e−2t + et . 6. Apply the Laplace transform to both sides of the equation. For the left hand side we get sY (s) + 2Y (s), while the right hand side is 1/(s + 2)2 . 1 1 2 −2t Solve for Y (s) to get Y (s) = (s+2) . 3 and therefore y(t) = 2 t e 7. Apply the Laplace transform to both sides. For the left hand side we get

1 Solutions

47

L {y !! + 3y ! + 2y} (s) = L {y !! } (s) + 3L {y ! } (s) + 2L {y} (s)

= s2 Y (s) − sy(0) − y ! (0) + 3(sY (s) − y(0)) + 2Y (s) = (s2 + 3s + 2)Y (s) − 3s − 3.

Since the Laplace transform of 0 is 0, we now get (s2 + 3s + 2)Y (s) − 3s − 3 = 0. Hence, Y (s) =

s2

3s + 3 3(s + 1) 3 = = , + 3s + 2 (s + 1)(s + 2) s+2

and therefore, y(t) = 3e−2t . 8. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 5y ! + 6y} (s) = L {y !! } (s) + 5L {y ! } (s) + 6L {y} (s) = s2 Y (s) − sy(0) − y ! (0) + 5(sY (s) − y(0)) + 6Y (s) = (s2 + 5s + 6)Y (s) − 2s − 4. Since the Laplace transform of 0 is 0, we now get (s2 + 5s + 6)Y (s) − 2s − 4 = 0. Hence, Y (s) =

s2

2s + 4 2(s + 2) 2 = = , + 5s + 6 (s + 3)(s + 2) s+3

and therefore, y(t) = 2e−3t . 9. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 25y} (s) = L {y !! } (s) + 25L {y} (s) = s2 Y (s) − sy(0) − y ! (0) + 25Y (s) = (s2 + 25)Y (s) − s + 1.

We now get Hence,

(s2 + 25)Y (s) − s + 1 = 0. Y (s) =

s−1 s 1 5 = 2 − , s2 + 25 s + 25 5 s2 + 25

and therefore, y(t) = cos 5t −

1 5

sin 5t.

10. Apply the Laplace transform to both sides. For the left hand side we get

48

1 Solutions

0 1 L y !! + a2 y (s) = L {y !! } (s) + a2 L {y} (s)

= s2 Y (s) − sy(0) − y ! (0) + a2 Y (s) = (s2 + 25)Y (s) − y0 s − y1 .

We now get

(s2 + a2 )Y (s) − y0 s − y1 = 0.

Hence, Y (s) =

y0 s + y1 s 1 a = y0 2 + y1 2 , s2 + a2 s + 25 a s + a2

and therefore, y(t) = y0 cos at + y1 a1 sin at. 11. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 8y ! + 16y} (s) = L {y !! } (s) + 8L {y ! } (s) + 16L {y} (s) = s2 Y (s) − sy(0) − y ! (0) + 8(sY (s) − y(0)) + 16Y (s) = (s2 + 8s + 16)Y (s) − s − 4.

We now get Hence,

(s + 4)2 Y (s) − (s + 4) = 0. Y (s) =

s+4 1 = (s + 4)2 s+4

and therefore y(t) = e−4t 12. Apply the Laplace transform to both sides. For the left hand side we get L {y !! − 4y ! + 4y} (s) = L {y !! } (s) − 4L {y ! } (s) + 4L {y} (s)

= s2 Y (s) − sy(0) − y ! (0) − 4(sY (s) − y(0)) + 4Y (s) = (s2 − 4s + 4)Y (s) + s.

0 1 Since L 4e2t (s) = 4/(s − 2) we get the algebraic equation (s − 2)2 Y (s) + s =

Hence,

4 . s−2

−s 4 + 2 (s − 2) (s − 2)3 −(s − 2) − 2 4 = + 2 (s − 2) (s − 2)3 1 2 2 = − − +2 s − 2 (s − 2)2 (s − 2)3

Y (s) =

1 Solutions

49

and therefore y(t) = −e2t − 2te2t + 2t2 e2t 13. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 4y ! + 4y} (s) = L {y !! } (s) + 4L {y ! } (s) + 4L {y} (s)

= s2 Y (s) − sy(0) − y ! (0) + 4(sY (s) − y(0)) + 4Y (s) = (s2 + 4s + 4)Y (s) − 1.

0 1 Since L e−2t = 1/(s + 2) we get the algebraic equation (s + 2)2 Y (s) − 1 =

1 . s+2

Hence, Y (s) =

1 1 1 1 2 + = + (s + 2)2 (s + 2)3 (s + 2)2 2 (s + 2)3

and therefore y(t) = te−2t + 12 t2 e−2t 14. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 4y} (s) = L {y !! } (s) + 4L {y} (s)

= s2 Y (s) − sy(0) − y ! (0) + 4Y (s) = (s2 + 4)Y (s) − 2s − 1.

Since L {8} = 8/s we get the algebraic equation (s2 + 4)Y (s) − 2s − 1 =

8 . s

Hence, 2s + 1 8 + s2 + 4 s(s2 + 4) 1 2 1 2 2 = 2 + = + 2 s +4 s 2s +4 s

Y (s) =

and therefore y(t) =

Section 2.2 1.

1 2

sin 2t + 2

50

1 Solutions

= = = = =

L {3t + 1} (s) * ∞ (3t + 1)e−st dt 0 * ∞ * ∞ 3 te−st dt + e−st dt 0 0 $∞ $∞ ' ( * t −st $$ 1 ∞ −st −1 −st $$ 3 e $ + e dt + e $ −s s 0 s 0 0 $∞ ( '' ( ' ( 1 −1 −st $$ 1 3 e $ + s s s 0 3 1 + . s2 s

2.

= = = = =

0 1 L 5t − 9et (s) * ∞ e−st (5t − 9et ) dt 0 * ∞ * ∞ −st 5 te dt − 9 e−st et dt 0 0 $∞ ' ( * * ∞ −t −st $$ 1 ∞ −st 5 e $ + e dt − 9 e−(s−1)t dt s s 0 0 0 ' ( 1 1 5 0+ 2 −9 s s−1 5 9 − . s2 s−1

3. 0 1 L e2t − 3e−t (s) * ∞ = e−st (e2t − 3e−t ) dt 0 * ∞ * ∞ −st 2t = e e dt − 3 e−st e−t dt 0 0 * ∞ * ∞ = e−(s−2)t dt − 3 e−(s+1)t dt 0

1 3 = − . s−2 s+1 4.

0

1 Solutions

51

= = = =

0 1 L te−3t (s) * ∞ e−st te−3t dt *0 ∞ te−(s+3)t dt 0 $∞ * ∞ te−(s+3)t $$ 1 + e−(s+3)t dt −(s + 3) $0 s+3 0 1 . (s + 3)2

0 1 0 1 5. L 5e2t = 5L e2t =

5 s−2

0 1 0 1 0 1 6. L 3e−7t − 7t3 = 3L e−7t − 7L t3 =

3 3! 3 42 −7 4 = − 4 s+7 s s+7 s

0 1 0 1 2 5 4 7. L t2 − 5t + 4 = L t2 − 5L {t} + 4L {1} = 3 − 2 + s s s 03 1 0 1 0 1 6 2 1 1 8. L t + t2 + t + 1 = L t3 + L t2 + L {t} + L {1} = 4 + 3 + 2 + s s s s 0 1 0 1 0 1 1 7 9. L e−3t + 7te−4t = L e−3t + 7L te−4t = + s + 3 (s + 4)2

0 1 10. L t2 e4t (s) =

2 (s − 4)3

11. L {cos 2t + sin 2t} = L {cos 2t}+L {sin 2t} =

s2

s 2 s+2 + 2 = 2 2 2 +2 s +2 s +4

12. L {et (t − cos 4t)} (s) = L {te−t } (s) − L {e−t cos 4t} (s) = s−1 . (s − 1)2 + 16

0 1 0 1 13. L (te−2t )2 (s) = L t2 e−4t (s) = √ 1 0 14. L e−t/3 cos 6t (s) =

1 − (s − 1)2

2 (s + 4)3

s + 13 (s + 13 )2 + 6

0 1 0 1 0 1 0 1 15. L (t + e2t )2 (s) = L t2 + 2te2t + e4t (s) = L t2 (s)+2L te2t (s)+ 0 1 2 2 1 L e4t (s) = 3 + + s (s − 2)2 s−4

52

1 Solutions

16. L {5 cos 3t − 3 sin 3t + 4} (s) = 5L {cos 3t} (s)−3L {sin 3t} (s)+L {4} (s) = s 3 4 5s − 9 4 5 2 −3 2 + = 2 + 2 2 s +3 s +3 s s +9 s 2 43 0 1 t 4! 24 17. L (s) = L t4 e−4t (s) = = 4t 5 e (s + 4) (s + 4)5 0 1 0 1 0 1 18. L e5t (8 cos 2t + 11 sin 2t) (s) = 8L e5t cos 2t (s)+11L e5t sin 2t (s) = 8(s − 5) 22 8s − 18 + = 2 (s − 5)2 + 4 (s − 5)2 + 4 s − 10s + 29 ' (! 0 1 % 0 1&! 1 1 19. L te3t (s) = − L e3t (s) = − = s−3 (s − 3)2 ' (! s s2 − 9 ! 20. L {t cos 3t} (s) = − (L {cos 3t}) = − 2 = 2 s +9 (s + 9)2 0 1 21. Here we use the transform derivative principle twice to get L t2 sin 2t (s) = ' (!! ' (! 2 −4s 12s2 − 16 !! (L {sin 2t}) = = = s2 + 4 (s2 + 4)2 (s2 + 4)3 , -! $ ' (! $ s s+1 ! −t −t $ 22. L {te cos t} (s) = −L {e cos t} (s) = − =− 2 = s2 + 1 $s'→s+1 s + 2s + 2 s2 + 2s (s2 + 2s + 2)2 ' ' 2 ((! s 2s 2 ! 23. L {tf (t)} (s) = −L {f (t)} (s) = − ln 2 = 2 − s +1 s +1 s 24. L

2

1 − cos 5t t

3

= 5L

2

1 − cos 5t 5t

3

=

1 (s/5)2 1 s2 ln = ln . 2 (s/5)2 + 1 2 s2 + 25

1 1 ln((s/6) + 1) ln(s + 6) − ln 6 L {Ei(t)} (s)|s'→s/6 = = 6 6 s/6 s ' ( 0 1 1 1 1 s s2 + 2b2 26. L cos2 bt (s) = L {1 + cos 2bt} (s) = + 2 = 2 2 s s + 4b2 s(s2 + 4b2 ) 25. L {Ei(6t)} (s) =

0 1 1 1 27. We use the identity sin2 θ = (1−cos 2θ). L sin2 bt (s) = L {1 − cos 2bt} (s) = 2 2 ' ( 1 1 s 2b2 − 2 = ; 2 s s + 4b2 s(s2 + 4b2 )

1 Solutions

53

1 28. We use the identity sin 2θ = 2 sin θ cos θ. L {sin bt cos bt} (s) = L {sin 2bt} (s) = 2 1 2b b = 2 ; 2 s2 + 4b2 s + 4b2 29. We use the identity sin at cos bt =

1 (sin(a + b)t + sin(a − b)t). 2

1 (L {sin(a + b)t} + L {sin(a − b)t}) 2' ( 1 a−b a+b = + 2 . 2 s2 + (a − b)2 s + (a + b)2

L {sin at cos bt} =

1& 1 1 % 0 bt 30. L {cosh bt} = L e + e−bt = 2 2 31. L {sinh bt} =

1& 1 1 % 0 bt L e − e−bt = 2 2

'

'

1 1 + s+b s−b 1 1 − s+b s−b

(

(

= =

s s2 − b2 s2

b − b2

32. Let f (t) = eat . Then f ! (t) = aeat and f (t)|t=0 = 1. Thus aL {eat } = 1 L {f ! (t)} = sL {eat } − 1. Solving for L {eat } gives L {eat } = . s−a

33. Let f (t) = sinh bt. Then f ! (t) = b cosh t and f !! (t) = b2 sinh t. Further, f (t)|t=0 = 0 and f ! (t)|t=0 = b. Thus b2 L {sinh bt} = L {f !! (t)} = s2 L {f (t)} − sf (0) − f ! (0) = s2 L {f (t)} − b. Solving for L {f (t)} gives b L {sinh bt} = 2 . s − b2

34. Let f (t) = cosh bt. Then f ! (t) = b sinh t and f !! (t) = b2 cosh t. Further, f (t)|t=0 = 1 and f ! (t)|t=0 = 0. Thus b2 L {cosh bt} = L {f !! (t)} = s2 L {f (t)} − sf (0) − f ! (0) = s2 L {f (t)} − s. Solving for L {f (t)} gives s L {cosh bt} = 2 . s − b2 )t )0 35. Let g(t) = 0 f (u) du and note that g ! (t) = f (t) and g(0) = 0 f (u) du = 0. Now apply the input derivative formula to g(t), to get F (s) = L {f (t)} (s) = L {g ! (t)} (s) = sL {g(t)} (s) − g(0) = sG(s). Solving for G(s) gives G(s) = F (s)/s. 36. For t ≥ 0, eat ≤ ebt . Thus |f (t)| ≤ Keat ≤ Kebt . So f is of exponential type of order b. 37. Suppose f is of exponential type of order a and g is of exponential type of order b. Then there are numbers K and L so that |f (t)| ≤ Keat and

54

1 Solutions

|g(t)| ≤ Lebt . Now |f (t)g(t)| ≤ Keat Lebt = KLe(a+b)t . If follows that f + g is of exponential type of order a + b. 38. Suppose f is of exponential type of order a in the sense given in the text. Then N can be chosen to be 0 and f satisfies the definition given in the statement of the problem. Now suppose f satisfies the definition given in the statement of the problem. I.e. there is a K ≥ 0 and N ≥ 0 so that |f | ≤ Keat for t ≥ N . Since |f | is continuous on the interval [0, N ] it has a maximum, K1 , say. It follows that |f | ≤ K1 ≤ K1 eat on [0, N ] and hence |f | ≤ (K + K1 )eat , for all t ≥ 0. It follows that f is of exponential type in the sense given in the text. 39. Suppose a and K are real and |y(t)| ≤ Keat . Then y(t)e−at is bounded by K. But 2

et e−at = et

2

2

−at+ a4

e−

a 2

a2 4

= e(t− 2 ) e− 2

= eu e−

a2 4

a2 4

, 2

where u = t− a2 . As t approaches infinity so does u. Since limu→∞ eu = ∞ 2 it is clear that limt→∞ et e−at = ∞, for all a ∈ R, and hence y(t)e−at is not bounded. It follows that y(t) is not of exponential type. )∞ 2 2 40. First of all, fix a ∈ R. Since et > 1 for all t it follows that a et dt > )∞ ) ∞ t2 1 dt = ∞ by the comparison test. Thus a e dt does not exist for a any real number a. Now let s be any real number. Then * ∞ 4 25 2 L et (2s) = e−2st et dt *0 ∞ 2 2 2 = et −2st+s −s dt 0 * ∞ 2 −s2 = e e(t−s) dt *0 ∞ 2 2 = e−s et dt. −s

But this last integral does not exist. Since the Laplace transform does not exist at 2s, for any s, the Laplace transform does not exist. 41. y(t) is of exponential type because it is continuous and bounded. On the 2 2 other hand, y ! (t) = cos(et )et (2t). Suppose there is a K and a so that |y ! (t)| ≤ Keat for all t ≥ 0. We need only show that there are some t for 2 which this inequality does not hold. Since cos et oscillates between −1 2 t2 and 1 let’s focus on those t for which cos e = 1. This happens when et

1 Solutions

55

! 2 is a multiple of 2π, i.e. et = 2πn for some n. Thus t = tn = ln(2πn). If the inequality |y ! (t)| ≤ Keat is valid for all t ≥ 0 it is valid for tn for 2 all n > 0. We then get the inequality 2tn etn ≤ Keatn . Now divide by atn e , combine, complete the square, and simplify to get the inequality 2 2 2tn e(tn −a/2) ≤ Kea /4 . Choose n so that tn > K and tn > a. Then this last inequality is not satisfied. It follows that y ! (t) is not of exponential )M type. Now consider the definite integral 0 e−st y ! (t) dt and compute by parts: We get *

M

e

−st !

y (t) dt =

0 2

$

M y(t)e−st $0

+s

*

M

e−st y(t) dt.

0

Since y(t) = sin(et ) is bounded and y(0) = 0 it follows that $M lim y(t)e−st $0 = 0.

M →∞

Taking limits as M → ∞ in the equation above gives L {y ! (t)} = sL {y(t)}. The righthand side exists because y(t) is bounded. (a) Show that Γ (1) = 1. (b) Show that Γ satisfies the recursion formula Γ (β + 1) = βΓ (β). (Hint: Integrate by parts.) (c) Show that Γ (n + 1) = n! when n is a nonnegative integer. )∞ ∞ 42. (a) Γ (1) = 0 e−x dx = −e−x |0 = 1. $∞ ) ∞ β −x )∞ (b) Γ (β + 1) = 0 x e dx = −xβ e−x $0 + β 0 xβ−1 e−x dx = βΓ (β). The second equality is obtained by integration by parts using u = xβ , dv = e−x dx. (c) Repeated use of (b) gives Γ (n+1) = nΓ (n) = n(n−1)Γ (n−1) = · · · = n(n−1) · · · 2·1Γ (1) = n!. 43. Using polar coordinates x = r cos θ, y = r sin θ. Then dx dy = r dr dθ and the domain of integration is the first quadrant of the plane, which in polar coordinates is given by 0 ≤ θ ≤ π/2, 0 ≤ r < ∞. Thus *

∞ 0

*

∞ 0

e−(x

2

+y 2 )

dx dy =

*

π/2 0

π = 2

*

*

∞

∞

2

e−r r dr dθ

0 2

e−r r dr 0 $ 2 ∞ π e−r $$ π = − $ = . 2 2 $ 4 0

Hence, I =

√

π/2.

56

1 Solutions

)∞ 44. Γ ( 12 ) = 0 x−1/2 e−x dx. Using the change of variables x = u2 , so dx = 2u du it follows that * ∞ * ∞ √ 2 2 1 Γ( ) = u−1 e−u 2u du = 2 e−u du = π. 2 0 0 Then using the recursion formula√gives: √ (a) Γ ( 32 ) = Γ ( 12 + 1) = 12 Γ ( 12 ) = π/2, (b) Γ ( 52 ) = 32 Γ ( 32 ) = 3 π/4. Now apply the general power formula (Formula 12) to get √ √ 0√ 1 Γ ( 32 ) 0 3/2 1 Γ ( 52 ) π 3 π (c) L t = 3/2 = 3/2 , (d) L t = 5/2 = 5/2 . s 2s s 4s

Section 2.3 The s − 1 -chain 1.

5s + 10 (s − 1)(s + 4) 2 s+4

3 s−1

The s − 2 -chain 2.

10s − 2 (s + 1)(s − 2) 4 (s + 1)

6 (s − 2)

The s − 5 -chain 3.

1 (s + 2)(s − 5) −1/7 (s + 2)

1/7 (s − 5)

1 Solutions

57

The s + 3 -chain 4.

5s + 9 (s − 1)(s + 3) 7/2 s−1

3/2 s+3

The s − 1 -chain 5.

3s + 1 (s − 1)(s2 + 1) −2s + 1 s2 + 1

2 s−1

The s + 1 -chain 2

6.

3s − s + 6 (s + 1)(s2 + 4) s−2 s2 + 4

2 (s + 1)

The s + 3 -chain

7.

s2 + s − 3 (s + 3)3 s−2 (s + 3)2 1 s+3 0

3 (s + 3)3 −5 (s + 3)2 1 s+3

58

1 Solutions

The s + 2 -chain 2

8.

5s − 3s + 10 (s + 1)(s + 2)2 5s + 23 (s + 1)(s + 2) 18 s+1

−36 (s + 2)2 −13 s+2

The s + 1 -chain s 9.

(s +

2)2 (s

1)2

+ s+4 (s + 2)2 (s + 1) −3s − 8 (s + 2)2

−1 (s + 1)2 3 s+1

The s − 1 -chain

10.

16s (s − 1)3 (s − 3)2 −4s + 36 (s − 1)2 (s − 3)2 −8s + 36 (s − 1)(s − 3)2 −7s + 27 (s − 3)2

4 (s − 1)3 8 (s − 1)2 7 (s − 1)

1 Solutions

59

The s − 5 -chain

11.

1 (s − 5)5 (s − 6) 1 (s − 5)4 (s − 6) 1 (s − 5)3 (s − 6) 1 (s − 5)2 (s − 6) 1 (s − 5)(s − 6) 1 s−6

−1 (s − 5)5 −1 (s − 5)4 −1 (s − 5)3 −1 (s − 5)2 −1 s−5

12. Use the technique of distinct linear factors (Example 5). 3/2 7/2 + s+3 s−1 13. Use the technique of distinct linear factors (Example 5). 13/8 5/8 − s−5 s+3 14.

−1 1 + s−1 s−2

15.

23 37 + 12(s − 5) 12(s + 7)

16.

1 2 + s s+1

17.

25 9 − 8(s − 7) 8(s + 1)

18.

25 9 15 + − 2(s − 3) 2(s − 1) s − 2

19.

1 1 1 − + 2(s + 5) 2(s − 1) s − 2

20. Use Theorem 1 to write

60

1 Solutions

s2 A1 = (s − 1)3 (s − 1)3

+

p1 (s) (s − 1)2

$ s2 $$ =1 1 $s=1 1 1 and p1 (s) = (s2 − (1)(1)) = (s2 − 1) = s + 1 s−1 s−1 where

A1 =

Continuing gives

s+1 A2 = (s − 1)2 (s − 1)2

+

p2 (s) s−1

$ s + 1 $$ where A2 = =2 1 $s=1 1 1 and p2 (s) = (s + 1 − (2)(1)) = (s − 1) = 1 s−1 s−1 Thus

s2 1 2 1 = + + . (s − 1)3 (s − 1)3 (s − 1)2 s−1

Alternate Solution: Write s = (s − 1) + 1 so that s2 ((s − 1) + 1)2 1 + 2(s − 1) + (s − 1)2 1 2 1 = = = + + 3 3 (s − 1) (s − 1) (s − 1)3 (s − 1)3 (s − 1)2 s − 1 21.

7 (s + 4)4

22. Use Theorem 1 to write s A1 = (s − 3)3 (s − 3)3

+

p1 (s) (s − 3)2

s $$ =3 $ 1 s=3 1 and p1 (s) = (s − (3)(1)) = 1. s−3 where A1 =

Thus,

s 3 1 = + 3 3 (s − 3) (s − 3) (s − 3)2

Alternate Solution: Write s = (s − 3) + 3 so that s (s − 3) + 3 3 1 = = + . (s − 3)3 (s − 3)3 (s − 3)3 (s − 3)2

1 Solutions

61

23. Use Theorem 1 to write s2 + s − 3 A1 = (s + 3)3 (s + 3)3 where A1 = and p1 (s) =

+

$ s2 + s − 3 $$ $ 1

p1 (s) (s + 3)2 =3

s=−3

1 1 (s2 + s − 3 − (3)(1)) = (s2 + s − 6) = s − 2 s+3 s+3

Continuing gives s−2 A2 = 2 (s + 3) (s + 3)2

+

p2 (s) s+3

$ s − 2 $$ = −5 1 $s=−3 1 1 and p2 (s) = (s − 2 − (−5)(1)) = (s + 3) = 1 s+3 s+3 where

Thus

A2 =

s2 + s − 3 3 5 1 = − + (s + 3)3 (s + 3)3 (s + 3)2 s+3

Alternate Solution: Write s = (s + 3) − 3 so that s2 + s − 3 ((s + 3) − 3)2 + ((s + 3) − 3) − 3 = 3 (s + 3) (s − 3)3 (s + 3)2 − 5(s + 3) + 3 = (s + 3)3 3 5 1 = − + . 3 2 (s + 3) (s + 3) s+3 24. Use Theorem 1 to compute the (s + 2)-chain: 5s2 − 3s + 10 A1 = 2 (s + 1)(s + 2) (s + 2)2

+

p1 (s) (s + 2)(s + 1)

$ 5s2 − 3s + 10 $$ = −36 $ s+1 s=−2 1 1 and p1 (s) = (5s2 − 3s + 10 − (−36)(s + 1)) = (5s2 + 33s + 46) = 5s + 23 s+2 s+2 where A1 =

62

1 Solutions

Continuing gives 5s + 23 A2 = (s + 2)(s + 1) s+2

+

p2 (s) s+1

$ 5s + 23 $$ where A2 = = −13 s + 1 $s=−2 1 1 and p2 (s) = (5s + 23 − (−13)(s + 1)) = (18s + 36) = 18 s+2 s+2 Thus 25.

5s2 − 3s + 10 −36 13 18 = − + (s + 1)(s + 2)2 (s + 2)2 s+2 s+1

s2 − 6s + 7 s2 − 6s + 7 = , so use Theorem 1 to compute the (s2 − 4s − 5)2 (s + 1)2 (s − 5)2 (s + 1)-chain: s2 − 6s + 7 A1 = (s + 1)2 (s − 5)2 (s + 1)2

+

p1 (s) (s + 1)(s − 5)2

$ s2 − 6s + 7 $$ 7 = (s − 5)2 $s=−1 18 1 and p1 (s) = (s2 − 6s + 7 − (7/18)(s − 5)2 ) s+1 1 1 = (11s2 − 38s − 49)/18 = (11s − 49) s+1 18 where

A1 =

Continuing gives 1 11s − 49 A2 = 2 18 (s + 1)(s − 5) s+1

+

p2 (s) (s − 5)2

$ 1 11s − 49 $$ where A2 = = −5/54 18 (s − 5)2 $s=−1 1 and p2 (s) = ((11s − 49)/18 − (−5/54)(s − 5)2 ) = (5s − 22)/54 s+1 s2 − 6s + 7 1/18 5/54 (5s − 22)/54 = − + Now either (s + 1)2 (s − 5)2 (s + 1)2 s+1 (s − 5)2 continue with Theorem 1 or replace s with s = (s−5)+5 in the numerator s2 − 6s + 7 of the last fraction to finish the calculation and get = (s + 1)2 (s − 5)2 ' ( 1 5 21 3 5 + + − 54 s − 5 (s + 1)2 (s − 5)2 s+1 Thus

1 Solutions

63

26. Use Theorem 1 to compute the (s + 9)-chain: 81 A1 = + 9) s+9

s3 (s

+

p1 (s) s3

$ 81 $$ where A1 = = −1/9 s3 $s=−9 1 1 and p1 (s) = (81 − (−1/9)(s3 )) = (s3 + 93 )/9 = (s2 − 9s + 81)/9 s+9 s+9 Thus

81 9 1 1 1 1 = 3− 2+ − + 9) s s 9s 9 s + 1

s3 (s

27. Use Theorem 1 to compute the (s + 2)-chain: s A1 = (s + 2)2 (s + 1)2 (s + 2)2

+

p1 (s) (s + 2)(s + 1)2

$ $ s $ where A1 = = −2 (s + 1)2 $s=−2 1 and p1 (s) = (s − (−2)(s + 1)2 ) s+2 2s2 + 5s + 2 (2s + 1)(s + 1) = = = 2s + 1 s+2 s+2 Continuing gives 2s + 1 A2 = (s + 2)(s + 1)2 s+2

+

p2 (s) (s + 1)2

$ 2s + 1 $$ where A2 = = −3 (s + 1)2 $s=−2 1 and p2 (s) = (2s + 1 − (−3)(s + 1)2 ) = 3s + 2 s+2 s −2 3 3s + 2 = − + . Now continue using (s + 2)2 (s + 1)2 (s + 2)2 s + 2 (s + 1)2 Theorem 1 or replace s by (s + 1) − 1 in the numerator of the last fraction s −2 3 1 3 to get = − − + 2 2 2 2 (s + 2) (s + 1) (s + 2) s + 2 (s + 1) s+1 Thus

28. Use Theorem 1 to compute the (s + 2)-chain:

64

1 Solutions

s2 A1 = (s + 2)2 (s + 1)2 (s + 2)2

+

p1 (s) (s + 2)(s + 1)2

$ s2 $$ =4 (s + 1)2 $s=−2 1 and p1 (s) = (s2 − (4)(s + 1)2 ) s+2 −3s2 − 8s − 4 −(3s + 2)(s + 2) = = = −(3s + 2) s+2 s+2 where A1 =

Continuing gives −3s − 2 A2 = (s + 2)(s + 1)2 s+2

+

p2 (s) (s + 1)2

$ −3s − 2 $$ =4 (s + 1)2 $s=−2 1 and p2 (s) = (−3s − 2 − (4)(s + 1)2 ) = −(4s + 3) s+2 where

A2 =

s2 4 −4s − 3 = + + . Now continue using (s + 2)2 (s + 1)2 (s + 2)2 s + 2 (s + 1)2 Theorem 1 or replace s by (s + 1) − 1 in the numerator of the last fraction s2 4 4 1 4 to get = + + − (s + 2)2 (s + 1)2 (s + 2)2 s + 2 (s + 1)2 s+1 Thus

29. Use Theorem 1 to compute the (s − 3)-chain: 8s A1 = (s − 1)(s − 2)(s − 3)3 (s − 3)3

+

p1 (s) (s − 1)(s − 2)(s − 3)2

$ $ 8s $ = 12 (s − 1)(s − 2) $s=3 1 and p1 (s) = (8s − (12)(s − 1)(s − 2)) s−3 −12s2 + 44s − 24 (−12s + 8)(s − 3) = = = −12s + 8 s−3 s−3 where A1 =

For the second step in the (s − 3)-chain:

1 Solutions

65

−12s + 8 A2 = 2 (s − 1)(s − 2)(s − 3) (s − 3)2

p2 (s) (s − 1)(s − 2)(s − 3)2

+

$ −12s + 8 $$ = −14 (s − 1)(s − 2) $s=3 1 and p2 (s) = (−12s + 8 − (−14)(s − 1)(s − 2)) s−3 14s2 − 54s + 36 (14s − 12)(s − 3) = = = 14s − 12 s−3 s−3 where A2 =

Continuing gives

14s − 12 A3 = (s − 1)(s − 2)(s − 3)2 s−3

p3 (s) (s − 1)(s − 2)

+

$ 14s − 12 $$ where A3 = = 15 (s − 1)(s − 2) $s=3 1 and p3 (s) = (14s − 12 − (15)(s − 1)(s − 2)) = −15s + 14 s−3 8s 12 14 15 −15s + 14 = − + + . (s − 1)(s − 3)(s − 3)3 (s − 3)3 (s − 3)2 s − 3 (s − 1)(s − 2) The last fraction has a denominator with distinct linear factors so we get 8s 12 −14 15 −16 1 = + + + + 3 3 2 (s − 1)(s − 3)(s − 3) (s − 3) (s − 3) s−3 s−2 s−1 Thus

30. Use Theorem 1 to compute the s-chain: s2 (s

25 A1 = − 5)(s + 1) s2

+

p1 (s) s(s − 5)(s + 1)

$ $ 25 $ = −5 (s − 5)(s + 1) $s=0 1 and p1 (s) = (25 − (−5)(s − 5)(s + 1)) s 5s2 − 20s = = 5s − 20 s where A1 =

Continuing gives

66

1 Solutions

5s − 20 A2 = s(s − 5)(s + 1) s

p2 (s) (s − 5)(s + 1)

+

$ 5s − 20 $$ =4 (s − 5)(s + 1) $s=0 1 and p2 (s) = (5s − 20 − (4)(s − 5)(s + 1)) = −4s + 16 s where A2 =

25 5 4 −4s + 16 = 2 − + . The last fraction has s2 (s − 5)(s + 1) s s (s − 5)(s + 1) 25 a denominator with distinct linear factors so we get 2 = s (s − 5)(s + 1) 2 1 20 1 5 4 − − − + 3s−5 3 s + 1 s2 s Thus

31. Use Theorem 1 to compute the (s − 2)-chain: s A1 = (s − 2)2 (s − 3)2 (s − 2)2

+

p1 (s) (s − 2)(s − 3)2

$ $ s $ where A1 = =2 (s − 3)2 $s=2 1 and p1 (s) = (s − (2)(s − 3)2 ) s−2 −2s2 + 13s − 18 (−2s + 9)(s − 2) = = = −2s + 9 s−2 s−2

Continuing gives

−2s + 9 A2 = (s − 2)(s − 3)2 s−3

+

p2 (s) (s − 3)2

$ −2s + 9 $$ =5 (s − 3)2 $s=2 1 and p2 (s) = (−2s + 9 − (5)(s − 3)2 ) = −5s + 18 s−2 where A2 =

2 5 −5s + 18 + + . Now continue 2 (s − − (s − 2) s−2 (s − 3)2 using Theorem 1 or replace s by (s − 3) + 3 in the numerator of the last s 2 5 3 5 fraction to get = + + − (s − 2)2 (s − 3)2 (s − 2)2 s − 2 (s − 3)2 s−3 Thus

s

2)2 (s

3)2

=

32. Use Theorem 1 to compute the (s − 1)-chain:

1 Solutions

67

16s A1 = 3 2 (s − 1) (s − 3) (s − 1)3

+

p1 (s) (s − 1)2 (s − 3)2

$ 16s $$ =4 (s − 3)2 $s=1 1 and p1 (s) = (16s − (4)(s − 3)2 ) s−1 −4s2 + 40s − 36 (−4s + 36)(s − 1) = = = −4s + 36 s−1 s−1 where A1 =

For the second step in the (s − 1)-chain: −4s + 36 A2 = 2 2 (s − 1) (s − 3) (s − 1)2

+

p2 (s) (s − 1)(s − 3)2

$ −4s + 36 $$ =8 (s − 3)2 $s=1 1 and p2 (s) = (−4s + 36 − (8)(s − 3)2 ) s−1 −8s2 + 44s − 36 (−8s + 36)(s − 1) = = = −8s + 36 s−1 s−1 where A2 =

Continuing gives

−8s + 36 A3 = (s − 1)(s − 3)2 s−1

+

p3 (s) (s − 3)2

$ −8s + 36 $$ where A3 = =7 (s − 3)2 $s=1 1 and p3 (s) = (−8s + 36 − (7)(s − 3)2 ) = −7s + 27 s−1 16s 4 8 7 −7s + 27 = + + + . Now (s − 1)3 (s − 3)2 (s − 1)3 (s − 1)2 s−1 (s − 3)2 continue using Theorem 1 or replace s by (s − 3) + 3 in the numerator of 16s 4 8 7 the last fraction to get = + + + (s − 1)3 (s − 3)2 (s − 1)3 (s − 1)2 s − 1 6 7 − 2 (s − 3) s−3 Thus

33. Apply the Laplace transform to both sides. For the left hand side we get

68

1 Solutions

L {y !! + 2y ! + y} = L {y !! } + 2L {y ! } + L {y}

= s2 Y (s) − sy(0) − y ! (0) + 2(sY (s) − y(0)) + Y (s) = (s2 + 2s + 1)Y (s).

0 1 Since L 9e2t =

9 we get s−2 Y (s) =

9 . (s + 1)2 (s − 2)

A partial fraction decomposition gives

The (s + 1) -chain 9 (s + 1)2 (s − 2) 3 (s + 1)(s − 2) 1 (s − 2)

It follows that Y (s) = and

−3 (s + 1)2 −1 s+1

−3 1 1 − + (s + 1)2 s+1 s−2

y(t) = −3te−t − e−t + e2t .

34. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 3y ! + 2y} = L {y !! } + 3L {y ! } + 2L {y} = s2 Y (s) − sy(0) − y ! (0) + 3(sY (s) − y(0)) + 2Y (s) = (s2 + 3s + 2)Y (s) − s − 2.

0 1 Since L 12e2t = 12/(s − 2) we get the algebraic equation (s2 + 3s + 2)Y (s) − s − 2 =

12 . s−2

Since s2 + 2s + 3 = (s + 1)(s + 2) we now solve to get

1 Solutions

69

12 s+2 + (s − 2)(s + 1)(s + 2) (s + 1)(s + 2) 12 1 = + . (s − 2)(s + 1)(s + 2) s + 1

Y (s) =

The first term has denominator that is a product of distinct linear terms. Thus $ $ A1 12 12 $ • For : A1 = = = 1. s−2 (s + 1)(s + 2) $s=2 12 $ $ A2 12 12 $ • For : A2 = = = −4. s+1 (s − 2)(s + 2) $s=−1 −3 $ $ A3 12 12 $ • For : A3 = = =3 s+2 (s − 2)(s + 1) $s=−2 4 and we get

12 1 4 3 = − + . (s − 2)(s + 1)(s + 2) s−2 s+1 s+2 It follows that Y (s) =

1 3 3 − + . s−2 s+1 s+2

Table 2.2 gives y(t) = e2t − 3e−t + 3e−2t

35. Apply the Laplace transform to both sides. For the left hand side we get L {y !! − 4y ! − 5y} = L {y !! } − 4L {y ! } − 5L {y}

= s2 Y (s) − sy(0) − y ! (0) − 4(sY (s) − y(0)) − 5Y (s) = (s2 − 4s − 5)Y (s) + s − 5.

Since L {150t} = 150/s2 we get the algebraic equation (s2 − 4s − 5)Y (s) + s − 5 =

150 . s2

Hence, −s + 5 150 + (s + 1)(s − 5) s2 (s + 1)(s − 5) −1 150 = + 2 . s + 1 s (s + 1)(s − 5)

Y (s) =

For the second term we start with the s-chain to get the following partial fraction decomposition

70

1 Solutions

The s -chain 150 + 1)(s − 5) 30(s − 4) s(s + 1)(s − 5) −244s + 124 (s + 1)(s − 5) 1 s−5

s2 (s

Thus Y (s) =

−30 s2 24 s −25 s+1

−30 24 26 1 + − + 2 s s s+1 s−5

and Table 2.2 gives y(t) = −30t + 24 − 26e−t + e5t .

36. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 4y ! + 4y} (s) = L {y !! } (s) + 4L {y ! } (s) + 4L {y} (s)

= s2 Y (s) − sy(0) − y ! (0) + 4(sY (s) − y(0)) + 4Y (s) = (s2 + 4s + 4)Y (s) − 1.

Since L {4 cos 2t} = 4s/(s2 + 4) we get the algebraic equation (s + 4)2 Y (s) − 1 = Hence, Y (s) =

4s . +4

s2

1 4s + 2 (s + 2)2 (s + 4)(s + 2)2

For the second term we start with the (s + 2)-chain to get the following partial fraction decomposition

1 Solutions

71

The s -chain 4s + 4)(s + 2)2 s+2 2 (s + 4)(s + 2) 1 2 s +4

(s2

−1 (s + 2)2 0 s+2

Thus 1 −1 1 + + 2 (s + 2)2 (s + 2)2 s +4 1 = 2 s +4

Y (s) =

and therefore y(t) =

1 2

sin 2t

37. Apply the Laplace transform to both sides. For the left hand side we get L {y !! − 3y ! + 2y} = L {y !! } − 3L {y ! } + 2L {y}

= s2 Y (s) − sy(0) − y ! (0) − 3(sY (s) − y(0)) + 2Y (s) = (s2 − 3s + 2)Y (s) − 2s + 3.

Since L {4} = 4/s we get the algebraic equation (s − 1)(s − 2)Y (s) − 2s + 3 = Hence, Y (s) =

4 . s

2s − 3 4 + . (s − 1)(s − 2) s(s − 1)(s − 2)

Each term has denominator a product of distinct linear terms. It is easy to see that 2s − 3 1 1 = + (s − 1)(s − 2) s−1 s−2 and

Thus

4 2 4 2 = − + . s(s − 1)(s − 2) s s−1 s−2 Y (s) =

2 3 3 + − s s−2 s−1

72

1 Solutions

and Table 2.2 gives y(t) = 2 + 3e2t − 3et . 38. Apply the Laplace transform to both sides. For the left hand side we get L {y !! − 3y ! + 2y} = L {y !! } − 3L {y ! } + 2L {y}

= s2 Y (s) − sy(0) − y ! (0) − 3(sY (s) − y(0)) + 2Y (s) = (s2 − 3s + 2)Y (s) + 3s − 9.

Since L {et } = 1/(s − 1) we get the algebraic equation (s − 1)(s − 2)Y (s) + 3s − 9 = Hence, Y (s) =

1 . s−1

−3s + 9 1 + . (s − 1)(s − 2) (s − 1)2 (s − 2)

The first term has partial fraction decomposition

−3s + 9 −6 3 = + . (s − 1)(s − 2) s−1 s−2 The (s − 1)-chain for the second term is

The (s − 1) -chain 1 (s − 1)2 (s − 2) 1 (s − 1)(s − 2) 1 s−2

Thus Y (s) =

−1 (s − 1)2 −1 s−1

−7 4 1 + − s − 1 s − 2 (s − 1)2

and Table 2.2 gives y(t) = −tet + 4e2t − 7et

1 Solutions

73

Section 2.4 1. Note that s = i is a root of s2 + 1. Applying Theorem 1 gives 1 (s2 + 1)2 (s2 + 2)

=

where B1 i + C1 = ⇒

and p1 (s) = =

B1 s + C1 p1 (s) + 2 (s2 + 1)2 (s + 1)(s2 + 2) $ $ 1 1 $ = 2 =1 $ 2 (s + 2) s=i i +2 B1 = 0 and C1 = 1 1 (1 − (1)(s2 + 2)) s2 + 1 −s2 − 1 = −1. s2 + 1

We now apply Theorem 1 on the remainder term

(s2

−1 + 1)(s2 + 2)

=

where B2 i + C2 = ⇒

and p2 (s) = =

−1 . (s2 + 1)(s2 + 2)

B2 s + C2 p2 (s) + 2 2 (s + 1) (s + 2) $ −1 $$ = −1 (s2 + 2) $s=i B2 = 0 and C2 = −1 1 (−1 − (−1)(s2 + 2)) 2 s +1 s2 + 1 = 1. s2 + 1

Thus the (s2 + 1)-chain is The s2 + 1 -chain 1 (s2 + 1)2 (s2 + 2) −1 2 (s + 1)(s2 + 2) 1 s2 + 2 2. Note that s =

√

1 (s2 + 1)2 −1 2 (s + 1)

2i is a root of s2 + 2. Applying Theorem 1 gives

74

1 Solutions

(s2

s3 + 2)2 (s2 + 3)

=

√

where B1 2i + C1

and p1 (s)

= ⇒ = =

B1 s + C1 p1 (s) + 2 (s2 + 2)2 (s + 2)(s2 + 3) √ $ √ s3 $$ ( 2i)3 √ = = −2 2i $ (s2 + 3) s=√2i ( 2i)2 + 3 B1 = −2 and C1 = 0 1 (s3 − (−2s)(s2 + 3)) s2 + 1 3s3 + 6s = 3s. s2 + 2

We now apply Theorem 1 on the remainder term

(s2

3s + 2)(s2 + 3) √

where B2 2i + C2

and p2 (s)

=

= ⇒ = =

3s . (s2 + 2)(s2 + 3)

B2 s + C2 p2 (s) + 2 2 (s + 2) (s + 3) $ √ 3s $$ = 3 2i $ 2 √ (s + 3) s= 2i B2 = 3 and C2 = 0 1 (3s − (3s)(s2 + 3)) 2 s +2 −3s(s2 + 2) = −3s. s2 + 2

Thus the (s2 + 2)-chain is The s2 + 2 -chain s3 (s2 + 2)2 (s2 + 3) 3s (s2 + 2)(s2 + 3) −3s s2 + 3 3. Note that s =

√

−2s (s2 + 2)2 3s s2 + 2

3i is a root of s2 + 3. Applying Theorem 1 gives

1 Solutions

75

(s2

8s + 8s2 + 3)3 (s2 + 1) √

where B1 3i + C1

and p1 (s)

B1 s + C1 p1 (s) + 2 (s2 + 3)3 (s + 3)2 (s2 + 1)

=

√ √ $ 8s + 8s2 $$ 8 3i + 8( 3i)2 √ = (s2 + 1) $s=√3i ( 3i)2 + 1 √ −4 3i + 12 B1 = −4 and C1 = 12 1 (8s + 8s2 − (−4s + 12)(s2 + 1)) 2 s +3 4s3 − 4s2 + 12s − 12 = 4(s − 1). s2 + 3

= = ⇒ = =

Apply Theorem 1 a second time on the remainder term 4s − 4 (s2 + 3)2 (s2 + 1)

=

√ where B2 3i + C2

=

and p2 (s)

⇒ = =

4s − 4 . (s2 + 3)2 (s2 + 1)

B2 s + C2 p2 (s) + 2 (s2 + 3)2 (s + 3)(s2 + 1) $ √ 4s − 4 $$ = −2 3i + 2 $ 2 √ (s + 1) s= 3i B2 = −2 and C2 = 2 1 (4s − 4 − (−2s + 2)(s2 + 1)) s2 + 3 2s3 − 2s2 + 6s − 6 = 2s − 2. s2 + 3

A third application of Theorem 1 on the remainder term gives (s2

2s − 2 + 3)(s2 + 1) √

where B3 3i + C3

and p3 (s)

=

= ⇒ = =

Thus the (s2 + 3)-chain is

(s2

2s − 2 + 3)(s2 + 1)

B3 s + C3 p3 (s) + 2 2 (s + 3) (s + 1) $ √ 2s − 2 $$ = − 3i + 1 $ 2 √ (s + 1) s= 3i B3 = −1 and C3 = 1 1 (2s − 2 − (−s + 1)(s2 + 1)) 2 s +3 s3 − s2 + 3s − 3 = s − 1. s2 + 3

76

1 Solutions

The s2 + 3 -chain 8s + 8s2 (s+ 3)3 (s2 + 1) 4(s − 1) (s2 + 3)2 (s2 + 1) 2(s − 1) 2 (s + 3)(s2 + 1) s−1 s2 + 1

12 − 4s (s2 + 3)3 2 − 2s (s2 + 3)2 1−s s2 + 3

4. Note that s = 2i is a root of s2 + 4. Applying Theorem 1 gives 4s4 (s2 + 4)3 (s2 + 6) where B1 2i + C1

and p1 (s)

=

= = ⇒ = =

B1 s + C1 p1 (s) + 2 2 4 (s + 4) (s + 4)3 (s2 + 6) $ 4s4 $$ 4(2i)4 = (s2 + 6) $s=2i (2i)2 + 6 32 B1 = 0 and C1 = 32 1 (4s4 − (32)(s2 + 6)) s2 + 4 4s4 − 32s2 − 192 = 4s2 − 48. s2 + 4

Apply Theorem 1 a second time on the remainder term 4s2 − 48 + 4)3 (s2 + 6)

=

where B2 2i + C2

=

(s2

and p2 (s)

⇒ = =

B2 s + C2 p2 (s) + 2 (s2 + 4)3 (s + 4)2 (s2 + 6) $ 4s2 − 48 $$ = −32 (s2 + 6) $s=2i B2 = 0 and C2 = −32 1 (4s2 − 48 − (−32)(s2 + 6)) s2 + 4 36s2 + 144 = 36. s2 + 4

A third application of Theorem 1 on the remainder term gives

4s2 − 48 . (s2 + 4)3 (s2 + 6)

(s2

36 + 4)2 (s2 + 6)

1 Solutions

77

36 + 4)2 (s2 + 6)

=

where B3 2i + C3

=

(s2

and p3 (s)

⇒ = =

B3 s + C3 p3 (s) + 2 2 2 (s + 4) (s + 4)(s2 + 6) $ 36 $$ = 18 (s2 + 6) $s=2i B3 = 0 and C3 = 18 1 (36 − (18)(s2 + 6)) 2 s +4 −18s2 − 72 = −18. s2 + 4

A fourth (and final) application of Theorem 1 on the remainder term −18 gives (s2 + 4)(s2 + 6) (s2 where

−18 + 4)(s2 + 6) B4 2i + C4

and p4 (s)

=

= ⇒ = =

B4 s + C4 p4 (s) + 2 2 (s + 4) (s + 6) $ −18 $$ = −9 (s2 + 6) $s=2i B4 = 0 and C4 = −9 1 (−18 − (−9)(s2 + 6)) 2 s +4 9s2 + 36 = 9. s2 + 4

Thus the (s2 + 4)-chain is The s2 + 4 -chain 4s4 (s2 + 4)4 (s2 + 6) 4s2 − 48 2 (s + 4)3 (s2 + 6) 36 (s2 + 4)2 (s2 + 6) −18 2 (s + 4)(s2 + 6) 9 s2 + 6

32 (s2 + 4)4 −32 2 (s + 4)3 18 (s2 + 4)2 −9 2 s +4

5. Note that s2 + 2s + 2 = (s + 1)2 + 1 so s = −1 ± i are the roots of s2 + 2s + 2. We will use the root s = −1 + i for the partial fraction algorithm. Applying Theorem 1 gives

78

1 Solutions

1 (s2

+ 2s +

2)2 (s2

+ 2s +

3)2

B1 s + C1 + 2s + 2)2

=

(s2

+ where B1 (−1 + i) + C1

$ $ 1 $ 2 2 (s + 2s + 3) $s=−1+i 1 =1 ((−1 + i)2 + 2)2 B1 = 0 and C1 = 1 1 − (1)(s2 + 2s + 3)2 s2 + 2s + 2 2 −(s + 2s + 4)(s2 + 2s + 2) s2 + 2s + 2 2 −(s + 2s + 4).

= = ⇒

and p1 (s)

= = =

Now apply Theorem 1 to the remainder term

(s2

−(s2 + 2s + 4) + 2s + 2)(s2 + 2s + 3)2

where B2 (−1 + i) + C2

−(s2 + 2s + 4) . (s2 + 2s + 2)(s2 + 2s + 3)2

B2 s + C2 p2 (s) + (s2 + 2s + 2) (s2 + 2s + 3)2

=

$ −(s2 + 2s + 4) $$ = −2 (s2 + 2s + 3) $s=−1+i B2 = 0 and C2 = −2 −(s2 + 2s + 4) − (−2)(s2 + 2s + 3)2 s2 + 2s + 2 2 (2(s + 1) + 5)((s + 1)2 + 1) s2 + 2s + 2 2 2s + 4s + 7.

= ⇒

and p2 (s)

p1 (s) (s2 + 2s + 2)(s2 + 2s + 3)2

= = =

Thus the (s2 + 2s + 2)-chain is The s2 + 2s + 2 -chain 1 (s2 + 2s + 2)2 (s2 + 2s + 3)2 −(s2 + 2s + 4) 2 (s + 2s + 2)(s2 + 2s + 3)2 2s2 + 4s + 7 (s2 + 2s + 3)2

1 (s2 + 2s + 2)2 −2 s2 + 2s + 2

1 Solutions

79

6. Note that s2 + 2s + 2 = (s + 1)2 + 1 so s = −1 ± i are the roots of s2 + 2s + 2. We will use the root s = −1 + i for the partial fraction algorithm. Applying Theorem 1 gives 1 (s2 + 2s + 2)2 (s2 + 4s + 5)

=

B1 s + C 1 (s2 + 2s + 2)2 +

where B1 (−1 + i) + C1

= = ⇒

and p1 (s)

= = =

$ $ 5s − 5 $ (s2 + 4s + 5) $s=−1+i 5i − 10 = 5i 2i + 1 B1 = 5 and C1 = 5 5s − 5 − (5s + 5)(s2 + 4s + 5) s2 + 2s + 2 −(5s + 15)(s2 + 2s + 2) s2 + 2s + 2 −(5s + 15).

Now apply Theorem 1 to the remainder term −(5s + 15) (s2 + 2s + 2)(s2 + 4s + 5)

=

where B2 (−1 + i) + C2

= ⇒

and p2 (s)

= = =

Thus the (s2 + 2s + 2)-chain is

p1 (s) (s2 + 2s + 2)(s2 + 4s + 5)

−(5s + 15) . (s2 + 2s + 2)(s2 + 4s + 5)

B2 s + C2 p2 (s) + (s2 + 2s + 2) (s2 + 4s + 5) $ −(5s + 15) $$ = −4 + 3i (s2 + 4s + 5) $s=−1+i B2 = 3 and C2 = −1 −(5s + 15) − (3s − 1)(s2 + 4s + 5) s2 + 2s + 2 2 (−3s − 5)(s + 2s + 2) s2 + 2s + 2 −3s − 5.

80

1 Solutions

The s2 + 2s + 2 -chain 5s − 5 + 2s + 2)2 (s2 + 4s + 5) −5s − 15 (s2 + 2s + 2)(s2 + 4s + 5) −3s − 5 2 s + 4s + 5

5s + 5 + 2s + 2)2 3s − 1 s2 + 2s + 2

(s2

(s2

7. Use Theorem 1 of Section 2.3 to compute the (s − 3)-chain: (s2

s A1 = + 1)(s − 3) s−3

p1 (s) s2 + 1

+

$ s $$ 3 = s2 + 1 $s=3 10 1 1 and p1 (s) = (s − (3/10)(s2 + 1)) = (−3s2 + 10s − 3) s−3 10(s − 3) −3s + 1 = 10 where A1 =

Since the remainder term we conclude

−3s + 1 is already a simple partial fraction, 10(s2 + 1)

s 1 = 2 (s + 1)(s − 3) 10

'

3 1 − 3s + 2 s−3 s +1

(

8. Use Theorem of Section 2.3 1 to compute the (s + 1)-chain: 4s A1 = (s2 + 1)2 (s + 1) s+1 where A1 =

+

p1 (s) (s2 + 1)2

$ $ 4s $ = −1 (s2 + 1)2 $s=−1

1 s4 + 2s2 + 4s + 1 (4s − (−1)(s2 + 1)2 ) = s+1 s+1 = s3 − s2 + 3s + 1

and p1 (s) =

s3 − s2 + 3s + 1 (s2 + 1)2 2 using Theorem 1. Since s = i is a root of s + 1, an application of this theorem gives Now compute the s2 + 1-chain for the remainder term

1 Solutions

81

s3 − s2 + 3s + 1 (s2 + 1)2 where

=

B1 s + C1 p1 (s) + 2 (s2 + 1)2 (s + 1)

$ s3 − s2 + 3s + 1 $$ = 2i + 2 $ 1 s=i ⇒ B1 = 2 and C1 = 2 1 and p1 (s) = (s3 − s2 + 3s + 1 − (2s + 2)(1)) s2 + 1 s3 − s2 + s − 1 = = s − 1. s2 + 1 B1 i + C1 =

s−1 is a simple partial fraction, we conclude s2 + 1 that the complete partial fraction decomposition is Since the remainder term

4s 2s + 2 s−1 1 = 2 + 2 − (s2 + 1)2 (s + 1) (s + 1)2 s +1 s+1 9. We compute the (s2 + 4)-chain: 9s2 (s2 + 4)2 (s2 + 1) where

=

B1 (2i) + C1 = ⇒ and p1 (s) = =

B1 s + C1 (s2 + 4)2

+

p1 (s) (s2 + 4)(s2 + 1)

$ 9s2 $$ = 12 s2 + 1 $s=2i B1 = 0 and C1 = 12 1 −3(s2 + 4) 2 2 (9s − 12(s + 1)) = s2 + 4 s2 + 4 −3

Now compute the next term in s2 + 4-chain. (s2

−3 + 4)(s2 + 1)

=

where B2 (2i) + C1 = ⇒

and p2 (s) =

=

B2 s + C2 p2 (s) + 2 2 s +4 s +1 $ −3 $$ =1 s2 + 1 $s=2i B2 = 0 and C2 = 1 1 (−3 − (s2 + 1)) s2 + 4 −(s2 + 1) = −1. s2 + 1

82

1 Solutions

−1 is a simple partial fraction, we conclude s2 + 1 that the complete partial fraction decomposition is Since the remainder term

9s2 12 1 1 = 2 + 2 − (s2 + 4)2 (s2 + 1) (s + 4)2 s +4 s+1 10. We compute the (s2 + 1)-chain: 9s (s2 + 1)2 (s2 + 4)

=

where B1 (i) + C1 = ⇒ and p1 (s) = =

B1 s + C1 (s2 + 1)2

+

p1 (s) (s2 + 1)(s2 + 4)

$ 9s $$ = 3i s2 + 4 $s=i B1 = 3 and C1 = 0 1 −3s(s2 + 1) (9s − 3s(s2 + 4)) = 2 s +1 s2 + 1 −3s

Now compute the next term in s2 + 1-chain. (s2

−3s + 1)(s2 + 4)

=

where B2 (i) + C2 = ⇒

and p2 (s) =

=

B2 s + C2 p2 (s) + 2 2 s +1 s +4 $ −3s $$ = −i s2 + 4 $s=i B2 = −1 and C2 = 0 1 (−3s + s(s2 + 4)) 2 s +1 s(s2 + 1) = s. s2 + 1

s is a simple partial fraction, we conclude +4 that the complete partial fraction decomposition is Since the remainder term

s2

9s 3s s s = 2 − 2 + (s2 + 1)2 (s2 + 4) (s + 1)2 s + 1 s2 + 4 11. Use Theorem of Section 2.3 1 to compute the (s − 3)-chain:

1 Solutions

(s2

83

2 A1 = − 6s + 10)(s − 3) s−3 where A1 =

+

(s2

p1 (s) − 6s + 10)

$ $ 2 $ =2 2 (s − 6s + 10) $s=3

1 −2s2 + 12s − 18 (2 − (2)(s2 − 6s + 10)) = s−3 s−3 = −2s + 6

and p1 (s) =

−2s + 6 is a simple partial fraction, we con− 6s + 10 2 clude that the complete partial fraction decomposition is 2 = (s − 6s + 10)(s − 3) 2 6 − 2s + s − 3 (s − 3)2 + 1 Since the remainder term

s2

12. Use Theorem of Section 2.3 1 to compute the (s − 1)-chain: (s2

30 A1 = − 4s + 13)(s − 1) s−1 where A1 =

+

(s2

p1 (s) − 4s + 13)

$ $ 30 $ =3 2 (s − 4s + 13) $s=1

1 −3s2 + 12s − 9 (30 − (3)(s2 − 4s + 13)) = s−1 s−1 = −3s + 9

and p1 (s) =

−3s + 9 is a simple partial fraction, we con− 4s + 13 30 clude that the complete partial fraction decomposition is 2 = (s − 4s + 13)(s − 1) 9 − 3s 3 + ((s − 2)2 + 9) s − 1 Since the remainder term

s2

13. Note that s2 −4s+8 = (s−2)2 +2 so s = 2±2i are the roots of s2 −4s+8. We will use the root s = 2+2i to compute the (s2 −4s+8)-chain. Applying Theorem 1 gives

84

1 Solutions

(s2

25 − 4s + 8)2 (s − 1)

B1 s + C1 − 4s + 8)2 p1 (s) + (s2 − 4s + 8)(s − 1)

=

(s2

$ 25 $$ s − 1) $s=2+2i 25 = 5 − 10i 2i + 1 B1 = −5 and C1 = 15 25 − (−5s + 15)(s − 1) s2 − 4s + 8 2 (5)(s − 4s + 8) s2 − 4s + 8 5.

where B1 (2 + 2i) + C1 = = ⇒ and p1 (s) = = =

Now apply Theorem 1 to the remainder term

(s2

5 − 4s + 8)(s − 1)

=

where B2 (2 + 2i) + C2 = ⇒ and p2 (s) = = = Thus the partial fraction expansion is −s + 3 1 + s2 − 4s + 8 s − 1

5 . (s2 − 4s + 8)(s − 1)

B2 s + C2 p2 (s) + 2 (s − 4s + 8) s − 1

$ 5) $$ = 1 − 2i s − 1 $s=2+2i B2 = −1 and C2 = 3 5 − (3 − s)(s − 1) s2 − 4s + 8 (1)(s2 − 4s + 8) s2 − 4s + 8 1.

25 −5s + 15 = 2 + (s2 − 4s + 8)2 (s − 1) (s − 4s + 8)2

14. Note that s2 + 6s + 10 = (s + 3)2 + 1 so s = −3 ± i are the roots of s2 +6s+10. We will use the root s = −3+i to compute the (s2 +6s+10)chain. Applying Theorem 1 gives

1 Solutions

85

(s2

s + 6s + 10)2 (s + 3)2

=

where B1 (−3 + i) + C1 = = ⇒ and p1 (s) = = =

B1 s + C1 + 6s + 10)2 p1 (s) + (s2 + 6s + 10)(s + 3)2

(s2

$ $ s $ 2 (s + 3) ) $s=−3+i 3−i B1 = −1 and C1 = 0 s − (−s)(s + 3)2 s2 + 6s + 10 (s)(s2 + 6s + 10) s2 + 6s + 10 s.

Now apply Theorem 1 to the remainder term s (s2 + 6s + 10)(s + 3)2

=

where B2 (−3 + i) + C2 = ⇒ and p2 (s) = = =

(s2

s . + 6s + 10)(s + 3)2 )

B2 s + C2 p2 (s) + (s2 + 6s + 10) (s + 3)2 $ s) $$ =3−i (s + 3)2 $s=−3+i B2 = −1 and C2 = 0 s − (−s)(s + 3)2 s2 + 6s + 10 (s)(s2 + 6s + 10) s2 + 6s + 10 s.

The remainder term is s (s + 3) − 3 −3 1 = = + , 2 2 2 (s + 3) (s + 3) (s + 3) s+3 so the partial fraction expansion of the entire rational function is s −s s = − 2 (s2 + 6s + 10)2 (s + 3)2 (s2 + 6s + 10)2 s + 6s + 10 3 1 − + 2 (s + 3) s+3 15. Note that s2 + 4s + 5 = (s + 2)2 + 1 so s = −2 ± i are the roots of s2 + 4s + 5. We will use the root s = −2 + i to compute the (s2 + 4s + 5)chain. Applying Theorem 1 gives

86

1 Solutions

(s2

s+1 + 4s + 5)2 (s2 + 4s + 6)2

=

B1 s + C1 + 4s + 5)2

(s2

+ where B1 (−2 + i) + C1 = = ⇒ and p1 (s) = = = =

p1 (s) (s2 + 4s + 5)(s2 + 4s + 6)2

$ $ s+1 $ 2 2 (s + 4s + 6) ) $s=−2+i −1 + i B1 = 1 and C1 = 1 s + 1 − (s + 1)(s2 + 4s + 6)2 s2 + 4s + 5 −(s + 1)((s2 + 4s + 6)2 − 1) s2 + 4s + 5 −(s + 1)(s2 + 4s + 7)(s2 + 4s + 5) s2 + 4s + 5 2 −(s + 1)(s + 4s + 7).

Now apply Theorem 1 to the remainder term −(s + 1)(s2 + 4s + 7) + 4s + 5)(s2 + 4s + 6)2

(s2

=

where B2 (−2 + i) + C2 = ⇒ and p2 (s) = = =

(s2

−(s + 1)(s2 + 4s + 7) . + 4s + 5)(s2 + 4s + 6)2 )

B2 s + C2 p2 (s) + 2 2 (s + 4s + 5) (s + 4s + 6)2 $ −(s + 1)(s2 + 4s + 7) $$ = 2 − 2i $ (s2 + 4s + 6)2 s=−2+i B2 = −2 and C2 = −2 −(s + 1)(s2 + 4s + 7) − (−2s − 2)(s2 + 4s + 6)2 s2 + 4s + 5 2 (s + 1)(2(s + 4s + 6) + 1)(s2 + 4s + 5) s2 + 4s + 5 2 (s + 1)(2(s + 4s + 6) + 1).

The remainder term is (s + 1)(2(s2 + 4s + 6) + 1) s+1 2s + 2 = 2 + 2 (s2 + 4s + 6)2 (s + 4s + 6)2 s + 4s + 6 so the partial fraction expansion of the entire rational function is s+1 s+1 2s + 2 = + 2 (s2 + 4s + 5)2 (s2 + 4s + 6)2 (s2 + 4s + 6)2 s + 4s + 6 s+1 2s + 2 + 2 − 2 (s + 4s + 5)2 s + 4s + 5

1 Solutions

87

u , (u + 5)3 (u + 6)2 which can be put in partial fraction form by the technique of Section 2.3. Use Theorem 1 of Section 2.3 to compute the (u + 5)-chain:

16. Using the hint, let u = s2 . Then the rational function becomes

u A1 = (u + 5)3 (u + 6)2 (u + 5)3

p1 (u) (u + 5)2 (u + 6)2

+

$ $ u $ where A1 = = −5 (u + 6)2 $u=−5 1 and p1 (u) = (u − (−5)(u + 6)2 ) u+5 5u2 + 61u + 180 (5u + 36)(u + 5) = = = 5u + 36. u+6 u+5 Continuing gives 5u + 36 A2 = (u + 5)2 )(u + 6)2 (u + 5)2 where A2 = and p2 (u) = = =

+

p2 (u) (u + 5)(u + 6)2

$ 5u + 36 $$ = 11 (u + 6)2 $u=−5 1 (5u + 36 − (11)(u + 6)2 ) u+5 −11u2 − 127u − 360 −(11u + 72)(u + 5) = u+5 u+5 −(11u + 72).

Another recursion gives −11u − 72 A3 = 2 (u + 5)(u + 6) (u + 5)

+

p3 (u) (u + 6)2

$ −11u − 72 $$ = −17 (u + 6)2 $u=−5 1 and p3 (u) = (−11u − 72 − (−17)(u + 6)2 ) u+5 17u2 + 193u + 540 (17u + 108)(u + 5) = = u+5 u+5 = 17u + 108. where A3 =

Thus we get

88

1 Solutions

u −5 11 17 = + − (u + 5)3 (u + 6)2 (u + 5)3 (u + 5)2 u+5 17u + 108 + (u + 6)2 −5 11 17 = + − 3 2 (u + 5) (u + 5) u+5 17 6 + + . (u + 6)2 u+6 Replacing u by s2 in the above expression, gives the following expression s2 for the partial fraction decomposition of 2 : (s + 5)3 (s2 + 6)2 (s2

−5 11 17 6 17 + 2 − 2 + 2 + 2 3 2 2 + 5) (s + 5) s + 5 (s + 6) s +6

17. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 4y ! + 4y} = L {y !! } + 4L {y ! } + 4L {y} = s2 Y (s) − sy(0) − y ! (0) + 4(sY (s) − y(0)) + 4Y (s) = (s2 + 4s + 4)Y (s) − 1.

Since L {4 cos 2t} = 4s/(s2 + 4) we get the algebraic equation (s + 4)2 Y (s) − 1 = Hence, Y (s) =

4s . s2 + 4

1 4s + 2 . 2 (s + 2) (s + 4)(s + 2)2

The (s2 + 4)-chain for the second term is

The (s2 + 4)-chain 4s (s2 + 4)(s + 2)2 −1 (s + 2)2

Thus

1 s2 + 4

1 Solutions

89

Y (s) = and Table 2.2 gives y(t) =

1 2

1 s2 + 4

sin 2t

18. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 6y ! + 9y} = L {y !! } + 6L {y ! } + 9L {y} = s2 Y (s) − sy(0) − y ! (0) + 6(sY (s) − y(0)) + 9Y (s) = (s2 + 6s + 9)Y (s) − 2. Since L {50 sin t} = 50/(s2 + 1) we get the algebraic equation (s + 3)2 Y (s) − 2 = Hence, Y (s) =

50 . s2 + 1

2 50 + 2 2 (s + 3) (s + 1)(s + 3)2

The partial fraction decomposition for the second term is

The (s2 + 1)-chain 50 (s2 + 1)(s + 3)2

−3s + 4 s2 + 1

3s + 14 (s + 3)2

5 (s + 3)2

3 s+3

Thus Y (s) =

3 −3s + 4 7 + 2 + s+3 s +1 (s + 3)2

and Table 2.2 give y(t) = 3e−3t + 7te−3t − 3 cos t + 4 sin t 19. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 4y} = L {y !! } + 4L {y} = s2 Y (s) − sy(0) − y ! (0) + 4Y (s) = (s2 + 4)Y (s) − 1.

90

1 Solutions

Since L {sin 3t} = 3/(s2 + 9) we get the algebraic equation (s2 + 4)Y (s) − 1 = Hence, Y (s) =

s2

3 . s2 + 9

1 3 + 2 . + 4 (s + 9)(s2 + 4)

Using quadratic partial fraction recursion we obtain the (s2 + 9)-chain

The (s2 + 9)-chain −3/5 s2 + 9

3 (s2 + 9)(s2 + 4) 3/5 +4

s2

Thus Y (s) =

8 1 3 1 4 2 1 3 − = − 5 s2 + 4 5 s2 + 9 5 s2 + 4 5 s2 + 9

and Table 2.2 gives y(t) =

4 5

sin 2t −

1 5

sin 3t

20. Apply the Laplace transform to both sides. For the left hand side we get L {y !! + 2y ! + 2y} = L {y !! } + 2L {y ! } + 2L {y} = s2 Y (s) + 2sY (s) + 2Y (s) = ((s + 1)2 + 1)Y (s). Since L{2 cos t + sin t} =

2s + 1 we get s2 + 1

Y (s) =

2s + 1 . ((s + 1)2 + 1)(s2 + 1)

We compute the (s2 + 1)-chain.

1 Solutions

91

The (s2 + 1)-chain 2s + 1 ((s + 1)2 + 1)(s2 + 1)

1 s2 + 1

−1 (s + 1)2 + 1

It follow that Y (s) = and hence

1 1 − s2 + 1 (s + 1)2 + 1

y(t) = sin t − e−t sin t.

Section 2.5 1. L−1 {−5/s} = −5L−1 {1/s} = −5 2. L−1 {3/(s − 4)} = 3L−1 {1/(s − 4)} = 3e4t 2 3 0 1 0 1 3 4 −1 3. L − 3 = 3L−1 1/s2 − 2L−1 2/s3 = 3t − 2t2 s2 s 2 3 2 3 4 1 −1 −1 4. L = 2L = 2e−3t/2 2s + 3 s + (3/2) 2 3 2 3 3s s −1 5. L−1 = 3L = 3 cos 2t s2 + 4 s2 + 22 / 6 √ 2 3 √ 2 2 −1 3 2 −1 √ 6. L =√ L = √ sin 3 t 2 s +3 3 s2 + ( 3)2 3 2s − 5 7. First, we have s2 + 6s + 9 = (s + 3)2 . Partial fractions gives 2 = s + 6s + 9 2 3 −11 2 2s − 5 + . So L−1 = −11te−3t + 2e−3t 2 (s + 3) s+3 (s + 3)2

92

1 Solutions

8. Partial fractions gives −11 2 −3t t e + 2te−3t 2

2s − 5 −11 2 = + . Thus L−1 3 3 (s + 3) (s + 3) (s + 3)2

2

2s − 5 (s + 3)3

3

=

2 3 6 6 −1 1 6 −1 9. 2 = = + . So L = s + 2s − 8 (s − 2)(s + 4) s+4 s−2 s2 + 2s − 8 2t −4t e −e 2 3 s s −2 3 s −1 10. 2 = = + . So L = s − 5s + 6 (s − 2)(s − 3) s−2 s−3 (s − 2)(s + 3) 3t 2t 3e − 2e 2 2 3 2s2 − 5s + 1 −1 3 2 2s − 5s + 1 −1 11. = + + . So L = (s − 2)4 (s − 2)4 (s − 2)3 (s − 2)2 (s − 2)4 −1 3 2t 3 2 2t t e + t e + 2te2t 6 2 2 3 2s + 6 2s + 6 −2 4 2s + 6 12. 2 = = + . So L−1 = s − 6s + 5 (s − 1)(s − 5) s−1 s−5 s62 − 6s + 5 4e5t − 2et 2 3 4s2 1 1 1 1 4s2 −1 13. = + + − . So L = (s − 1)2 (s + 1)2 (s − 1)2 s − 1 (s + 1)2 s + 1 (s − 1)2 (s + 1)2 tet + et + te−t − e−t 2 3 27 9 3 1 1 27 9t2 −1 14. 2 = 3− 2+ − . So L = −3t+1−e−3t s (s + 3) s s s s−3 s3 (s + 3) 2 2 3 8s + 16 4 1 s 2 8s + 16 −1 15. 2 = − + − . So L = (s + 4)(s − 2)2 (s − 2)2 s − 2 s2 + 4 s2 + 4 (s2 + 4)(s − 2)2 4te2t − e2t + cos 2t − sin 2t 2 3 5s + 15 2 s 3 5s + 15 16. 2 = −2 2 + 2 . So L−1 = (s + 9)(s − 1) s−1 s +9 s +9 (s2 + 9)(s − 1) −2 cos 3t + sin 3t + 2et 2 3 12 −6 3 4 1 12 17. 2 = 2+ − + . So L−1 = s (s + 1)(s − 2) s s s+1 s−2 s2 (s + 1)(s − 2) 3 − 6t + e2t − 4e−t 18.

2s 6 14 22 8 22 = + + + − . So (s −23)3 (s − 4)2 (s3− 3)3 (s − 3)2 s−3 (s − 4)2 s−4 2s L−1 = 3t2 e3t + 14te3t + 22e3t + 8te4t − 22e4t (s − 3)3 (s − 4)2

1 Solutions

19. First we have s2 + 2s + 5 = (s + 1)2 + 4. So

93

2s 2s = = s2 + 2s + 5 (s + 1)2 + 4

2(s + 1) − 2 s+1 2 =2 − . The First Translation prin2 2 2+4 (s + 1) + 4 2 (s + 1) + 43 (s + 1)2 3 2 3 2s s+1 2 −1 −1 ciple gives L−1 = 2L −L = s2 + 2s + 5 (s + 1)2 + 4 (s + 1)2 + 4 2e−t cos 2t − e−t sin 2t 2 3 1 1 1 −1 20. 2 = . Thus L = e−3t sin t s + 6s + 10 (s + 3)2 + 1 (s + 3)2 + 1 2 3 s−1 s−4 1 s−1 −1 21. 2 = +3 . Thus L = s − 8s + 17 (s − 4)2 + 1 (s − 4)2 + 1 s2 − 8s + 17 e4t cos t + 3e4t sin t √ 2 3 √ 2s + 4 s−2 8 2s + 4 √ + 8 √ . Thus L−1 22. 2 =2 = 2 + ( 8)2 2 + ( 8)2 s − 4s + 12 s2 − 4s + 12 (s − 2) (s − 2) √ √ √ 2e2t cos 8 t + 8 e2t sin 8 t 2 3 s−1 s−1 s−1 −1 23. 2 = . Thus L = et cos 3t s − 2s + 10 (s − 1)2 + 32 s2 − 2s + 10 2 3 s−5 s−3 2 s−5 −1 24. 2 = − . Thus L = s − 6s + 13 (s − 3)2 + 22 (s − 3)2 + 22 s2 − 6s + 13 e3t cos 2t − e3t sin 2t 2 3 2 3 8s s 8 −1 25. L−1 = 8L = (2t sin 2t) = 2t sin 2t 2 2 2 2 2 (s + 4) (s + 2 ) 2 · 22 2 3 2 3 9 3 3 −1 −1 26. L = 3L = (sin 3t − 3t cos 3t) = (s2 + 9)2 (s2 + 32 )2 2 · 32 1 1 sin 3t − t cos 3t 6 2 27. We first complete the square s2 + 4s + 53= (s + 2)22+ 1. By the transla2 3 2s (s + 2) − 2 −1 tion principle we get L−1 = 2L = 2 2 ((s ' + 2)2 + 1)2 ' 2 3 (s + 4s 2 + 5) 3( ( s 1 1 1 −1 −2t 2e−2t L−1 − 2L = 2e t sin t − 2( (sin t − t cos t) = (s2 + 1)2 (s2 + 1)2 2 2 2te−2t cos t + (t − 2)e−2t sin t 28. We first complete the 2 square s2 −6s+103= (s−3)2 +1. 2 By the translation 3 2s + 2 (s − 3) + 3 + 1 −1 −1 principle we get L = 2L = 2 2 2 2 ((s ' 2 3 (s − 6s2+ 10) 3( ' − 3) + 1) ( s 1 1 1 3t −1 −1 3t 2e L + 4L = 2e t sin t + 4( (sin t − t cos t) = (s2 + 1)2 (s2 + 1)2 2 2 3t 3t −4te cos t + (t + 4)e sin t

94

1 Solutions

29. We first complete the 2 square s2 +8s+173= (s+4)2 +1. 2 By the translation 3 2s (s + 4) − 4 −1 principle we get L−1 = 2L = 2 2 2 2 ((s + ' 2 3(s + 8s + 2 17) 3( ' 4) + 1) ( s 1 1 1 −4t −1 −1 −4t 2e L − 4L = 2e t sin t − 4( (sin t − t cos t) = (s2 + 1)2 (s2 + 1)2 2 2 −4t −4t 4te cos t + (t − 4)e sin t 2 2 30. We first complete the square 2 s + 2s + 2 = 3 (s + 1) 2+ 1. By the transla3 s + 1 s+1 −1 −1 tion principle we get L =L = (s2 + 2s + 2)3 ((s + 1)2 + 1)3 ' 2 3( & 1 s 1% e−t L−1 = e−t t sin t − t2 cos t = (te−t sin t−t2 e−t cos t) 2 3 (s + 1) 8 8 2 2 2 31. We first complete the square 2 s − 2s + 5 = 3 (s − 1) 2+ 2 . By the transla3 1 1 −1 tion principle we get L−1 = L = 2 3 2 2 3 ' 2 3( (s − ' 2s +25) 3(((s − 1) + 2 ) 1 1 2 et L−1 = et L−1 (s2 + 22 )3 2 (s2 + 22 )3 % & 1 = et (3 − (2t)2 ) sin 2t − 6t cos 2t 2 · 8 · 24 & 1 % = (3 − 4t2 )et sin 2t − 6tet cos 2t 256

32. We first complete the 2 square s2 −6s+103= (s−3)2 +1. 2 By the translation 3 8s (s − 3) + 3 −1 principle we get L−1 = 8L = 2 3 ((s − 3)2 + 1)3 ' 2 3 (s − 6s2+ 10) 3( s 1 8e3t L−1 + 3L−1 2 + 1)3 2 + 1)3 (s (s ' ( & 1% 1 3t 2 2 = 8e t sin t − t cos t + 3 ((3 − t ) sin t − 3t cos t) 8 8 = (−3t2 + t + 9)e3t sin t − (t2 + 9t)e3t cos t

2 33. We first complete the square = (s − 4)22+ 1. By the transla2 s − 8s + 17 3 3 s − 4 s−4 −1 tion principle we get L−1 = L = 2 4 ((s − 4)2 + 1)4 ' 2 3( (s − 8s + 17) & s 1 % e4t L−1 = e4t (3t − t3 ) sin t − 3t2 cos t 2 4 (s + 1) 48 & 1 % 3 4t 2 4t = (−t + 3t)e cos t − 3t e cos t 48 2 2 2 34. We first complete the square 2 s + 4s + 8 = 3 (s + 2) 2+ 2 . By the transla3 2 2 −1 −1 tion principle we get L =L = 2 3 ((s + 2)2 + 22 )3 ' 2 3( (s + 4s + 8) & 2 1 % e−2t L−1 = e−2t (3 − (2t)2 ) sin 2t − 6t cos 2t 2 2 3 4 (s + 2 ) 8·2 & 1 % 2 −2t −2t = (3 − 4t )e sin 2t − 6te cos 2t 128

1 Solutions

95

35. Apply the Laplace transform to get s2 Y (s) − s + 1 + Y (s) =

4 . s2 + 1

Solving for Y (s) we get Y (s) =

s−1 4 + . s2 + 1 (s2 + 1)2

We use Table 2.5 to get y(t) = cos t − sin t + 2(sin t − t cos t) = cos t + sin t − 2t cos t. 36. Apply the Laplace transform and use Table 2.5 to get (s2 + 9)Y (s) − 3 = and hence Y (s) =

216s + 9)2

(s2

3 216s + . s2 + 9 (s2 + 9)3

Now use Table 2.5 again to determine the Laplace inversion: y(t) = sin 3t +

& 216 % 3t sin 3t − 9t2 cos 3t 8(34 )

= sin 3t + t sin 3t − 3t2 cos 3t 37. Apply the Laplace transform to get 2

(s − 3)Y (s) = 4

' '

s 2 s +1

(!!

1 − s2 (s2 + 1)2 8s(s2 − 3) = (s2 + 1)3 = 4

It follows that Y (s) =

(s2

(!

8s . Table 2.5 now gives + 1)3

y(t) = t sin t − t2 cos t. 38. Apply the Laplace transform to get

96

1 Solutions

s2 Y (s) − 2 + 4Y (s) = 32

s2 − 4 . (s2 + 4)2

Solving for Y (s) we get 2 s2 + 4 − 8 + 32 2 +4 (s + 4)3 2 1 1 = 2 + 32 2 − 256 2 2 s +4 (s + 4) (s + 4)3 2 2 2 = 2 + 16 2 − 128 2 s +4 (s + 4)2 (s + 4)3

Y (s) =

s2

We use Table 2.5 to get & 16 128 % (sin 2t − 2t cos 2t) − (3 − 4t2 ) sin 2t − 6t cos 2t 8 128 = 4t2 sin 2t + 2t cos 2t.

y(t) = sin 2t +

39. Compute the partial fraction Then 2 L−1

1 (s − a)(s − b)

3

= L−1

2

1 1/(a − b) 1/(b − a) = + . (s − a)(s − b) s−a s−b 1/(a − b) 1/(b − a) + s−a s−b

3

=

eat ebt + . a−b b−a

40. Apply the inverse Laplace transform to the partial fraction expansion s a/(a − b) b/(b − a) = + . (s − a)(s − b) s−a s−b 41. Apply the inverse Laplace transform to the partial fraction expansion 1 1 1 1 1 1 1 = + + . (s − a)(s − b)(s − c) (a − b)(a − c) s − a (b − a)(b − c) s − b (c − a)(c − b) (s − c) 42. Apply the inverse Laplace transform to the partial fraction expansion s a 1 b 1 c 1 = + + . (s − a)(s − b)(s − c) (a − b)(a − c) s − a (b − a)(b − c) s − b (c − a)(c − b) (s − c) 43. Apply the inverse Laplace transform to the partial fraction expansion s2 a2 1 b2 1 c2 1 = + + . (s − a)(s − b)(s − c) (a − b)(a − c) s − a (b − a)(b − c) s − b (c − a)(c − b) (s − c)

1 Solutions

97

44. Apply the inverse Laplace transform to the partial fraction expansion sk A1 An = + ··· + (s − r1 ) · · · (s − rn ) s − r1 s − rn where

$ $ sk rik $ Ai = = . q(s)/(s − ri ) $s=ri q ! (ri )

The last equality is true since the product rule for derivatives implies that q ! (ri ) = (r1 − r1 ) · · · (ri − ri−1 )(ri − ri+1 ) · · · (ri − rn ),

that is, the derivative of q(s) evaluated at one of the roots ri is obtained by deleting the term s − ri from q(s) and then evaluating at ri and this is the same expression which is evaluated to put in the denominator of the coefficient Ai . 45. This is directly from Table 2.4. 46. Apply the inverse Laplace transform to the partial fraction expansion s (s − a) + a 1 a = = + . (s − a)2 (s − a)2 s − a (s − a)2 47. This is directly from Table 2.4. 48. Apply the inverse Laplace transform to the partial fraction expansion s (s − a) + a 1 a = = + . (s − a)3 (s − a)3 (s − a)2 (s − a)3 49. Apply the inverse Laplace transform to the partial fraction expansion s2 ((s − a) + a)2 1 2a a2 = = + + . (s − a)3 (s − a)3 s − a (s − a)2 (s − a)3 50. To compute the partial fraction write s = (s − a) + a and compute sk = ((s − a) + a)k by the binomial theorem: sk = ((s − a) + a)k = Thus,

k ' ( 7 k k−l a (s − a)l . l l=0

98

1 Solutions k ' ( 7 sk k k−l 1 = a . n (s − a) l (s − a)n−l l=0

Now apply the inverse Laplace transform.

Section 2.6 0 1 1. The root of q(s) is 4 with multiplicity 1. Thus Bq = e4t 0 1 2. The root of q(s) is −6 with multiplicity 1. Thus Bq = e−6t

3. q(s) = s2 + 5s = s(s + 5), hence the 0 1 roots of q(s) are 0 and −5 each with multiplicity 1. Thus Bq = 1, e−5t

4. q(s) = s2 − 3s − 4 = (s − 4)(s + 1), hence the1roots of q(s) are 4 and −1 0 each with multiplicity 1. Thus Bq = e4t , e−t

2 5. q(s) = s2 − 6s0+ 9 = (s − 1 3) , hence the root of q(s) is 3 with multiplicity 2. Thus Bq = e3t , te3t

6. q(s) = s2 − 9s + 14 = (s − 7)(s − 2),0hence the 1 roots of q(s) are 7 and 2 each with multiplicity 1. Thus Bq = e2t , e7t 7. q(s) = s2 − s − 6 = (s − 3)(s + 2), hence the 1root of q(s) are 3 and −2 0 each with multiplicity 1. Thus Bq = e3t , e−2t 8. q(s) = s2 + 9s + 18 = (s + 6)(s + 3), hence the0 root of q(s) 1 are −3 and −6 each with multiplicity 1. Thus hence Bq = e−3t , e−6t

9. q(s) = 6s2 − 11s + 4 = (3s − 4)(2s0 − 1) so the1 roots are 4/3 and 1/2, each with multiplicity 1. Hence Bq = et/2 , e4t/3

10. q(s) does √ not easily factor. However the quadratic formula gives roots 4 5 √ √ √ −2 ± 8 s= = −1 ± 2. Hence Bq = e(−1+ 2)t , e(−1− 2)t 2 √ √ 4 ± 12 11. The quadratic formula gives roots = 2 ± 3. Hence Bq = 2 4 5 √ √ e(2+ 3)t , e(2− 3)t

2 12. q(s) = s2 −10s+25 , hence the root of q(s) is 5 with multiplicity 0 5t =5t(s−5) 1 2. Thus Bq = e , te

13. q(s) = 4s2 + 12s0+ 9 = (2s + 3)21; so the root is −3/2 with multiplicity 2 and hence Bq = e−3t/2 , te−3t/2

1 Solutions

99

14. q(s) has complex roots ±3i. Hence Bq = {cos 3t, sin 3t} 15. q(s) = 4s2 + 25 = 4(s2 + (5/2)2 ). Therefore q(s) has complex roots ± 52 i. Hence Bq = {cos(5t/2), sin(5t/2)} 16. q(s) = s2 + 4s + 13 = s2 + 4s + 4 + 90 = (s + 2)2 + 32 . Therefore q(s) has 1 complex roots −2 ± 3i. Hence Bq = e−2t cos 3t, e−2t sin 3t 17. q(s) = s2 − 2s + 5 = s2 − 2s + 1 + 4 = (s − 1)2 + 22 . Therefore q(s) has complex roots 1 ± 2i. Hence Bq = {et cos 2t, et sin 2t} √ 2 18. q(s) = s2 −s+1 = s2 −s+(1/4)+(3/4) = (s−(1/2)) +( 3/2)2 . Therefore 4 √ √ √ 5 q(s) has complex roots 12 ± 23 i. Hence Bq = et/2 cos 23 t, et/2 sin 23 t 19. q(s) has 0 root −3 with multiplicity 1 4. Hence Bq = e−3t , te−3t , t2 e−3t , t3 e−3t . 20. q(s) has 0 root 2 with multiplicity 1 5. Hence Bq = e2t , te2t , t2 e2t , t3 e2t , t4 e2t .

21. q(s) = − 1)3 has 0 (s 1 root 1 with multiplicity 3. Hence t Bq = e , tet , t2 et . 22. q(s) has 0 root −1 with multiplicity 6. Hence 1 Bq = e−t , te−t , t2 e−t , t3 e−t , t4 e−t , t5 e−t .

23. We complete the square to get q(s) = ((s + 2)2 + 1)2 . Thus q(s) has complex 0 root −2 ± i with multiplicity 2 It follows 1 that Bq = e−2t cos t, e−2t sin t, te−2t cos t, te−2t sin t . 24. We complete the square to get q(s) = ((s − 4)2 + 22 )4 . Thus q(s) has complex 0 roots 4 ± 2i with multiplicity 3. If follows that 1 Bq = e4t cos 2t, e4t sin 2t, te4t cos 2t, te4t sin 2t, t2 e4t cos 2t, t2 e4t sin 2t 25. 0The complex roots of q(s) are ±i with multiplicity14. Thus Bq = cos t, sin t, t cos t, t sin t, t2 cos t, t2 sin t, t3 cos t, t3 sin t

Section 2.7 1. Yes. 2. No; t−2 is not a polynomial. 3. Yes;

t = te−t . et

100

1 Solutions

1 is not a polynomial. t √ √ π 2 2 5. Yes; t sin(4t − ) = t( sin 4t − cos 4t). 4 2 2 4. No;

6. Yes; (t + et )2 = t2 + 2tet + e2t . 7. No. 8. Yes. 1

9. No; t 2 is not a polynomial. 10. Yes;

sin 2t = e−2t sin 2t. e2t

11. No.

1 is not in E. sin 2t

12. s3 +s = s(s2 +1); The roots are 1 and ±i each with multiplicity 1. Hence Bq = {1, cos t, sin t}. 13. s4 − 1 = (s2 − 1)(s2 + 1) = (s − 1)(s + 1)(s2 + 1); The roots ate 1, −1, and ±i each with multiplicity 1. Hence Bq = {et , e−t , cos t, sin t}. 14. The roots are 0 with multiplicity 3 and −1 with multiplicity 2. Hence 0 1 Bq = 1, t, t2 , e−t , te−t . 15. The roots are 1 with multiplicity 3 and −7 with multiplicity 2. Hence 0 1 Bq = et , tet , t2 et , e−7t , te−7t . 16. The roots are −8 with multiplicity 2 and ±3i with multiplicity 3.1 Hence 0 Bq = e−8t , te−8t , cos 3t, sin 3t, t cos 3t, t sin 3t, t2 cos 3t, t2 sin 3t . 17. The roots are −2 with multiplicity 3 and ±2i with multiplicity 2. Hence 0 1 Bq = e−2t , te−2t , t2 e−2t , cos 2t, sin 2t, t cos 2t, t sin 2t . 18. The roots are −5 with multiplicity 2, 4 with multiplicity 2, and 0 1 −3 with multiplicity 2. Hence Bq = e−5t , te−5t , e4t , te4t , e−3t , te−3t .

19. We must gather the roots together to get the correct multiplicity. Thus q(s) = (s − 2)2 (s + 3)3 . The0 roots are 2 with multiplicity 1 2 and −3 with multiplicity 3. Hence Bq = e2t , te2t , e−3t , te−3t , t2 e−3t . 20. The roots at 1 with0 multiplicity 1, 2 with multiplicity 2, and 3 with 1 multiplicity 3. Bq = et , e2t , te2t , e3t , te3t , t2 e3t .

21. By completing the square we may write q(s) = (s+4)2 ((s+3)2 +4)2 The roots 0are −4 with multiplicity 2 and −3 ± 2i with multiplicity 2.1 Hence Bq = e−4t , te−4t , e−3t cos 2t, e−3t sin 2t, te−3t cos 2t, te−3t sin 2t .

1 Solutions

101

22. By completing the square we may write q(s) = (s + 5)((s + 2)2 + 1)2 . The roots 0are −5 with multiplicity 1 and −2 ± i with multiplicity 2. Hence 1 Bq = e−5t , e−2t cos t, e−2t sin t, te−2t cos t, te−2t sin t .

23. First observe that s2 +2s+10 = (s+1)2 +32 and hence q(s) = (s−3)3 ((s+ 1)2 +32 )2 . The0roots are 3 with multiplicity 3 and −1±3i with multiplicity1 2. Thus Bq = e3t , te3t , t2 e3t , e−t cos 3t, e−t sin 3t, te−t cos 3t, te−t sin 3t .

2 24. s3 + 8√ = (s + 2)(s2 − 2s + = (s + 2)(s √ −t 2s +√1 + 0 4) 1 3) = (s + 2)((s − 2 2 −2t t 1) + ( 3) ); hence Bq = e , e cos 3t, e sin 3t 0 1 25. 2s3 − 5s2 + 4s − 1 = (2s − 1)(s − 1)2 ; hence Bq = et/2 , et , tet 0 1 26. s3 + 2s2 − 9s − 18 = (s + 2)(s − 3)(s + 3); hence Bq = e−2t , e3t t, e−3t √ √ √ √ 1 0 27. s4 +5s2 +6 = (s2 +3)(s2 +2); hence Bq = cos 3t, sin 3t, cos 2t, sin 2t 0 1 28. s4 −8s2 +16 = (s2 −4)2 = (s−2)2 (s+2)2 ; hence Bq = e2t , e−2t , te2t , te−2t

p1 (s) p2 (s) with deg p1 (s) < deg q1 (s) and r2 (s) = with q1 (s) q2 (s) p1 (s)p2 (s) deg p2 (s) < deg q2 (s). Thus, r1 (s)r2 (s) = and q1 (s)q2 (s)

29. r1 (s) =

deg(p1 (s)p2 (s) = deg p1 (s)+deg p2 (s) < deg q1 (s)+deg q2 (s) = deg(q1 (s)q2 (s)). p(s) with deg p(s) < deg q(s). But deg p(s) = deg p(s − a) since, q(s) for any natural number n, (s − a)n = sn + lower degree terms. Thus translating the highest power of s found in p(s) will produce the same highest power of s in p(s − a). Since the same is also true for q(s), it follows that deg p(s − a) = deg p(s) < deg q(s) = deg q(s − a). Thus, p(s − a) r(s − a) = is a proper rational function. q(s − a)

30. r(s) =

31. If r(s) ∈ R then r(s) =

p(s) where deg p(s) = m < n = deg q(s). Then q(s)

r! (s) =

q(s)p! (s) − q ! (s)p(s) , (q(s))2

and deg(q(s)p! (s) − q ! (s)p(s)) ≤ max(deg(q(s)q ! (s)), deg(q ! (s)p(s))) = max(n + (m − 1), (n − 1) + m) = n + m − 1 < 2n = deg(q(s))2 . Hence r! (s) is a proper rational function. 32. Suppose r(s) ∈ Rqn but not in Rqn−1 . Then we can write r(s) = qp(s) n (s) 2 2 where p(s) does not have a factor of q(s) = (s − a) + b . By the quotient

102

1 Solutions

rule we have p! q n − nq n−1 2(s − a)p q 2n p! q − 2n(s − a)p = . q n+1

r! (s) =

Thus r! (s) ∈ Rqn+1 . Suppose r! (s) ∈ Rqn . Then the numerator p! q − 2n(s − a)p has a factor of q. But this implies p has a factor of q. But this is impossible since r(s) ∈ / Rqn−1 . Hence r! (s) ∈ / Rq n . 33. By exercise 32 this is true for n = 1. Now apply induction. If n is given and we assume the result is true for derivatives of order n − 1, then r(n−1) ∈ Rqn but not in Rqn−1 . Another application of exercise 32 then % &! shows that r(n) = r(n−1) ∈ Rqn+1 but not in Rqn . 34. First note the following trigonometric identities:

1 (cos(b + d)t + cos(b − d)t) 2 1 sin bt sin dt = (cos(b − d)t − cos(b + d)t) 2 1 sin bt cos dt = (sin(b + d)t + sin(b − d)t) . 2

cos bt cos dt =

Thus if t1 (t) and t2 (t) are two basic trig functions ( sin or cos) then t1 (t)t2 (t) is a linear combination of basic trig functions. Now if f (t) = tn eat t1 (t) and g(t) = tm ect t2 (t) then f (t)g(t) = tn+m e(a+c)t t1 (t)t2 (t). Since t1 (t)t2 (t) is a linear combinations of basic trig functions then f (t)g(t) is a linear combination of exponential polynomial. Now if f (t) and g(t) is any exponential polynomial then they are each made up of linear combinations of simple exponential polynomials and their product is a sum of terms of products of simple exponential polynomials, which we have shown is the sum of possibly two simple exponential polynomials. It now follows that f (t)g(t) is an exponential polynomial. 35. Observe that et−t0 = e−t0 et . So the translate of an exponential function is a multiple of an exponential function. Also, if f (t) is a polynomial so is f (t−t0 ). By the addition rule for cos we have cos b(t−t0 ) = cos bt cos bt0 − sin bt sin bt0 and similarly for sin. It follows that all these translates remain in E. By Exercise 34 the result follows. 36. Since f ∈ E is a linear combination of terms of the form tn eat cos bt and tn eat sin bt it is enough to show that the derivative of each of these terms is again in E. But, by the product rule we have (tn eat cos bt)! = ntn−1 eat cos bt + atn eat cos bt − btn eat sin bt,

1 Solutions

103

a linear combination of simple exponential polynomials and hence back in E. A similar calculation holds for tn eat sin bt. 37. By linearity of integration it is enough to show this result for fn (t) = t)n eat (c1 cos bt + c2 sin bt), where c1 and c2 are scalars. Let In (t) = fn (t) dt. First assume n = 0. Then a standard trick using integration by parts twice gives * I0 (t) : = eat (c1 cos bt + c2 sin bt) dt =

1 ((c1 a − c2 b) cos bt + (c1 b + c2 a) sin bt)eat . a 2 + b2

Clearly, I0 is an exponential polynomial. Observe that I0 is of the same n form as f0 . Now assume n > 0. Using integration ) nby parts with u = t and dv = (c1 cos)bt + c2 sin bt) dt we have In (t) = t (c1 cos bt + c2 sin bt) dt = tn I0 (t) − n t)n−1 I0 (t) dt. Since I0 ∈ E so are tn I0 and tn−1 I0 . By induction we have tn−1 I0 (t) dt ∈ E. It now follows that In ∈ E.

38. All exponential polynomials are continuous functions on R. If f (t) = t then f is an exponential polynomial. However, f −1 (t) = 1/t is not continuous at 0. Thus f −1 is not an exponential polynomial. 39. It is enough to show this for each f ∈ Bq since differentiation is linear and Eq = Span Bq . Suppose f (t) = tn eat cos bt. Then f ! (t) = ntn−1 eat cos bt − btn eat sin bt + atn eat cos bt. The derivative f ! (t) is a linear combination of the simple exponential polynomials tn−1 eat cos bt, tn eat sin bt, and tn eat cos bt each of which are in Bq . Hence f ! (t) ∈ Eq . A similar argument applies to tn eat sin bt. 40. This is seen by repeated applications of Exercise 39. 41. Observe that et−t0 = e−t0 et . So the translate of an exponential function is a multiple of an exponential function. Also, if p(t) is a polynomial of degree n the binomial theorem implies that p(t − t0 ) is a polynomial of degree n. By the addition rule for cos we have cos b(t−t0 ) = cos bt cos bt0 − sin bt sin bt0 and similarly for sin. Thus if f (t) = tn eat cos bt ∈ Bq then f (t − t0 ) is a linear combination of terms in Bq . Since Eq = Span Bq it follows that all translates remain in Eq .

Section 2.8 1.

104

1 Solutions

t∗t = =

* *

t

x(t − x) dx

0 t 0

(tx − x2 ) dx

' 2 ($x=t x x3 $$ = t − 2 3 $x=0 = t

t2 t3 t3 − = 2 3 6

2. 3

3

t∗t =t ∗t = = =

* *

t

x3 (t − x) dx

0 t 0

(tx3 − x4 ) dx

' 4 ($x=t x x5 $$ t − 4 5 $x=0

t4 t5 − 4 5 t5 = . 20 = t

3. 3 ∗ sin t = sin t ∗ 3 =

*

t

(sin x)(3) dx 0 x=t

= −3 cos x|x=0 = −3(cos t − cos 0) = −3 cos t + 3 4. (3t + 1) ∗ e4t =

*

t

(3x + 1)e4(t−x) dx ' * t ( * t = e4t 3 xe−4x dx + e−4x dx 0 ' '0 ( ( 1 1 1 1 1 = e4t 3 − te−4t − e−4t + − e−4t + 4 16 16 4 4 4t 7e − 12t − 7 = . 16 0

1 Solutions

105

5. From the Convolution table we get 1 (2e3t − 2 cos 2t − 3 sin 2t) 32 + 22 1 = (2e3t − 2 cos 2t − 3 sin 2t). 13

sin 2t ∗ e3t =

6. From linearity and the Convolution table we get (2t + 1) ∗ cos 2t = 2t ∗ cos 2t + 1 ∗ cos 2t * t 1 − cos 2t = 2 + ( cos 2x dx) 22 0 1 − cos 2t sin 2t = + 2 2 7. From the Convolution table we get 2 (e−6t − (−6 − 6t + (36t2 )/2)) (−6)3 1 = (18t2 − 6t − 6 + e−6t ). 108

t2 ∗ e−6t =

8. From the Convolution table we get cos t ∗ cos 2t = =

12

1 (sin t − 2 sin 2t) − 22

1 (2 sin 2t − sin t). 3

9. From the Convolution table we get e2t ∗ e−4t =

e2t − e−4t 2 − (−4)

1 2t (e − e−4t ). 6 10.

106

1 Solutions

Thus

L {t ∗ tn } (s) = L {t} (s)L {tn } (s) 1 n! = 2 n+1 s s n! = n+3 s 1 (n + 2)! = (n + 1)(n + 2) sn+3 0 1 1 = L tn+2 . (n + 1)(n + 2) t ∗ tn =

tn+2 . (n + 1)(n + 2)

11. 0 1 L eat ∗ sin bt (s) =

Thus

1 b s − a s2 + b2 ' ( 1 b bs + ba = 2 − s + b2 s − a s2 + b2 % 0 at 1 & 1 = 2 bL e − (bL {cos bt} + aL {sin bt}) 2 s +b

eat ∗ sin bt =

a2

1 (beat − b cos bt − a sin bt). + b2

12. 0 1 L eat ∗ cos bt (s) =

Thus

1 s s − a s2 + b2 ' ( 1 a as − b2 = 2 − 2 a + b2 s − a s + b2 % 0 at 1 & 1 = 2 aL e − (aL {cos bt} − bL {sin bt}) a + b2

eat ∗ sin bt = 13. First assume a &= b. Then

1 (aeat − a cos bt + b sin bt). a 2 + b2

1 Solutions

107

a b 2 2 + a s + b2 ' ( 1 ab ab = 2 − b − a2 s2 + a2 s2 + b2

L {sin at ∗ sin bt} =

s2

From this it follows that sin at ∗ sin bt = Now assume a = b. Then

1 (b sin at − a sin bt). b2 − a 2

L {sin at ∗ sin at} =

(s2

a2 + a2 )2

By Table 2.5 in Section 2.5 we get L {sin at ∗ sin at} =

1 (sin at − at cos at). 2a

14. First assume a &= b. Then a s 2 2 + a s + b2 ' ( 1 as as = 2 − 2 b − a 2 s2 + a 2 s + b2

L {sin at ∗ cos bt} =

s2

From this it follows that sin at ∗ cos bt = Now assume a = b. Then

b2

1 (a cos at − a cos bt). − a2

L {sin at ∗ cos at} =

as (s2 + a2 )2

By Table 2.5 in Section 2.5 we get L {sin at ∗ cos at} = 15. First assume a &= b. Then

1 (at sin at). 2a

108

1 Solutions

s s s2 + a 2 s2 + b2 ' ( 1 −a2 b2 = 2 + 2 b − a2 s2 + a2 s + b2

L {cos at ∗ cos bt} =

From this it follows that cos at ∗ cos bt =

b2

Now assume a = b. Then

1 (−a sin at + b sin bt). − a2 s2 (s2 + a2 )2 1 a2 = 2 − . s + a2 (s2 + a2 )2

L {cos at ∗ cos at} =

By Table 2.5 we get L {cos at ∗ cos at} =

1 1 1 sin at− (sin at−at cos at) = (sin at+at cos at). a 2a 2a

16. The key is to recognize the integral defining f (t) as the convolution integral of two functions. Thus f (t) = (cos 2t) ∗ t so that F (s) = s 1 1 L {(cos 2t) ∗ t} = L {cos 2t} L {t} = 2 = . 2 2 s +4s s(s + 4) 17. f (t) = t2 ∗ sin 2t so F (s) =

2 2 4 = 3 2 . 3 2 s s +4 s (s + 4)

18. f (t) = t3 ∗ e−3t so F (s) =

6 1 6 = 4 4 s s+3 s (s + 3)

19. f (t) = t3 ∗ e−3t so F (s) =

6 1 6 = 4 4 s s+3 s (s + 3) 2 s 2s = 2 2 +4s +1 (s + 4)(s2 + 1)

20. f (t) = sin 2t ∗ cos t so F (s) =

s2

21. f (t) = sin 2t ∗ sin 2t so F (s)

2 2 4 = 2 2 2 2 +2 s +2 (s + 4)2

22.

s2

1 Solutions

109

L−1

2

1 (s − 2)(s + 4)

3

= L−1

2

1 s−2

3

∗ L−1

= e2t ∗ e−4t e2t − e−4t = 2 − −4 1 2t = (e − e−4t ). 6

2

1 s+4

3

23. L−1

2

1 2 s − 6s + 5

3

3 1 (s − 1)(s − 5) 2 3 2 3 1 1 = L−1 ∗ L−1 s−1 s−5 = L−1

2

= et ∗ e5t et − e5t = 1−5 1 = (−et + e5t ) 4 24. L−1

2

1 2 (s + 1)2

3

−1

2

s 2 (s + 1)2

3

3 2 3 1 1 −1 ∗ L s2 + 1 s2 + 1 = sin t ∗ sin t 1 = (sin t − t cos t) 2 = L−1

2

25. L

2

1 = L 2 s +1 = sin t ∗ cos t 1 = t sin t 2 −1

3

∗L

−1

2

s 2 s +1

3

26. L

−1

2

1 (s + 6)s3

3

3 2 3 1 1 −1 = L ∗L s+6 s3 t2 = e−6t ∗ 2 1 = (−e−6t + 1 − 6t + 18t2 ) 216 −1

2

110

1 Solutions

27. L−1

2

2 (s − 3)(s2 + 4)

3

= L−1

2

s (s − 4)(s2 + 1)

3

= L

−1

3

= L−1

2

1 s−3

3

∗ L−1

2

1 s−4

3

∗L

−1

2

1 s−a

3

∗ L−1

2

2 s2 + 4

2

s s2 + 1

2

1 s−b

= e3t ∗ sin 2t 1 = (2e3t − 2 cos 2t − 3 sin 2t) 13

3

28. L

−1

= e4t ∗ cos t 1 = (4e4t − 4 cos t + sin t) 17

3

29. L−1

2

1 (s − a)(s − b)

= eat ∗ ebt eat − ebt = . a−b

30. L−1

2

G(s) s+2

3

= L−1 {G(s)} ∗ L−1

2

1 s+2

= g(t) ∗ e−2t * t = g(x)e−2(t−x) dx

3

0

31. L

−1

2

G(s) s2 + 2

3

= L

−1

{G(s)} ∗ L

−1

/

s √ 2 s2 + 2

√ = g(t) ∗ cos( 2)t * t √ = g(x) cos 2(t − x) dx 0

32. Hint: use the Input Integral Principle repeatedly.

6

3

1 Solutions

111

33. We apply the input integral principle twice: 2 3 * t 1 L−1 = sin x dx s(s2 + 1) 0 t

= − cos x|0

= − cos t + 1 L

−1

2

1 s2 (s2 + 1)

3

34. The inverse Laplace transform of e2t − e−2t . Thus 4 L−1

2

1 s(s2 − 4)

3

*

=

t 0

1 − cos x dx

= t − sin t dx 1 1 1 = is e2t ∗ e−2t = s2 − 4 s−2s+2

1 4

*

t

(e2x − e−2x ) dx ,0 $t $t 1 e2x $$ e−2x $$ = + 4 2 $0 2 $0 =

1 2t (e − 1 + e−2t − 1) 8 1 1 1 = − + e2t + e−2t 4 8 8 =

Repeating the input integral principle we get 2 3 ( * t' 1 1 1 2x 1 −2x −1 L = − + e + e dx s2 (s2 − 4) 4 8 8 0 t 1 1 = − + (e2t − 1) − (e−2t − 1) 4 16 16 −t 1 1 = + e2t − e−2t 4 16 16 35. We apply the input integral principle three times: 2 3 * t 1 −1 L = e−3x dx s(s + 3) 0 $t e−3x $$ = −3 $ 0

1 = (1 − e−3t ). 3

112

1 Solutions

L−1

L

−1

2

2

1 s2 (s + 3)

1 3 s (s + 3)

3

3

* 1 t 1 − e−3x dx 3 0 ' ( 1 1 − e−3t = t− 3 3 1 = (3t − 1 + e−3t ). 9 =

* 1 t = 3x − 1 + e−3x dx 9 0 ' 2 ( 1 t e−3t − 1 = 3 −t− 9 2 3 1 = (2 − 6t + 9t2 − 2e−3t ) 54

36. We apply the input integral principle twice: 2 3 * t 1 −1 L = xe2x dx s(s − 2)2 0

$t $ 1 (2xe2x − e2x )$$ 4 0 1 2t 2t = (2te − e + 1) 4 =

L

−1

2

1 2 s (s − 2)2

3

=

1 4

*

t 0

2xe2x − e2x + 1 dx

$t $ 1 (xe2x − e2x + x)$$ 4 0 1 2t 2t = (te − e + t + 1) 4 =

37. First, L−1

2

1 (s2 + 9)2

3

=

1 (sin 3t − 3t cos 3t). Thus 54

1 Solutions

L−1

113

2

1 s(s2 + 9)2

3

38. First, from Table 2.5, principle twice: L

1 54

=

*

t 0

(sin 3x − 3x cos 3x) dx

' ' (($t $ 1 cos 3x cos 3x $ = − − x sin 3x + $ 54 3 3 0 ' ( 1 2 cos 3t 2 = − − t sin 3t + 54 3 3 1 = (−2 cos 3t − 3t sin 3t + 2). 162 1 1 = 2 . We apply the input integral s3 + s2 s (s + 1)

−1

2

1 s(s + 1)

3

*

=

t

e−x dx

0

$t = −e−x $0

= 1 − e−t . L−1

2

1 2 s (s + 1)

3

= =

*

t 0

%

(1 − e−x ) dx

&$t x + e−x $0

= t + e−t − 1.

Section 3.1 1. no, not linear. 2. yes; (D 2 − 3D)(y) = et , q(s) = s2 − 3s, nonhomogeneous 3. no, third order. 4. no, not linear. 5. no, not constant coefficient. 6. yes; (D 2 + 2D + 3)(y) = e−t , q(s) = s2 + 2s + 3, nonhomogeneous 7. yes; (D 2 − 7D + 10)(y) = 0, q(s) = s2 − 7s + 10, homogeneous 8. no, first order.

114

1 Solutions

9. yes; D 2 (y) = −2 + cos t, q(s) = s2 , nonhomogeneous 10. yes; (2D 2 − 12D + 18)(y) = 0, q(s) = 2s2 − 12s + 18, homogeneous 11. (a) Let = et + 3et + 2et = 6et (b) Le−t = e−t + 3(−e−t ) + 2e−t = 0 (c) L sin t = − sin t + 3(cos t) + 2 sin t = sin t − 3 cos t 12. (a) L(4et ) = 4et − 2(4et ) + 4et = 0 (b) L cos t = − cos t − 2(− sin t) + cos t = 2 sin t (c) L(−e2t ) = −4e2t − 2(−2e−2t ) + −e−2t = −e2t 13. (a) L(−4 sin t) = 4 sin t + −4 sin t = 0 (b) L(3 cos t) = 3(− cos t) + 3 cos t = 0 (c) L1 = 0 + 1 = 1 14. (a) Le2t = 4e2t − 4(2e2t ) + 8e2t = 4e2t (b) Le2t sin 2t = 8e2t cos 2t − 4e2t (2 sin 2t + 2 cos 2t) + 8e2t sin 2t = 0 (c) Le2t cos 2t = −8e2t sin 2t − 4e2t (2 cos 2t − 2 sin 2t) + 8e2t cos 2t = 0 15. et and e4t are homogeneous solution so yh = c1 et +c2 e4t are homogeneous solutions for all scalars c1 and c2 . A particular solution is yp = cos 2t. Thus y(t) = yp (t) + yh (t) = cos 2t + c1 et + c2 e4t where c1 , c2 are arbitrary constants. 16. e3t and e−2t are homogeneous solution so yh = c1 e3t + c2 e−2t are homogeneous solutions for all scalars c1 and c2 . A particular solution is yp = te3t . Thus y(t) = yp (t) + yh (t) = te3t + c1 e3t + c2 e−2t where c1 , c2 are arbitrary constants. 17. From Exercise 15 we have y(t) = cos 2t + c1 et + c2 e4t . Since y ! = −2 sin 2t + c1 et + 4c2 e4t we have 1 = y(0) = 1 + c1 + c2 −3 = y ! (0) = c1 + 4c2 , from which follows that c1 = 1 and c2 = −1. Thus y(t) = cos 2t + et − e4t . 18. From Exercise 16 we have y(t) = te3t + c1 e3t + c2 e−2t . Since y ! = (1 + 3t)e3t + 3c1 e3t − 2c2 e−2t we have −1 = y(0) = c1 + c2

8 = y ! (0) = 1 + 3c1 − 2c2

from which follows c1 = 1 and c2 = −2. Thus y(t) = te3t + e3t − 2e−2t . 19. L(ert ) = a(ert )!! + b(ert )! + cert = ar2 ert + brert + cert = (ar2 + br + c)ert

1 Solutions

115

20. Let t = a be the point on the t-axis where φ is tangent. Let y1 (t) = φ(t) and y2 (t) = 0. Both y1 and y2 are solutions to y !! + ay ! + by = 0. Further y1 (a) = 0 = y2 (a) and y1! (a) = 0 = y2! (a). By the existence and uniqueness theorem y1 = y2 . Hence φ = 0. 21. Let t = a be the point φ1 and φ2 are tangent. Then φ1 (a) = φ2 (a) and φ!1 (a) = φ!2 (a). By the existence and uniqueness theorem φ1 = φ2 .

Section 3.2 1. Suppose c1 t + c2 t2 = 0. Evaluating at t = 1 and t = 2 gives c1 + c2 = 0 and 2c 0 1 +14c2 = 0. The simultaneous solution is c1 = c2 = 0. It follows that t, t2 is linearly independent. 2. We compute the Wronskian:

9 et e2t w(e , e ) = det t = 2e3t − e3t = e3t . e 2e2t 0 1 Since the Wronskian is nonzero it follows that et , e2t is linearly independent. 0 1 3. Since et+2 = et e2 is a multiple of et it follows that et , et+2 is linearly dependent. t

2t

8

4. We compute the Wronskian: 8 9 1 ln 2 − ln 5 ln 2t ln 5t w(ln 2t, ln 5t) = det = (ln 2t − ln 5t) = . 1/t 1/t t t Since the Wronskian is nonzero it follows that {ln 2t, ln 5t} is linearly independent. 5. Since ln t2 = 2 ln t and ln t5 = 5 ln t they are multiples of each other and hence linearly dependent 6. A trig identity gives cos(t+π) = cos(t−π). Hence, {cos(t + π), cos(t − π)} is linearly dependent. 7. Suppose c1 t + c2 (1/t) = 0 Evaluating at t = 1 and t = 2 gives c1 + c2 = 0 and 2c1 + c2 /2 = 0. The simultaneous solution is c1 = c2 = 0. It follows that {t, 1/t} is linearly independent. 8. Suppose c1 + c2 t + c3 t2 = 0. Evaluating at t = 1, t = 2, and t = 3 gives the following system:

116

1 Solutions

c1 + c2 + c 3 = 0 c1 + 2c2 + 4c3 = 0 c1 + 3c2 + 9c3 = 0 Subtracting the first equation from the second and third gives c2 + 3c3 = 0 2c2 + 8c3 = 0 Subtracting twice the first equation from the second give 2c3 = 0 and hence c13 = 0. Now back substitution implies c2 = 0 and c1 = 0. Hence 0 1, t, t2 is linearly independent.

9. Suppose c1 + c2 (1/t) + c3 (1/t2 ) = 0. Evaluating at t = 1, t = 1/2, and t = 1/3 gives the same system as in the solution to Exercise 8 and hence 0 1 c1 , c2 and c3 are zero. It follows that 1, 1/t, 1/t2 on I = (0, ∞) is linearly independent. 0 1 10. Since cos2 t + sin2 t − 1 = 0 it follows that cos2 t, sin2 t, 1 is linearly dependent. 11. Let q(s) = s(s − 1)(s + 1). Then Bq = {1, et , e−t } which is linearly independent.

12. Let q(s) = (s − 1)((s − 1)2 + 4). Then Bq = {et , et cos 2t, et sin 2t} ⊃ {et , et sin 2t}. This implies linear independence. 0 1 13. Let q(s) = (s0− 1)5 . Then B1q = et , tet , t2 et , t3 et . Linear independence follows since t2 et , t3 et , t4 et ⊂ Bq . 14.

w(e3t , e5t ) = det

15.

8

9 e3t e5t = 5e8t − 3e8t = 2e8t 3e3t 5e5t

8

9 t t ln t w(t, t ln t) = det = t ln t + t − t ln t = t 1 ln t + 1

16. w(t cos(3 ln t), t sin(3 ln t)) 8 t cos(3 ln t) = det cos(3 ln t) − 3 sin(3 ln t)

t sin(3 ln t) sin(3 ln t) + 3 cos(3 ln t)

= 3t cos2 (3 ln t) + t cos(3 ln t) sin(3 ln t) −t sin(3 ln t) cos(3 ln t) + 3t sin2 (3 ln t) = 3t

9

1 Solutions

17.

117

9 t10 t20 w(t , t ) = det = 20t29 − 10t29 = 10t29 10t9 20t19 10

20

8

18. 

e2t 2t 3t 4t  w(e , e , e ) = det 2e2t 4e2t  1 9t  = e det 2 4

 e3t e4t 3e3t 4e4t  9e3t 16e4t  1 1 3 4 9 16

= e9t (1(48 − 36) − 1(32 − 16) + 1(18 − 12)) = 2e9t .

19. w(er1 t , er2 t , er3 t )  rt e1 e r2 t r t = det r1 e 1 r2 er2 t r12 er1 t r22 er2 t  1 = e(r1 +r2 +r3 )t det r1 r12

 er3 t r3 er3 t  r32 er3 t  1 1 r2 r3  r22 r32

= e(r1 +r2 +r3 )t ((r2 r32 − r3 r22 ) − (r1 r32 − r3 r12 ) + (r1 r22 − r2 r12 )) = e(r1 +r2 +r3 )t (r3 − r1 )(r3 − r2 )(r2 − r1 ). The last line requires a little algebra. 20.

 1 t t2 w(1, t, t2 ) = det 0 1 2t = 2 0 0 2

21.





1 0 2 3 w(1, t, t , t ) = det  0 0

 t t 2 t3 1 2t 3t2   = 12 0 2 6t 0 0 6

22. Let q(s) = s2 + 4. Then Bq = {sin 2t, cos 2t} is linearly independent. We can thus equate coefficients to get a + b = 2 a − b = −3

118

1 Solutions

The solution is a = −1/2 and b = 5/2. 0 1 23. Let q(s) = (s − 2)2 . Then Bq = e2t , te2t is linearly independent. We can equation coefficients to get 25c1 + 10c2 = 0 25c2 = 25 We thus get c2 = 1 and then c1 = −10/25 = −2/5. 24. We first rewrite this equation as 3a1 t − 3a2 t ln t = (a2 + 1)t + 5(a1 − a2 )t ln t. By Exercise 15 the set {t, t ln t} is linearly independent. Thus we get 3a1 = a2 + 1 and −3a2 = 5(a1 − a2 ). Solving this system gives a1 = 2 and a2 = 5. 0 1 25. If q(s) = s3 then Bq = 1, t, t2 is linearly independent. Thus we can equate coefficients to get a1 = a2 3 = a1 −a2 = −3 It follows that a1 = 3 and a2 = 3 is the solution. $ $ 26. Suppose at3 + b $t3 $ = 0 on (−∞, ∞). Then for t = 1 and t = −1 we get a+b = 0

−a + b = 0. These equations imply a = b = 0. So y1 and y2 are linearly independent. / −3t2 if t < 0 ! 2 ! 27. Observe that y1 (t) = 3t and y2 (t) = If t < 0 2 3t if t ≥ 0. ' 3 ( t −t3 then w(y1 , y2 )(t) = = 0. If t ≥ 0 then w(y1 , y2 )(t) = 2 3t −3t2 ' 3 ( t t3 = 0. It follows that the Wronskian is zero for all t ∈ 3t2 3t2 (−∞, ∞).

1 Solutions

119

Section 3.3 1. The polynomial is q(s) = s2 − s − 2 = (s − 2)(s + 1) so Bq = 0 2t characteristic 1 −t e ,e and the general solution takes the form y(t) = c1 e2t + c2 e−t , c1 , c 2 ∈ R 2. The polynomial is q(s) = s2 +s−12 = (s+4)(s−3) so Bq = 0 −4tcharacteristic 1 3t e ,e and the general solution takes the form y(t) = c1 e−4t + c2 e3t , c1 , c 2 ∈ R 3. The characteristic polynomial is q(s) = s2 + 10s + 24 = (s + 6)(s + 4) 0 −6t −4t 1 so Bq = e , e and the general solution takes the form y(t) = c1 e−6t + c2 e−4t , c1 , c2 ∈ R 4. The polynomial is q(s) = s2 −4s−12 = (s−6)(s+2) so Bq = 0 6t characteristic 1 −2t e ,e and the general solution takes the form y(t) = c1 e6t + c2 e−2t , c1 , c 2 ∈ R 5. The polynomial is q(s) = s2 + 8s + 16 = (s + 4)2 so Bq = 0 −4tcharacteristic 1 −4t e , te and the general solution takes the form y(t) = c1 e−4t + −4t c2 te , c1 , c2 ∈ R 6. The polynomial is q(s) = s2 −3s−10 = (s−5)(s+2) so Bq = 0 5t characteristic 1 e , e−2t and the general solution takes the form y(t) = c1 e5t + c2 e−2t , c1 , c 2 ∈ R 7. The characteristic polynomial is q(s) = s2 + 2s + 5 = (s + 1)2 + 4 so Bq = {e−t cos 2t, e−t sin 2t} and the general solution takes the form y(t) = c1 e−t cos 2t + c2 e−t sin 2t, c1 , c2 ∈ R 8. The polynomial is q(s) = 2s2 −12s+18 = 2(s−3)2 so Bq = 0 3t characteristic 1 3t e , te and the general solution takes the form y(t) = c1 e3t + c2 te3t , c1 , c 2 ∈ R 9. The characteristic polynomial is q(s) = s2 + 13s + 36 = (s + 9)(s + 4) 0 −9t −4t 1 so Bq = e , e and the general solution takes the form y(t) = c1 e−9t + c2 e−4t , c1 , c2 ∈ R 10. The characteristic polynomial1 is q(s) = s2 + 8s + 25 = (s + 4)2 + 9 0 −4t so Bq = e cos 3t, e−4t sin 3t and the general solution takes the form −4t y(t) = c1 e cos 3t + c2 e−4t sin 3t, c1 , c2 ∈ R 2 2 11. The 0 characteristic 1 polynomial is q(s) = s + 10s + 25 = (s + 5)−5tso −5t −5t Bq = e , te and the general solution takes the form y(t) = c1 e + c2 te−5t , c1 , c2 ∈ R

120

1 Solutions

12. 0The characteristic polynomial is q(s) = s2 −4s−21 = (s−7)(s+3) so Bq = 1 7t −3t e ,e and the general solution takes the form y(t) = c1 e7t + c2 e−3t , c1 , c 2 ∈ R 13. The characteristic polynomial is q(s) = s2 − 1 = (s − 1)(s + 1) so Bq = {et , e−t } and the general solution takes the form y(t) = c1 e2t + c2 e−t . The initial conditions imply that c1 + c2 = 0 and c1 − c2 = 1. Solving t −t gives c1 = 1/2 and c2 = −1/2. Thus y = e −e 2 14. 0The characteristic polynomial is q(s) = s2 −3s−10 = (s−5)(s+2) so Bq = 1 e5t , e−2t and the general solution takes the form y(t) = c1 e5t + c2 e−2t . The initial conditions imply that c1 + c2 = 5 and 5c1 − 2c2 = 4. Solving gives c1 = 2 and c2 = 3. Thus y = 2e5t + 3e−2t 15. 0The characteristic polynomial is q(s) = s2 − 10s + 25 = (s − 5)2 so Bq = 1 5t 5t e , te and the general solution takes the form y(t) = c1 e5t + c2 te5t . The initial conditions imply that c1 = 0 and 5c1 + c2 = 1. Solving gives c1 = 0 and c2 = 1. Thus y = te5t 16. The characteristic polynomial1 is q(s) = s2 + 4s + 13 = (s + 2)2 + 9 0 −2t so Bq = e cos 3t, e−2t sin 3t and the general solution takes the form −2t y(t) = c1 e cos 3t+c2 e−2t sin 3t. The initial conditions imply that c1 = 1 and −2c1 + 3c2 = −5. Solving gives c1 = 1 and c2 = −1. Thus y = e−2t cos 3t − e−2t sin 3t 0 1 17. Let q(s) = (s − 83)(s + 7) = s92 + 4s − 21. Then Bq = e3t , e−7t . e3t e−7t w(e3t , e−7t ) = det = −10e−4t . So K = −10. 3t 3e −7e−7t 2 r1 t r2 t 18. Let q(s) = (s−r1 )(s−r 8 r 2t) = s r−(r 9 1 +r2 )s+r1 r2 . Then Bq = {e , e }. t e1 e2 w(er1 t , er2 t ) = det = (r2 − r1 )e(r1 +r2 )t . So K = r2 − r1 . r1 e3t r2 e−7t 0 1 19. Let8q(s) = (s − 3)2 =9 s2 − 6s + 9. Then Bq = e3t , te3t . w(e3t , te3t ) = e3t te3t det = (1 + 3t)e6t − 3te6t = e6t . So K = 1. 3t 3e (1 + 3t)e3t

20. Let8q(s) = (s − r)2 =9s2 − 2rs + r2 . Then Bq = {ert , tert }. w(ert , tert ) = ert tert det = (1 + rt)e2rt − rte2rt = e2rt . So K = 1. rt re (1 + rt)ert 21. Let q(s) = (s − 1)2 + 22 = s2 − 2s + 5. Then Bq = {et cos 2t, et sin 2t}.

1 Solutions

121

w(et cos 2t, et sin 2t) = det

8

et cos 2t et sin 2t t t e (cos 2t − 2 sin 2t) e (sin 2t + 2 cos 2t)

= e2t (sin 2t cos 2t + 2 cos2 2t)

9

−e2t (cos 2t sin 2t − 2 sin2 2t) = 2e2t . So K = 2. 22. Let q(s) = (s−a)2 +b2 = s2 −2as+a2 +b2 . Then Bq = {eat cos bt, eat sin bt}. w(eat cos bt, eat sin bt) 8 9 eat cos bt eat sin bt = det at e (a cos bt − b sin bt) eat (a sin bt + b cos bt) = e2at (a sin bt cos bt + b cos2 bt) −e2at (a cos bt sin bt − b sin2 bt) = be2at . So K = b.

Section 3.4 0 1 1. q(s)v(s) − 2)(s − 3) so Bqv = e−t , e2t , e3t while Bq = 0 −t 2t 1= (s + 1)(s e , e . Since e3t is the only function in the first set but not in the second yp (t) = a1 e3t . 0 1 2. q(s)v(s) =1 (s + 2)(s + 4)(s + 3) so Bqv = e−2t , e−4t , e−3t while Bq = 0 −2t −4t e ,e . Since e−3t is the only function in the first set but not in the second yp (t) = a1 e−3t . 0 1 0 1 3. q(s)v(s) = (s − 2)2 (s − 3) so Bqv = e2t , te2t , e3t while Bq = e2t , e3t . Since te2t is the only function in the first set but not in the second yp (t) = a1 te2t . 0 1 3 4. q(s)v(s) Bqv = e4t , te4t , t2 e4t , e3t while Bq = 0 4t 3t 1= (s − 4)4t(s − 3)2 so e , e . Since te and t e4t are the only functions in the first set but not in the second yp (t) = a1 te4t + a2 t2 e4t . 0 1 5. q(s)v(s) − 5)2 (s2 + 25) so Bqv = e5t , te5t , cos 5t, sin 5t while 0 5t= (s 1 Bq = e , te5t . Since cos 5t and sin 5t are the only functions in the first set that are not in the second yp (t) = a1 cos 5t + a2 sin 5t.

122

1 Solutions

6. q(s)v(s) = (s2 + 1)(s2 + 4) so Bqv = {cos t, sin t, cos 2t, sin 2t} while Bq = {cos t, sin t}. Since cos 2t and sin 2t are the only functions in the first set that are not in the second yp (t) = a1 cos 2t + a2 sin 2t. 7. q(s)v(s) = (s2 + 4)2 so Bqv = {cos 2t, sin 2t, t cos 2t, t sin 2t} while Bq = {cos 2t, sin 2t}. Since t cos 2t and t sin 2t are the only functions in the first set that are not in the second yp (t) = a1 t cos 2t + a2 t sin 2t. 0 1 2 8. q(s)v(s) = (s + 4s + 5)(s − 1)21 so Bqv = et , tet , e−2t cos t, e−2t sin t 0 −2t while Bq = e cos t, e−2t sin t . Since et and tet are the only functions in the first set that are not in the second yp (t) = a1 et + a2 tet . 0 1 9. q(s)v(s) = (s2 + 4s + 5)(s − 1)2 so Bqv = et , tet , e−2t cos t, e−2t sin t while Bq = {et , tet }. Since e−2t cos t and e−2t sin t are the only functions in the first set that are not in the second yp (t) = a1 e−2t cos t+a2 e−2t sin t. 10. The characteristic polynomial is q(s) = s2 + 3s − 4 = (s + 4)(s − 1). 0 2t 1 Since L e = 1/(s − 2), we 0 set v(s) =1 s − 2. Then 0 q(s)v(s) 1 = (s + 4)(s − 1)(s + 2). Since Bqv = e−4t , et , e2t and Bq = e−4t , et we have yp = a1 e2t , a test function. Substituting yp into the differential equation gives 6a1 e2t = e2t . It follows that a1 = 1/6. The general solution is y = 16 e2t + c1 et + c2 e−4t . 2 11. The characteristic 0 −2t 1 polynomial is q(s) = s − 3s − 10 = (s − 5)(s + 2). Since L 7e = 7/(s + 2),0 we set v(s) = 1 s + 2. Then 0 q(s)v(s) 1 = (s − 5)(s + 2)2 . Since Bqv = e5t , e−2t , te−2t and Bq = e5t , e−2t we have yp = a1 te−2t , a test function. Substituting yp into the differential equation gives −a1 e−2t = 7e−t . It follows that a1 = −1. The general solution is y = −te−2t + c1 e−2t + c2 e5t .

12. The characteristic polynomial is q(s) = s2 + 2s + 1 = (s + 1)2 . Since L {et } = 1/(s − 1), we set v(s) = s − 1. Then q(s)v(s) = (s + 1)2 (s − 1). Since Bqv = {e−t , te−t , et } and Bq = {e−t , te−t } we have yp = a1 et , a test function. Substituting yp into the differential equation gives 4a1 et = et . It follows that a1 = 1/4. The general solution is y = 14 et + c1 e−t + c2 te−t . 13. The characteristic polynomial is q(s) = s2 + 2s + 1 = (s + 1)2 . Since L {e−t }0= 1/(s + 1), we1set v(s) = s + 1. Then q(s)v(s) = (s + 1)3 . Since Bqv = e−t , te−t , t2 e−t and Bq = {e−t , te−t } we have yp = a1 t2 e−t , a test function. Substituting yp into the differential equation gives 2a1 e−t = e−t . It follows that a1 = 1/2. The general solution is y = 12 t2 e−t +c1 e−t + c2 te−t . 14. The characteristic polynomial is q(s) = s2 + 3s + 2 = (s + 1)(s + 2). Since L {4} = = (s + 1)(s + 2)s. 0 4/s, we set 1 v(s) = s.0 Then q(s)v(s) 1 Since Bqv = e−t , e−2t , 1 and Bq = e−t , e−2t we have yp = a1 , a test

1 Solutions

123

function. Substituting yp into the differential equation gives 2a1 = 4. It follows that a1 = 2. The general solution is y = 2 + c1 e−t + c2 e−2t . 15. The characteristic polynomial is1q(s) = s2 + 4s + 5 = (s + 2)2 + 1, an irre0 −3t ducible quadratic. Since L e = 1/(s + 3),0we set v(s) = s + 3. Then 1 −2t −2t −3t q(s)v(s) = 0 ((s + 2)2 + 1)(s + 3). Since B = e cos t, e sin t, e qv 1 and Bq = e−2t cos t, e−2t sin t we have yp = a1 e−3t , a test function. Substituting yp into the differential equation gives 2a1 e−3t = e−3t . It follows that a1 = 1/2. The general solution is y = 12 e−3t + c1 e−2t cos t + c2 e−2t sin t. 16. We use the principle of superposition and break this problem into two separate parts: y !! + 4y = 1 and y !! + 4y = et . For y !! + 4y = 1: The characteristic polynomial is q(s) = s2 + 4. Since L {1} = 1/s, we set v(s) = s. Then q(s)v(s) = (s2 + 4)s. Since Bqv = {cos 2t, sin 2t, 1} and Bq = {cos 2t, sin 2t} we have yp1 = a1 , a test function. Substituting yp1 into the differential equation gives 4a1 = 1. It follows that a1 = 1/4 and yp1 = 1/4. For y !! + 4y = et : Since L {et } = 1/(s − 1) we set v(s) = s − 1. Then q(s)v(s) = (s2 +4)(s−1) and Bqv = {cos 2t, sin 2t, et }. It follows that yp2 = a1 et is a test function. Substituting yp2 into the differential equation gives 5a1 et = et from which we get a1 = 1/5. Thus yp2 = (1/5)et . The general solution is y = 14 + 15 et + c1 cos(2t) + c2 sin(2t). 17. The polynomial is q(s) = s2 − 1 = (s − 1)(s + 1). Since 0 2 1characteristic 3 L t 0= 2/s , we set1v(s) = s3 . Then q(s)v(s) = (s − 1)(s + 1)s3 . Since Bqv = et , e−t , 1, t, t2 and Bq = {et , e−t } we have yp = a1 + a2 t + a3 t2 , a test function. Substituting yp into the differential equation gives 2a3 − a1 − a2 t − a3 t2 = t2 . Using linear independence we equate the coefficients to get 2a3 − a1 = 0 −a2 = 0 −a3 = 1 It follows that a3 = −1, a2 = 0, and a1 = −2. The general solution is y = −t2 − 2 + c1 et + c2 e−t .

18. The characteristic polynomial is q(s) = s2 − 4s + 4 = (s − 2)2 . Since 2 L {et } = 1/(s0− 1), we set 1 v(s) = s −01.2tThen 1q(s)v(s) = (s − 2) t(s − 1). 2t 2t t 2t Since Bqv = e , te , e and Bq = e , te we have yp = a1 e , a test function. Substituting yp into the differential equation gives a1 et = et . It follows that a1 = 1. The general solution is y = et + c1 e2t + c2 te2t . 19. The polynomial is q(s) = s2 − 4s + 4 = (s − 2)2 . Since 0 2tcharacteristic 1 L e 0= 1/(s − 2), we set v(s) =0s − 2. Then q(s)v(s) = (s − 2)3 . Since 1 1 Bqv = e2t , te2t , t2 e2t and Bq = e2t , te2t we have yp = a1 t2 e2t , a test function. Substituting yp into the differential equation gives 2a1 e2t = e2t . It follows that a1 = 1/2. The general solution is y = 12 t2 e2t +c1 e2t +c2 te2t

124

1 Solutions

20. The characteristic polynomial is q(s) = s2 + 1, an irreducible quadratic. Since L {2 sin t} = 2/(s2 + 1), we set v(s) = s2 + 1. Then q(s)v(s) = (s2 + 1)2 . Since Bqv = {cos t, sin t, t cos t, t sin t} and Bq = {cos t, sin t} we have yp = a1 t cos t + a2 t sin t, a test function. Substituting yp into the differential equation gives −2a1 sin t + 2a2 cos t = 2 sin t. Linear independence implies −2a1 = 2 and 2a2 = 0. Thus a1 = −1 and a2 = 0. The general solution is y = −t cos t + c1 sin t + c2 cos t. 21. The polynomial is q(s) = s2 + 6s + 9 = (s + 3)2 . Since 0 characteristic 1 2t 2 L 25te = 25/(s − 2)2 , 0we set v(s) = (s − 2) = (s + 1 . Then q(s)v(s) 0 1 2 2 −3t −3t 2t 2t 3) (s − 2) . Since Bqv = e , te , e , te and Bq = e−3t , te−3t we have yp = a1 e2t + a2 te2t , a test function. Substituting yp into the differential equation gives (25a1 + 10a2 )e2t + 25a2 te2t = 25te2t . Linear independence implies 25a1 + 10a2 = 0 and 25a2 = 25. We get a2 = 1 and a1 = −2/5 The general solution is y = te2t − 25 e2t + c1 e−3t + c2 te−3t 22. The polynomial is q(s) = s2 + 6s + 9 = (s + 3)2 . Since 0 characteristic 1 −3t 2 L 25te = , we set v(s) =1 (s+3)2 . Then q(s)v(s) 1 = (s+3)4 . 0 25/(s+3) 0 Since Bqv = e−3t , te−3t , t2 e−3t , t3 e−3t and Bq = e−3t , te−3t we have yp = a1 t2 e−3t + a2 t3 e−3t , a test function. Substituting yp into the differential equation gives (2a1 + 6a2 t)e−3t = 25te−3t . Linear independence implies 2a1 = 0 and 6a2 = 25. We get a1 = 0 and a2 = 25/6 The general 3 −3t solution is y = 25 + c1 e−3t + c2 te−3t . 6 t e 2 2 23. The characteristic polynomial is q(s) 0 −3t= s +16s + 13 = (s + 3) +2 4, an irreducible quadratic. Since L e cos 2t = (s + 3)/((s + 3) + 2 2 2 4), we set v(s) = (s + 3) + 4. Then q(s)v(s) = ((s + 0 −3t 1 3) + 4) . −3t −3t −3t Since cos12t, e sin 2t, te cos 2t, te sin 2t and Bq = 0 −3t Bqv = −3te e cos 2t, e sin 2t we have yp = a1 te−3t cos 2t + a2 te−3t sin 2t, a test function. Substituting yp into the differential equation gives (after a long calculation) −4a1 e−3t sin 2t + 4a2 e−3t cos 2t = e−3t cos 2t. It follows that −4a1 = 0 and 4a2 = 1. Thus a1 = 0 and a2 = 1/4. The general solution is y = 14 te−3t sin(2t) + c1 e−3t cos(2t) + c2 e−3t sin(2t).

24. The characteristic polynomial is q(s) = s2 − 8s + 25 = (s − 4)2 + 9, an irreducible quadratic. Since L {104 sin 3t} = 312/(s2 + 9), we set 2 2 v(s) 9. Then q(s)v(s) = 1((s − 4)2 + 9)(s Bqv = 0 −4t= s + −4t 0 −4t+ 9). Since 1 e cos 3t, e sin 3t, cos 3t, sin 3t and Bq = e cos 3t, e−4t sin 3t we have yp = a1 cos 3t + a2 sin 3t, a test function. Substituting yp into the differential equation gives (16a1 − 24a2 ) cos 3t + (16a2 + 24a1 ) sin 3t = 104 sin 3t. Linear independence implies 16a1 −24a2 = 0 and 16a2 +24a1 = 104. Solving these equations gives a1 = 3 and a2 = 2. The general solution is y = 3 cos 3t + 2 sin 3t + te4t sin(3t) + c1 e4t cos(3t) + c2 e4t sin 3t. 25. The characteristic polynomial is q(s) = s2 − 5s − 6 = (s − 6)(s + 1). 0 3t 1 Since L e = 1/(s − 3), we set v(s) = s − 3. Then q(s)v(s) = (s −

1 Solutions

125

0 1 0 1 6)(s + 1)(s − 3). Since Bqv = e6t , e−t , e3t and Bq = e6t , e−t we have yp = a1 e3t , a test function. Substituting yp into the differential equation gives −12a1 e3t = e3t . It follows that a1 = −1/12. The general solution is −1 3t 3t 6t −t ! 6t −t y = −1 the initial 12 e + c1 e + c2 e . Since y = 4 e + 6c1 e − c2 e condition imply −1 c1 + c2 = 2 12 + −1 4

+ 6c1 − c2 = 1

It is easy to calculate that c1 = 10/21 and c2 = 135/84. Thus y = −1 3t 10 6t 135 −t 12 e + 21 e + 84 e . 26. The characteristic polynomial is q(s) = s2 + 2s + 5 = (s + 1)2 + 4, an irreducible quadratic. Since L {8e−t } = 8/(s + 1), we set v(s) = s + 1. Then q(s)v(s) = ((s + 1)2 + 4)(s + 1). Since Bqv = {e−t cos 2t, , e−t sin 2t, e−t } and Bq = {e−t cos 2t, e−t sin 2t} we have yp = a1 e−t , a test function. Substituting yp into the differential equation gives 4a1 e−t = 8e−t . It follows that a1 = 2. The general solution is y = 2e−t + c1 e−t cos 2t + c2 e−t sin 2t. Since y ! = −2e−t +c1 (−e−t cos 2t−2e−t sin 2t)+c2 (−e−t sin 2t+2e−t cos 2t) the initial conditions imply 2 + c1 = 0 −2 − c1 + 2c2 = 8 Solving this system gives c1 = −2 and c2 = 4. Thus y = 2e−t − 2e−t cos 2t + 4e−t sin 2t. 0 1 27. The characteristic polynomial is q(s) = s2 + 1. Since L 10e2t = 2 10/(s−2), we set 0 1 v(s) = s−2. Then q(s)v(s) = (s +1)(s−2). Since Bqv = cos t, sin t, e2t and Bq = {cos t, sin t} we have yp = a1 e2t , a test function. Substituting yp into the differential equation gives 5a1 e2t = 10e2t and hence a1 = 2. The general solution is y = 2e2t + c1 cos t + c2 sin t. Since y ! = 4e2t − c1 sin t + c2 cos t the initial conditions imply 2 + c1 4 +

= 0 c2 = 0

and so c1 = −2 and c2 = −4. Thus y = 2e2t − 2 cos t − 4 sin t. 28. The characteristic polynomial is q(s) = s2 − 4 = (s − 2)(s + 2). Since L {2 − 8t} = (2/s) − (8/s2 ) = (2s − 8)/s2 , 0we set v(s) 1= s2 . Then 0q(s)v(s) 1= (s − 2)(s + 2)s2 . Since Bqv = e2t , e−2t , 1, t and Bq = e2t , e−2t we have yp = a1 + a2 t, a test function. Substituting yp into the differential equation gives −4(a1 + a2 t) = 2 − 8t. It follows that a1 = −1/2 and a2 = 2. The general solution is y = 2t − 12 + c1 e−2t + c2 e2t . Since y ! = 2 − 2c1 e−2t + 2c2 e2t the initial conditions imply

126

1 Solutions −1 2

2

+ c1 + c2 = 0 − 2c1 + 2c2 = 5

The solution to this system is c1 = −1/2 and c2 = 1. Thus y = 2t − 1 −2t + e2t . 2e

1 2

−

Section 3.5 1. The characteristic polynomial is q(s) = s2 − 4 = (s − 2)(s + 2) and 0 −6t 1 L e = 1/(s + 6). Thus L {y} =

1 1 p(s) = 32 + . (s − 2)(s + 2)(s + 6) s + 6 (s − 2)(s + 2)

A particular solution is yp = 1 −6t + c1 e2t + c2 e−2t 32 e

1 −6t 32 e

and the general solution is y =

2. The characteristic polynomial is q(s) = s2 + 2s − 15 = (s + 5)(s − 3) and L {16et } = 16/(s − 1). Thus L {y} =

16 −1 p(s) = + . (s − 1)(s + 5)(s − 3) s − 1 (s + 5)(s − 3)

A particular solution is yp = −1et and the general solution is y = −et + c1 e−3t + c2 e5t 3. The characteristic polynomial is q(s) = s2 + 5s + 6 = (s + 2)(s + 3) and 0 −2t 1 L e = 1/(s + 2). Thus L {y} =

1 1 p(s) = + . (s + 2)2 (s + 3) (s + 2)2 (s + 2)(s + 3)

A particular solution is yp = te−2t and the general solution is y = te−2t + c1 e−2t + c2 e−3t 4. The characteristic polynomial is q(s) = s2 + 3s + 2 = (s + 1)(s + 2) and L {4} = 4/s. Thus L {y} =

4 2 p(s) = + . s(s + 1)(s + 2) s (s + 1)(s + 2)

A particular solution is yp = 2 and the general solution is y = 2 + c1 e−t + c2 e−2t

1 Solutions

127

5. The polynomial is q(s) = s2 + 2s − 8 = (s − 2)(s + 4) and 0 characteristic 1 −4t L 6e = 6/(s + 4). Thus L {y} =

6 −1 p(s) = + . 2 2 (s − 2)(s + 4) (s + 4) (s − 2)(s + 4)

A particular solution is yp = −te−4t and the general solution is y = −te−4t + c1 e2t + c2 e−4t 6. The characteristic polynomial is q(s) = s2 + 3s − 10 = (s + 5)(s − 2) and L {sin t} = 1/(s2 + 1). Thus L {y} =

1 1 −3s − 11 p(s) = + . (s2 + 1)(s + 5)(s − 2) 130 s2 + 1 (s + 5)(s − 2)

1 A particular solution is yp = 130 (−3 cos t − 11 sin t) and the general so3 11 lution is y = − 130 cos t − 130 sin t + c1 e−5t + c2 e2t

7. The polynomial is q(s) = s2 + 6s + 9 = (s + 3)2 and 0 characteristic 1 2t L 25e = 25/((s − 2)2 ). Thus L {y} =

25 1 2 1 p(s) = − + . (s − 2)2 (s + 3)2 (s − 2)2 5 s − 2 (s + 3)2

A particular solution is yp = te2t − 25 e2t and the general solution is y = te2t − 25 e2t + c1 e−3t + c2 te−3t 8. The polynomial is q(s) = s2 − 5s − 6 = (s − 6)(s + 1) and 0 characteristic 1 4t L 10te = 10/(s − 4)2 . Thus L {y} =

(s −

10 −1 3 1 p(s) = − + . 2 − 6)(s + 1) (s − 4) 10 s − 4 (s − 6)(s + 1)

4)2 (s

A particular solution is yp = −te4t − 3 4t y = −te4t − 10 e + c1 e−t + c2 e6t

3 4t 10 e

and the general solution is

2 2 9. The 0 characteristic 1 polynomial is q(s) = s − 8s + 25 = (s − 4) + 9 and L 36te4t sin 3t = 216(s − 4)/((s − 4)2 + 9)2 . Thus

L {y} =

216(s − 4) . ((s − 4)2 + 9)3

This is a partial fraction. Table 2.5 gives y = −3t2 e4t cos 3t + te4t sin 3t. A particular solution is yp = −3t2 e4t cos 3t + te4t sin 3t and the general solution is y = −3t2 e4t cos 3t + te4t sin 3t + c1 e4t cos 3t + c2 e4t sin 3t 10. The polynomial is q(s) = s2 − 4s + 4 = (s − 2)2 and 0 2tcharacteristic 1 L te = 1/(s − 2)2 . Thus

128

1 Solutions

L {y} =

1 (s − 2)4

This is a partial fraction. A particular solution is yp = general solution is y = 16 t3 e2t + c1 e2t + c2 te2t

t3 2t 6e

and the

11. The characteristic polynomial is q(s) = s2 + 2s + 1 = (s + 1)2 and L {cos t} = s/(s2 + 1). Thus L {y} =

s 1 1 p(s) = + . (s + 1)2 (s2 + 1) 2 s2 + 1 (s + 1)2

A particular solution is yp = 1 −t + c2 te−t 2 sin t + c1 e

1 2

sin t and the general solution is y =

12. The characteristic polynomial is q(s) = s2 + 2s + 1 = (s + 1)2 and L {et cos t} = s − 1/((s − 1)2 + 1). Thus ' ( s−1 1 3(s − 1) + 4 p(s) L {y} = = + . ((s − 1)2 + 1)(s + 1)2 25 (s − 1)2 + 1 (s + 1)2 1 A particular solution is yp = 25 (3 cos t+4 sin t)et and the general solution 1 t is y = 25 (3 cos t + 4 sin t)e + c1 e−t + c2 te−t .

Section 3.6 1. The force is 16 lbs. A length of 6 inches is 1/2 ft. The spring constant is k = 16/(1/2) = 32 lbs/ft. 2. The equation for the spring constant gives 20 = lbs.

x 9/12 .

So x = 20 34 = 15

3. The force exerted by the mass is 40 · 9.8 = 392 N. Thus k = 392/.8 = 490 N/m. 4. The force exerted by the mass is 20 · 9.8 = 196 N. Let x be the distance stretched. Thus 784 = 196 x . Solving for x gives x = 196/784 = 1/4 m or 25 centimeters. 5. The force is 4 lbs and the velocity is 1/2 ft per second. So µ = 4 Force/velocity = 1/2 = 8 lbs s/ft. 6. The force is 40 Newtons and the velocity is 30 centimeters per second which is .3 m per second. So µ = 40/.3 = 133.33 N s/m.

1 Solutions

129

7. Let x be the force. Then 100 = x/4 so x = 400 lbs. 8. Let v be the velocity of deceleration. Then 200 = 40/v and so v = 1/5 m/s or 20 centimeters per second. 9. The mass is m = 6. The spring constant is given by k = 2/.1 = 20. The damping constant is µ = 0. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) = .1 m and y ! (0) = 0. The following equation 6y !! + 20y = 0,

y(0) = .1,

y ! (0) = 0

represents the model for the motion of the body. The characteristic poly! ! 2 2 2 nomial !is q(s) = 6s + 20 = 6(s + 10/3 ). Thus y = c1 cos 10/3 t + c2 sin 10/3 t. The initial conditions imply c1 = 1/10 and c2 = 0. Thus y=

! 1 cos 10/3 t. 10

The motion is undamped free or simple harmonic motion. Since y is written in the form y = A cos ωt + φ we can read!off the amplitude, frequency, and phase shift; they are A = 1/10, β = 10/3, and φ = 0.

10. The mass is m = 4. The spring constant k is given by k = 4 · 9.8/.8 = 49. The damping constant is given by µ = 98/2 = 49. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) = 0 and y ! (0) = −.1. The equation 4y !! +49y ! +49y = 0, y(0) = 0, y ! (0) = −.1 models the motion of the body. The roots of the characteristic polynomial, √ √ 33 33 q(s) = 4s2 + 49s + 49 are s = −49±7 which are r1 = −49+7 ≈ −1.1 8 8 √ −49−7 33 r1 t r2 t and r2 = ≈ −11.15. Thus y = c1 e + c2 e . The initial 8 conditions imply c1 + c2 = 0 and r1 c1 + r2 c2 = −.1. Thus c1 = − r1 .1 −r2 = 2 .1 2 √ √ − 35 33 and c2 = r1 −r2 = 35 33 . Thus y=−

2 √

35 33

e

−49+7 8

√ 33

t

+

√ −49−7 33 2 t 8 √ e . 35 33

Since the discriminant of the characteristic equation is D = 492 −4·4·49 = 1617 > 0 the motion is overdamped free motion. To cross equilibrium there must be a t > 0 so the y(t) = 0. But this would imply e(r1 −r2 )t = 1 which cannot happen. 11. The mass is m = 16/32 = 1/2 slugs. The spring constant k is given by k = 16/(6/12) = 32. The damping constant is given by µ = 4/2 = 2. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) = 1 and y ! (0) = 1. The following equation 1 !! ! ! 2 y + 2y + 32y = 0, y(0) = 1, y (0) = 1 models the motion of the body.

130

1 Solutions

The characteristic polynomial is q(s) = 12 s2 + 2s + 32 = 12 (s2 + 4s + 64) = √ 2 1 2 60 ). Thus 2 ((s + 2) + √ 60t + c2 e−2t sin 60t. √ The initial conditions imply c1 = 1 and c2 = 3/ 60. Thus y = c1 e−2t cos

y = e−2t cos

√

√

√ 3 60t + √ e−2t sin 60t. 60

The discriminant of the characteristic equation is D = 22 − 4 · (1/2) · 32 −60 < 0 so the motion is underdamped free motion. Let A = @ = " #2 ! √ √ 1 + √360 = 23/20. If tan φ = 3/ 60 = 60/20 then φ ≈ .3695. We can write

y=

@

√ 23 −2t e cos( 60t + φ). 20

12. The mass is m = 32/32 = 1 slug. The spring constant k is given by k = 32/2 = 16. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) = −1 and y ! (0) = 0. The equation y !! + 8y ! + 16y = 0, y(0) = −1, y ! (0) = 0 models the motion of the body. The characteristic polynomial is q(s) = (s + 4)2 . Thus y = c1 e−4t + c2 te−4t . The initial conditions imply c1 = 1 and c2 = 4. Thus y = (1 + 4t)e−4t . The discriminant of the characteristic equation is D = 82 − 4 · 1 · 16 = 0 so the motion is critically damped free motion. The equation y(t) = 0 implies t = −1/4. So there is no positive time where the body crosses equilibrium. 13. The mass is m = 2/32 = 1/16 slug. The spring constant k is given by k = 2/(4/12) = 6 and the damping constant is µ = 0. The initial conditions 1 !! are y(0) = 0 and y ! (0) = 8/12 = 2/3. The equation 16 y + 6y = 0 or !! equivalently y + 96y = 0 with initial conditions y(0) = 0, y ! (0) = 2/3 models the motion of the characteristic polynomial is q(s) = √ body. The √ s2 + 96 so y = c1√cos 96t + c2 √ sin 96t. The initial conditions imply √ √ %√ & c1 = 0 and c2 = 366 . Thus y = 366 sin 96t = 366 cos 96t − π2 . The motion is undamped free or simple harmonic motion so the mass crosses equilibrium.

1 Solutions

131

14. The mass is m = 2. The spring constant k is given by k = 5/1 = 5. The damping constant is given by µ = 6/1 = 6. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) = .1 and y ! (0) = .1. The equation 2y !! + 6y ! + 5y = 0, y(0) = .1, y ! (0) = .1 models the motion The characteristic polynomial is q(s) = "% of the& body. % 1 &2 # 3 2 2 2s + 6s + 5 = 2 s + 2 + 2 . Thus 3 3 1 1 y = c1 e 2 t cos t + c2 e 2 t sin t. 2 2

The initial conditions imply c1 = y=

1 10

and c2 = 12 . So

1 3t 1 1 3 1 e 2 cos t + e 2 t sin t. 10 2 2 2

The discriminant of the characteristic equation is D = 62 −4·2·5 = −4 < 0 so the motion is underdamped free motion and crosses equilibrium. If √ % 1 &1 A = 100 + 14 2 = 1026 and φ = arctan 5 = 1.3734 then we can write √ %1 & 3 y = 1026 e 2 t cos 10 t + 1.3734 .

15. By the quadratic formula the roots of q(s) = ms2 + µs + k are @" ! −µ ± µ2 − 4mk −µ µ #2 k s= = ± − . 2m 2m 2m m

If the discriminant D = µ2 − 4mk is negative then the roots are complex and the real part is −µ 2m which is negative. If the discriminant is zero then −µ is a negative double root. If the discriminant is positive then both 2m roots are real and distinct. It is enough to show that+ the larger of the +% & % µ &2 −µ µ 2 µ k k two, r = 2m + − m is negative. Let p = 2m + −m and 2m 2m observe that it is positive. Further, , -, @" @" −µ µ #2 k µ µ #2 k rp = + − + − 2m 2m m 2m 2m m µ2 µ2 k + − 4m2 4m2 m k = − 0 it follows the r < 0. It follows that a solution to my !! + µy ! + ky = 0 is of the following form 1. y = c1 er1 t + c2 er2 t where r1 and r2 are negative. 2. y = (c1 + c2 t)ert where r is negative

132

1 Solutions

3. y = c1 eαt cos βt + c2 eαt sin βt where α is negative In each case limt→∞ y(t) = 0. 16. For an overdamped system the solution is of the form y = c1 er1 t + c2 er2 t , where both r1 and r2 are negative. The equation y(t) = 0 gives −c1 = e(r2 −r1 )t . c2 If c1 and c2 have the same sign there are no solutions. If c1 and c2 have 1 1 opposite signs then t = r2 −r ln −c c2 . If t > 0 the solution crosses equilib1 rium only at this point. For a critically damped system the solution is of the form y = (c1 +c2 t)ert , 1 where r is negative. The equation y(t) = 0 implies c1 +c2 t = 0 so t = −c c2 . It t > 0 the solution crosses equilibrium only at this point.

Section 3.7 1. We first solve for the charge. The initial conditions are q(0) = 0 and q ! (0) = I(0) = 0. The differential equation is 0.25q !! + 10q ! + (5/1000)

−1

q=6

which is equivalent to q !! + 40q ! + 800q = 24. The characteristic polynomial is s2 + 40s + 800 = (s + 20)2 + 400 so the homogeneous solution is qh (t) = c1 e−20t cos 20t+c2 e−20t sin 20t. It is easy to see that the particular solution is qp (t) = 24/800 = 3/100. Thus q(t) = c1 e−20t cos 20t + c2 e−20t sin 20t + 3/100. The initial conditions imply 3 = 0 100 −20c1 + 20c2 = 0 c1 +

which gives c1 = −3/100 and c2 = −3/100. It follows that q(t) = −

3 −20t 3 −3t 3 e cos 20t − e sin 20t + . 100 100 100

1 Solutions

133

Differentiating we get I(t) =

6 −20t 6 e sin 20t = e−20t cos(20t − π/2). 5 5

2. We first solve for the charge. The initial conditions are q(0) = 0.01 and q ! (0) = I(0) = 0. The differential equation is 0.1q !! + 5q ! + (0.025)

−1

q=0

which is equivalent to q !! + 50q ! + 400q = 0. The characteristic polynomial is s2 + 50s + 400 = (s + 40)(s + 10) and hence q(t) = c1 e−40t + c2 e−10t . c1 + c2 = 0.01 −40c1 − 10c2 = 0 which gives c1 = −1/300 and c2 = 4/300. It follows that q(t) = −

1 −40t 4 −10t e + e 300 300

Differentiating we get I(t) =

2 −40t 2 e − e−10t . 15 15

3. We first solve for the charge. The initial conditions are q(0) = 0 and q ! (0) = I(0) = 2. The differential equation is 0.2q !! + 4q ! + (0.02)

−1

q = 25 sin 5t

which is equivalent to q !! + 20q ! + 100q = 125 sin 5t. The characteristic polynomial is s2 + 20s + 100 = (s + 10)2 and hence qh (t) = c1 e−10t + c2 te−10t . We use the method of undetermined coefficients to get the test function qp (t) = a1 cos 5t + a2 sin 5t. Substitution leads to the equations 3a1 + 4a2 = 0 −4a1 + 3a2 = 5

134

1 Solutions

and therefore a1 = −4/5 and a2 = 3/5. It follows that qp = 3 5 sin 5t and q(t) = c1 e−10t + c2 te−10t −

−4 5

cos 5t +

4 3 cos 5t + sin 5t. 5 5

The initial conditions lead to the equations 4 5 = −1

c1 = −10c1 + c2

which gives c1 = 4/5 and c2 = 7. It follows that q(t) =

4 −10t 4 3 e + 7te−10t − cos 5t + sin 5t. 5 5 5

Differentiating we get I(t) = −e−10t − 70te−10t + 4 sin 5t + 3 cos 5t. 4. We first solve for the charge. The initial conditions are q(0) = 1 and q ! (0) = I(0) = 2. The differential equation is q !! + 11q ! + 30q = 10e−5t . The characteristic polynomial is s2 + 11s + 30 = (s + 5)(s + 6) and hence qh (t) = c1 e−5t + c2 e−6t . We use the incomplete partial fraction method with Q(s) = (s+5)10 2 (s+6) to determine qp .

Incomplete s + 5 -chain 10 (s + 5)2 (s + 6)

10 (s + 5)2

p(s) (s + 5)(s + 6) It follows that qp (t) = L−1

4

10 (s+5)2

5

= 10te−5t . From this we get

q(t) = c1 e−5t + c2 e−6t + 10te−5t . The initial conditions lead to the equations

1 Solutions

135

c1 + c 2 = 1 5c1 + 6c2 = 8 which gives c1 = −2 and c2 = 3. It follows that q(t) = −2e−5t + 3e−6t + 10te−5t . Differentiating we get I(t) = 20e−5t − 18e−6t − 50te−5t . 5. The differential equation that models the charge is given by 0.1q !! + 10q = 0.1 cos 5t,

q(0) = 0, q ! (0) = 0

which is equivalent to q !! + 100q = cos 5t q(0) = 0, q ! (0) = 0. Use the Laplace transform method to get Q(s) =

(s2

s . + 100)(s2 + 25)

Quadratic partial fraction recursion gives ' ( 1 s s Q(s) = − . 75 s2 + 25 s2 + 100 Hence q(t) =

1 (cos 25t − cos 100t). 75

Observe that 1 |cos 25t − cos 100t| 75 1 ≤ (|cos 25t| + |cos 100t|) 75 2 ≤ . 75

|q(t)| =

In this case the maximum charge on the capacitor is 2/75 C. It will not fail. 6. The differential equation that models the charge is given by 0.1q !! + 10q = 0.1 cos 10t,

q(0) = 0, q ! (0) = 0

136

1 Solutions

which is equivalent to q !! + 100q = cos 10t q(0) = 0, q ! (0) = 0. The Laplace transform method gives Q(s) =

(s2

s + 100)2

and therefore

1 t sin 10t. 20 The factor of t tell us that q(t) is unbounded and will eventually hit the 2 C threshold where capacitor failure occurs. q(t) =

Section 4.1 1. yes; (D 3 − 3D)y = et , order 3, q(s) = s3 − 3s, nonhomogeneous 2. yes; (D 4 + D + 4)y = 0, order 4, q(s) = s4 + s + 4, homogeneous 3. no, because of the presence of y 4 4. no, because of the presence of ty !! 5. (a) L(e2t ) = 8e2t − 4(2e2t ) = 0 (b) L(3−2t = −8e−2t − 4(−2e−2t ) = 0 (c) L(2) = 0 − 4(0) = 0 6. (a) L(−2e−2t ) = −2e−2t − 2(e−2t = −4e−2t (b) L(3e2t = 3(2e2t ) − 2(3e2t ) = 0 (c) L tan t = sec2 t − 2 tan t 7. (a) Le−t = e−t + 5e−t + 4e−t = 10e−t (b) L cos t = cos t + 5(− cos t) + 4 cos t = 0 (c) L sin 2t = 16 sin 2t + 5(−4 sin 2t) + 4 sin 2t = 0 8. (a) (b) (c) (d)

Let = et − et + et − et = 0 L(tet ) = (3 + t)et − (2 + t)et + (1 + t)et − tet = 2et L(cos t) = sin t − (− cos t) + (− sin t) − cos t = 0 L(sin t) = − cos t − (− sin t) + cos t − sin t = 0

9. e2t , e−2t , and 1 are homogeneous solution so yh = c1 e2t + c2 e−2t + c3 are homogeneous solutions for all scalars c1 , c2 , and c3 . A particular solution is yp = te2t . Thus y(t) = yp (t) + yh (t) = te2t + c1 e2t + c2 e−2t + c3 where c1 , c2 , and c3 are arbitrary constants

1 Solutions

137

10. e2t , e−2t , sin 2t, and cos 2t are homogeneous solution so yh = c1 e2t + c2 e−2t + c3 sin 2t + c4 cos 2t are homogeneous solutions for all scalars c1 , . . . , c4 . A particular solution is yp = sin t. Thus y(t) = yp (t) + yh (t) = sin t + c1 e2t + c2 e−2t + c3 sin 2t + c4 cos 2t where c1 , . . . , c4 are arbitrary constants. 11. From Exercise 9 we have y(t) = yp (t) + yh (t) = te2t + c1 e2t + c2 e−2t + c3 . Since y = te2t + c1 e2t + c2 e−2t + c3 y ! = (1 + 2t)e2t + 2c1 e2t − 2c2 e−2t

y !! = (4 + 4t)e2t + 4c1 e2t + 4c2 e−2t we have 2 = y(0) = c1 + c2 + c3 −1 = y ! (0) = 1 + 2c1 − 2c2 16 = y !! (0) = 4 + 4c1 + 4c2 , from which follows that c1 = 1, c2 = 2, and c3 = −1. Thus y(t) = te2t + e2t + 2e−2t − 1 12. From Exercise 10 we have y(t) = sin t+c1 e2t +c2 e−2t +c3 sin 2t+c4 cos 2t. Since y(t) = sin t + c1 e2t + c2 e−2t + c3 sin 2t + c4 cos 2t y ! (t) = cos t + 2c1 e2t − 2c2 e−2t + 2c3 cos 2t − 2c4 sin 2t

y !! (t) = − sin t + 4c1 e2t + 4c2 e−2t − 4c3 sin 2t − 4c4 cos 2t y !!! (t) = − cos t + 8c1 e2t − 8c2 e−2t − 8c3 cos 2t + 8c4 sin 2t we have 0 = y(0) = c1 + c2 + c4 3 = y ! (0) = 1 + 2c1 − 2c2 + 2c3 −16 = y !! (0) = 4c1 + 4c2 − 4c4

−9 = y !!! (0) = −1 + 8c1 − 8c2 − 8c3 .

from which follows c1 = −1, c2 = −1, c3 = 1, and c4 = 2. Thus y(t) = sin t − e2t − e−2t + sin 2t + 2 cos 2t.

138

1 Solutions

Section 4.2 1. The characteristic polynomial is q(s) 4= s3 − 1 = (s − 1)(s2 + s + 5 1) = √ √ 3 3 2 t − 12 t − 12 t (s − 1)((s + 1/2) + 3/4). Thus Bq = e , e cos 2 t, e sin 2 t . It 1

follows that y(t) = c1 e−t + c2 e− 2 t cos

√ 3 2 t

1

+ c3 e− 2 t sin

√ 3 2 t

2. The characteristic is q(s) = s3 − 6s2 + 12s − 8 = (s − 2)3 . 0 2t 2tpolynomial 1 2 2t Thus Bq = e , te , t e . It follows that y(t) = (c1 + c2 t + c3 t2 )e2t

3. The characteristic polynomial is q(s) = s4 − 1 = (s2 − 1)(s2 + 1) = (s − 1)(s + 1)(s2 + 1). Thus Bq = {et , e−t , cos t, sin t} . It follows that y(t) = c1 et + c2 e−t + c3 sin t + c4 cos t 4. The characteristic polynomial is q(s) = s3 + 2s2 + s = s(s + 1)2 . Thus Bq = {1, e−t , te−t } . It follows that y(t) = c1 + c2 e−t + c3 te−t 2 2 5. The characteristic polynomial is q(s) =0s4 − 5s2 + 4 = (s 1 − 1)(s − 4) = t −t 2t −2t (s − 1)(s + 1)(s − 2)(s + 2). Thus Bq = e , e , e , e . It follows that y(t) = c1 et + c2 e−t + c3 e2t + c4 e−2t 2 6. The characteristic (s−2)(s−5)(s+ 0 2tpolynomial 1 is q(s) = (s−2)(s −25) = 5t −5t 5). Thus Bq = e , e , e . It follows that y(t) = c1 e2t +c2 e5t +c3 e−5t .

7. The 2)(s2 + 25). Thus Bq = 0 −2tcharacteristic1polynomial is q(s) = (s +−2t e , cos 5t, sin 5t . It follows that y(t) = c1 e + c2 cos 5t + c3 sin 5t. 0 8. The characteristic polynomial is q(s) = (s2 +9)3 . Thus Bq = cos 3t, sin 3t, t cos 3t, t sin 3t, t2 cos 3t, t2 sin It follows that y(t) = c1 cos 3t+c2 sin 3t+c3 t cos 3t+c4 t sin 3t+c5 t2 cos 3t+ c6 t2 sin 3t. 9. The characteristic polynomial is q(s) = (s + 13)(s − 1)(s + 3)2 = (s − 0 t −3t 3 1)(s + 3) . Thus Bq = e , e , te−3t , t2 e−3t . It follows that y(t) = c1 et + c2 e−3t + c3 te−3t + c4 t2 e−3t . 10. The characteristic polynomial is q(s) = s3 + s2 − s − 1 = (s − 1)(s + 1)2 . Thus Bq = {et , e−t , te−t } . It follows that y(t) = c1 et + c2 e−t + c3 te−t . Since y(t) = c1 et + c2 e−t + c3 te−t y ! (t) = c1 et − c2 e−t + c3 (1 − t)e−t

y !! (t) = c1 et + c2 e−t + c3 (−2 + t)e−t we have

1 Solutions

139

1 = y(0) = c1 + c2 4 = y ! (0) = c1 − c2 + c3 −1 = y !! (0) = c1 + c2 − 2c3 and hence c1 = 2, c2 = −1, and c3 = 1. Thus y = 2et − e−t + te−t . 11. The characteristic polynomial is q(s) = s4 − 1 = (s2 − 1)(s2 + 1) = (s − 1)(s + 1)(s2 + 1). Thus Bq = {et , e−t , cos t, sin t}. It follows that y(t) = c1 et + c2 e−t + c3 cos t + c4 sin t. Since y(t) = c1 et + c2 e−t + c3 cos t + c4 sin t y ! (t) = c1 et − c2 e−t − c3 sin t + c4 cos t

y !! (t) = c1 et + c2 e−t − c3 cos t − c4 sin t y !!! (t) = c1 et − c2 e−t + c3 sin t − c4 cos t we have −1 = y(0) = c1 + c2 + c3 6 = y ! (0) = c1 − c2 + c4 −3 = y !! (0) = c1 + c2 − c3 2 = y !!! (0) = c1 − c2 − c4

from which we get c1 = 1, c2 = −3, c3 = 1, and c4 = 2. Hence y = et − 3e−t + cos t + 2 sin t

Section 4.3 1. Since q(s) = s3 − s = s(s − 1)(s + 1) we have Bq = {1, et , e−t } and since q(s)v(s) = s(s − 1)(s + 1)2 we have Bqv = {1, et , e−t , te−t }. Thus Bqv \ Bq = {te−t } and y = cte−t is the test function. 2. Since q(s) = s3 − s2 − s + 1 = (s − 1)2 (s + 1) we have 0Bq = {et , tet , e−t1} and since q(s)v(s) (s1− 1)3 (s + 1) we have Bqv = et , tet , t2 et , e−t . 0 2= −t Thus Bqv \ Bq = t e and y = ct2 e−t is the test function. 3. q(s) = s(s − 1)(s + 1) we have Bq0 = {1, et , e−t1} and since q(s)v(s) 0 = 1 s(s−1)(s+1)(s−2) we have Bqv = 1, et , e−t , e2t . Thus Bqv \Bq = e2t and y = ce2t is the test function. 4 2 2 2 4. q(s) = + 9)(s 0 s − 81 = (s 1 − 9) = (s + 9)(s −2 3)(s 2+ 3) we have 3t −3t Bq = cos 3t, sin03t, e , e and since q(s)v(s) = (s 1+ 9) (s − 3)(s + 3) we have Bqv = t cos 3t, t sin 3t, cos 3t, sin 3t, e3t , e−3t . Thus Bqv \ Bq = {t cos 3t, t sin 3t} and y = c1 t cos 3t + c2 t sin 3t is the test function.

140

1 Solutions

1 5. We have q(s) = s3 − s = s(s − 1)(s + 1) and L {et } = s−1 . Let v(s) = 2 t −t s − 1. Then q(s)v(s) = s(s − 1) (s + 1), Bq = {1, e , e } and Bqv = {1, et , tet , e−t } and Bqv \ Bq = {tet }. It follows that y = ctet is the test function. Since

y = ctet y ! = c(1 + t)et y !! = c(2 + t)et y !!! = c(3 + t)et we have c(3 + t)et − c(1 + t)et = et . Simplifying we get 2cet = et which implies c = 1/2. It follows that y = 12 tet + c1 e−t + c2 et + c3 6. We have q(s) = s3 − s2 − s − 1 = (s − 1)(s2 + 1) and L {4 cos t} = s24s +1 . Let v(s) = s2 + 1. Then q(s)v(s) = (s − 1)(s2 + 1)2 , Bq = {et , cos t, sin t} and Bqv = {et , cos t, sin t, t cos t, t sin t}. Thus Bqv \ Bq = {t cos t, t sin t}. It follows that y = c1 t cos t + c2 t sin t is the test function and we have y = c1 t cos t + c2 t sin t y ! = c1 (cos t − t sin t) + c2 (sin t + t cos t)

y !! = c1 (−2 sin t − t cos t) + c2 (2 cos t − t sin t) y !!! = c1 (−3 cos t + t sin t) + c2 (−3 sin t − t cos t) Substitution into the differential equation and simplifying gives (−2c1 − 2c2 ) cos t + (2c1 − 2c2 ) sin t = 4 cos t. Thus −2c1 − 2c2 = 4

2c1 − 2c2 = 0.

It follows that c1 = c2 = −1 and that y = −t(sin t + cos t) + c1 et + c2 cos t + c3 sin t is the general solution. 4 2 2 2 7. We have q(s) 0 2t 1 = s1 −5s +4 = (s −1)(s −4) = (s−1)(s+1)(s−2)(s+2)) and L e = s−2 . Let v(s) = s − 2. Then q(s)v(s) = (s − 1)(s + 1)(s − 0 1 0 1 2)2 (s + 2), Bq =0 et ,1e−t , e2t , e−2t and Bqv = et , e−t , e2t , te2t , e−2t . Thus Bqv \ Bq = te2t . It follows that y = cte2t is the test function and

y = cte2t

y ! = c(1 + 2t)e2t y !! = c(4 + 4t)e2t y !!! = c(12 + 8t)e2t y (4) = c(32 + 16t)e2t .

1 Solutions

141

Substituting into the differential equation and simplifying gives 12ce2t = 1 e2t . We thus get c = 1/12. It follows that y = 12 te2t + c1 et + c2 e−t + 2t −2t c3 e + c4 e 8. We use the principle of superposition and solve separately y (4) − y = et and y (4) − y = e−t . We have q(s) = s4 − 1 = (s2 + 1)(s − 1)(s + 1) and 1 L {et } = s−1 . Let v(s) = s − 1. Then q(s)v(s) = (s2 + 1)(s − 1)2 (s + 1), Bq = {cos t, sin t, et , e−t }, Bqv = {cos t, sin t, et , tet , e−t }, and Bqv \ Bq = {tet }. It follows that y = ctet is the test function and y = ctet y ! = c(1 + t)et y !! = c(2 + t)et y !!! = c(3 + t)et y (4) = c(4 + t)et . Substituting into the differential equation and simplifying gives 4cet = et which implies c = 1/4. It follows that y1 = 4t et is a particular solution for 1 y (4) − y = et . For y (4) − y = e−t we have L {e−t } = s+1 . Let v(s) = s + 1. 2 2 Then q(s)v(s) = (s + 1)(s − 1)(s + 1) , Bqv = {cos t, sin t, et , e−t , te−t }, and Bqv = {te−t }. It follows that y = cte−t is the test function and y = cte−t y ! = c(1 − t)e−t y !! = c(−2 + t)e−t y !!! = c(3 − t)e−t y (4) = c(−4 + t)e−t . Substituting into the differential equation and simplifying gives −4ce−t = −t e−t which implies c = −1/4. Thus y2 = −t 4 e . Let yp = y1 + y2 = t t −t 4 (e − e ). Then by the principle of superposition yp is a particular solution to the given differential equation. Thus y = 4t (et − e−t ) + c1 et + c2 e−t + c3 cos t + c4 sin t 1 9. We have q(s) = s3 − s = s(s − 1)(s + 1) and L {et } = s−1 . Thus Y (s) = 1 . One iteration of the partial fraction decomposition algorithm s(s+1)(s−1)2 gives

142

1 Solutions

Incomplete (s − 1)-chain 1 s(s + 1)(s − 1)2

1/2 (s − 1)2

p(s) s(s + 1)(s − 1) It follows that yp = y=

1 t 2 te

+ c1 e

−t

1 −1 2L t

4

1 (s−1)2

+ c2 e + c 3

5

=

1 t 2 te

and the general solution is

0 1 4 10. We have q(s) = s3 − 4s2 + 4s = s(s − 2)2 and L e2t = (s−2) 2 . Thus 4 Y (s) = s(s−2)4 . The partial fraction decomposition algorithm gives

Incomplete s − 2-chain 4 s(s − 2)4 −2 s(s − 2)3 p(s) s(s − 2)2

2 (s − 2)4 −1 (s − 2)3

It follows that yp = 13 t3 e2t − 12 t2 e2t and the general solution is y = 1 3 2t 1 2 2t t t 3 t e − 2 t e + c1 + c2 e + c3 te . 11. We have q(s) = s(s2 + 4) and L {t} = s12 . Thus Y (s) = partial fraction decomposition algorithm gives

1 s3 (s2 +4) .

The

1 Solutions

143

Incomplete s-chain 1 s3 (s2 + 4) −s/4 + 4)

s2 (s2

1/4 s3 0

−1/4 s(s2 + 4) It follows that yp = c3 sin 2t.

t2 8

and the general solution is y =

t2 8

+ c1 + c2 cos 2t +

4 2 2 2 12. We have q(s) 0 1= s −15s + 4 = (s − 1)(s − 4) = 1(s − 1)(s + 1)(s − 2)(s + 2) and L e2t = s−2 . Thus Y (s) = (s−1)(s+1)(s−2)2 (s+2) . The partial fraction decomposition algorithm gives

Incomplete s − 2-chain 1 (s − 1)(s + 1)(s − 2)2 (s + 2) p(s) (s − 1)(s + 1)(s − 2)(s + 2)

1/12 (s − 2)2

1 It follows that yp = 12 te2t and the general solution is y = c2 e−t + c3 e2t + c4 e−2t .

1 2t 12 te

+ c 1 et +

13. We have q(s) = s3 − s2 + s − 1 = (s − 1)(s2 + 1) and L {4 cos t} = s24s +1 . 4s Thus Y (s) = (s−1)(s 2 +1)2 . The partial fraction decomposition algorithm gives

Incomplete s2 + 1-chain 4s (s − 1)(s2 + 1)2 p(s) (s − 1)(s2 + 1)

−2s + 2 (s2 + 1)2

144

1 Solutions

It follows from Table 2.9 that yp = (−t sin t + sin t − t cos t). But since sin t is a homogeneous solution we can write the general solution as y = −t(sin t + cos t) + c1 et + c2 cos t + c2 sin t. 14. We use the principle of superposition and solve y (4) −y = et and y (4) −y = e−t separately and add the results. We have q(s) = s4 − 1 = (s − 1)(s + 1 1 1)(s2 + 1) and L {et } = s−1 . Thus Y1 (s) = (s−1)2 (s+1)(s 2 +1) . The partial fraction decomposition algorithm gives

Incomplete s − 1-chain 1 (s − 1)2 (s + 1)(s2 + 1) p(s) (s − 1)(s + 1)(s2 + 1)

1/4 (s − 1)2

1 It follows that y1 = 14 tet . On the other hand, L {e−t } = s+1 Thus Y2 (s) = 1 (s−1)(s+1)2 (s2 +1) . The partial fraction decomposition algorithm gives

Incomplete s + 1-chain 1 (s − 1)(s + 1)2 (s2 + 1)

−1/4 (s + 1)2

p(s) (s − 1)(s + 1)(s2 + 1)

−t It follows that y2 = −1 and a particular solution to the given dif4 te ferential equation is yp = y1 + y2 = 4t (et − e−t ). The general solution is y = 4t (et − e−t ) + c1 et + c2 e−t + c3 cos t + c4 sin t.

Section 4.4 1. Here L1 = D − 6 and L2 = D. It is easy to see that L = q(D), where q(s) = s2 − 6s + 8 =0(s − 2)(s 1 − 4). Therefore y1 and y2 are linear combinations of Bq = e2t , e4t . Next we recursively extend the initial values to derivatives of order 1 to get

1 Solutions

145

y1 (0) = 2 y1! (0) = 16

y2 (0) = −1 y2! (0) = 4

If y = c1 e2t + c2 e4t then c1 + c2 = y(0) 2c1 + 4c2 = y ! (0) For y1 we get c1 + c2 = 2 2c1 + 4c2 = 16 which gives c1 = −4 and c2 = 6. Thus y1 (t) = −4e2t + 6e4t . For y2 we get c1 + c2 = −1 2c1 + 4c2 = 4 which gives c1 = −4 and c2 = 3. Thus y2 (t) = −4e2t + 3e4t . 2. Here L1 = D − 3 and L2 = D + 1. It is easy to see that L = q(D), where q(s) = s2 − 2s + 1 = (s − 1)2 . Therefore y1 and y2 are linear combinations of Bq = {et , tet }. Next we recursively extend the initial values to derivatives of order 1 to get y1 (0) = 1 y1! (0) = −1

y2 (0) = 1 y2! (0) = 0

If y = c1 et + c2 tet then c1 = y(0) c1 + c2 = y ! (0) For y1 we get c1 = 1 c1 + c2 = −1

which gives c1 = 1 and c2 = −2. Thus y1 (t) = et − 2tet . For y2 we get c1 = 1 c1 + c2 = 0 which gives c1 = 1 and c2 = −1. Thus and y2 (t) = et − tet . 3. Here L1 = D and L2 = D. It is easy to see that L = q(D), where q(s) = s2 +4. Therefore y1 and y2 are linear combinations of Bq = {cos 2t, sin 2t}. Next we recursively extend the initial values to derivatives of order 1 to get y1 (0) = 1 y2 (0) = −1 y1! (0) = −2 y2! (0) = 2

146

1 Solutions

If y = c1 cos 2t + c2 sin 2t then c1

= y(0) 2c2 = y ! (0)

For y1 we get c1

= 1 2c2 = −2

which gives c1 = 1 and c2 = −1. Thus y1 (t) = cos 2t − sin 2t. For y2 we get c1 = −1 2c2 = 2 which gives c1 = −1 and c2 = 1. Thus and y2 (t) = − cos 2t + sin 2t. 4. Here L1 = D − 2 and L2 = D 2 + 2D + 1. It is easy to see that L = q(D), 2 where q(s) = (s − 2)(s2 + 2s + 1) − −40= (s + 2)(s − 1 1) . Therefore y1 and −2t t t y2 are linear combinations of Bq = e , e , te . Next we recursively extend the initial values to derivatives of order 2 to get y1 (0) = 3 y1! (0) = 6 y1!! (0) = 18

y2 (0) = 0 y2! (0) = 3 y2!! (0) = −12.

If y = c1 e−2t + c2 et + c3 tet then c1 + c2 = y(0) −2c1 + c2 + c3 = y ! (0) 4c1 + c2 + 2c3 = y !! (0). For y1 we get c1 + c 2 = 3 −2c1 + c2 + c3 = 6 4c1 + c2 + 2c3 = 18. which gives c1 = 1, c2 = 2, and c3 = 6. Thus y1 (t) = e−2t + 2et + 6tet . For y2 we get c1 + c2 = 0 −2c1 + c2 + c3 = 3 4c1 + c2 + 2c3 = −12. which gives c1 = −2, c2 = 2, and c3 = −3. Thus and y2 (t) = −2e−2t + 2et − 3tet

5. Here L1 = D +4 and L2 = D 2 −6D +23. It is easy to see that L = q(D), where q(s) = (s+4)(s2 −6s+23)−90 = (s2 −1)(s−2) =0(s+1)(s−1)(s−2). 1 Therefore y1 and y2 are linear combinations of Bq = e−t , et , e2t . Next we recursively extend the initial values to derivatives of order 2 to get

1 Solutions

147

y1 (0) = 0 y1! (0) = 20 y1!! (0) = −60

y2 (0) = 2 y2! (0) = 2 y2!! (0) = −34.

If y = c1 e−t + c2 et + c3 e2t then c1 + c2 + c3 = y(0) −c1 + c2 + 2c3 = y ! (0) c1 + c2 + 4c3 = y !! (0). For y1 we get c1 + c2 + c3 = 0 −c1 + c2 + 2c3 = 20 c1 + c2 + 4c3 = −60.

which gives c1 = −20, c2 = 40, and c3 = −20. Thus y1 (t) = −20e−t + 40et − 20e2t . For y2 we get c1 + c2 + c3 = 2 −c1 + c2 + 2c3 = 2 c1 + c2 + 4c3 = −34. which gives c1 = −6, c2 = 20, and c3 = −12. Thus and y2 (t) = −6e−t + 20et − 12e2t . 6. Here L1 = D − 2 and L2 = D 2 + D + 6. It is easy to see that L = q(D), where q(s) = (s − 2)(s2 + s + 6) − −8 = s3 − s2 + 4s − 4 = (s − 1)(s2 + 4). Therefore y1 and y2 are linear combinations of Bq = {et , cos 2t, sin 2t}. Next we recursively extend the initial values to derivatives of order 2 to get y1 (0) = 1 y2 (0) = 5 y1! (0) = −8 y2! (0) = 4 !! y1 (0) = −24 y2!! (0) = −30. If y = c1 et + c2 cos 2t + c3 sin 2t then

c 1 + c2 = y(0) c1 + 2c3 = y ! (0) c1 + −4c2 = y !! (0). For y1 we get c1 + c 2 = 1 c1 + 2c3 = −8 c1 + −4c2 = −24.

which gives c1 = −4, c2 = 5, and c3 = −2. Thus y1 (t) = −4et + 5 cos 2t − 2 sin 2t. For y2 we get

148

1 Solutions

c1 + c 2 = 5 c1 + 2c3 = 4 c1 + −4c2 = −30. which gives c1 = −2, c2 = 7, and c3 = 3. Thus and y2 (t) = −2et + 7 cos 2t + 3 sin 2t. 7. Here L1 = D 2 + 2D + 6 and L2 = D 2 − 2D + 6. It is easy to see that L = q(D), where q(s) = (s2 + 2s + 6)(s2 − 2s + 6) − 45 = s4 + 8s2 − 9 = (s2 − 1)(s2 + 9) = (s − 1)(s + 1)(s2 + 9). Therefore y1 and y2 are linear combinations of Bq = {et , e−t , cos 3t, sin 3t}. Next we recursively extend the initial values to derivatives of order 3 to get y1 (0) y1! (0) y1!! (0) y1!!! (0)

= = = =

0 0 30 −30

y2 (0) y2! (0) y2!! (0) y2!!! (0)

= = = =

6 6 −24 −84

If y = c1 et + c2 e−t + c3 cos 3t + c4 sin 3t then c1 c1 c1 c1

+ − + −

c2 + c3 c2 + 3c4 c2 − 9c3 c2 − 27c4

= y(0) = y ! (0) = y !! (0) = y !!! (0)

For y1 we get c1 c1 c1 c1

+ − + −

c2 + c3 c2 + 3c4 c2 − 9c3 c2 − 27c4

= 0 = 0 = 30 = −30

which gives c1 = 0, c2 = 3, c3 = −3, and c4 = 1. Thus y1 (t) = 3e−t − 3 cos 3t + sin 3t. For y2 we get c1 c1 c1 c1

+ − + −

c2 + c3 c2 + 3c4 c2 − 9c3 c2 − 27c4

= 6 = 6 = −24 = −84

which gives c1 = 0, c2 = 3, c3 = 3 and c4 = 3. Thus and y2 (t) = 3e−t + 3 cos 3t + 3 sin 3t. 8. Here L1 = D 2 +2 and L2 = D 2 +2D −9. It is easy to see that L = q(D), where q(s) = (s2 +2)(s2 +2s−9)+18 = s4 +2s3 −7s2 +4 0 = s(s−1)( s+4). 1 Therefore y1 and y2 are linear combinations of Bq = 1, et , tet , e−4t . Next we recursively extend the initial values to derivatives of order 3 to

1 Solutions

149

get y1 (0) y1! (0) y1!! (0) y1!!! (0)

= = = =

−1 −4 −1 2

y2 (0) y2! (0) y2!! (0) y2!!! (0)

= = = =

If y = c1 + c2 et + c3 tet + c4 e−4t then c1 + c2 c2 + c3 c2 + 2c3 c2 + 3c3

+ c4 − 4c4 + 16c4 − 64c4

1 2 −1 −4

= y(0) = y ! (0) = y !! (0) = y !!! (0)

For y1 we get c1 + c2 c2 + c3 c2 + 2c3 c2 + 3c3

+ c4 − 4c4 + 16c4 − 64c4

= −1 = −4 = −1 = 2

which gives c1 = 6, c2 = −7, c3 = 3, and c4 = 0. Thus y1 (t) = 6 − 7et + 3tet . For y2 we get c1 + c2 c2 + c 3 c2 + 2c3 c2 + 3c3

+ c4 − 4c4 + 16c4 − 64c4

= 1 = 2 = −1 = −4

which gives c1 = −4, c2 = 5, c3 = −3 and c4 = 0. Thus and y2 (t) = −4 + 5et − 3tet . 9. Here a = 2, b = 1, and c = 2 and the coupled system that describes the motion is given by y1!! + 3y1 = y2 y2!! + 2y2 = 2y1 . Let L1 = D 2 + 3 and L2 = D 2 + 2. Then y1 and y2 a solutions to q(D)y = 0, where q(s) = (s2 +3)(s2 +2)−2 = s4 +5s2 +4 = (s2 +1)(s2 +4). Thus y1 and y2 a linear combinations of Bq = {cos t, sin t, cos 2t, sin 2t}. Next we recursively extend the initial values to derivatives of order 3 to get y1 (0) = 3 y2 (0) = 0 y1! (0) = 3 y2! (0) = 0 !! y1 (0) = −9 y2!! (0) = 6 !!! y1 (0) = −9 y2!!! (0) = 6 If y = c1 cos t + c2 sin t + c3 cos 2t + c4 sin 2t then

150

1 Solutions

c1

+

c3

= y(0) + 2c4 = y ! (0) = y !! (0) − 8c4 = y !!! (0)

c2 −c1

− 4c3

−c2

For y1 we get c1

+

c3

= 3 + 2c4 = 3 = −9 − 8c4 = −9

c2 −c1

− 4c3

−c2

which gives c1 = 1, c2 = 1, c3 = 2, and c4 = 1. Thus y1 (t) = cos t + sin t + 2 cos 2t + sin 2t. For y2 we get c1

+

c3

c2 −c1

− 4c3

−c2

= + 2c4 = = − 8c4 =

0 0 6 6

which gives c1 = 2, c2 = 2, c3 = −2 and c4 = −1. Thus y2 (t) = 2 cos t + 2 sin t − 2 cos 2t − sin 2t. 10. Here a = 2, b = 3, and c = 4 and the coupled system that describes the motion is given by y1!! + 5y1 = 3y2 y2!! + 4y2 = 4y1 . Let L1 = D 2 +5 and L2 = D 2 +4. Then y1 and y2 a solutions to q(D)y = 2 0, where q(s) = (s2 +5)(s2 +4)−12 = s40+9s2 +8 = (s2 +1)(s +8). y1 √ √ Thus 1 and y2 a linear combinations of Bq = cos t, sin t, cos 8t, sin 8t . Next we recursively extend the initial values to derivatives of order 3 to get y1 (0) y1! (0) y1!! (0) y1!!! (0)

= = = =

1 0 13 0

y2 (0) y2! (0) y2!! (0) y2!!! (0)

√

8t + c4 sin

If y = c1 cos t + c2 sin t + c3 cos c1

+

c3

c2 −c1 For y1 we get

−c2

+ − 8c3

√

√

= = = =

6 0 −20 0

8t then

8c4

√ − 8 8c4

= y(0) = y ! (0) = y !! (0) = y !!! (0)

1 Solutions

151

c1

+

c3

c2 −c1

−c2

+ − 8c3

√

8c4

√ − 8 8c4

= 1 = 0 = 13 = 0

which√gives c1 = 3, c2 = 0, c3 = −2, and c4 = 0. Thus y1 (t) = 3 cos t − 2 cos 8t. For y2 we get c1

+

c3

c2 −c1

−c2

+ − 8c3

√

8c4

√ − 8 8c4

= 6 = 0 = −20 = 0

which√gives c1 = 4, c2 = 0, c3 = 2 and c4 = 0. Thus y2 (t) = 4 cos t + 2 cos 8t. 11. 1. We begin by taking the Laplace transform of each equation above to get q1 (s)Y1 (s) − p1 (s) = λ1 Y2 (s) q2 (s)Y2 (s) − p2 (s) = λ2 Y1 (s) which can be rewritten: q1 (s)Y1 (s) − λ1 Y2 (s) = p1 (s)

q2 (s)Y2 (s) − λ2 Y1 (s) = p2 (s). In matrix form this becomes ' (' ( ' ( q1 (s) −λ1 Y1 (s) p1 (s) = −λ2 q2 (s) Y2 (s) p2 (s) 2. The inverse of the coefficient matrix is ' (−1 ' ( 1 q1 (s) −λ1 q2 (s) λ1 = −λ2 q2 (s) λ2 q1 (s) q1 (s)q2 (s) − λ1 λ2 and therefore ' ( ' (' ( 1 Y1 (s) q2 (s) λ1 p1 (s) = Y2 (s) λ2 q1 (s) p2 (s) q1 (s)q2 (s) − λ1 λ2 ' ( 1 p1 (s)q2 (s) + λ1 p2 (s) = . q1 (s)q2 (s) − λ1 λ2 p2 (s)q1 (s) + λ2 p1 (s) 12. We first take the Laplace transform of each equation to get

152

1 Solutions

sY1 (s) − 1 = −Y2 (s)

sY2 (s) − (−1) − 2Y2 (s) = Y1 (s). We associate the Y1 and Y2 . In matrix form we get ' (' ( ' ( s 1 Y1 (s) 1 = . −1 s − 2 Y2 (s) −1 By matrix inversion we get '

Y1 (s) Y2 (s)

(

(−1 ' ( s 1 1 −1 s − 2 −1 ' (' ( 1 s − 2 −1 1 = 1 s −1 (s − 1)2 ' ( ' 1 ( 1 s−1 s−1 = = −1 (s − 1)2 1 − s (s−1) =

'

We now get by Laplace inversion ' ( ' t( y1 (t) e = . y2 (t) −et 13. We first take the Laplace transform of each equation to get sY1 (s) − 2 − Y1 = −2Y2 (s) sY2 (s) − (−2) − Y2 (s) = 2Y1 (s). We associate the Y1 and Y2 . In matrix form we get ' (' ( ' ( s−1 2 Y1 (s) 2 = . −2 s − 1 Y2 (s) −2 By matrix inversion we get '

Y1 (s) Y2 (s)

(

(−1 ' ( s−1 2 2 −2 s − 1 −2 ' (' ( 1 s−1 −2 2 = 2 s−1 −2 (s − 1)2 + 4 ' ( , 2(s−1)+4 1 2s + 2 (s−1)2 +22 = = −2(s−1)+4 2 −2s + 6 (s − 1) + 4 (s−1)2 +22 =

'

We now get by Laplace inversion

1 Solutions

153

' ( ' ( y1 (t) 2et cos 2t + 2et sin 2t = . y2 (t) −2et cos 2t + 2et sin 2t 14. We first take the Laplace transform of each equation to get sY1 (s) − 2Y1 = −Y2 (s) s2 Y2 (s) + s − 2 − (sY2 (s) + 1) + Y2 (s) = Y1 (s). We associate the Y1 and Y2 . In matrix form we get ' (' ( ' ( s−2 1 Y1 (s) 0 = . −1 s2 − s + 1 Y2 (s) −s + 3 By matrix inversion we get '

Y1 (s) Y2 (s)

(

' (−1 ' ( s−2 1 0 = −1 s2 − s + 1 −s + 3 ' 2 (' ( 1 s −s+1 −1 0 = 1 s−2 −s + 3 (s − 1)3 ' ( , 1 2 1 s−3 (s−1)2 − (s−1)3 = = −1 3 2 (s − 1)3 −s2 + 5s − 6 s−1 + (s−1)2 − (s−1)3

We now get by Laplace inversion ' ( ' ( y1 (t) tet − t2 et = . y2 (t) −et + 3tet − t2 et 15. We first take the Laplace transform of each equation to get 2

sY1 (s) − 1 + 2Y1 = 5Y2 (s)

s Y2 (s) − 3 − 2(sY2 (s)) + 5Y2 (s) = 2Y1 (s). We associate the Y1 and Y2 . In matrix form we get ' (' ( ' ( s+2 −5 Y1 (s) 1 = . −2 s2 − 2s + 5 Y2 (s) 3 By matrix inversion we get

154

1 Solutions

'

Y1 (s) Y2 (s)

(

' (−1 ' ( s+2 −5 1 = −2 s2 − 2s + 5 3 ' 2 (' ( 1 s − 2s + 5 5 1 = 3 2 s+2 3 s +s ' 2 ( ' 20 ( − 19 s2s+1 − 2 s21+1 1 s − 2s + 20 s = = 8 s 1 3s + 8 s(s2 + 1) s − 8 s2 +1 + 3 s2 +1

We now get by Laplace inversion ' ( ' ( y1 (t) 20 − 19 cos t − 2 sin t = . y2 (t) 8 − 8 cos t + 3 sin t 16. We first take the Laplace transform of each equation to get s2 Y1 (s) − 10s + 2Y1 = −3Y2 (s) s2 Y2 (s) − 10s + 2(sY2 (s) − 10) − 9Y2 (s) = 6Y1 (s). We associate the Y1 and Y2 . In matrix form we get ' 2 (' ( ' ( s +2 3 Y1 (s) 10s = . −6 s2 + 2s − 9 Y2 (s) 10s + 20 By matrix inversion we get '

Y1 (s) Y2 (s)

(

' 2 (−1 ' ( s +2 3 10s −6 s2 + 2s − 9 10s + 20 ' 2 (' ( 1 s + 2s − 9 −3 10s = 4 6 s2 + 2 10s + 20 s + 2s3 − 7s2 + 4s ' 3 ( 2 1 10s + 20s − 120s − 60 = s(s − 1)2 (s + 4) 10s3 + 20s2 + 80s + 40 , 26 30 1 − 15 s + s−1 − (s−1)2 − s+4 = 10 6 30 6 s − s−1 + (s−1)2 + s+4 =

We now get by Laplace inversion ' ( ' ( y1 (t) −15 + 26et − 30tet − e−4t = . y2 (t) 10 − 6et + 30tet + 6e−4t

1 Solutions

155

Section 4.5 1. The only characteristic mode is e−5t . Thus the zero-input response is y(t) = ce−5t . The initial condition a = y(0) = 10 implies c = 10. Thus y(t) = 10e−5t , The characteristic value is −5, to the left of the imaginary axis. Hence the system is stable. 2. The only characteristic mode is e2t . Thus y(t) = ce2t . The initial condition a = y(0) = 2 implies c = 2. Thus the zero-input response is y(t) = 2e2t , The characteristic value is 2, to the right of the imaginary axis. Hence the system is unstable. 2 3. The characteristic polynomial 3)(s − 1). The 0 t 3tis1 q(s) = s − 4s + 3 t= (s − 3t characteristic modes is e , e . Thus y(t) = c1 e + c2 e . The initial condition a = (2, 4) = (y(0), y ! (0)) implies c1 = 1 and c2 = 1. Thus the zero-input response is y(t) = et +e3t . The characteristic value are 1, 3 and both lie to the right of the imaginary axis. Hence the system is unstable. 2 4. The characteristic polynomial q(s) + 1)(s + 4). The 0 −t is −4t 1 = s + 5s + 4 = (s−t characteristic modes are e , e . Thus y(t) = c1 e + c2 e−4t . The initial condition a = (0, 3) = (y(0), y ! (0)) implies c1 = 1 and c2 = −1. Thus the zero-input response is y(t) = e−t − e−4t . The characteristic value are −1, −4 and both lie to the left of the imaginary axis. Hence the system is stable.

5. The characteristic polynomial = s2 +14s + 5 = (s + 2)2 + 1. The 0 −2t is q(s)−2t characteristic modes are e cos t, e sin t . Thus y(t) = c1 e−2t cos t+ −2t c2 e sin t. The initial condition a = (0, 1) = (y(0), y ! (0)) implies c1 = 0 and c2 = 1. Thus the zero-input response is y(t) = e−2t sin t. The characteristic value are −2 + i, −2 − i and both lie to the left of the imaginary axis. Hence the system is stable. 6. The characteristic polynomial is q(s) = s2 + 9. The characteristic modes are {cos 3t, sin 3t}. Thus y(t) = c1 cos 3t + c2 sin 3t. The initial condition a = (1, 1) = (y(0), y ! (0)) implies c1 = 1 and c2 = 1/3. Thus the zeroinput response is y(t) = cos 3t + 13 sin 3t. The characteristic value are ±3i and lie on the imaginary axis. Hence the system is marginally stable. 2 2 7. The characteristic polynomial is q(s) 0 1 = s + 6s + 9 = (s + 3) . The characteristic modes are e−3t , te−3t . Thus y(t) = c1 e−3t + c2 te−3t . The initial condition a = (1, 1) = (y(0), y ! (0)) implies c1 = 1 and c2 = 4. Thus the zero-input response is y(t) = e−3t + 4te−3t . The characteristic value is −3 with multiplicity 2 lies to the left of the imaginary axis. Hence the system is stable. 2 8. The characteristic polynomial (s + 2)(s − 1). The 0 −2t ist q(s) 1 =s +s−2= characteristic modes are e , e . Thus y(t) = c1 e−2t + c2 et . The initial

156

1 Solutions

condition a = (1, −2) = (y(0), y ! (0)) implies c1 = 1 and c2 = 0. Thus the zero-input response is y(t) = e−2t . The characteristic value are {−2, 1}. Since one of the characteristic values lies to the right of the imaginary axis the system is unstable. 9. The characteristic polynomial is q(s) = s2 − 2s + 2 = (s − 1)2 + 1. The characteristic modes are {et cos t, et sin t}. Thus y(t) = c1 et cos t + c2 et sin t. The initial condition a = (1, 2) = (y(0), y ! (0)) implies c1 = 1 and c2 = 1. Thus the zero-input response is y(t) = et cos t + et sin t. The characteristic values are {1 + i, 1 − i} and lie to the right of the imaginary axis. Hence the system is unstable. 10. The characteristic polynomial is q(s) = s3 + s2 = s2 (s + 1). The characteristic modes are {1, t, e−t }. Thus y(t) = c1 + c2 t + c3 e−t . The initial condition a = (1, −1, 1) = (y(0), y ! (0), y !! (0)) implies c1 = 0, c2 = 0, and c3 = 1. Thus the zero-input response is y(t) = e−t . The characteristic value are 0 with multiplicity 2 and −1 with multiplicity 1. Since 0 lies on the imaginary axis and has multiplicity 2 the system is unstable. 11. The characteristic polynomial is q(s) = (s+1)(s2 +1). The characteristic modes are {e−t , cos t, sin t}. Thus y(t) = c1 e−t + c2 cos t+c3 sin t. The initial condition a = (1, −1, 1) = (y(0), y ! (0), y !! (0)) implies c1 = 1, c2 = 0, and c3 = 0. Thus the zero-input response is y(t) = e−t . The characteristic value are −1, i, and −i. The system is then marginally stable. 12. The characteristic polynomial is q(s) = s4 − 1 = (s − 1)(s + 1)(s2 + 1). The characteristic modes are {et , e−t , cos t, sin t}. Thus y(t) = c1 et + c2 e−t + c3 cos t + c4 sin t. The initial condition a = (0, 1, 0, −1) = (y(0), y ! (0), y !! (0), y !!! (0)) implies c1 = 0, c2 = 0, c3 = 0, and c4 = 1. Thus the zero-input response is y(t) = sin t. The characteristic values are 1, −1, i, −i. Since one of the characteristic values lies to the right of the imaginary axis the system is unstable. 13. The characteristic mode is e−t so y(t) = ce−t . For the unit impulse we have y(0) = 1 and this implies c = 1. Thus y(t) = e−t 14. The characteristic polynomial is q(s) = s2 + 4 and hence the characteristic modes are {cos 2t, sin 2t}. Hence, y(t) = c1 cos 2t + c2 sin 2t. For the unit impulse we have y(0) = 0 and y ! (0) = 1 and this implies c1 = 0 and c2 = 12 . Thus y(t) = 12 sin 2t. 15. The characteristic polynomial0is q(s) =1s2 − 4 = (s − 2)(s + 2) and hence the characteristic modes are e2t , e−2t . Hence, y(t) = c1 e2t + c2 e−2t . For the unit impulse we have y(0) = 0 and y ! (0) = 1 and this implies c1 = 1/4 and c2 = −1/4 . Thus y(t) = 14 e2t − 14 e−2t .

1 Solutions

157

16. The characteristic polynomial is q(s) = s2 + 2s + 5 = (s + 1)2 + 4. The characteristic modes are {e−t cos 2t, e−t sin 2t}. For the unit impulse we have y(0) = 0 and y ! (0) = 1 and this implies c1 = 0 and c2 = 1/2 . Thus y(t) = 12 e−t sin 2t. 17. The characteristic polynomial is q(s) = s3 +s = s(s2 +1). The characteristic modes are {1, cos 2t, sin 2t}. For the unit impulse we have y(0) = 0, y ! (0) = 0, and y !! (0) = 1 and this implies c1 = 1, c2 = −1 and c3 = 0. Thus y(t) = 1 − cos(t). 19. Since f is bounded there is an M such that |f (t)| ≤ M for all t ≥ 0. We then have $* t $ $ k αt $ $ $ k λx $t e cos βt ∗ f (t)$ = $ x e cos βxf (t − x) dx$$ $ *

0 t

xk eαx |f (t − x)| dx * t ≤ M xk eαx dx ≤

0

0

= M (C + p(t)eαt ),

where C and p(t) are as in Exercise 18, which also implies that tk eαt cos βt∗ f is bounded. The argument for tk eαt sin βt ∗ f is the same. 20. Since the system is asymptotically stable each characteristic mode is of the form tk eαt cos βt or tk eαt sin βt, with α negative. The unit impulse function h is a linear combination of such modes and the output function for an input function f is h ∗ f . By linearity and Exercise 19 h ∗ f is a bounded function if f is bounded. There the system is BIBO-stable.

Section 5.1 1. No, it is not linear because of the presence of the product y ! y. 2. yes, homogeneous, yes 3. yes, nonhomogeneous, yes 4. no, it is not linear because of the presence of y 2 . 5. yes, nonhomogeneous, no 6. No, it is not linear because of the presence of 7. yes, nonhomogeneous, no

√

y! .

158

1 Solutions

8. This can be written y !! + (−2 − t)y = 0. So it is linear but not constant coefficient. It is homogeneous. 9. No, it is not linear because of the presence of sin y. 10. yes, homogeneous, no 11. yes, homogeneous, no 12. 1. L(1) = t(0) + 1 = 1 2. L(t) = t(0) + t = t 3. L(e−t ) = te−t + e−t = (t + 1)e−t 4. L(cos 2t) = t(−4 cos 2t) + cos 2t = (−4t + 1) cos 2t 13. 1. L( 1t ) = t2 (2t−3 ) + t(−t−2 ) − t−1 = (2 − 1 − 1)t−1 = 0 2. L(1) = t2 (0) + t(0) − 1 = −1 3. L(t) = t2 (0) + t(1) − t = 0 4. L(tr ) = t2 r(r − 1)tr−2 + t(rtr−1 ) − tr = (r2 − 1)tr 14. yp! = 12 Ct

−1 2

and yp!! =

−3 −1 2 . 4 Ct

Thus

( ' ( 1 −1 −3 1 −1 Ct 2 + t Ct 2 − Ct 2 4 2 ' ( 1 −1 1 = C + − 1 t2 4 2 −3 1 = Ct 2 4

t2 yp!! + typ! − yp = t2

The equation

1 −3 2 4 Ct

'

1

= t 2 implies C =

−4 3 .

15. yp! = C(2t − t2 )e−t and yp!! = C(t2 − 4t + 2)e−t . Thus t2 yp!! + typ! − yp = C(t3 − 4t2 + 2t)e−t + C(−t3 + 3t2 − 2t)e−t + C(−t2 )e−t = C(−2t2 )e−t

The equation C(−2t2 )e−t = t2 e−t implies C =

−1 2 .

16. If y = t then y ! = 1 and y !! = 0 so that Ly = (1 + t2 )(0) − 4t(1) + 6t = 2t. Parts (1) follows. If y = 1 − 3t2 then y ! = −6t and y !! = −6 so that Ly = (1 + t2 )(−6) − 4t(−6t) + 6(1 − 3t2 ) = 0. It follows that y = 1 − 3t2 is 3 a solution to Ly = 0. The case where y = t − t3 is similar. Part (2) now 2 follows. By " # linearity every function of the form y(t) = t + c1 (1 − 3t ) + 3 c2 t − t3 is a solution to Ly = 2t, where c1 and c2 are constants. If we

want a solution to L(y) = 2t with y(0) = a and y ! (0) =" b, then # we need t3 2 to solve for c1 and c2 : Since y(t) = t + c1 (1 − 3t ) + c2 t − 3 we have

1 Solutions

159

y ! (t) = 1 + c1 (−6t) + c2 (1 − t2 ). Hence, a = y(0) = c1 b = y ! (0) = 1 + c2 These equations give c1 = a, c2 = b − 1. Particular choices of a and b give the answers for Part (3). " (3)a. y(t) = t + (1 − 3t2 ) − t −

t3 3

#

3

= 1 − 3t2 + t3 " # 3 (3)b. y(t) = t + (0)(1 − 3t2 ) + (0) t − t3 = t " # 3 (3)c. y(t) = t − 1(1 − 3t2 ) + 3 t − t3 = −1 + 4t + 3t2 − t3 " # 3 (3)d. y(t) = t + a(1 − 3t2 ) + (b − 1) t − t3 17. If y = e−t then y ! = −e−t and y !! = e−t so that Ly = (t − 1)e−t − t(−e−t ) + e−t = 2te−t . Parts (1) follows. If y = et then Ly = (t − 1)(et ) − t(et )+(et ) = 0. It follows that y = et is a solution to Ly = 0. If y = t then y ! = 1 and y !! = 0. Thus Ly = (t−1)(0)−t(1)+t = 0. Part (2) now follows. By linearity every function of the form y(t) = e−t +c1 et +c2 t is a solution to Ly = 2te−t , where c1 and c2 are constants. If we want a solution to L(y) = 2te−t with y(0) = a and y ! (0) = b, then we need to solve for c1 and c2 : Since y(t) = e−t + c1 et + c2 t we have y ! (t) = −e−t + c1 et + c2 . Hence, a = y(0) = 1 + c1 b = y ! (0) = −1 + c1 + c2 . These equations give c1 = a − 1 and c2 = b − a + 2. Particular choices of a and b give the answers for Part (3). (3)a. y(t) = e−t − et + 2t

(3)b. y(t) = e−t + (0)et + (1)t = e−t + t (3)c. y(t) = e−t + −et + 3t

(3)d. y(t) = e−t + (a − 1)et + (b − a + 2)t % & 3 2 20 3 18. If %y =& 16 t5 %then& y ! = 56 t4 and y !! = 20 − 6 t so that Ly = t 6 t 4t 56 t4 + 6 16 t5 = t5 . Parts (1) follows. If y = t2 then Ly = t2 (2) − 4t(2t) + 6(t2 ) = 0. It follows that y = t2 is a solution to Ly = 0. If y = t3 then Ly = t2 (6t) − 4t(3t2 ) + 6t3 = 0. Part (2) now follows. By linearity every function of the form y(t) = 16 t5 +c1 t2 +c2 t3 is a solution to Ly = t5 , where c1 and c2 are constants. If we want a solution to L(y) = t5

160

1 Solutions

with y(0) = a and y ! (0) = b, then we need to solve for c1 and c2 : Since y(t) = 16 t5 c1 t2 + c2 t3 we have y ! (t) = 56 t5 + 2c1 t + 3c2 t2 . Hence, 1 + c1 + c2 6 5 b = y ! (1) = + 2c1 + 3c2 . 6

a = y(1) =

These equations give c1 = 3a − b + 13 and c2 = b − 2a − 12 . Particular choices of a and b give the answers for Part (3). (3)a. y = 16 t5 + (3)b. y = 16 t5 − (3)c. y = 16 t5 −

(3)d. y = 16 t5 +

10 2 5 3 3 t − 2t 2 2 1 3 3t + 2t 17 2 t + 92 t3 %31 & 2 3 + 3a − b t

% & + − 12 − 2a + b t3

19. Write the equation in the standard form:

3 1 y !! + y ! − 2 y = t2 . t t The forcing function is continuous on R while the coefficient functions, 3 1 t and − t2 , are continuous except at t = 0. Thus the largest intervals of common continuity are (0, ∞) and (−∞, 0). Since the initial conditions are given at t0 = −1 it follows from Theorem 6 that the interval (−∞, 0) is the largest interval with a unique solution. 20. This differential equation is in standard form with forcing function f (t) = 1+t2 1−t2 , which is continuous except at t = ±1. The intervals continuity are therefore (−∞, −1), (−1, 1), (1, ∞). Since t0 = 2 it follows that the maximal interval for a unique solution is (1, ∞). 21. Write the equation in the standard form: y !! +

y cos t = . sin t sin t

The intervals of continuity are of the form (kπ, (k + 1)π), k ∈ Z. Since t0 = π2 it follows that the maximal interval for a unique solution is (0, π). 22. Write the equation is standard form: y !! −

t t2 cos t ! y + = . 1 + t2 1 + t2 1 + t2

Since the coefficient functions and the forcing function is continuous on R the maximal interval for a unique solution is (−∞, ∞)

1 Solutions

161

23. The common interval of continuity of the coefficient functions is (3, ∞) and t0 = 10 is in this interval. 24. Write the equation is standard form: y !! +

1 et y = . t(t2 − 4) t(t2 − 4)

The intervals of continuity are (−∞, −2), (−2, 0), (0 2), (2, ∞). Since t0 = 1 the maximal interval for a unique solution is (0, 2). 25. The initial condition occurs at t = 0 which is precisely where a2 (t) = t2 has a zero. Theorem 6 does not apply. 26. In this case y(t0 ) = 0 and y ! (t0 ) = 0. The function y(t) = 0, t ∈ I is a solution to the initial value problem. By the uniqueness part of Theorem 6 y = 0 is the only solution. 27. The assumptions say that y1 (t0 ) = y2 (t0 ) and y1! (t0 ) = y2! (t0 ). Both y1 and y2 therefore satisfies the same initial conditions. By the uniqueness part of Theorem 6 y1 = y2 .

Section 5.2 1. dependent; 2t and 5t are multiples of each other. 2. independent; If c1 y1 + c2 y2 = 0 then evaluating at t = 0 gives c1 + c2 = 0 and evaluating at t = 1 gives 2c1 + 5c2 = 0. These two equations imply that c1 = 0 and c2 = 0. Hence, {2t , 5t } is linearly independent. 3. independent; If c1 ln t + c2 t ln t = 0 then evaluating at t = e and t = e2 gives c1 + ec2 = 0 and 2c1 + 2e2 c2 = 0. These equations imply that c1 and c2 are both zero so {ln t, t ln t} is linearly independent. 4. dependent; e2t+1 = e1 e2t and e2t−3 = e−3 e2t , they are multiples of each other. 5. independent, If c1 ln 2t + c2 ln 5t = 0 then evaluating at t = 1 and t = e gives (ln 2)c1 + (ln 5)c2 = 0 and (1 + ln 2)c1 + (ln 5 + 1)c2 = 0. These equations imply that c1 and c2 are both zero so {ln t, t ln t} is linearly independent. 6. dependent; ln t2 = 2 ln t and ln t5 = 5 ln t, they are multiples of each other.

162

1 Solutions

7. f1! (t) = et − 1 and f2!! (t) = et . Thus (t − 1)f1!! − tf1! + f1 = (t − 1)(et ) − t(et − 1) + et − t = 0. Similarly, f2! (t) = 1 and f2!! (t) = 0. Thus (t − 1)f2!! − tf2! + f2 = −t(1) + t = 0. Now, $ t $ $ e − t t$ $ $ = et − t − (et − 1)t = (1 − t)et . w(t) = $ t e − 1 1$

On the other hand the coefficient function of y ! in the standard form of t 1 the differential equation is a1 (t) = − t−1 = −1 − t−1 Integrating gives )t 1 t −1 − x−1 dx = −x − ln |x − 1| |0 = −t + ln(1 − t), (since x − 1 < 0) and 0 !t

e− 0 a1 (x) dx = et (1 − t). At t = 0 we have w(1) = 1 so Abel’s formula is verified. It follows from Proposition 4 that f1 and f2 are linearly independent. By Theorem 2 the solution set is {c1 (et − t) + c2 t : c1 , c2 ∈ R} 8. f1! (t) = −2t and f2!! (t) = −2. Thus (1 + t2 )f1!! − 2tf1! + 2f1 = (1 + t2 )(−2) − 2t(−2t) + 2(1 − t2 ) = 0. Similarly, f2! (t) = 1 and f2!! (t) = 0. Thus (1 + t2 )f2!! − 2tf2! + 2f2 = −2t(1) + 2t = 0. Now, $ $ $1 − t 2 t $ $ $ = 1 − t2 + 2t2 = 1 + t2 . w(t) = $ −2t 1$

On the other hand the coefficient function of y ! in the standard ) t −2xform of −2t the differential equation is a1 (t) = 1+t dx = 2 Integrating gives 0 1+x2 !t

− ln(1 + x2 )|t0 = − ln(1 + t2 ) and e− 0 a1 (x) dx = 1 + t2 . At t = 0 we have w(1) = 1 so Abel’s formula is verified. It follows from Proposition 4 that By Theorem 2 the solution set 0 f1 and f2 are linearly independent. 1 is c1 (1 + t2 ) + c2 t : c1 , c2 ∈ R

ln t) cos(2 ln t) ln t) 9. f1! (t) = −2 sin(2 , f1!! (t) = 2 sin(2 ln t)−4 , f2! (t) = 2 cos(2 , and t t2 t −4 sin(2 ln t)−2 cos(2 ln t) !! 2 !! ! f2 (t) = . Thus t f1 + tf1 + 4f1 = 2 sin(2 ln t) − t2 4 cos(2 ln t) − 2 sin(2 ln t) + 4 cos(2 ln t) = 0. Similarly, t2 f2!! + tf2! + 4f2 = −4 sin(2 ln t) − 2 cos(2 ln t) + 2 cos(2 ln t) + 4 sin(2 ln t) = 0. Now,

$ $ $ cos(2 ln t) sin(2 ln t)$ 4 $ $ w(t) = $ −2 sin(2 ln t) 2 cos(2 ln t) $ = . $ $ t t t

On the other hand the coefficient function of y ! in the)standard form of t the differential equation is a1 (t) = 1t Integrating gives 1 x1 dx = ln t and !t e− 0 a1 (x) dx = 1/t. At t = 1 we have w(1) = 1 so Abel’s formula is verified. It follows from Proposition 4 that f1 and f2 are linearly independent. By Theorem 2 the solution set is {c1 cos(2 ln t) + c2 sin(2 ln t) : c1 , c2 ∈ R} 10. Suppose f1 and f2 are solutions to

1 Solutions

163

y !! + a1 (t)y ! + a0 (t)y = 0,

(1)

where a0 and a1 are continuous functions on an interval I. Let t0 ∈ I. First observe that since f1 is a solution to Equation (1) we have f1!! = −a1 f1! − a0 f1 and similarly for f2 . To simplify the notation let w = w(f1 , f2 )(t) = f1 f2! − f2 f1! . Then the product rule gives w! = f1! f2! + f1 f2!! − (f2! f1! + f2 f1!! )

= f1 f2!! − f2 f1!! = f1 (−a1 f2! − a2 f2 ) − f2 (−a1 f1! − a2 f1 ) = −a1 (f1 f2! − f2 f1! ) = −a1 w.

Therefore w satisfies the differential equation w! + a1 w = 0. By Corollary 1.4.8 there is a constant K so that w(t) = Ke

!t

t0

−a1 (x) dx

.

Evaluating at t = t0 gives w(t0 ) = K. $ $ 11. 1. Suppose at3 + b $t3 $ = 0 on (−∞, ∞). Then for t = 1 and t = −1 we get a+b = 0

−a + b = 0.

2.

3.

4. 5.

These equations imply a = b = 0. So y1 and y2 are linearly independent. / −3t2 if t < 0 Observe that y1! (t) = 3t2 and y2! (t) = If t < 0 2 3t if t ≥ 0. ' 3 ( t −t3 then w(y1 , y2 )(t) = = 0. If t ≥ 0 then w(y1 , y2 )(t) = 2 3t −3t2 ' 3 ( 3 t t = 0. It follows that the Wronskian is zero for all t ∈ 3t2 3t2 (−∞, ∞). The condition that the coefficient function a2 (t) be nonzero in Theorem 2 and Proposition 4 is essential. Here the coefficient function, t2 , of y !! is zero at t = 0, so Proposition 4 does not apply on (−∞, ∞). The largest open intervals on which t2 is nonzero are (−∞, 0) and (0, ∞). On each of these intervals y1 and y2 are linearly dependent. Consider the cases t < 0 and t ≥ 0. The verification is then straightforward. Again the condition that the coefficient function a2 (t) be nonzero is essential. The Uniqueness and Existence theorem does not apply.

164

1 Solutions

Section 5.3 1. The indicial polynomial is Q(s) = s2 + s − 2 = (s +02)(s −11). There are two distinct roots 1 and −2. The fundamental set is t, t−2 . The general solution is y(t) = c1 t + c2 t−2 . 2. The indicial polynomial is Q(s) = 2s2 −7s+3 = (2s−1)(s−3). There are 4 1 5 1 3 2 two distinct roots 2 and 3. The fundamental set is t , t . The general 1

solution is y(t) = c1 t 2 + c2 t3 .

3. The indicial polynomial is Q(s) = 9s2 − 6s + 1 = (3s −41)2 . There5is one 1 1 root, 1/3, with multiplicity 2. The fundamental set is t 3 , t 3 ln t . The 1

1

general solution is y(t) = c1 t 3 + c2 t 3 ln t.

√ √ 4. The indicial polynomial is Q(s) = s2 − 2 = (s − 2)(s +4 2). There √ √ 5 √ √ are two distinct roots 2 and − 2. The fundamental set is t 2 , t− 2 . The general solution is y(t) = c1 t

√ 2

√

+ c 2 t−

2

.

2 5. The indicial polynomial is Q(s) = 4s2 − 4s + 14= (2s − 1) 5 . The root is 1 1 1 2 2 2 with multiplicity 2. The fundamental set is t , t ln t . The general 1

1

solution is y(t) = c1 t 2 + c2 t 2 ln t.

6. The indicial polynomial is Q(s) = s2 − 4s 0 − 21 = 1 (s − 7)(s + 3). The roots are 7 and −3. The fundamental set is t7 , t−3 . The general solution is y(t) = c1 t7 + c2 t−3 . 7. The indicial polynomial is Q(s) = s2 + 6s + 90= (s + 3)2 . 1 The root is −3 with multiplicity 2. The fundamental set is t−3 , t−3 ln t . The general solution is y(t) = c1 t−3 + c2 t−3 ln t. 8. The indicial polynomial is Q(s) = s2 − s + 1 = s2 − s + 1/4 + 3/4√= √ √ 1+i 3 (s − 1/2)2 + ( 3/2)2 . There are two complex roots, and 1−i2 3 . 2 4 5 1

The fundamental set is 1 2

y(t) = c1 t sin

√ 3 2 t

t 2 sin

1 2

+ c2 t cos

√ 1 3 2 2 t, t

√ 3 2 t.

cos

√ 3 2 t

. The general solution is

9. The indicial polynomial is Q(s) = s2 − 4 = (s − 2)(s 0 + 2).1There are two distinct roots, 2 and −2. The fundamental set is t2 , t−2 . The general solution is y(t) = c1 t2 + c2 t−2 . 10. The indicial polynomial is Q(s) = s2 + 4. There are two complex roots, 2i and −2i. The fundamental set is {cos(2 ln t), sin(2 ln t)}. The general solution is y(t) = c1 cos(2 ln t) + c2 sin(2 ln t).

1 Solutions

165

11. The indicial polynomial is Q(s) = s2 − 4s + 13 = (s − 2)2 + 9. There are roots, 12 + 3i and 2 − 3i. The fundamental set is 0 2 two complex t cos(3 ln t), t2 sin(3 ln t) . The general solution is y(t) = c1 t2 cos(3 ln t)+ c2 t2 sin(3 ln t). 12. The indicial polynomial is Q(s) = s2 + s − 2 = (s + 2)(s − 1). There are 0 1 two distinct roots, r1 = −2 and r2 = 1. The fundamental set is t−2 , t . The general solution is y(t) = c1 t−2 + c2 t. The initial conditions imply c1 + c 2 = 0 −2c1 + c2 = 1. Thus c1 = −1/3 and c2 = 1/3. Hence y =

1 (t − t−2 ) 3

13. The indicial polynomial is Q(s) = 4s2 − 4s + 1 4 = (2s − 1)(2s 5 − 1). There 1

1

is a double root, r = 12 . The fundamental set is t 2 , t 2 ln t . The general 1

1

solution is y(t) = c1 t 2 + c2 t 2 ln t. The initial conditions imply c1 = 2 1 c1 + c2 = 0. 2 Thus c1 = 2 and c2 = −1. Hence y = 2t1/2 − t1/2 ln t

14. The indicial polynomial is Q(s) = s2 + 4. There are two complex roots, ±2i. The fundamental set is {cos(2 ln t), sin(2 ln t)}. The general solution is y(t) = c1 cos(2 ln t) + c2 sin(2 ln t). The initial conditions imply c1 = −3 2c2 = 4. Thus c1 = −3 and c2 = 2. Hence y = −3 cos(2 ln t) + 2 sin(2 ln t) 15. The coefficient functions for the given equation in standard form are a1 (t) = −4/t and a2 (t) = 6/t2 both of which are not defined at the initial condition t0 = 0. Thus the uniqueness and existence theorem does not guarantee a solution. In fact, the condition that y ! (0) exist presupposes that y is defined near t = 0. For t positive the indicial polynomial is Q(s) = s2 − 5s + 6 = (s − 6)(s + 1) and therefore y(t) = c1 t6 + c2 t−1 . The only way that y can be extended to t = 0 is that c2 = 0. In this case y(t) = c1 t6 cannot satisfy the given initial conditions. Thus, no solution is possible.

166

1 Solutions

Section 5.4 bt

at

1. By L’Hospital’s rule limt→0 e −e = b − a. So Theorem 4 applies and t gives 2 bt 3 * ∞ e − eat 1 1 L (s) = − dσ t σ−b σ−a s ' ( ' ( M −b s−b = lim ln − ln M →∞ M −a s−a ' ( s−a = ln s−b at 2. By L’Hospital’s rule limt→0 2 cos bt−cos = 0. So Theorem 4 applies and t gives 2 3 * ∞ cos bt − cos at 2σ 2σ L 2 (s) = − 2 dσ 2 2 t σ +b σ + a2 s ' 2 ( ' 2 ( M + b2 s + b2 = lim ln − ln M →∞ M 2 + a2 s2 + a2 ' 2 ( s + a2 = ln s2 + b2

cos bt − cos at 3. Apply L’Hospital’s rule twice to get lim 2 = a2 − b2 . Now t→∞ t2 use Exercise 2 to get 2 3 ( * ∞ ' 2 cos bt − cos at σ + a2 L 2 (s) = ln dσ t2 σ 2 + b2 s ,* * M M 2 2 2 2 = lim ln(σ + a ) dσ − ln(σ + b ) dσ . M →∞

s

s

We now use two facts from calculus: ) 1. ln(x2 +"a2 ) dx #= x ln(x2 + a2 ) − 2x + 2a tan−1 (x/a) + C 2. lim x ln x→∞

x2 +a2 x2 +b2

=0

The first fact is shown by integration by parts and the second fact is shown by L’Hospitals rule. We now get (after some simplifications) 2 3 ' 2 ( ' ( " # cos bt − cos at s + b2 b −1 a −1 L 2 (s) = s ln 2 + 2a tan − 2b tan t2 s + a2 s s

1 Solutions

167

4. We have limt→0 sint at = a. So Theorem 4 applies and gives 2 3 * ∞ sin at a L (s) = dσ 2 t σ + a2 s = =

lim tan−1 (M/a) − tan−1 s/a

M →∞

π − tan−1 (s/a) = tan−1 (a/s) 2

5. Applying the Laplace transform we get Y!+

3s + 2 2y0 Y = 2 . 2 s +s s +s

The integrating factor is I = s2 (s + 1); we get Y (s) = Laplace inversion gives

y0 s+1

+

C s2 (s+1) .

y(t) = y0 e−t + C(t − 1 + e−t ) = (y0 + C)e−t + C(t − 1). Let c1 = C and c2 = y0 + C to get y(t) = c1 e−t + c2 (t − 1). 6. Applying the Laplace transform we get Y ! (s) + integrating factor is I = s + 1; we get Y (s) =

1 Y (s) = 0. The s+1

C and y(t) = Ce−t s+1

−y0 7. Applying the Laplace transform we get Y ! (s) = and therefore (s + 2)2 y0 Y (s) = + C. However, since lims→∞ Y (s) = 0 we must have C = 0. s+2 Hence y(t) = y0 e−2t . 4s 3y0 Y (s) = 2 . An +1 s +1 3 y0 (s + 3s) integrating factor is I = (s2 + 1)2 . Solving gives Y (s) = + (s2 + 1)2 C and y(t) = (C/2)(sin t − t cos t) + y0 (t sin t + cos t). 2 (s + 1)2

8. Applying the Laplace transform we get Y ! (s) +

s2

6s Y (s) = 0. An +1 C integrating factor is I = (s2 + 1)3 . We then get Y (s) = 2 , and (s + 1)3 % & y(t) = (C/8) (3 − t2 ) sin t − 3t cos t

9. Applying the Laplace transform we get Y ! (s) +

s2

168

1 Solutions

10. Applying the Laplace transform we get Y ! (s) = and y(t) = y0 e−t .

−y0 y0 , Y (s) = , (s + 1)2 s+1

' ( −y0 1 1 11. Apply the Laplace transform to get Y (s) = = y0 − . s(s − 1) s s−1 ' ( s Then Y (s) = y0 ln + C. Take C = 0 since lims→∞ Y (s) = 0. s−1 et − 1 Hence y(t) = y0 . t " # −y0 1 1 12. Apply the Laplace transform to get Y ! (s) = 2 = (y0 /2) s+1 − s−1 . s −1 ' ( y0 s+1 Hence Y (s) = ln + C. Take C = 0 since lims→∞ Y (s) = 0. 2 s−1 y0 et − e−t Then y(t) = . 2 t !

−y0 13. Apply the Laplace transform, simplify, and get Y ! (s) = 2 = (s − 5s + 6) ' ( ' ( 1 1 s−2 y0 − . Then Y (s) = y0 ln + C. Take C = 0. Then s − 2' s − 3 ( s−3 e3t − e2t y(t) = y0 . t −y0 14. Apply the Laplace transform to get Y ! (s) = 2 . Then Y (s) = s +9 −y0 π (tan−1 (s/3) + C). Since lims→∞ Y (s) = 0 we have C = − . Hence 3 2 y0 y0 sin 3t −1 3 Y (s) = tan ( ) and y(t) = . 3 s 3 t −sy0 15. Apply the Laplace transform, simplify, and get Y ! (s) = − s(s2 + 1) ' ( 2y1 1 1 s = −y0 2 − 2y1 − 2 . Integrating gives Y (s) = s(s2 + 1 s + 1 s s +1 ' 2 ( s +1 −y0 tan−1 (s) + y1 ln + C. Since lims→∞ Y (s) = 0 we must have s2 ' ( ' 2 ( π 1 s +1 −1 C = y0 and hence Y (s) = y0 tan + y1 ln . Therefore 2 s s2 sin t 1 − cos t y(t) = y0 + 2y1 t t

1 Solutions

169

' ( −y0 1 1 16. Apply the Laplace transform to get Y ! (s) = 2 = y0 − . s +s s+1 s ' ( s+1 Hence y(t) = y0 ln + C. But C = 0 since lims→∞ Y (s) = 0. Thus s 1 − e−t y(t) = y0 . t 17. We use the formula n ' ( k 7 dn n d dn−k (f (t)g(t)) = f (t) · n−k g(t). n k dt k dt dt k=0

Observe that

dk −t e = (−1)k e−t dtk

and

dn−k n t = n(n − 1) · · · (k + 1)tk . dtn−k

It now follows that 1 t dn −t n e (e t ) n! dtn ' ( n 1 t 7 n dk −t dn−k n = e e t n! k dtk dtn−k k=0 n ' ( 7 n n(n − 1) · · · (k + 1) k t = e (−1)k e−t t k n! k=0 ' ( k n 7 n t = (−1)k k k! k=0

= +n (t).

18. This follows in a similar manner to the proof of Equation (2) given in Theorem 9. 20.

170

1 Solutions

L {+n (at)} (s) 2 k3 n ' ( 7 n t k k = (−1) a L k k! k=0 n ' ( 7 n 1 = (−1)k ak k+1 k s k=0 ' ( n 1 7 n = n+1 (−a)k sn−k s k k=0 n

=

(s − a) . sn+1

The last line follows from the binomial theorem. Note: the dilation principle gives the same formula for a > 0. 21. Hint: Take the Laplace transform of each side. Use the previous exercise and the binomial theorem. )t 22. We have that +n ∗ 1(t) = 0 +n (x) dx. By the convolution theorem 1 (s − 1)n s sn+1 ( ' s − 1 (s − 1)n = 1− s sn+1 (s − 1)n (s − 1)n+1 = − n+1 s sn+2 = L {+n } (s) − L {+n+1 } (s).

L {+n ∗ 1} (s) =

The result follows by inversion. 23. We compute the Laplace transform of both sides. We’ll do a piece at a time. L {(2n + 1)+n } (s) (s − 1)n = (2n + 1) n+1 s (s − 1)n−1 = (2n + 1)(s(s − 1)). sn+2 L {−t+n } (s) ' (! (s − 1)n = sn+1 (s − 1)n−1 = (n + 1 − s). sn+2

1 Solutions

171

−nL {+n−1 } (s) (s − 1)n−1 = −n sn n−1 (s − 1) = (−ns2 ). sn+2 We have written each so that the common factor is coefficients are

(s − 1)n−1 . The sn+2

n + 1 − s + (2n + 1)(s(s − 1)) − ns2 = (n + 1)(s2 − 2s + 1) = (n + 1)(s − 1)2 The right hand side is now ' ( n−1 1 2 (s − 1) (n + 1)(s − 1) n+1 sn+2 n+1 (s − 1) = sn+2 = L {+n+1 } (s). Taking the inverse Laplace transform completes the verification. )t 24. We have that +n ∗ +m (t) = 0 +n (x)+m (t − x) dx. By the convolution theorem

= = = =

L {+n ∗ +m } (s) (s − 1)m (s − 1)n n+1 sm+1 s' ( m+n (s − 1) s−1 1 − sm+n+1 s m+n (s − 1) (s − 1)m+n+1 − m+n+1 s sm+n+2 L {+m+n (s)} − L {+m+n+1 } (s).

The result follows by inversion. )∞ 25. First of all 0 e−t +n (t) dt = L {+n } (1) = 0. Thus

172

1 Solutions

*

∞ t

*

e−x +n (x) dt t

e−x +n (x) dx * ∞ = −e−t et−x +n (x) dx = −

0

0

= −e−t (et ∗ +n (t)).

By the convolution theorem 0 1 L et ∗ +n (s) 1 (s − 1)n = s − 1 sn+1 (s − 1)n−1 = sn+1 ' ( (s − 1)n−1 s−1 = 1− sn s n−1 (s − 1) (s − 1)n = − sn sn+1 −1 = L {+n−1 (t)} − L−1 {+n (t)} . It follows by inversion that et ∗ +n = +n−1 − +n and substituting this formula into the previous calculation gives the needed result.

Section 5.5 1. Let y2 (t) = t2 u(t). Then t4 u!! + t3 u! = 0, which gives u! = t−1 and u(t) = ln t. Substituting gives y2 (t) = t2 ln t. The general solution can be written y(t) = c1 t2 + c2 t2 ln t. 2. Let y2 (t) = tu(t). Then t3 u!! + 4t2 u! = 0, which gives u! = t−4 and −3 −2 −1 u(t) = t−3 . Substituting gives y2 (t) = − t 3 = 3t 2 . The general solution 1 can be written y(t) = c1 t + c2 t2 . 1

5

3

2 u!! + 4t 2 u! = 0 leads to u! = 1/t and hence 3. Let y2 (t) = t 2 u(t). Then 4t√ u(t) = ln√t. Thus √ y2 (t) = t ln t. The general solution can be written y(t) = c1 t + c2 t ln t.

4. Observe that we can reduce the order of this differential equation by letting v = y ! . Then t2 v ! + 2tv = 0. Separating variables and solving gives y ! = v = kt−2 . Integrating (and setting c1 = −k) gives y = ct1 + c2 . Since

1 Solutions

173

1/t is the given solution it follows that y2 (t) = 1 is a second independent solution. The general solution can be written y(t) = ct1 + c2 . 5. Let y2 (t) = tu(t). Then u satisfies t3 u!! − t3 u! = 0. Thus u! = et and u = et . It follows that y2 (t) = tet is a second independent solution. The general solution can be written y(t) = c1 t + c2 tet . 6. Let y2 (t) = t2 cos t u(t). Then t4 cos t u!! − 2t4 sin t u! = 0 which gives u! (t) = sec2 t and hence u(t) = tan t. Thus y2 (t) = t2 sin t. The general solution can be written y(t) = c1 t2 cos t + c2 t2 sin t. 7. Let y2 (t) = u(t) sin t2 . Then u(t) satisfies t sin t2 u!! +(4t2 cos t2 −sin t2 )u! = u!! 1 cos t2 0 and hence ! = − 4t . It follows that u! = t csc2 t2 and therefore u t sin t2 −1 2 2 u(t) = −1 2 cot t . We now get y2 (t) = 2 cos t . The general solution can 2 2 be written y(t) = c1 sin t + c2 cos t . 8. Let y2 = ue−2t Then tu!! + (2t − 2)u! = 0 which gives u! = t2 e−2t and 2 −2t 2 u = −1 . Therefore y2 (t) = −1 4 (2t + 2t + 1)e 4 (2t + 2t + 1). The general 2t 2 solution may be written y(t) = c1 e + c2 (2t + 2t + 1). 9. Let y2 (t) = u(t) tan t. Then u!! tan t + 2u! sec2 t = 0 which gives u! = cot2 t = csc2 t − 1. Hence u = − cot t − t and y2 (t) = −1 − t tan t. The general solution can be written y(t) = c1 tan t + c2 (1 + t tan t). 10. Let y2 (t) = u(t)e−t . Then u(t) satisfies te−t u!! − (e−t + te−t )u! = 0 from which we get u! (t) = tet and u(t) = tet − et . Thus y2 (t) = t − 1. The general solution can be written y(t) = c1 e−t + c2 (t − 1). 11. The functions tan t and sec t are continuous except at points of the form π2 + 2nπ, n ∈ Z. We will work in the interval (−π/2, π/2). Let y2 (t) = u(t) tan t. Then u!! tan t + u! (tan2 t + 2) = 0 and hence u## ! u# = − tan t − 2 cot t. It follows that ln |u | = ln |cos t| − 2 ln |sin t| and −2 −1 ! thus u = cos t sin t. Further u(t) = sin t and we have y2 (t) = − sec t. The general solution can be written y(t) = c1 tan t + c2 sec t. 2

12. Let y2 (t) = tu(t). Then (t3 + t)u!! + 2u! = 0 which gives u! = t t+1 and 2 u = t − 1t . Thus y2 (t) = t2 − 1. The general solution can be written y(t) = c1 t + c2 (t2 − 1). ##

sin 2t u !! ! 13. Let y2 = u 1+cos 2t . Then u(t) satisfies u sin 2t + 4u = 0 and hence u# = ! ! −4 csc 2t. We now get ln u = 2 ln |csc 2t + cot 2t|. Thus u = (csc 2t + cot 2t)2 = csc2 2t + 2 csc 2t cot 2t + cot2 2t = 2 csc2 2t + 2 csc 2t cot 2t − 1. 2t By integrating we get u = − cot 2t − csc 2t − t = − 1+cos sin 2t − t. It now t sin 2t follows that y2 = −1 − 1+cos 2t . The general solution can be written " # sin 2t t sin 2t y(t) = c1 1+cos 2t + c2 1 + 1+cos 2t .

174

1 Solutions

14. Let y2 (t) = u(t)t cos t. Then u satisfies t3 (cos t)u!! − 2t3 (sin t)u! = 0 and ## sin t ! 2 ! 2 hence uu# = 2cos t . We now get ln u = ln sec t and hence u = sec t. It follows that u = tan t and y2 (t) = t sin t. The general solution can be written y(t) = c1 t cos t + c2 t sin t. 15. Let y2 (t) = (1 − t2 )u(t). Substitution gives (1 − t2 )2 u!! − 4t(1 − t2 )u! = 0 ## −2t 1 ! ! and hence uu# = −2 1−t 2 . From this we get u = (1−t2 )2 . Integrating u by " # t 1 1+t partial fractions give u = 12 1−t and hence 2 + 4 ln 1−t y2 (t) =

1 1 t + (1 − t2 ) ln 2 4

'

1+t 1−t

(

.

The general solution can be written ' ' (( 1 1 1+t 2 2 y = c1 (1 − t ) + c2 t + (1 − t ) ln . 2 4 1−t 16. Let y2(t) = tu(t). Substitution gives t(1 − t)2 u!! + (2 − 4t2 )u! = 0 and ## 2−4t2 1 1 1 thus uu# = t(t−1(t+1) = −2 u! (t) = t2 (t−1)(t+1) = t − t−1 − t+1 . Hence $ $ $ $ 1/2 1/2 −1 1 1 t−1 t2 + t−1 − t+1 . Therefore u(t) = t + 2 ln $ t+1 $. It now follows that y2 (t) = 1 + 2t ln 1−t . The general solution can be written y(t) = c1 t + " # t+1 t 1−t c2 1 + 2 ln t+1 .

Section 5.6 1. sin t and cos t form a fundamental set for the homogeneous solutions. Let y'p (t) = u1 cos ( t +'u2 ( sin t.'Then(the matrix equation cos t sin t u!1 0 = implies u!1 (t) = − sin2 t = 12 (cos 2t − 1) − sin t cos t u!2 sin t and u!2 (t) = cos t sin t = 12 (sin 2t). Integration give u1 (t) = 14 (sin(2t) − −1 2t) = 12 (sin t cos t − t) and u2 (t) = −1 4 cos 2t = 4 (2 cos 2t − 1). This 1 1 1 implies yp (t) = 4 sin t − 2 t cos t. Since 4 sin t is a homogeneous solution we can write the general solution in the form y(t) = −1 2 t cos t + c1 cos t + c2 sin t. We observe that a particular solution is the imaginary part of a solution to y !! + y = eit . We use the incomplete partial fraction method and p(s) 1 1 get Y (s) = (s−i)12 (s+i) . This can be written Y (s) = 2i (s−i)2 + (s−i)(s+i) . " # 1 −1 1 −1 it From this we get yp (t) = Im 2i L { (s−i) = Im −i 2} 2 te = 2 t cos t. The general solution is y(t) =

−1 2 t cos t

+ c1 cos t + c2 sin t.

1 Solutions

175

0 1 2. A fundamental set for y !! − 4y = 0 is e2t , e−2t . Let yp (t) = u1 (t)e2t + −2t u . Then '2 (t)e ( the ' ! matrix ( ' equation ( 2t 4t e e−2t u1 0 = implies u!1 (t) = 14 and u!2 (t) = −e4 and 2e2t −2e−2t u!2 e2t 4t t 2t e2t e2t hence u1 (t) = 4t and u2 (t) = −e 16 . Now yp (t) = 4 e − 16 . Since 16 is a homogeneous solution we can write the general solution as y(t) = t 2t 2t −2t . On the other hand, the incomplete partial fraction 4 e + c1 e + c 2 e 1

p(s) 4 method gives Y (s) = (s−2)12 (s+2 = (s−2) 2 + (s−2)(s+2) . From this we see that a particular solution is yp (t) = 14 te2t . The general solution is y(t) = 14 te2t + c1 e2t + c2 e−2t .

3. The functions et cos 2t and et sin 2t form a fundamental set. Let yp (t) = c1 et cos 2t + c2 et sin 2t. equation ' Then ( the ' matrix ( ! u 0 W (et cos 2t, et sin 2t) 1! = implies that u!1 (t) = −1 2 sin 2t and u2 et 1 1 1 ! u2 (t) = 2 cos 2t. Hence, u1 (t) = 4 cos2t and u2 (t) = 4 sin 2t. From this we get yp (t) = 14 et cos2 2t + 14 et sin2 2t = 14 et . On the other hand, the method of undetermined coefficients implies that a particular solution is of the form yp (t) = Cet . Substitution gives 4Cet = et and hence C = 14 . It follows that yp (t) = 14 et . Furthermore, the general solution is y(t) = 14 et + c1 et cos 2t + c2 et sin 2t. 0 −3t 1 4. A set(is 1, ' fundamental ( ' 'e (. The matrix equation 1 e−3t u!1 0 = implies u!1 (t) = 13 e−3t and u!2 (t) = −1 3 . 0 −3e−3t u!2 e−3t −3t −e−3t Thus u1 (t) = −e9 , u2 (t) = −t − 3t e−3t . Observe 3 , and yp (t) = 9 −e−3t though that 9 is a homogeneous solution and so the general solution can be written y(t) = − 3t e−3t + c1 + c2 e−3t . The incomplete partial −1

p(s) 1 3 fraction method gives Y (s) = (s+3) 2 s = (s+3)2 + (s+3)s which implies −3t that −1 is a particular solution. The general solution is as above. 3 te 0 1 5. A set is' et ,(e2t . The matrix equation ' fundamental ( ' ( et e2t u!1 0 = implies u!1 (t) = −e2t and u!2 (t) = et . Hence t 2t e 2e u!2 e3t −1 2t t 2t t t 2t u1 (t) = −1 = 12 e3t . The 2 e , u2 (t) = e , and yp (t) = 2 e e + e e 1 3t t 2t general solution is y(t) = 2 e +c1 e +c2 e . The method of undetermined coefficients implies that a particular solution is of the form yp = Ce3t . Substitution gives 2Ce3t = 3e3t and hence C = 12 . The general solution is as above.

6. sin t and cos t form a fundamental set for the homogeneous solutions. Let y'p (t) = u1 cos ( t +'u2 ( sin t. ' Then ( the matrix equation cos t sin t u!1 0 = implies that u!1 (t) = cos t − sec t and − sin t cos t u!2 tan t

176

1 Solutions

u!2 (t) = sin t. From this we get u1 (t) = sin t − ln |sec t + tan t| and u2 (t) = − cos t. Therefore yp (t) = − cos t ln |sec t + tan t|. The general solution is thus y(t) = − cos t ln |sec t + tan t| + c1 cos t + c2 sin t. 7. A is {et', tet(}. The matrix equation ' fundamental ( 'set! ( t t 0 e te u1 = et implies u!1 (t) = −1 and u!2 (t) = 1t . Hence, et et + tet u!2 t u1 (t) = −t, u2 (t) = ln t, and yp (t) = −tet + t ln tet . Since −tet is a homogeneous solution we can write the general solution as y(t) = t ln tet + c1 et + c2 tet . 8. A t}. The matrix equation ' fundamental(set ' !is({cos't, sin ( cos t sin t u1 0 = implies u!1 (t) = − tan t and u!2 (t) = 1. − sin t cos t u!2 sec t Hence u1 (t) = ln(cos t), u2 (t) = t, and yp (t) = cos t ln(cos t) + t sin t. The general solution is y(t) = cos t ln(cos t) + t sin t + c1 cos t + c2 sin t. 9. The associated homogeneous equation is Cauchy-Euler 0 1 with indicial equation s2 −3s+2 = (s−2)(s−1). It follows that t, t2 forms a fundamental set. We put the given equation is standard form to get y !! − 2t y ! + t22 y = t2 . Thus f( (t)'= ( t2 . The'matrix equation ' ( 2 ! t t u1 0 = implies u!1 (t) = −t2 and u!2 (t) = t. Hence 1 2t u!2 t2 3 2 3 2 4 u1 (t) = −t3 , u2 (t) = t2 , and yp (t) = −t3 t + t2 t2 = t6 . It follows that the 4 general solution is y(t) = t6 + c1 t + c2 t2 . 1 ! 1 !! 10. 'In standard ( ' ! (form ' we get ( y − t y = 3t − t . The matrix equation 2 0 1 t u1 1 3 1 2 ! = implies u!1 (t) = −3 2 t + 2 and u2 (t) = 2 − 2t2 . 0 2t u!2 3t − 1t 1 3 1 3 3 Hence u1 (t) = −1 2 t + 2 t, u2 (t) = 2 t + 2t , and yp (t) = t + t. The general 3 2 solution is y(t) = t + t + c1 + c2 t .

11. The homogeneous equation is Cauchy-Euler with indicial equation s2 − 2s + 1 = (s − 1)2 . It follows that {t, t ln t} is a fundamental set. After writing in standard form we see the forcing function f (t) is 1t . The matrix equation ' (' ! ( ' ( 0 t t ln t u1 t = 1 implies u!1 (t) = − ln and u!2 (t) = 1t . Hence t 1 ln t + 1 u!2 t 2 2 2 2 t u1 (t) = − ln2 t , u2 (t) = ln t, and yp (t) = −t 2 ln t + t ln t = 2 ln t. The 2 general solution is y(t) = 2t ln t + c1 t + c2 t ln t. 0 1 12. A fundamental set is e2t , te2t . The matrix equation ' 2t ( ' ( ' ( 0 e te2t u!1 ! = implies u!1 (t) = t2−t e2t +1 and u2 (t) = 2e2t e2t (1 + 2t) u!2 t2 +1 1 −1 −1 2 t, and yp (t) = t2 +1 . Hence u1 (t) = 2 ln(t + 1), u2 (t) = tan

1 Solutions

177

−1 2t 2 e

ln(t2 + 1) + t tan−1 e2t . The general solution is y(t) = 1) + t tan−1 e2t + c1 e2t + c2 te2t .

−1 2t 2 e

ln(t2 +

13. 'The matrix equation (' ! ( ' ( tan t sec t u1 0 = implies u!1 (t) = t and u!2 (t) = −t sin t. sec2 t sec t tan t u!2 t 2 2 Hence u1 (t) = t2 , u2 (t) = t cos t − sin t, and yp (t) = t2 tan t + (t cos t − 2 sin t) sec t = t2 tan t + t − tan t. Since tan t is a homogeneous solution we 2 can write the general solution as y(t) = t2 tan t + t + c1 tan t + c2 sec t. 14. When put in standard form one sees that f (t) = te−t . The matrix equation ' (' ! ( ' ( t−1 e−t u1 0 = implies u!1 (t) = e−t and u!2 (t) = 1 − t. 1 −e−t u!2 te−t 2 2 Hence u1 (t) = −e−t , u2 (t) = t− t2 , and yp (t) = −(t−1)e−t +(t− t2 )e−t = −t2 −t −t −t is a homogeneous solution we can write the general 2 e +e . Since e −t2 −t solution as y(t) = 2 e + c1 (t − 1) + c2 e−t . 15. After put in standard form the forcing function f is 4t4 . The matrix equation ' (' ! ( ' ( cos t2 sin t2 u1 0 = implies u!1 (t) = −2t3 sin t2 and −2t sin t2 2t cos 2t u!2 4t4 u!2 (t) = 2t3 cos t2 . Integration by parts gives u1 (t) = t2 cos t2 − sin t2 and u2 (t) = t2 sin t2 +cos t2 . Hence yp (t) = t2 cos2 t2 −cos t2 sin t2 +t2 sin2 t2 + cos t2 sin t2 = t2 . The general solution is y(t) = t2 + c1 cos t2 + c2 sin t2 . t −t 16. 'A fundamental ( ' ! set ( is {e ' , e }.(The matrix equation t −t 0 e e u1 1 ! = implies u!1 (t) = 12 1+e t and u2 (t) = 1 t −t ! e −e u2 1+e−t −1 et 1 −1 t t t 2 1+et . Hence u1 (t) = 2 (t − ln(1 + e ), u2 (t) = 2 (e − ln(1 + e )), and yp (t) = 12 (tet − 1 − (et − e−t ) ln(1 + et )). (Note: in the integrations of u!1 and u!2 use the substitution u = et .) The general solution can now be written y(t) = 12 (tet − 1 − (et − e−t ) ln(1 + et )) + c1 et + c2 e−t .

17. Let a and t be in the interval I. Let z1 and z2 be the definite integrals defined as follows: * t −y2 (x)f (x) z1 (t) = dx w(y 1 , y2 )(x) a z2 (t) =

*

t a

y1 (x)f (x) dx. w(y1 , y2 )(x)

These definite integrals determine the constant of integration in Theorem 1 so that z1 (a) = z2 (a) = 0. It follows that

178

1 Solutions

yp (t) = z1 (t)y1 (t) + z2 (t) + y2 (t) * t * t −y2 (x)y1 (t)f (x) y1 (x)y2 (t)f (x) = dx + w(y , y )(x) w(y1 , y2 )(x) 1 2 a a * t (y1 (x)y2 (t) − y2 (x)y1 (t)) = f (x) dx w(y1 , y2 )(x) a $ $ $y (x) y2 (x)$ $ * t $$ 1 y1 (t) y2 (t)$ $ $ f (x) dx. = $ $ a $y1 (x) y2 (x)$ $y1! (x) y2! (x)$

18. Let y1 (t) = cos at and y2 (t) = sin at. Then {y1 , y2 } is a fundamental set. We have $ $ $y1 (x) y2 (x)$ $ $ $ y1 (t) y2 (t)$ = cos ax sin at − sin ax cos at = sin(a(t − x)) and

Thus

$ $ w(y1 , y2 )(x) = $$ *

yp (t) =

$ cos at sin at$$ = a. −a sin at a cos at$

t 0

sin a(t − x) f (x) dx a

1 = f (t) ∗ sin at. a Applying the Laplace transform to y !! + a2 y = f , with initial conditions y(0) = y ! (0) = 0, gives s2 Y (s) + a2 Y (s) = F (s). Solving for Y (s) we get Y (s) =

F (s) 1 a = F (s). 2 2 +a a s + a2

s2

The convolution theorem gives a particular solution yp (t) =

1 sin at ∗ f (t). a

19. Let y1 (t) = e−at and y2 (t) = eat . Then {y1 , y2 } is a fundamental set. We have $ $ $y1 (x) y2 (x)$ −ax at $ $ e − eax e−at = ea(t−x) − e−a(t−x) = 2 sinh(a(t − x)) $ y1 (t) y2 (t)$ = e and

1 Solutions

Thus

179

$ $ $ e−at eat $ $ $ = 2a. w(y1 , y2 )(x) = $ −ae−at aeat $ yp (t) =

*

t 0

2 sinh a(t − x) f (x) dx 2a

1 = f (t) ∗ sinh at. a Applying the Laplace transform to y !! − a2 y = f , with initial conditions y(0) = y ! (0) = 0, gives s2 Y (s) − a2 Y (s) = F (s). Solving for Y (s) we get Y (s) =

F (s) 1 a = F (s). s2 − a 2 a s2 − a 2

The convolution theorem gives a particular solution yp (t) =

1 sinh at ∗ f (t). a

20. Let y1 (t) = eat and y2 (t) = teat . Then {y1 , y2 } is a fundamental set. We have $ $ $y1 (x) y2 (x)$ ax at ax at a(x+t) $ $ − xea(x+t) = (t − x)ea(x+t) $ y1 (t) y2 (t)$ = e te − xe e = te and

$ ax $ $ e xeax $$ w(y1 , y2 )(x) = $$ ax ax = e2ax + axe2ax − axe2ax = e2ax . ae e + axeax $

Thus

yp (t) = =

* *

t 0

(t − x)ea(x+t) f (x) dx e2ax

t 0

(t − x)ea(t−x) f (x) dx

= f (t) ∗ (teat ). Applying the Laplace transform to y !! − 2ay ! + a2 y = f , with initial conditions y(0) = y ! (0) = 0, gives s2 Y (s) − 2asY (s) + a2 Y (s) = F (s). Solving for Y (s) we get Y (s) =

F (s) 1 = F (s). (s − a)2 (s − a)2

The convolution theorem gives a particular solution

180

1 Solutions

yp (t) = (teat ) ∗ f (t). 21. Let y1 (t) = eat and y2 (t) = ebt . Then {y1 , y2 } is a fundamental set. We have $ $ $y1 (x) y2 (x)$ ax bt bx at ax+bt $ $ − ebx+at $ y1 (t) y2 (t)$ = e e − e e = e and

Thus

$ ax $ $ e ebx $$ w(y1 , y2 )(x) = $$ ax = (b − a)e(a+b)x . ae bebx $ *

t

eax+bt − ebx+at f (x) dx (a+b)x 0 (b − a)e * t 1 = (eb(t−x) − ea(t−x) )f (x) dx b−a 0 1 = f (t) ∗ (ebt − eat ). b−a

yp (t) =

Applying the Laplace transform to y !! − (a + b)y ! + aby = f , with initial conditions y(0) = y ! (0) = 0, gives s2 Y (s) − (a + b)sY (s) + abY (s) = F (s). Solving for Y (s) we get ' ( F (s) 1 1 1 Y (s) = = − F (s). (s − a)(s − b) a−b s−a s−b The convolution theorem gives a particular solution yp (t) =

Section 6.1 1. Graph (c) 2. Graph (g) 3. Graph (e) 4. Graph (a) 5. Graph (f)

1 (eat − ebt ) ∗ f (t). a−b

1 Solutions

181

6. Graph (d) 7. Graph (h) 8. Graph (b) % &$2 )5 )2 )3 )5 9. 0 f (t) dt = 0 (t2 − 4) dt + 2 0 dt + 3 (−t + 3) dt = t3 /3 − 4t $0 + 0 + % 2 &$5 −t /2 + 3t $3 = (8/3 − 8) + (−25/2 + 15) − (−9/2 + 9) = −22/3.

$1 $2 )1 )2 f (u) du = 0 (2 − u) du + 1 u3 du = (2u − u2 /2)$0 + u4 /4$1 = 3/2 + 4 − 1/4 = 21/4. ) 2π )π ) 2π π 2π 11. 0 |sin x| dx = 0 sin x dx + π − sin x dx = − cos x|0 + cos x|π = 4. 10.

12. 13. 14. 15. 16.

)2 0

)3 0

)5 2

)6 0

)6 0

)6 0

f (w) dw = f (t) dt = f (t) dt =

0

)3 2

)2 0

f (u) du = f (t) dt =

)1

(1 − t) dt +

0

0

1 1 w

(3 − t) dt +

)1

)2

)2

w dw +

u du +

t2 dt +

)2 1

)3 2

)4 3

)4 2

dw +

)3

1 2 2

dw = 1/2 + ln 2 + 1/2 = 1 + ln 2

2(t − 3) dt + (3 − t) dt +

(2 − u) du +

4 dt +

)6 3

)6 2

)6 4

)6 4

2 dt = 1/2 + 1 + 4 = 11/2

(5 − t) dt = 0 + 0 + 0 = 0.

1 du = 1/2 + 1/2 + 4 = 5.

(7 − t) dt = 8/3 + 4 + 15/2 = 85/6

17. A is true since y(t) satisfies the differential equation on each subinterval. B is true since the left and right limits agree at t = 2. C is not true since y(0) = 1 &= 2. 18. A is true since y(t) satisfies the differential equation on each subinterval. B is true since limt→2− y(t) = 1 + e−8 = limt→2+ y(t). C is true since y(0) = 2. 19. A is true since y(t) satisfies the differential equation on each subinterval. B is false since limt→2− y(t) = 1 + e−8 while limt→2+ y(t) = 1. C is false since B is false. 20. A is false since 2e−4t does not satisfy the differential equation y ! +4y = 4 on the interval [0, 2). Since A is false, B and C are necessarily false. 21. A is true since y(t) satisfies the differential equation on each subinterval. B is true since limt→1− y(t) = −2e + e2 = limt→1+ y(t). C is false since limt→1− y ! (t) = −3e + 2e2 while limt→1+ y ! (t) = 3e2 − 2e. D is false since C is false. 22. A is true since y(t) satisfies the differential equation on each subinterval. B is false since limt→1− y(t) = −2e+e2 while limt→1+ y(t) = (1/2)e2 −3e.

182

1 Solutions

C and D are false since B is false. You cannot have a continuous derivative if the function is not continuous. 23. A is true since y(t) satisfies the differential equation on each subinterval. B is true since limt→1− y(t) = −2e + e2 = limt→1+ y(t). C is true since limt→1− y ! (t) = −3e + 2e2 = limt→1+ y ! (t). D is true since y(0) = y ! (0) = 0. 24. A is true since y(t) satisfies the differential equation on each subinterval. B is true since limt→1− y(t) = −e2 = limt→1+ y(t). C is true since limt→1− y ! (t) = −e − 2e2 = limt→1+ y ! (t). D is false since y(0) = −2 &= 0. 25. The general solution of y ! − y = 1 on the interval [0, 2) is found by using the integrating factor e−t . The general solution is y(t) = −1 + cet and the initial condition y(0) = 0 gives c = 1, so that y(t) = −1 + et for t ∈ [0, 2). Continuity of y(t) at t = 2 will then give y(2) = limt→2− y(t) = −1 + e2 , which will provide the initial condition for the next interval [2, 4). The general solution of y ! − y = −1 on [2, 4) is y(t) = 1 + ket . Thus −1 + e2 = y(2) = 1 + ke2 and solve for k to get k = −2e−2 + 1, so that y(t) = 1 + (−2e−2 + 1)et for t ∈ [2, 4). Continuity will then give y(4) = 1 + (−2e−2 + 1)e4 , which will provide the initial condition for the next interval [4, ∞). The general solution to y ! − y = 0 on [4, ∞) is y(t) = bet and the constant b is obtained from the initial condition be4 = y(4) = 1 + (−2e−2 + 1)e4 , which gives b = e−4 − 2e−2 + 1, so that y(t) = (e−4 − 2e−2 + 1)et for t ∈ [4, ∞). Putting these three pieces together, we find that the solution is  t  if 0 ≤ t < 2, −1 + e y(t) = 1 − 2et−2 + et if 2 ≤ t < 4   t−4 t−2 t e − 2e + e if 4 ≤ t < ∞. 26. The general solution of y ! +3y = t on the interval [0, 1) is found by using the integrating factor e3t . The general solution is y(t) = 13 t − 19 + ce−3t and the initial condition gives c = 19 . By continuity we must have y(1) = 1 1 1 −3 = 29 + 19 e−3 , which will provide the initial condition for 3 − 9 + 9e the next interval [1, ∞). The general solution of y ! + 3y = 1 on (1, ∞) is y(t) = 13 + ke−3t . Setting y(1) = 13 − 19 + 19 e−3 = 29 + 19 e−3 found from the interval [0, 1) equal to y(1) = 13 + ke−3 found from the interval [1, ∞) and solving for k gives k = 19 − 19 e3 . Thus the complete solution is / 1 t − 19 + 19 e−3t if 0 ≤ t < 1, y(t) = 31 1 1 3 −3t if 1 ≤ t < ∞. 3 + ( 9 − 9 e )e

1 Solutions

183

27. The general solution of y ! − y = f (t) on any interval is found by using the integrating factor e−t . The general solution on the interval [0, 1) is y(t) = aet and since the initial condition is y(0) = 0, the solution on [0, 1) is y(t) = 0. Continuity then given y(1) = 0, which will be the initial condition for the interval [1, 2). The general solution of y ! − y = t − 1 on the interval [1, 2) is y(t) = −t + bet and the initial condition y(1) = 0 gives 0 = −1+be1 so that b = e−1 . Thus y(t) = −t+e−1 et = −t+et−1 for t ∈ [0, 2). Continuity of y(t) at t = 2 will then give y(2) = limt→2− y(t) = −2 + e1 , which will provide the initial condition for the next interval [2, 3). The general solution of y ! − y = 3 − t on [2, 3) is y(t) = t − 2 + cet . Thus −2 + e1 = y(2) = ce2 and solve for c to get c = −2e−2 + e−1 , so that y(t) = t − 2 + (−2e−2 + e−1 )et = t − 2 − 2et−2 + et−1 for t ∈ [2, 3). Continuity will then give y(3) = 1−2e1 +e2 , which will provide the initial condition for the next interval [3, ∞). The general solution to y ! − y = 0 on [4, ∞) is y(t) = ket and the constant k is obtained from the initial condition ke3 = y(3) = 1 − 2e1 + e2 , which gives c = e−3 − 2e−2 + e−1 , so that y(t) = (e−3 − 2e−2 + e−1 )et = et−3 − 2et−2 + et−1 for t ∈ [3, ∞). Putting these three pieces together, we find that the solution is  0 if 0 ≤ t < 1,    −t + et−1 if 1 ≤ t < 2, y(t) = t−2 t−1  t − 2 − 2e +e if 2 ≤ t < 3    t−3 e − 2et−2 + et−1 if 3 ≤ t < ∞.

28. The general solution of y ! +y = f (t) on any interval is found by using the integrating factor et . Using this integrating factor the general solution of y ! + y = sin t on the interval [0, π) is found to be y(t) = (sin t − cos t)/2 + ce−t . The initial condition actually occurs at the end of this interval, but by continuity we can substitute t = π in this formula to get −1 = y(π) = 1/2 + ce−π so c = −(3/2)eπ . Hence on the interval [0, π] the solution is y(t) = (sin t − cos t)/2 − (3/2)eπ e−t = (sin t − cos t)/2 − (3/2)e−(t−π) . The general solution of y ! + y = 0 on the interval [π, ∞) is y(t) = ke−t and y(π) = −1 this gives −1 = ke−π so k = −eπ and y(t) = −e−(t−π) on [π, ∞). Putting these two pieces together, we find that the solution is / (sin t − cos t)/2 − (3/2)e−(t−π) if 0 ≤ t < π, y(t) = −e−(t−π) if π ≤ t < ∞. 29. The characteristic polynomial of the equation y !! − y = f (t) is s2 − 1 = (s − 1)(s + 1) so the homogeneous equation has the solution yh (t) = aet + be−t for constants a and b. On the interval [0, 1] the equation y !! − y = t has a particular solution yp (t) = −t so the general solution has the form y(t) = −t + aet + be−t . The initial conditions give 0 = y(0) = a + b and

184

1 Solutions

1 = y ! (0) = −1+ a − b. Solving gives a = 1, b = −1 so y(t) = −t+ et − e−t on [0, 1). By continuity it follows that y(1) = −1 + e1 − e−1 and y ! (1) = −1 + e1 + e−1 and these constitute the initial values for the equation y !! − y = 0 on the interval [1, ∞). The general solution on this interval is y(t) = aet +be−t and at t = 1 we get y(1) = ae1 +be−1 = −1+e1 −e−1 and y ! (1) = ae1 − be−1 = −1 + e1 + e−1 . Solving for a and b gives a = 1 − e−1 and b = −1 so that y(t) = (1 − e−1 )et − e−t = et − et−1 − e−1 . Putting the two pieces together gives / −t + et − e−t if 0 ≤ t < 1, y(t) = et − et−1 − e−1 1 ≤ t < ∞. 30. The characteristic polynomial of the equation y !! − 4y ! + 4y = f (t) is s2 − 4s + 4 = (s − 2)2 so the homogeneous equation has the solution yh (t) = ae2t + bte2t for constants a and b, which is valid on the interval [0, 2). The initial conditions y(0) = 1 and y ! (0) = 0 imply that c1 = 1 and c2 = −2. So y(t) = e2t − 2te2t on [0, 2). By continuity it follows that y(2) = e4 − 4e4 = −3e4 and y ! (2) = −8e4 and these constitute the initial values for the equation y !! − 4y ! + 4y = 4 on the interval [2, ∞). The general solution on this interval is y(t) = 1 + ae2t + bte2t and at t = 2 we get y(2) = 1 + ae4 + 2be4 = −3e4 and y ! (1) = (2a + b)e4 + 4be4 = −8e4 . Solving for a and b gives a = 1 − 5e−4 and b = −2 + 2e−4 so that y(t) = 1 + (1 − 5e−4 )e2t + (−2 + 2e−4 )te2t = 1 + e2t − 5e2(t−2) − 2te2t + 2te2(t−2) . Putting the two pieces together gives / e2t − 2te2t if 0 ≤ t < 2, y(t) = 2t 2(t−2) 2t 2(t−2) 1 + e − 5e − 2te + 2te 2 ≤ t < ∞. 33. 1. |f (t)| = |sin(1/t)| ≤ 1 for all t &= 0, while |f (0)| = |0| = 0 ≤ 1. 2. It is enough to observe that limt→0+ does not exist. But letting tn = 1 nπ gives f (tn ) = sin nπ = 0 for all positive integers n, while letting 2 tn = (4n+1)π gives f (tn ) = sin(1/tn ) = sin((4n + 1)π/2) = sin(2nπ + π ) = 1 so there is one sequence tn → 0 with f (tn ) → 0 while another 2 sequence tn → 0 with f (tn ) → 1 so f (t) cannot be continuous at 0. 3. To be piecewise continuous, f (t) would have to have a limit at t approaches 0 from above, and this is not true as shown in part 2.

1 Solutions

185

Section 6.2   0 if t < 2, 1. f (t) = 3h(t − 2) − h(t − 5) = 3 if 2 ≤ t < 5, Thus, the graph is   2 if t ≥ 5. 3 2 1 0 yt 0 1 2 3 4 5 6 7  0    2 2. f (t) = 2h(t − 2) − 3h(t − 3) + 4h(t − 4) =   −1  3 graph is

8 if if if if

t < 2, 2 ≤ t < 3, Thus, the 2 ≤ t < 4, t ≥ 4.

3 2 1 yt −1

1 2 3 4 5 6 7

3. This function is g(t − 1)h(t − 1) where g(t) = t, so the graph of f (t) is the graph of g(t) = t translated 1 unit to the right and then truncated at t = 1, with the graph before t = 1 replaced by the line y = 0. Thus the graph is 2 1 0 yt 0

1

2

3

4. This function is g(t − 2)h(t − 2) where g(t) = t2 , so the graph of f (t) is the graph of g(t) = t2 translated 1 unit to the right and then truncated at t = 2, with the graph before t = 2 replaced by the line y = 0. Thus the graph is

186

1 Solutions

2 1 0 yt 0

1

2

3

5. This function is just t2 truncated at t = 2, with the graph before t = 2 replaced by the line y = 0. Thus the graph is 8 6 4 2 0 yt 0 1 2 3 where the dashed line is the part of the t2 graph that has been truncated. It is only shown for emphasis and it is not part of the graph. 6. This is the sin t function, truncated at t = π. The graph is 1 0 yt

π

2π

3π

−1 7. This is the function cos 2t shifted π units to the right and then truncated at t = π. The graph is 1 0 yt −1

π

2π

3π

1 Solutions

187

8. f (t) = t2 χ[0, 1) (t) + (2 − t)χ[1, 3) (t) + 3χ[3, ∞) (t)  2  if 0 ≤ t < 1, t = 2 − t if 1 ≤ t < 3, Thus, the graph is   3 if t ≥ 3. 3 2 1 yt

1

2

3

4

5

−1 9. (a) (t − 2)χ[2, ∞) (t); (b) (t − 2)h(t − 2); (c) L {(t − 2)h(t − 2)} = e−2s L {t} = e−2s /s2 . 10. (a) tχ[2, ∞) (t); (b) th(t − 2); (c) L {th(t − 2)} = e

−2s

L {t + 2} = e

−2s

'

( 1 2 + . s2 s

11. (a) (t + 2)χ[2, ∞) (t); (b) (t + 2)h(t − 2); (c) L {(t + 2)h(t − 2)} = e−2s L {(t + 2) + 2} = e−2s

'

( 1 4 + . s2 s

12. (a) (t − 4)2 χ[4, ∞) (t); (b) (t − 4)2 h(t − 4); 0 1 0 1 2 (c) L (t − 4)2 h(t − 4) = e−2s L t2 = e−4s 3 . s

2 13. (a) t20χ[4, ∞) (t); (b) h(t −04); 1 t −4s 1 0 1 2 (c) L t'h(t − 4) = e ( L (t + 4)2 = e−4s L t2 + 8t + 16 2 8 16 = e−4s + 2+ . s3 s s

14. (a) (t02 − 4)χ[4, ∞) (t); (b) (t2 − 4)h(t 1 0 − 4); 1 02 1 2 −4s 2 −4s (c) L (t ' − 4)h(t − 4) (= e L (t + 4) − 4 = e L t + 8t + 12 2 8 12 = e−4s + 2+ . s3 s s 15. (a) (t0− 4)2 χ[2, ∞) (t); (b) (t − 4)20h(t − 2); 1 1 02 1 2 −2s 2 −2s (c) L (t − 4) h(t − 2) = e L ((t + 2) − 4) = e L t − 4t + 4 ' ( 2 4 4 = e−2s − + . s3 s2 s

188

16. (a) et−4 χ[4, ∞) (t); (b) et−4 h(t − 4); 0 1 (c) L et−4 h(t − 4) = e−4s L {et } = e−4s

1 Solutions

1 . s−1

17. (a) et χ[4, ∞) (t); (b) et h(t −04); 1 (c) L {et h(t − 4)} = e−4s L et+4 = e−4s e4 L {et } 1 = e−4(s−1) . s−1

18. (a) et−4 χ[6, ∞) (t); (b) et−4 h(t − 6); 0 1 0 1 (c) L et−4 h(t − 6) = e−6s L e(t+6)−4 = e−6s e2 L {et } 1 = e−6s+2 . s−1

19. (a) tet χ[4, ∞) (t); (b) tet h(t − 0 4); 1 −4s t+4 (c) L {tet h(t − 4)} = e L (t = e−4s e4 L {tet + 4et } ' ( + 4)e 1 4 = e−4(s−1) + . (s − 1)2 s−1 20. (a) χ[0,4) (t) − χ[4,5) (t); (b) 1 − 2h(t − 4) + h(t − 5); 1 e−4s e−5s (c) L {1 − 2h(t − 4) + h(t − 5)} = − 2 + . s s s 21. (a) tχ[0,1) (t) + (2 − t)χ[1,∞) (t); (b) t + (2 − 2t)h(t − 1); (c) L {t + (2 − 2t)h(t − 1)} = L {t} + e−s L {(2 − 2(t + 1))} 1 2e−s = L {t} + e−s L {−2t} = 2 − 2 . s s 22. (a) tχ[0,1) (t) + (2 − t)χ[1,2) (t) + χ[2,∞) (t); (b) t − (2 − 2t)h(t − 1) + (t − 1)h(t − 2); (c) L {t + (2 − 2t)h(t − 1) + (t − 1)h(t − 2)} = L {t} + e−s L {(2 − 2(t + 1))} + e−2s L {(t + 2) − 1} −2s = L {t} + e−s L {−2t} '+ e L({t + 1} 1 2e−s 1 1 = 2 − 2 + e−2s + . 2 s s s s 23. (a) t2 χ[0, 2) (t) + 4χ[2, 3) (t) + (7 − t)χ[3, ∞) (t); (b) t20+ (4 − t2 )h(t − 2) + (3 − t)h(t − 3); 1 2 (c) L0 t1 + (4 − t20 )h(t − 2) + (31− t)h(t − 3) 2 = L t + e−2s L 4 − (t + 2)2 + e−3s L {3 − (t + 3)} ' ( 2 2 4 e−3s −2s = 3 −e + − . s s3 s2 s2 24. (a) χ[0,2) (t) + (3 − t)χ[2,3) (t) + 2(t − 3)χ[3,4) (t) + 2χ[4,∞) (t); (b) 1 + (2 − t)h(t − 2) + (3t − 9)h(t − 3) − (2t − 4)h(t − 4); (c) L {1 + (2 − t)h(t − 2) + (3t − 9)h(t − 3) − (2t − 4)h(t − 4)}

1 Solutions

189

−3s −4s = L {1}+e−2s L {2 − (t + 2)}+e L {3(t ' ( + 3) − 9}−e L {2(t + 4) − 4} −2s −3s 1 e 3e 2 4 = − 2 + − e−4s + . s s s2 s2 s .∞ 25. (a) n=0 (t − n)χ[n,n+1) (t); .∞ (b) t − n=1 .∞h(t − n); .∞ (c) L {t − n=1 h(t − n)} = L {t} − n=1 L {h(t − n)} .∞ e−ns 1 1 1 .∞ n = 2 − n=1 = 2− (e−s ) s s s s n=1 1 e−s = 2− . s s(1 − e−s ) .∞ .∞ 26. (a) n=0 χ[2n,2n+1) (t); (b) n=0 (−1)n h(t − n); 1 (c) . s(1 + e−s ) .∞ .∞ 27. (a) n=0 (2n + 1 − t)χ[2n,2n+2) (t); (b) −(t + 1) + 2 n=0 h(t − 2n); 1 1 2 (c) − 2 − + . s s s(1 − e−2s ) 2 −3s 3 2 3$ $ e 1 $ 28. L−1 = h(t − 3) L−1 s−1 s − 1 $t→t−3 / 0 if 0 ≤ t < 3, = h(t − 3) (et )|t→t−3 = et−3 h(t − 3) = t−3 e if t ≥ 3.

29. L

−1

2

e−3s s2

3

= h(t − 3) L

−1

2

3$ 1 $$ s2 $t→t−3 /

= h(t − 3) (t)|t→t−3 = (t − 3)h(t − 3) =

if 0 ≤ t < 3, if t ≥ 3.

3 2 3$ $ e−3s 1 −1 $ = h(t − 3) L 3 3 (s − 1) (s − 1) $t→(t−3) % &$ = h(t − 3) 12 t2 et $t→t−3 = 12 (t − 3)2 et−3 h(t − 3) / 0 if 0 ≤ t < 3, = 1 2 t−3 (t − 3) e if t ≥ 3. 2

30. L−1

2

0 t−3

3 2 3$ $ e−πs 1 −1 $ 31. L = h(t − π) L s2 + 1 s2 + 1 $t→t−π = h(t − π) (sin t)|t→t−π = h(t − π) sin(t − π) / / 0 if 0 ≤ t < π, 0 if 0 ≤ t < π, = = sin(t − π) if t ≥ π − sin t if t ≥ π. −1

2

190

1 Solutions

3 2 3$ $ se−3πs s −1 $ = h(t − 3π) L 2 2 s +1 s + 1 $t→t−3π = h(t − 3π) (cos t)|t→t−3π = h(t − 3π) cos(t − 3π) / / 0 if 0 ≤ t < 3π, 0 if 0 ≤ t < 3π, = = cos(t − 3π) if t ≥ 3π − cos t if t ≥ 3π.

32. L−1

2

3 2 3$ $ e−πs 1 −1 $ 33. L = h(t − π) L s2 + 2s + 5 s2 + 2s + 5 $t→t−π 2 3$ $ $ 1 $ = h(t − π) L−1 = h(t − π) ( 12 e−t sin 2t)$t→t−π $ 2 2 (s + 1) + 2 t→t−π / 0 if 0 ≤ t < π, = 12 e−(t−π) sin 2(t − π)h(t − π) = 1 −(t−π) e sin 2t if t ≥ π. 2 −1

2

3 e−s e−2s 34. L + 3 s2 (s2− 1)3$ 2 3$ $ $ 1 1 −1 $ $ = h(t − 1) L−1 + h(t − 2) L $ $ 2 3 s (s − 1) t→t−1 t→t−2 $ 1 2 t $ = h(t − 1) (t)|t→t−1 + h(t − 2) ( 2 t e ) t→t−2 = (t − 1)h(t − 1) + 12 (t − 2)2 et−2 h(t − 2)   if 0 ≤ t < 1, 0 = t−1 if 1 ≤ t < 2,   t − 1 + 12 (t − 2)2 et−2 if t ≥ 2. 2 −2s 3 2 3$ $ e 1 −1 −1 $ 35. L = h(t − 2) L 2+4 $ s2 + 4 s t→t−2 $ = h(t − 2) ( 12 sin 2t)$t→t−2 = 12 h(t − 2) sin 2(t − 2) / 0 if 0 ≤ t < 2, = 1 sin 2(t − 2) if t ≥ 2. 2 −1

2

3$ $ 1 $ 36. L = h(t − 2) L s2 − 4 $t→t−2 2 ' (3$ $ 1 1 1 $ = h(t − 2) L−1 − $ 4 s−2 s+2 t→t−2% $ & 1 2t 1 = h(t − 2) ( 4 (e − e−2t )$t→t−2 = 4 h(t − 2) e2(t−2) − e−2(t−2) / 0 if 0 ≤ t < 2, & = 1 % 2(t−2) −2(t−2) e − e if t ≥ 2. 4 −1

2

3

−1

2

3 2 3$ $ se−4s s −1 $ = h(t − 4) L 2 2 s + 3s + 2 s + 3s + 2 $t→t−4 2 3$ $ 2 1 $ = h(t − 4) L−1 − s + 2 s + 1 $t→t−4

37. L−1

2

e−2s s2 − 4

1 Solutions

191

$ % & = h(t − 4) (2e−2t − e−t )$t→t−4 = h(t − 4) 2e−2(t−4) − e−(t−4) / 0 if 0 ≤ t < 4, = −2(t−4) −(t−4) 2e −e if t ≥ 4.

3 2 3$ 2 3$ $ $ e−2s + e−3s 1 1 −1 −1 $ $ 38. L = h(t−2) L +h(t−3) L $ 2 2 2 s − 3s + 2 s − 3s + 2 t→t−2 s − 3s + 2 $t→t−3 2 3$ $ 1 1 −1 $ = h(t − 2) L − s − 2 s − 1 $t→t−2 2 3$ $ 1 1 $ + h(t − 3) L−1 − s −$ 2 s − 1 $t→t−3 $ = h(t − 2) (e2t − et )$t→t−2 + h(t − 3) (e2t − et )$t→t−3 % & % & = h(t − 2) e2(t−2) − et−2 + h(t − 3) e2(t−3) − et−3 2 3 2 3 2 3$ 1 − e−5s 1 1 $$ −1 −1 −1 39. L =L − h(t − 5) L 2 2 s s s2 $t→t−5 / t if 0 ≤ t < 5, = t − h(t − 5) (t)|t→t−5 = t − (t − 5)h(t − 5) = 5 if t ≥ 5. −1

2

3 2 3 2 3$ 1 + e−3s 1 1 $$ −1 −1 = L + h(t − 3) L 4 4 s s s4 $t→t−3 % 1 3 &$ 1 3 1 1 = 6 t + h(t − 3) 6 t $t→t−3 = 6 t3 + 6 (t − 3)3 h(t − 3) / 1 3 t if 0 ≤ t < 3, = 61 3 1 3 t + (t − 3) if t ≥ 3. 6 6 2 3 2 3$ $ 2s + 1 2s + 1 $ 41. L−1 e−πs 2 = h(t − π)L−1 2 s + 6s + 13 s + 6s + 13 $t→t−π 2 3$ 2(s + 3) − 5 $$ = h(t − π) L−1 (s + 3)2 + 22 $t→t−π 2 3$ $ 2(s + 3) $ = h(t − π) L−1 2 2 (s + 3) + 2 $t→t−π 2 3$ $ −5 $ + h(t − π) L−1 2 2 (s + 3) + 2 $t→t−π$ = h(t − π) (2e−3t cos 2t − 52 e−3t sin 2t)$t→t−π % & −3(t−π) = h(t 2 cos 2(t − π) − 52 sin 2(t − π) / − π)e 40. L−1

=

2

0 % e−3(t−π) 2 cos 2t −

5 2

sin 2t

&

if 0 ≤ t < π, if t ≥ π.

42. This is the same calculation as the previous exercise with the additional term added: 2 3 2s + 1 5 −1 L = 2e−3t cos 2t − e−3t sin 2t. s2 + 6s + 13 2

192

1 Solutions

2 3 2s + 1 Combining this with the previous calculation gives L−1 (1 − e−πs ) 2 = s + 6s + 13 2 3 2s + 1 L−1 s22 + 6s + 13 3 2s + 1 −1 −πs −L e s2 + 6s + 13& % −3t =e 2 cos 2t − 52 sin 2t % & − h(t − π)e−3(t−π) 2 cos 2(t − π) − 52 sin 2(t − π)

43. Let b > 0. Since f1 and f2 are piecewise continuous on [0, ∞) they only have finitely many jump discontinuities on [0, b). It follows that f1 + cf2 have only finitely many jump on [0, b). Thus f1 + cf2 is piecewise continuous on [0, ∞). 44. By Exercise 43 f1 + cf2 is piecewise continuous on [0, ∞). There are constants K1 , K2 , a1 and a2 so that f1 (t) ≤ K1 ea1 t and f2 (t) ≤ K2 ea2 t . Now let K = K1 + |c| K2 and let a be the larger of a1 and a2 . Then |f1 (t) + cf2 (t)| ≤ |f2 (t)| + |c| |f2 (t)| ≤ K1 eat + K2 |c| eat = Keat .

It follows that f1 + cf2 is of exponential type. Thus H is closed under addition and scalar multiplication, i.e. is a linear space. 0 /

Section 6.3 1. We write the forcing function as f (t) = 3h(t − 1). Applying the Laplace transform, partial fractions, and simplifying gives ' ( −3 −3 1 1 −s Y (s) = e = − e−s . s(s + 2) 2 s s+2 Laplace inversion now gives " # 3 y = − h(t − 1) 1 − e−2(t−1) = 2

/

0 % & − 32 1 − e−2(t−1)

if 0 ≤ t < 1 . if 1 ≤ t < ∞

2. We write the forcing function as f (t) = −5χ[0,1) +5χ[1,∞) = −5+10h(t− 1). Applying the Laplace transform, partial fractions, and simplifying gives

1 Solutions

193

1 5 10 − + e−s s + 5 s(s + 5) s(s + 5) ' ( 2 1 2 2 = − + − e−s . s+5 s s s+5

Y (s) =

Laplace inversion now gives " # y = 2e−5t − 1 + 2 − 2e−5(t−1) h(t − 1) / 2e−5t − 1 if 0 ≤ t < 1 = . 2e−5t + 1 − 2e−5(t−1) if 1 ≤ t < ∞ 3. We write the forcing function as f (t) = 2χ[2,3) = 2h(t − 2) − 2h(t − 3). Applying the Laplace transform, partial fractions, and simplifying gives % −2s & 2 e − e−3s s(s − 3) ' ( & 2 1 1 % −2s = − e − e−3s . 3 s−3 s

Y (s) =

Laplace inversion now gives

# " # # 2 "" 3(t−2) e − 1 h(t − 2) − e3(t−3) − 1 h(t − 3) 3   if 0 ≤ t < 2 0 % & 2 3(t−2) = −1 if 2 ≤ t < 3 . 3 e  &  2 % 3(t−2) 3(t−3) −e if 3 ≤ t < ∞ 3 e

y =

4. We write the forcing function as f (t) = t−th(t−1). Applying the Laplace transform, partial fractions, and simplifying gives 1 1 1 − 2 e−s − e−s + 2) s (s + 2) s(s + 2) ' ( ' ( 1 2 1 1 1 2 1 1 = − + − + − e−s . 4 s2 s s+2 4 s2 s s+2

Y (s) =

s2 (s

Laplace inversion now gives # & 1" 1% 2t − 1 + e−2t − 2(t − 1) + 1 − e−2(t−1) h(t − 1) 4/ 4 −2t if 0 ≤ t < 1 1 2t − 1 + e = 4 e−2t + e−2(t−1) if 1 ≤ t < ∞.

y =

194

1 Solutions

5. We write the forcing function as f (t) = 12et χ[0,1) + 12eχ[1,∞) = 12et − 12(et − e)h(t − 1). Applying the Laplace transform, partial fractions, and simplifying gives ' ( 2 12 12e 12e Y (s) = + − e−s − s − 4 (s − 1)(s − 4) (s − 1)(s − 4) s(s + 4) ' ( 6 4 −4 1 3 = − − e−s e + + . s−4 s−1 s−1 s−4 s Laplace inversion now gives " # y = 6e4t − 4et − e −4et−1 + e4(t−1) + 3 h(t − 1)

= 6e4t − 4et + 4et h(t − 1) − e4t−3 h(t − 1) − 3eh(t − 1) / 6e4t − 4et if 0 ≤ t < 1 = . 4t 4t−3 6e − e − 3e if 1 ≤ t < ∞

6. We write the forcing function as f (t) = 10 sin tχ[0,π) = 10 sin t − 10(sin t)h(t − π). Applying the Laplace transform, partial fractions, and simplifying gives ' ( −1 10 10 −πs Y (s) = + +e s + 3 (s + 3)(s2 + 1) (s + 3)(s2 + 1) ' ( −s 3 1 s 3 −πs = 2 + +e − + . s + 1 s2 + 1 s + 3 s2 + 1 s2 + 1 Laplace inversion now gives " # y = − cos t + 3 sin t + e−3(t−π) − cos(t − π) + 3 sin(t − π) h(t − π) " # = − cos t + 3 sin t + e−3(t−π) + cos t − 3 sin t / − cos t + 3 sin t if 0 ≤ t < π = . −3(t−π) e if π ≤ t < ∞ 7. Applying the Laplace transform, partial fractions, and simplifying gives e−3s s(s2 + 9) ' ( 1 1 s = − e−3s . 9 s s2 + 9

Y (s) =

1 Solutions

195

Laplace inversion now gives 1 y = (1 − cos 3(t − 3))h(t − 3) = 9

/

0 1 9 (1

− cos 3(t − 3))

if 0 ≤ t < 3 . if 3 ≤ t < ∞

8. Write the forcing function as f (t) = χ[0,5) = 1 − h(t − 5). Now apply the Laplace transform, partial fractions, and simplify to get 1 1 1 + − e−5s (s − 1)(s − 4) s(s − 1)(s − 4) s(s − 1)(s − 4) ' ( 11 5 1 2 1 11 1 1 1 1 = + − − + − e−5s 4 s 12 s − 4 3 s − 1 4 s 12 s − 4 3 s − 1

Y (s) =

Laplace inversion now gives ' ( 1 5 4t 2 t 1 1 4(t−5) 1 t−5 y = + e − e − + e − e h(t − 5) 4 12 3 4 12 3 " # # 1 " = 3 + 5e4t − 8et − 3 + e4(t−5) − 4et−5 h(t − 5) 12 / if 0 ≤ t < 5 1 3 + 5e4t − 8et = . 12 5e4t − 8et + e4(t−5) − 4et−5 if 5 ≤ t < ∞ 9. Write the forcing function as f (t) = 6χ[1,3) = 6h(t − 1) − 6h(t − 3). Now apply the Laplace transform, partial fractions, and simplify to get % −s & 6 e − e−3s s(s + 2)(s + 3) ' ( % −s & 1 3 2 = − + e − e−3s s s+2 s+3

Y (s) =

Now we take the inverse Laplace transform and simplify " # y = 1 − 3e−2(t−1) + 2e−3(t−1) h(t − 1) " # − 1 − 3e−2(t−3) + 2e−3(t−3) h(t − 3)   0 = 1 − 3e−2(t−1) + 2e−3(t−1)   −2(t−3) 3e − 3e−2(t−1) − 2e−3(t−3) + 2e−3(t−1)

to get

if 0 ≤ t < 1 if 1 ≤ t < 3 if 3 ≤ t < ∞

10. Observe that L {h(t − 2π) sin t} = e−2πs L {sin(t + 2π)} = e−2πs s21+1 , since sin(t + 2π) = sin t. Apply the Laplace transform, partial fractions,

196

1 Solutions

and simplify to get s 1 + e−2πs s2 + 9 (s2 + 9)(s2 + 1) ' ( s −1 3 1 1 = 2 + + e−2πs s +9 24 s2 + 32 8 s2 + 1

Y (s) =

Laplace inversion gives 1 1 sin(3(t − 2π))h(t − 2π) + sin(t − 2π)h(t − 2π) 24 8 1 1 = cos 3t − sin(3t)h(t − 2π) + sin(t)h(t − 2π) 24 8 / cos 3t if 0 ≤ t < 2π = 1 cos 3t − 24 sin 3t + 18 sin t if 2π ≤ t < ∞

y = cos 3t −

11. Apply the Laplace transform, partial fractions, and simplify to get 1 1 + e−3s (s + 1)2 s(s + 1)2 ' ( 1 1 1 1 −3s = + − − + e (s + 1)2 (s + 1)2 s+1 s

Y (s) =

Laplace inversion gives " # y = te−t + −(t − 3)e−(t−3) − e−(t−3) + 1 h(t − 3) " # = te−t + 1 − (t − 2)e−(t−3) h(t − 3) / te−t if 0 ≤ t < 3 = . −t −(t−3) 1 + te − (t − 2)e if 3 ≤ t < ∞ 12. Write the forcing function as f (t) = e−t χ[0,4) = e−t − e−t h(t − 4) and 0 1 1 1 1 then L {f (t)} = s+1 − e−4s L e−(t+4) = s+1 − s+1 e−4 e−4s . Apply the Laplace transform, partial fractions, and simplify to get Y (s) =

1 1 − e−4 e−4s . (s + 1)3 (s + 1)3

Laplace inversion then gives

1 Solutions

197

t2 −t (t − 4)2 −(t−4) −4 e − e e h(t − 4) 2 2 t2 −t (t − 4)2 −t = e − e h(t − 4) 2 2 / 2 t −t if 0 ≤ t < 4 2e = . −t (4t − 8)e if 4 ≤ t < 0

y =

gal lb 13. For the first three minutes, source one adds salt at a rate of 1 gal · 2 min = lbs 2 min . and after that source two takes over and adds salt at a rate of gal lb lbs 5 gal · 2 min = 10 min . Thus the rate at which salt is being added is given by the function / 2 if 0 ≤ t < 3 f (t) = 10 if 3 ≤ t < ∞.

= 2χ[0,3) + 10χ[3,∞)

= 2(1 − h(t − 3)) + 10(h(t − 3)) = 2 + 8h(t − 3). The output rate of salt is given by led to the differential equation

y(t) 4

·2 =

1 y ! + y(t) = 2 + 8h(t − 3), 2

y(t) 2

lbs/min. We are thus

y(0) = 0.

We take the Laplace transform of both sides and use partial fractions to get 2 8e−3s + s(s + 1/2) s(s + 1/2) ' ( 4 4 16 16 = − + e−3s − . s s + 1/2 s s + 1/2

Y (s) =

Laplace inversion now gives −t

−(t−3)

y(t) = 4 − 4e 2 + 16h(t − 3) − 16e 2 h(t − 3) / −t 4 − 4e 2 if 0 ≤ t < 3 = . −(t−3) −t 20 − 4e 2 − 16e 2 if t ≥ 3. L 14. For the first five minutes, source one adds salt at a rate of 2 kg L · 4 min = kg 8 min and after that source two takes over and adds salt at a rate of kg kg L 3 L · 4 min = 12 min . Thus the rate at which salt is being added is given

198

1 Solutions

by the function f (t) =

/

8 12

if 0 ≤ t < 5 if 5 ≤ t < ∞.

= 8χ[0,5) + 12χ[5,∞)

= 8(1 − h(t − 5)) + 12(h(t − 5)) = 8 + 4h(t − 5). The output rate of salt is given by to the differential equation

y(t) 4

· 4 = y(t) kg/min. We are thus led

y ! + y(t) = 8 + 4h(t − 5),

y(0) = 1.

We take the Laplace transform of both sides and use partial fractions to get 1 8 4e−5s + + s + 1 s(s + 1) s(s + 1) ' ( 8 7 4 4 = − + − e−5s . s s+1 s s+1

Y (s) =

Laplace inversion now gives y(t) = 8 − 7e−t + (4 − 4e−(t−5) )h(t − 5) / 8 − 7e−t if 0 ≤ t < 5 = . −t −(t−5) 12 − 7e − 4e if 5 ≤ t < ∞ L 15. For the first two minutes, source one adds salt at a rate of 1 kg L · 3 min = kg 3 min . Thereafter source two takes over for two minutes but the input rate of salt is 0. Thereafter source on take over again and adds salt to the tank kg at a rate of 3 min . Thus the rate at which salt is being added is given by the function   if 0 ≤ t < 2 3 f (t) = 0 if 2 ≤ t < 4   3 if 4 ≤ t < ∞. = 3χ[0,2) + 3χ[4,∞) = 3(1 − h(t − 2) + h(t − 4)).

The output rate of salt is given by led to the differential equation

y(t) 10

·3 =

3 10 y(t)

kg/min. We are thus

1 Solutions

199

y! +

3 y(t) = 3(1 − h(t − 2) + h(t − 4)), 10

y(0) = 2.

We take the Laplace transform of both sides, simplify, and use partial fractions to get 2 3 + (1 − e−2s + e−4s ) s + 3/10 s(s + 3/10) ' ( 10 8 10 10 = − + − (e−4s − e−2s ). s s + 3/10 s s + 3/10

Y (s) =

Laplace inversion now gives " # y(t) = 10 − 8e−3t/10 − 10 − 10e−3(t−2)/10 h(t − 2) " # + 10 − 10e−3(t−4)/10 h(t − 4)  −3t/10  10 − 8e = 10e−3(t−2)/10 − 8e−3t/10   10 − 8e−3t/10 + 10e−3(t−2)/10 − 10e−3(t−4)/10

if 0 ≤ t < 2 if 2 ≤ t < 4 . if 4 ≤ t < ∞

16. We apply the Laplace transform to each side of y ! + ay = Aχ[α,β) to get sY (s) − y0 + aY (S) =

A −αs (e − e−βs ). s

Now simplify, solve for Y (s), and apply partial fractions to get y0 A + (e−αs − e−βs ) s + a s(s + a) ' ( y0 A 1 1 = + − (e−αs − e−βs ). s+a a s s+a

Y (s) =

Laplace inversion now gives # A" (1 − e−a(t−α) )h(t − α) − (1 − e−a(t−β) )h(t − β) a 0 if 0 ≤ t < α A = y0 e−at + 1 − e−a(t−α) if α ≤ t < β a  −a(t−β) −a(t−α) e −e if β ≤ t < ∞.

y(t) = y0 e−at +

200

1 Solutions

Section 6.4 1. Take the Laplace transform, solve for Y (s), and simplify to get Y (s) = e−s s+2 . Laplace inversion then gives y = e−2(t−1) h(t − 1) / 0 if 0 ≤ t < 1 = . e−2(t−1) if 1 ≤ t < ∞ 2. Take the Laplace transform, solve for Y (s), and simplify to get −1 3 e−2s + + s − 3 s(s − 3) s − 3 −1 e−2s = + . s s−3

Y (s) =

. Laplace inversion then gives

y = −1 + e3(t−2) h(t − 2) / −1 if 0 ≤ t < 2 = . −1 + e3(t−2) if 2 ≤ t < ∞ 3. Take the Laplace transform, solve for Y (s), and simplify to get Y (s) = 2 e−4s s−4 + s−4 . Laplace inversion then gives y = 2e4t + e4(t−4) h(t − 4) / 2e4t if 0 ≤ t < 4 = . 4t 4(t−4) 2e + e if 4 ≤ t < ∞ 4. Take the Laplace transform, solve for Y (s), and simplify to get Y (s) = e−s e−3s s+1 − s+1 . Laplace inversion then gives y = e−(t−1) h(t − 1) − e−(t−3) h(t − 3)   if 0 ≤ t < 1 0 −(t−1) = e if 1 ≤ t < 3 .   −(t−1) e − e−(t−3) if 3 ≤ t < ∞

5. We begin by taking the Laplace transform of each side and simplifying −πs to get Y (s) = s21+4 + se2 +4 . Laplace inversion then gives

1 Solutions

201

1 1 sin 2t + sin 2(t − π)h(t − π) 2 2 1 1 = sin 2t + sin(2t) h(t − π) 2 2 / sin 2t if 0 ≤ t < π 2 = . sin 2t if π ≤ t < ∞

y =

6. Apply the Laplace transform, take partial fractions, and simplify to get e−s e−2s − 2 −1 s −1 ' ( ' ( 1 1 1 1 1 1 −s = − e − − e−2s . 2 s−1 s+1 2 s−1 s+1

Y (s) =

s2

Laplace inversion now gives # # 1 " t−1 1 " t−2 e − e−(t−1) h(t − 1) − e − e−(t−2) h(t − 2) 2 2  0 if 0 ≤ t < 1 1  t−1 = e − e−(t−1) if 1 ≤ t < 2 . 2  t−1 −(t−1) t−2 −(t−2) e −e −e +e if 2 ≤ t < ∞

y =

7. Apply the Laplace transform, partial fractions, and simplify to get s+3 2e−2s + (s + 1)(s + 3) (s + 1)(s + 3) ' ( 1 1 1 = + − e−2s . s+1 s+1 s+3

Y (s) =

Laplace inversion gives " # y = e−t + e−(t−2) − e−3(t−2) h(t − 2) / e−t if 0 ≤ t < 2 = . −t −(t−2) −3(t−2) e +e −e if 2 ≤ t < ∞ 8. Apply the Laplace transform to get Y (s) = Laplace inversion gives

s 1 1 + e−πs − 2 e−2πs . s2 + 4 s2 + 4 s +4

202

1 Solutions

1 1 sin(2(t − π))h(t − π) − sin(2(t − 2π))h(t − 2π) 2 2 1 1 = cos 2t + (sin 2t)h(t − π) − (sin 2t)h(t − 2π) 2 2   if 0 ≤ t < π cos 2t 1 = cos 2t + 2 sin 2t if π ≤ t < 2π .   cos 2t if 2π ≤ t < ∞

y = cos 2t +

9. Take the Laplace transform, apply partial fractions, and simplify to get s+1 3 + e−s (s + 2)2 (s + 2)2 1 1 3 = − + e−s . 2 (s + 2) s + 2 (s + 2)2

Y (s) = −

Laplace inversion now gives y = te−2t − e−2t + 3(t − 1)e−2(t−1) h(t − 1) / te−2t − e−2t if 0 ≤ t < 1 = . te−2t − e−2t + 3(t − 1)e−2(t−1) if 1 ≤ t < ∞ 10. Take the Laplace transform, complete the square, and simplify to get 1 3 + e−πs s2 + 4s + 5 s2 + 4s + 5 1 3 = + e−πs . (s + 2)2 + 1 (s + 2)2 + 1

Y (s) =

Laplace inversion gives y = e−2t sin t + 3e−2(t−π) sin(t − π)h(t − π) = e−2t sin t − 3e2π e−2t (sin t)h(t − π) / 1 if 0 ≤ t < π = e−2t sin t . 1 − 3e2π if π ≤ t < ∞

11. The input rate of salt is 6 + 4δ3 while the output rate is 3 y(t) 12 . We thus have the differential equation y ! + 14 y = 6 + 4δ3 , y(0) = 0. Take the Laplace transform, apply partial fractions, and simplify to get

1 Solutions

203

6 4 + e−3s s(s + 1/4) s + 1/4 24 24 4 = − + e−3s . s s + 1/4 s + 1/4

Y (s) =

Laplace inversion now gives 1

1

y = 24 − 24e− 4 t + 4e− 4 (t−3) h(t − 3) / 1 24 − 24e− 4 t if 0 ≤ t < 3 = . − 14 t − 14 (t−3) 24 − 24e + 4e if 3 ≤ t < ∞ 12. There is 1 kg of salt per liter , so y(0) = 10. The input rate of salt is 1+δ2 while the output rate is 2 y(t) 10 . We thus have the differential equation y ! + 15 y = 1 + δ2 , y(0) = 10. Take the Laplace transform, apply partial fractions, and simplify to get 10 1 e−2s + + s + 1/5 s(s + 1/5) s + 1/5 5 5 1 = − + e−2s . s s + 1/5 s + 1/5

Y (s) =

Laplace inversion now gives 1

1

y = 5 + 5e− 5 t + e− 5 (t−2) h(t − 2) / 1 5 + 5e− 5 t if 0 ≤ t < 2 = . − 15 t − 15 (t−2) 5 + 5e +e if 2 ≤ t < ∞ 13. Clearly, y(0) = 0. The input rate is δ0 +δ2 +δ4 +δ6 while the output rate is .3 y. We are thus led to the differential equation y ! +y = k=0 δ2k , y(0) = 0. Take the Laplace transform and solve for Y (s) to get Y (s) =

3 7 e−2ks

k=0

Laplace inversion gives

s+1

.

204

1 Solutions

y =

3 7

k=0

e−(t−2k) h(t − 2k)

 −t  e  e−t + e−(t−2) = e−t + e−(t−2) + e−(t−4)    −t e + e−(t−2) + e−(t−4) + e−(t−6)

Using the formula 1 + r + r2 + · · · + rn = y(6) =

3 7

k=0

=

e−(6−2k) = %

& −2 4

1− e 1 − e−2

3 7

1−r n+1 1−r

e−2k =

k=0

if if if if

0≤t2 ). and therefore the differential equation that models this system is 1 y ! + y = 5δ0 (< t >2 ), 2

y(0) = 0.

By Proposition 6.6.6 the Laplace transform gives Y (s) =

5 s+

1 2

1 . 1 − e−2s

By Theorem 6.6.7 Laplace inversion gives

228

1 Solutions

y(t) = 5

,

∞ 7

N =0

N 7

e

n=0 ∞ N +1 7

e

1

= 5e− 2 t

− 12 (t−2n)

N =0 − 12 t e

= 5e

-

χ[2N,2(N +1))

−1 χ[2N,2(N +1)) e−1

1 2 [t]2 +1

−1 . e−1

The solution is sandwiched in between a lower and upper curve. The upper curve, u(t), is obtained by setting t = 2m to be an even integer in the formula for the solution and then continuing it to all reals. We obtain 1

1

u(2m) = 5e− 2 2m Thus

1

1 e 2 [2m]2 +1 − 1 e 2 2m+1 − 1 = 5e− 2 2m . e−1 e−1 1

1

u(t) = 5e− 2 t

1

e 2 t+1 − 1 e − e− 2 t =5 . e−1 e−1

In a similar way, the lower curve, l(t), is obtained by setting t = 2(m+1)− ( slightly less than the even integer 2(m + 1)) and continuing to all reals. We obtain 1 1 − e− 2 t l(t) = 5 . e−1 An easy calculation gives e limt→∞ u(t) = 5 e−1 1 7.91

1 and limt→∞ l(t) = 5 e−1 1 2.91.

This means that the salt fluctuation in the tank varies between 2.91 and 7.91 pounds for large values of t. 4. Let y(t) be the number of allegators at time t measured in months. We assume the Malthusian growth model y ! = ry. Thus y(t) = y(0)ert = 3000ert . To determine the growth rate r we know y(−12) = 2500 (12 months earlier there were 2500 allegators). Thus 2500 = 3000e−12r and 1 hence r = 12 ln 65 . The rate at which allegators are taken out of the swamp is 80 per month in odd months and 0 per month in even months. This can be modeled by 80 sw1 . The hunting policy implies that the rate of growth y ! is the difference between the natural growth ry and the rate at which allegators are taken out of the swamp. Thus y ! = ry − 80 sw1 , where r = get

1 12

y(0) = 3000,

ln 65 . We apply the Laplace transform and solve for Y (s) to Y (s) =

3000 1 1 1 − 80 . s−r s s − r 1 + e−s

1 Solutions

229

Partial fractions gives 1 1 = s(s − r) r

'

1 1 − s−r s

(

.

Hence we can write Y (s) = Let

3000 80 1 1 80 1 1 + − . s−r r s 1 + e−s r s − r 1 + e−s

Y1 (s) =

3000 s−r

Y2 (s) =

80 1 1 r s 1 + e−s

Y3 (s) =

80 1 1 . r s − r 1 + e−s

Then Y (s) = Y1 (s) + Y2 (s) − Y3 (s). Using Example 6.6.2 and Exercise 6.6.19 we get y1 (t) = 3000ert y2 (t) =

80 sw1 r

80 ert y3 (t) = r 1 + e−r =

,

1 + e−r

/

e−rN −e−rN

if t ∈ [N, N + 1) if t ∈ [N, N + 1)

N even N odd

-

80 ert (1 + e−r (−1)[t]1 e−r[t]1 ). r 1 + e−r

It follows now that y(t) = y1 (t) + y2 (t) − y3 (t) 80 80 ert = 3000ert + sw1 − (1 + e−r (−1)[t]1 e−r[t]1 ). r r 1 + e−r To determine the population at the beginning of 5 years = 60 months we compute y1 (60) = 3000e60r = 7464.96 80 y2 (60) = = 5265.42 r 80 e60r y3 (60) = (1 + e−61 r ) = 9213.43. r 1 + e−r

230

1 Solutions

Therefore y(60) ≈ 3517. 5. Let y(t) be the number of allegators at time t measured in months. We assume the Malthusian growth model y ! = ry. Thus y(t) = y(0)ert = 3000ert . To determine the growth rate r we know y(−12) = 2500 (12 months earlier there were 2500 allegators). Thus 2500 = 3000e−12r and 1 hence r = 12 ln 65 . The elite force of Cajun allegator hunters instantaneously remove 40 allegators at the beginning of each month. This can be modeled by 40(δ0 +δ1 +· · · ) = 40δ0 (< t >1 ). The mathematical model is thus y ! = ry − 40δ0 (< t >1 ), y(0) = 3000,

1 where r = 12 ln 65 . We apply the Laplace transform and use Proposition 6.6.6 to get 3000 1 1 Y (s) = − 40 . s−r s − r 1 − e−s Let

Y1 (s) =

3000 s−r

Y2 (s) = 40

1 1 s − r 1 − e−s

Then Y (s) = Y1 (s) + Y2 (s). We use Theorem 6.6.7 to get y1 (t) = 3000ert ,N ∞ 7 7 r(t−n) y2 (t) = 40 e χ[N,N +1) =

N =0 n=0 ∞ 7 1 − e−r(N +1) 40ert χ[N,N +1) 1 − e−r N =0 rt −r([t]1 −t+1)

= 40 It follows now that

e −e 1 − e−r

y(t) = y1 (t) − y2 (t) = 3000ert − 40

.

ert − e−r([t]1 −t+1) . 1 − e−r

To determine the population at the beginning of 5 years = 60 months we compute

1 Solutions

231

e60r − e−r 1 − e−r = 7464.96 − 3988.16 ≈ 3477

y(60) = 3000e60r − 40

Section 6.8 1. Since cβ =

√

2 is not an odd multiple of π we get " # y(t) = 2 2 sw1 (t) − (−1)[t]1 (cos < t >1 −α sin < t >1 ) −2 (cos t + α sin t)) ,

where α =

√ − sin √2 . 1+cos 2

Since

βc π

=

2 π

is irrational the motion is non periodic

2. Since cβ = 2 is not an odd multiple of π we get ' ( 1 1 2 [t/2]1 y(t) = π 2 sw2 (t) − (−1) (cos < t >2 −α sin < t >2 ) π π ' ( 1 1 −π 2 cos t + α sin t . π π Since

βc π

= 2/π is irrational the motion is non periodic.

3. Since cβ = 2π is not an odd multiple of π we get 1 π2 1 = π2

y(t) =

"

2 sw2 (t) − (−1)[t/2]1 cos π < t >2 − cos πt " " ## 2 sw2 (t) − cos πt (−1)[t/2]1 + 1 ,

where we have used the identity cos π < t >2 = cos πt. Since rational the motion is periodic.

# βc π

= 2 is

4. Since cβ = π/2 is not an odd multiple of π we get y(t) =

Since

βc π

# 8 " π π 2 sw1 (t) − (−1)[t]1 (cos < t >1 + sin < t >1 ) 2 π 2 2 8 " π π # − 2 cos t − sin t . π 2 2

= 1/2 is rational the motion is periodic.

232

1 Solutions

5. Since cβ = π is an odd multiple of π we get y(t) = Resonance occurs.

2 π2

(sw1 (t) − [t]1 cos πt − cos πt).

6. Since cβ = 5π is an odd multiple of π we get y(t) = 3 (sw5w (t) − [t/5π]1 cos t − cos t). Resonance occurs. 7. Since cβ = π is not a multiple of 2π and γ = 0 we get y(t) = sin t + sin < t >π + (1 + (−1)[t/π]1 ) sin t, where we have used sin < t >π = (−1)[t/π]1 sin t. Since the motion is periodic.

βc 2π

=

1 2

is rational

8. Since cβ = π/2 is not a multiple of 2π and γ = −1 we get y(t) = Since

βc 2π

1 4

=

# 2" π π π π sin t − cos t + sin < t >2 + cos < t >2 π 4 4 4 4

is rational the motion is periodic.

9. Since cβ = 1 is not a multiple of 2π and γ =

− sin 1 1−cos 1

we get

y(t) = sin t + γ cos t + sin < t >1 −γ cos < t >1 , where γ = periodic. 10. Since cβ = y(t) = where γ = periodic.

− sin 1 1−cos 1 . √ π 2 2

Since

βc 2π

=

1 2π

is not rational the motion is non

is not a multiple of 2π and γ =

− sin

1−cos

√

2 2 π √ 2 2 π

we get

# 2" π π π π sin t + γ cos t + sin < t >√2 −γ cos < t >√2 , π 2 2 2 2

√ 2 2 π √ 1−cos 22 π

− sin

. Since

βc 2π

=

√

2 4

is not rational the motion is non

11. Since βc = 2π we get y(t) = 2(sin t)(1 + [t/2π]1 ). Resonance occurs. 12. Since βc = 4π we get y(t) = Resonance occurs.

2 (sin π)t(1 + [t/4]1 ). π

1 Solutions

233

13. First we have N 7 %

n=0

eiθ

&n

=

N 7

einθ

n=0

=

N 7

cos nθ + i sin nθ

n=0

=

N 7

n=0

cos nθ + i

N 7

sin nθ.

n=0

On the other hand, N 7 %

n=0

eiθ

&n

=

ei(N +1)θ − 1 eiθ − 1

cos(N + 1)θ − 1 + i sin(N + 1)θ cos θ − 1 + i sin θ cos(N + 1)θ − 1 + i sin(N + 1)θ cos θ − 1 − i sin θ = · cos θ − 1 + i sin θ cos θ − 1 − i sin θ =

The product of the denominators simplifies to 2 − 2 cos θ. The product of the numerators has a real and imaginary part. Call them R and I, respectively. Then R = (cos((N + 1)θ) − 1)(cos θ − 1) + sin((N + 1)θ) sin θ = cos((N + 1)θ) cos θ + sin((N + 1)θ) sin θ − cos(N + 1)θ − cos θ + 1 = cos(N θ) − cos(N θ) cos θ + sin(N θ) sin θ − cos θ + 1 = (cos(N θ) + 1)(1 − cos θ) + sin(N θ) sin θ

and I = sin((N + 1)θ)(cos θ − 1) − sin θ(cos((N + 1)θ) − 1)

= sin((N + 1)θ) cos θ − sin((N + 1)θ) + sin θ − cos((N + 1)θ) sin θ = sin(N θ) − sin(N θ) cos θ − cos(N θ) sin θ + sin θ = sin(N θ)(1 − cos θ) + sin θ(1 − cos(N θ)).

Equating real and imaginary parts and simplifying now gives

234

1 Solutions N 7

cos nθ =

n=0

R 2 − 2 cos θ

(cos(N θ) + 1)(1 − cos θ) + sin(N θ) sin θ 2 − 2 cos θ 1 = (1 + cos N θ + γ sin N θ) 2 =

and N 7

sin nθ =

n=0

I 2 − 2 cos θ

sin(N θ)(1 − cos θ) + sin θ(1 − cos(N θ)) 2 − 2 cos θ 1 = (sin N θ + γ(1 − cos N θ)) 2 =

14. Observe that cos n(θ + π) = (−1)n cos nθ sin n(θ + π) = (−1)n sin nθ sin(θ + π) − sin θ γ(θ + π) = = = −α(θ). 2 − 2 cos(θ + π) 2 + 2 cos θ Now replace θ by θ + π in the summation formulas given in Exercise 13. We get N 7

n

(−1) cos nθ =

n=0

N 7

cos(n(θ + π))

n=0

1 (1 + cos(N (θ + π)) + γ(θ + π) sin(N (θ + π))) 2 & 1% = 1 + (−1)N cos(N θ) − α(θ)(−1)N sin(N θ) 2 =

and N 7

n

(−1) sin nθ =

n=0

N 7

sin(n(θ + π))

n=0

1 (sin(N (θ + π)) + γ(θ + π)(1 − cos(N (θ + π)))) 2 & 1% = (−1)N sin(N θ) − α(θ)(1 − (−1)N cos(N θ)) 2 & 1% = −α + (−1)N (sin N θ + α cos N θ) 2 =

1 Solutions

235

15. Let R(v) =

N 7

cos nv =

n=0

I(v) =

N 7

sin nv =

n=0

N 7 1 (1 + cos N v + γ sin N v) = Re einv 2 n=0

N 7 1 (sin N v + γ(1 − cos N v)) = Im einv , 2 n=0

as in Exercise 13. Now N 7

cos(u + nv) = Re

n=0

N 7

ei(u+nv)

n=0

,

= Re eiu

N 7

einv

n=0

-

= Re ((cos u + i sin u)(R(v) + iI(v))) = (cos u)R(v) − (sin u)I(v) 1 = (cos u + cos u cos N v + γ cos u sin N v) 2 1 − (sin u sin N v + γ(sin u − sin u cos N v)) 2 1 = (cos u + cos(u + N v) + γ(− sin u + sin(u + N v)) . 2 Similarly, N 7

sin(u + nv) = Im

n=0

N 7

ei(u+nv)

n=0

,

= Im e

iu

N 7

n=0

e

inv

-

= Im ((cos u + i sin u)(R(v) + iI(v))) = (sin u)R(v) + (cos u)I(v) 1 = (sin u + sin u cos N v + γ sin u sin N v) 2 1 + (cos u sin N v + γ(cos u − cos u cos N v)) 2 1 = (sin u + sin(u + N v) + γ(cos u − cos(u + N v))) . 2 16. Observe that

236

1 Solutions

cos n(θ + π) = (−1)n cos nθ sin n(θ + π) = (−1)n sin nθ sin(θ + π) − sin θ γ(θ + π) = = = −α(θ). 2 − 2 cos(θ + π) 2 + 2 cos θ Now replace v by v + π in Exercise 15. We then get N 7

(−1)n cos(u + nv)

n=0

=

N 7

cos(u + nv + nπ)

n=0

1 (cos u − γ(v + π) sin u) 2 1 + (cos(u + N v + N π) + γ(v + π) sin(u + N v + N π)) 2 1 = (cos u + α(v) sin u) 2 & 1% + (−1)N cos(u + N v) − α(v)(−1)N sin(u + N v) 2 1 = (cos u + α sin u) 2 (−1)N + (cos(u + N v) − α sin(u + N v)) 2 =

and N 7

(−1)n sin(u + nv)

n=0

=

N 7

sin(u + nv + nπ)

n=0

1 (sin u + γ(v + π) cos u) 2 1 + (sin(u + N v + N π) − γ(v + π) cos(u + N v + N π)) 2 1 = (sin u − α(v) cos u) 2 & 1% + (−1)N sin(u + N v) + α(v)(−1)N cos(u + N v) 2 1 = (sin u − α cos u) 2 (−1)N + (sin(u + N v) + α cos(u + N v)) 2 =

1 Solutions

237

Section 7.1 1. The ratio test gives

(n+1)2 n2

2. The ratio test gives

n n+1

3. The ratio test gives

2n n! 2n+1 (n+1)!

4. The ratio test gives

3n+1 n+1 n+2 3n

5. The ratio test gives

(n+1)! n!

6. Let u = t2 to get

∞ .

n=0

→ 1. R = 1.

→ 1. R = 1 =

=

1 2(n+1)

3(n+1) n+2

→ 0. R = ∞.

→ 3. R = 13 .

= n + 1 → ∞. R = 0.

(−1)n un (2n+1)!

$ $ n+1 $ (2n+1)! $ The ratio test gives $ (−1) n (−1) (2n+3)! $ =

→ 0. It follows that the radius of convergence in u is ∞ and hence the radius of convergence in t is ∞. 1 (2n+2)(2n+3)

7. t is a factor in this series which we factor out to get t

∞ .

n=0

(−1)n t2n (2n)! .

Since

t is a polynomial its presence will not change the radius of convergence. ∞ . (−1)n un Let u = t2 in the new powers series to get (2n)! . The ratio test n=0 $ $ $ (−1)n+1 (2n)! $ 1 gives $ (−1) n (2n+2)! $ = (2n+1)(2n+2) → 0. The radius of convergence in u and hence t is ∞. % &n (n+1) n! 8. The ratio test gives (n+1) = n+1 → e. Thus R = 1e . nn (n+1)! n

1·3·5···(2n+1) = 1·2·3·4···(2n+1) 1 2·4···2n ∞ 2 n n . (n!) 2 t is . The ratio (2n+1)! n=0 1 2 . R = 2.

9. The expression in the denominator can be written (2n+1)! 2n (1·2·3···n)

test gives

=

(2n+1)! 2n n!

and the given power series

((n+1)!)2 2n+1 (2n+1)! (n!)2 2n (2n+3)!

=

10. Use the geometric series to get

(n+1)2 2 (2n+3)(2n+2)

1 1+t2

11. Use the geometric series to get −

∞ .

n=0

=

→

1 1−(−t2 )

1 t−a

=

=

∞ .

(−t2 )n =

n=0

−1 1 a 1− at

(−1)n t2n .

n=0

=

−1 a

tn an+1 .

12. Replace t by at in the power series for et to get eat = ∞ a n tn . . n=0 n!

∞ .

∞ % &n . t

n=0

a

=

∞ (at)n . = n=0 n!

=

238

13. 14.

1 Solutions sin t t

et −1 t

1 t

=

n=0

1 t

=

∞ .

'

(−1)n t2n+1 (2n+1)!

∞ .

n=0

tn n!

=

∞ .

(−1)n t2n (2n+1)! .

n=0

( ∞ . −1 =

n=1

tn−1 n!

=

∞ .

n=0

tn (n+1)!

) 1 15. Recall that tan−1 t = 1+t 2 dt. Using the result of Exercise 10 we get ∞ ) ∞ . . 2n+1 tan−1 t = (−1)n t2n dt + C = (−1)n t2n+1 + C. Since tan−1 0 = 0 n=0

n=0

it follows that C = 0. Thus tan

−1

t=

∞ .

n=0

16. It is easy to verify that ln(1 + t2 ) = Exercise 10 it is easy to see that ln(1 + t2 ) = 2

∞ .

Thus ln(1 + t ) = 2 2

ln(1 + t ) = 2

∞ .

2t 1+t2

∞ .

2n+2

(−1)n t2n+2 . An index shift n → n − 1 gives

n=0

2n

(−1)n−1 t2n =

∞ .

n=1

2n

(−1)n−1 tn .

17. Since tan t is odd we can write tan t = ∞ .

dt. Using the result of ∞ . = 2 (−1)n t2n+1 . Hence

2t 1+t2

n=0

2n+2

n=1

cos t

)

(−1)n t2n+2 + C. Evaluating at t = 0 give C = 0.

n=0 2

2n+1

(−1)n t2n+1 .

∞ .

d2n+1 t2n+1 and hence sin t =

n=0

d2n+1 t2n+1 . Writing out a few terms gives t −

n=0 2 4 (1 − t2! + t4!

t3 3!

+

t5 5!

− ··· =

− · · · )(d1 t + d3 t3 + d5 t5 · · · ). Collecting like powers of t gives the following recursion relations d1 d1 d3 − 2! d3 d1 d5 − + 2! 4! d5 d3 d1 d7 − + − 2! 4! 6! Solving these equations gives d1 = 1 1 d3 = 3 2 d5 = 15 17 d7 = . 315

= 1 −1 = 3! 1 = 5! −1 = . 7!

1 Solutions

239

Thus tan t = 1 + 13 t3 +

2 5 15 t

+

17 7 315 t

+ ···.

18. Since sec t is even we can write its power series in the form The equation sec t = 2

4

1 cos t

=

∞ .

d2n t2n implies 1 =

n=0

6

∞ .

n=0

∞ .

d2n t2n .

n=0 ∞ 2n .

(−1)n t (2n)!

d2n t2n =

n=0

(1 − t2 + t4! − t6! + · · · ) · (d0 + d2 t2 + d4 t4 + d6 t6 + · · · ). Collecting like powers of t gives the following recursion relations d0 d2 d2 − 2 d2 d2 d4 − + 2! 4! d4 d2 d0 d6 − + − 2! 4! 6!

= 1 = 0 = 0 = 0.

Solving these equations gives d0 = 1 1 d2 = 2 5 d4 = 4! 69 d6 = . 6! Thus sec t = 1 + 12 t2 + 2

5 4 4! t 3

+

61 6 6! t

+ ···.

4

3

5

19. et sin t = (1 + t + t2! + t3! + t4! + · · · )(t − t3! + t5! − · · · ) = t + (1)t2 + ( −1 3! + 1 3 −1 1 4 1 1 1 1 5 1 3 1 5 2 )t + ( + )t + ( − + )t · · · = t + t + t − t . 2! 3! 3! 5! 2! 3! 4! 3 30 2

3

2

4

1 20. et cos t = (1 + t + t2! + t3! + · · · )(1 − t2! + t4! − · · · ) = 1 + (1)t + ( 2! − 1 2 1 1 3 1 1 1 1 4 1 3 1 4 )t + ( − )t + ( − + )t · · · = 1 + t − t − t + · · · . 2! 3! 2! 4! 2! 2! 4! 3 6

21. The ratio test gives infinite radius of convergence. Let f (t) be the function defined by the given power series. Then

240

1 Solutions ∞ 7

n+1 n t n! n=0 ' ( ∞ 7 n 1 n = (−1) + tn n! n! n=0

f (t) =

=

∞ 7

(−1)n

(−1)n

n=1 ∞ 7

= t

(−1)n+1

n=0 −t

= −te

∞ 7 1 1 tn + (−1)n tn (n − 1)! n! n=0

+ e−t

∞

tn 7 (−t)n + n! n=0 n!

22. The radius of convergence in infinite. Let f (t) the be function defined by the given power series. Then ∞ ∞ n 7 7 3 n tn t f (t) = −2 n! n! n=0 n=0

= e3t − 2et .

23. It is easy to check that the interval of convergence is (−1, 1). Let f (t) be the function defined by the given power series. Then *

f (t) dt =

∞ 7

(n + 1)

n=0 ∞ 7

= t

tn+1 +c n+1

tn + c

n=0

= Differentiation gives

t + c. 1−t

f (t) =

1 . (1 − t)2

24. It is not hard to see that the interval of convergence is (−1, 1). Let f (t) be the function defined by the given power series. Then f (t) = −t+t2 g(t), ∞ t2n−1 . where g(t) = . Observe that g(0) = 0 and n=1 2n − 1

1 Solutions

241 !

g (t) = =

∞ 7

n=1 ∞ 7

t2n−2 t2n

n=0

=

1 1 1 1 1 = + 1 − t2 21−t 21+t

where the last line is the partial fraction decomposition of the previous line. Integration gives 1 1 g(t) = − ln(1 − t) + ln(1 + t) 2 2 1 1+t = ln + c. 2 1−t Since 0 = g(0) = 0 + c we get c = 0. Therefore f (t) = −t +

t2 1 + t ln . 2 1−t

25. It is not hard to see that the interval of convergence is (−1, 1). Let f (t) be the given power series. A partial fraction decomposition gives ' ( 1 1 1 1 = − . (2n + 1)(2n − 1) 2 2n − 1 2n + 1 Therefore f (t) = Let f1 (t) = t2 2

ln

1+t 1−t

.∞

t2n+1 n=0 2n−1

∞ ∞ 1 7 t2n+1 1 7 t2n+1 − . 2 n=0 2n − 1 2 n=0 2n + 1

and f2 (t) =

.∞

t2n+1 n=0 2n+1 .

Then f1 (t) = −t +

by Exercise 24. Observe that f2 (0) = 0 and f2! (t) =

∞ 7

t2n

n=0

=

1 1 1 1 1 = + . 2 1−t 21+t 21−t

Integration and the fact that f2 (0) = 0 gives f2 (t) = now that

1 2

1+t ln 1−t . It follows

242

1 Solutions

1 (f1 (t) − f2 (t)) 2' ( 1 t2 1 + t 1 1 + t = −t + ln − ln 2 2 1−t 2 1−t 2 −t t − 1 1 + t = + ln . 2 4 1−t

f (t) =

26. et eit = e(1+i)t =

∞ 7 (1 + i)n tn . n! n=0

Now let’s compute a few powers of (1 + i). n=0 n=1 n=2 n=3 n=4 n=5 .. .

(1 + i)0 (1 + i)1 (1 + i)2 (1 + i)3 (1 + i)4 (1 + i)5 .. .

=1 =1+i = 2i = −2 + 2i = −4 = −4 − 4i

It now follows that et cos t = Re e(1+i)t ∞ 7 tn = (Re(1 + i)n ) n! n=0

t3 t4 t5 − 4 − 4 + ··· 3! 4! 5! t3 t4 t5 = 1+t− − − + ··· 3 6 30 = 1 + t + 0t2 − 2

Similarly, et sin t = Im e(1+i)t ∞ 7 tn = (Im(1 + i)n ) n! n=0

t2 t3 t4 t5 + 2 + 0 − 4 + ··· 2! 3! 4! 5! t3 t5 2 = t+t + − + ··· 3 30 = 0+t+2

27. The binomial theorem: (a + b)n =

.n

k=0

%n& k

ak bn−k .

1 Solutions

243

28. The root test gives

+ n

1 nn

=

1 n

→ 0. R = ∞.

$ $ $ $ if n is even and 2 if n is odd. Thus lim $ cn+1 $ does n→∞ cn √ n not exist. The ratio √ test does not apply. The root test gives that cn is 1 if n is odd and n 2 if n is even. As n approaches ∞ both even and odd terms approach 1. It follows that the radius of convergence is 1. & & −1 % −1 % 30. y ! (t) = e t t12 and y !! (t) = e t t14 − t23 . Here p1 (x) = x2 and p2 (x) = x4 − 2x3 . cn+1 cn

29. The ratio

is

1 2

−1

31. Suppose f (n) (t) = e t pn ( 1t ) where pn is a polynomial. Then f (n+1) (t) = & % & −1 −1 % −1 1 2 t = e t p e t t12 pn ( 1t )+p!n ( 1t ) −1 n+1 ( t ), where pn+1 (x) = x (pn (x)− t2 e % & −1 p!n (x)). By mathematical induction it follows that f (n) (t) = e t pn 1t for all n = 1, 2, . . .. pn (u) eu .

32. f (n) (u) = 0.

Repeated applications of L’Hospitals rule gives lim

u→∞

pn (u) eu

33. Since f (t) = 0 for t ≤ 0 clearly lim− f (n) (t) = 0 The previous problems t→0

imply that the right hand limits are also zero. Thus f (n) (0) exist and is 0.

34. Since all derivatives at t = 0 are zero the Taylor series for f at t = 0 gives the zero function and not f . Hence f is not analytic at 0.

Section 7.2 1. Let y(t) =

∞ .

cn tn . Then cn+2 =

n=0

cases to get c2n = c1

∞ .

n=0

t2n+1 (2n+1)!

c0 (2n)!

cn (n+2)(n+1) .

and c2n+1 =

c1 (2n+1)! .

Consider even and odd

Thus y(t) = c0

∞ .

n=0

t2n (2n)!

+

= c0 cosh t + c1 sinh t. (see Example 7.1.7) We observe that

the characteristic polynomial is s2 − 1 = (s − 1)(s + 1) so {et , e−t } is t −t t −t a fundamental set. But cosh t = e +e and sinh t = e −e ; the set 2 2 {cosh t, sinh t} is also a fundamental set. 2. Let y(t) =

∞ .

n=0

cn tn . Then cn+2 =

terms. After simplifying we get:

cn 1 n+2 (2cn+1 − n+1 ).

We write out several

=

244

1 Solutions

n=0 n=1 n=2 n=3 .. .

c2 c3 c4 c5 .. .

= = = =

1 2! (2c1 1 3! (3c1 1 4! (4c1 1 5! (5c1

− c0 ) − 2c0 ) − 3c0 ) − 4c0 )

A clear pattern emerges that can be verified by induction: cn =

1 (nc1 − (n − 1)c0 ). n!

It follows now that ∞ 7

c n tn =

n=0 ∞ 7

c1

∞ 7 1 n 1 nt − c0 (n − 1)tn . n! n! n=0 n=0

The first series simplifies to ∞ ∞ 7 7 1 n 1 nt = tn n! (n − 1)! n=0 n=1

=

∞ n+1 7 t n! n=0

= t

∞ n 7 t = tet . n! n=0

The second series simplifies to ∞ 7 1 (n − 1)tn n! n=0

=

∞ ∞ 7 n n 7 tn t − n! n! n=0 n=0

= tet − et . It follows that

3. Let y(t) = 2

∞ .

y(t) = c0 (1 − t)et + c1 tet . cn tn . Then cn+2 (n + 2)(n + 1) + k 2 cn = 0 or cn+2 =

n=0

k cn − (n+2)(n+1) . We consider first the even case.

1 Solutions

245 2

c0 c2 = − k(2·1

n=0

2

c4 = − k4·3c2 =

n=2

k 4 c0 4!

6

c6 = − k 6!c0 .. .

n=4 .. .

2n

c0 From this it follows that c2n = (−1)n k(2n)! . The odd case is similar. We 2n+1

k get c2n+1 = (−1)n (2n+1)! . The power series expansion becomes

y(t) =

∞ 7

c n tn

n=0 ∞ 7

= c0 + c1

n=0 ∞ 7

(−1)n

k 2n t2n (2n)!

(−1)n

k 2n+1 t2n+1 (2n + 1)!

n=0

= c0 cos kt + c1 sin kt. 4. Let y(t) =

∞ .

cn tn . Then cn+2 (n + 2)(n + 1) − 3(n + 1)cn+1 + 2cn = n=0 1 (n+2)(n+1) (3(n + 1)cn+1 − 2cn ). We write out several terms

cn+2 = simplify:

n=0 n=1 n=2 n=3 n=4 .. .

c2 c3 c4 c5 c6 .. .

= = = = =

0 or and

1 2! (3c1 − 2c0 ) 1 3! (7c1 − 6c0 ) 1 4! (15c1 − 14c0 ) 1 5! (31c1 − 30c0 ) 1 6! (63c1 − 62c0 )

After careful observation we get cn =

1 ((2n − 1)c1 − (2n − 2)c0 ). n!

(Admittedly, not easy to notice. However, the remark below shows why the presence of 2n is not unexpected.). From this we get ∞ 7 (2n − 1)tn y(t) = c1 n! n=0

−c0

∞ 7 (2n − 2)tn . n! n=0

246

1 Solutions

The first series can be written ∞ ∞ 7 2n tn 7 tn − = e2t − et . n! n! n=0 n=0

Similarly, the second series is e2t − 2et . The general solution is y(t) = c0 (2et − e2t ) + c1 (e2t − 2et ). The characteristic polynomial is s2 − 3s + 2 =0 (s − 1)(s − 2). It 1 follows 0 t 2t 1 that e , e is a fundamental set and so is 2et − e2t , e2t − 2et . The presence of e2t in the general solution implies that 2n tn appears in a power series solution. Thus the presence of 2n in the formula for cn is necessary. 5. Let y(t) =

∞ .

cn tn . Then the recurrence relation is

n=0

(n + 2)(n + 1)cn+2 − (n − 2)(n + 1)cn = 0

or

n−2 cn . n+2 Since there is a difference of two in the indices we consider the even and odd case. We consider first the even case. cn+2 =

c2 = −c0 c4 = 04 c2 = 0 c6 = 26 c4 = 0 .. .

n=0 n=2 n=4 .. .

It follows that c2n = 0 for all n = 2, 3, . . .. Thus ∞ 7

n=0

c2n t2n = c0 + c2 t2 + 0t4 + · · · = c0 (1 − t2 )

and hence y0 (t) = 1 − t2 . We now consider the odd case. n=1 n=3 n=5 n=7 .. .

c3 c5 c7 c9 .. .

= −1 3 c1 1 1 = 5 c3 = − 5·3 c1 3 1 = 7 c5 = − 7·5 c1 5 1 = 9 c5 = − 9·7 c1

1 Solutions

247

From this we see that c2n+1 = ∞ 7

n=0

−c1 (2n+1)(2n−1) .

c2n+1 t2n+1 = −c1

Thus

∞ 7

t2n+1 (2n + 1)(2n − 1) n=0

∞ . t2n+1 and hence y1 (t) = − (2n+1)(2n−1) . By Exercise 7.1.25 we can write y1 n=0 as ' ( t t2 − 1 1+t y1 (t) = − ln . 2 4 1−t

The general solution is

2

y(t) = c0 (1 − t ) − c1

'

t t2 − 1 + ln 2 4

'

1−t 1+t

((

.

(See also Exercise 5.5.15.) 6. Let y(t) =

∞ .

cn tn . Then the recurrence relation is cn+2 =

n=0

index step is two so we consider we get n=0 n=2 n=4 n=4 .. . In general, c2n =

−1 2n−1 c0 ∞ 7

n=0

The

the even and odd cases. In the even case c2 c4 c6 c8 .. .

= −c0 = 13 c2 = − 13 c0 = 35 c4 = − 15 c0 = 57 c6 = − 17 c0

and c2n t2n = −c0

Thus y0 (t) = − Now observe that

n−1 n+1 cn .

∞ 7

∞ 7

t2n . 2n − 1 n=0

t2n . 2n − 1 n=0

248

1 Solutions ∞ 7 y0 (t) t2n−1 = − t 2n − 1 n=0 ' (! ∞ 7 y0 (t) = t2n−2 t n=0

1 1 t2 1 − t2 ' ( y0 (t) 1 1 1−t = + ln t t 2 1+t ' ( t 1−t y0 (t) = 1 + ln . 2 1+t =

For the odd terms we get n=1 n=3 n=5 .. .

c3 = 02 c1 c5 = 24 c3 = 0 c7 = 0 .. .

Thus ∞ 7

n=0

c2n+1 t2n+1 = c1 t + 0t3 + · · · = c1 t

and hence y1 (t) = t. The general solution is ' ' (( t 1−t y(t) = c0 1 + ln + c1 t. 2 1+t 7. Let y(t) =

∞ .

cn tn . Then the recurrence relation is

n=0

cn+2 =

2 n−1 cn+1 − cn . n+2 (n + 2)(n + 1)

For the first several terms we get n=0 n=1 n=2 n=3 .. .

c2 c3 c4 c5 .. .

1 = 0c1 + 12 c0 = 2! c0 1 1 = 2 c2 − 0 = 3! c0 1 1 = 24 c3 − 4·3 c2 = 4! c0 3 2 3 = 5 c4 − 5·4 c3 = 5! c0 −

2 5! c0

=

1 5! c0

1 Solutions

249

In general, cn =

1 c0 , n!

n = 2, 3, . . . .

We now get y(t) =

∞ 7

cn tn

n=0

= c0 + c1 t +

∞ 7

c n tn

n=2

= (c1 − c0 )t + c0 + c0 t + c0 = (c1 − c0 )t + c0 et = c0 (et − t) + c1 t. 8. Let y(t) =

∞ .

∞ n 7 t n! n=2

cn tn . Then the recurrence relation is

n=0

cn+2 = −

(n − 2)(n − 1) cn . (n + 2)(n + 1)

We separate into even and odd cases. The even case gives n=0 n=2 n=4 .. .

c2 = − 22 c0 = −c0 c4 = 0 c6 = 0 .. .

Generally, c2n = 0

n = 2, 3, . . . .

Thus ∞ 7

n=0

For the odd indices

Generally,

c2n t2n = c0 + c2 t2 = c0 (1 − t2 ). n=1 n=3 n=5 .. .

c3 = 0 c5 = 0 c7 = 0 .. .

250

1 Solutions

c2n+1 = 0,

n = 1, 2, . . . .

Thus ∞ 7

c2n+1 t2n+1 = c1 t.

n=0

It follows that

9. Let y(t) =

∞ .

y(t) = c0 (1 − t2 ) + c1 t. cn tn . Then the recurrence relation is

n=0

cn+2 = −

(n − 2)(n − 3) cn . (n + 2)(n + 1)

The even case gives: n=0 n=2 n=4 .. .

c2 = − 62 c0 = −3c0 c4 = 0 c6 = 0 .. .

Hence ∞ 7

n=0

c2n t2n = c0 + c2 t2 = c0 (1 − 3t2 ).

The odd case gives n=1 n=3 n=5 .. .

c3 = − 13 c1 c5 = 0 c7 = 0 .. .

Hence ∞ 7

n=0

c2n+1 t2n+1 = c1 t + c3 t3 = c1 (t −

The general solution is y(t) = c0 (1 − 3t2 ) + c1 (t −

t3 ). 3

t3 ). 3

1 Solutions

10. Let y(t) =

251 ∞ .

cn tn . Then the recurrence relation is

n=0

cn+2 =

(n + 4) cn . (n + 2)

The even case gives n=0 n=2 n=4 .. .

c2 = 42 c0 = 2c0 c4 = 64 c2 = 3c0 c6 = 86 c4 = 4c0 .. .

More generally, c2n = (n + 1)c0 and ∞ 7

c2n t

2n

= c0

n=0

Let y0 (t) =

∞ .

∞ 7

(n + 1)t2n .

n=0

(n + 1)t2n and observe that

n=0

ty0 (t) = *

ty0 (t) dt = =

∞ 7

(n + 1)t2n+1

n=0 ∞ 7

1 t2n+2 + c 2 n=0 ∞ t2 7 2n t +c 2 n=0

t2 1 +c 2 1 − t2 ' 2 ( d t 1 ty0 (t) = dt 2 1 − t2 t = (1 − t2 )2 1 y0 (t) = . (1 − t2 )2 =

The odd case gives

252

1 Solutions

n=1 n=3 n=5 .. .

c3 = 53 c1 c5 = 75 c3 = 73 c1 c7 = 97 c5 = 93 c1 .. .

More generally, c2n+1 =

2n + 3 c1 3

and ∞ 7

c2n+1 t2n+1 = c1

n=0

Let y1 (t) =

∞ .

n=0

2n+3 2n+1 3 t

and observe that

3ty1 (t) = *

∞ 7 2n + 3 2n+1 t . 3 n=0

3ty1 (t) dt =

∞ 7

(2n + 3)t2n+2

n=0 ∞ 7

t2n+3 + c

n=0 ∞ 7 3 2n

= t

t

+c

n=0 3

t +c 1 − t2 d t3 3ty1 (t) = dt (1 − t2 ) t2 (3 − t2 ) = (1 − t2 )2 t(3 − t2 ) y1 (t) = . 3(1 − t2 )2 =

The general solution is y(t) = c0

1 t(3 − t2 ) + c . 1 (1 − t2 )2 3(1 − t2 )2

11. Let n be an integer. Then einx = (eix )n . By Euler’s formula this is (cos x + i sin x)n = cos nx + i sin nx.

1 Solutions

253

12. By de Moivre’s formula cos nx is the real part of (cos x + i sin x)n . The binomial theorem gives cos nx + i sin nx = (cos x + i sin x)n n ' ( 7 n = cosn−k xik sink x k k=0

Only the even powers of i contribute to the real part. It follows that [n/2] '

7

cos nx =

j=0

( n cosn−2j x(i2j ) sin2j x 2j

[n/2]

7

=

(−1)j

j=0

' ( n cosn−2j x(1 − cos2 x)j , 2j

where we use the greatest integer function 2x3 to denote the greatest integer less than or equal to x. Now replace t = cos x and using the .2 n2 3 % n & n−2k definition Tn (cosx) = cos nx to get Tn (t) = k=0 (1 − t2 )k . It 2k t follows that cos nx is a polynomial in cos x. 13. By de Moivre’s formula sin(n + 1)x is the imaginary part of (cos x + i sin x)n+1 . The binomial theorem gives cos(n + 1)x + i sin(n + 1)x = (cos x + i sin x)n+1 n+1 7 'n + 1( = cosn+1−k xik sink x k k=0

Only the odd powers of i contribute to the imaginary part. It follows that sin(n + 1)x = Im

n 27 ( 23' n+1

j=0

273 n 2

=

j=0

2j + 1

(−1)j

cosn+1−(2j+1) x(i2j+1 ) sin2j+1 x

'

( n+1 cosn−2j x(1 − cos2 x)j sin x, 2j + 1

where we use the greatest integer function 2x3 to denote the greatest integer less than or equal to x. Now replace t = cos x and using the definition % n+1 & n−2j .2 n2 3 sin xUn (cosx) = sin(n + 1)x to get Un (t) = j=0 (−1)j 2j+1 t (1 − 2 j t ) . It follows that sin nx is a product of sin x and a polynomial in cos x. 14. We have

254

1 Solutions

cos x cos(n + 1)x = cos xTn+1 (cos x) sin x sin(n + 1)x = sin2 xUn (cos x). Subtracting gives cos(n + 2)x = cos xTn+1 (cos x) − (1 − cos2 x)Un (cos x) and from this we get Tn+2 (t) = tTn+1 (t) − (1 − t2 )Un (t). 15. We have sin((n + 2)x) = sin((n + 1)x + x) = sin((n + 1)x) cos x + cos((n + 1)x) sin x and hence (sin x)Un+1 (cos x) = (sin x)Un (x) cos x + (sin x)Tn+1 (cos x). Now divide by sin x and let t = cos x. We get Un+1 (t) = tUn (t) + Tn+1 (t).

16. By using the sum and difference formula it is easy to verify the following trigonometric identity: 2 cos a cos b = cos(a + b) + cos(a − b). Let a = x and b = nx. Then 2 cos x cos nx = cos((n + 1)x) + cos((n − 1)x) and hence 2(cos x)Tn (cos x) = Tn+1 (cos x) + Tn−1 (cos x). Now let t = cos x. 17. By using the sum and difference formula it is easy to verify the following trigonometric identity: 2 sin a cos b = sin(a + b) + sin(a − b). Let a = (n + 1)x and b = x. Then 2 cos x sin(n + 1)x = sin((n + 2)x) + sin(nx)

1 Solutions

255

and hence 2(cos x)Un (cos x)/ sin x = Un+1 (cos x)/ sin x + Un−1 (cos x)/ sin x. Now cancel out sin x and let t = cos x. 18. By using the sum and difference formula it is easy to verify the following trigonometric identity: 2 sin a cos b = sin(a + b) − sin(b − a). Let a = x and b = nx. Then 2 sin x cos nx = sin((n + 1)x) − sin((n − 1)x) and hence 2(sin x)Tn (cos x) = (sin x)Un (cos x) − (sin x)Un−2 (cos x). Now divide by 2 sin x and let t = cos x. 19. By using the sum and difference formula it is easy to verify the following trigonometric identity: 2 sin a sin b = cos(b − a) − cos(a + b). Let a = x and b = nx. Then 2 sin x sin nx = cos((n − 1)x) − cos((n + 1)x) and hence 2Un−1 (cos x) = Tn−1 (cos x) − Tn+1 (cos x).

Now let t = cos x, replace n by n + 1, and divide by 2.

Section 7.3 t 1. The function 1−t 2 is analytic except at t = 1 and t = −1. The function 1 t = 1 and t = −1 are 1+t is analytic except at t = −1. It follows that " # t −t the only singular points. Observe that (t − 1) 1−t2 = 1+t is analytic " # 1 at 1 and (t − 1)2 1+t is analytic at t = 1. It follows that 1 is a regular " # t t singular point. Also observe that (t + 1) 1−t = 1−t is analytic at −1 2 " # 1 and (t + 1)2 1+t = (1 + t) is analytic at t = −1. It follows that −1 is a regular singular point. Thus 1 and −1 are regular points.

256

1 Solutions

1−cos t 2. The functions 1−t are analytic t and t3 % 1−t &except at t = 0. So t = 0 is the% only singular point. Observe that t = 1 − t is analytic at 0 and t & t 1−cos t t2 1−cos = is analytic at t = 0. It follows that 0 is a regular t3 t singular point. t

3. Both 3t(1 − t) and 1−e are analytic. There are no singular points and t hence no regular points. % & 4. Clearly, t% =&0 is the only singular point. Observe that t 1t = 1 is analytic while t2 t13 = 1t is not. Thus t = 0 is an irregular singular point.

! 5. We first write it in standard form: y !! + 1−t t y + 4y = 0. While the coeffi1−t ! cient of y is analytic the coefficient of y is t is analytic%except & at t = 0. It follows that t = 0 is a singular point. Observe that t 1−t = 1 − t is t analytic and 4t2 is too. It follows that t = 0 is a regular point.

6. In standard form the equation is 2t2 y !! + ty ! + t2 y = 0. The indicial equation is q(s) = 2s(s − 1) + s = 2s2 − s = s(2s − 1) The exponents of singularity are 1 and 12 . Theorem 2 guarantees two Frobenius solutions. 7. The indicial equation is q(s) = s(s − 1) + 2s = s2 + s = s(s + 1) The exponents of singularity are 0 and −1. Theorem 2 guarantees one Frobenius solution but there could be two. 8. The indicial equation is q(s) = s(s − 1) + s + 4 = s2 + 4 The exponents of singularity are ±2i. Theorem 2 guarantees that there are two complex Frobenius solutions. 9. In standard form the equation is t2 y !! + t(1 − t)y ! + λty = 0. The indicial equation is q(s) = s(s − 1) + s = s2 The exponent of singularity is 0 with multiplicity 2. Theorem 2 guarantees that there is one Frobenius solution. The other has a logarithmic term. 10. The indicial equation is p(s) = s(s−1)+3s+1 = s2 +2s+1 = (s+1)2 The exponent of singularity is −1 with multiplicity 2. Theorem 2 guarantees that there is exactly Frobenius solution. There is a logarithmic solution as well. 11. y1 (t) =

∞ .

n=0

(−1)n t2n (2n+1)!

=

1 t

∞ .

n=0

(−1)n t2n (2n)!

=

1 t

sin t. y2 is done similarly. n

12. The entries in the table in Example 8 confirm cn = in! for n = 0, . . . 5. Assume the formula is true for all k ≤ n. Let’s consider the recursion relation with n replaced by n + 1 to get the following formula for cn+1 :

1 Solutions

257

((i + n + 1)2 + 1)cn+1

= −((i + n)(i + n − 1) + 3)cn + 3cn−1 + cn−2

(n + 1)(2i + n + 1)cn+1

= −(n2 − n + 2in − i + 2)

(n + 1)(2i + n + 1)cn+1

= −

(n + 1)(2i + n + 1)cn+1

=

in 3in−1 in−2 + + n! (n − 1)! (n − 2)!

" in−2 ! 2 2 i (n − n + 2in − i + 2) + 3in + n(n − 1) n!

in (i(n + 1 + 2i)). n!

On the second line we use the induction hypothesis. Now divide both sides by (n + 1 + 2i)(n + 1) in the last line to get cn+1 =

in+1 . (n + 1)!

We can now conclude by Mathematical Induction that cn = n ≥ 0.

in n! ,

for all

13. Let y(t) = t−2 v(t). Then y ! (t) = −2t−3 v(t) + t−2 v ! (t) and y !! (t) = 6t−4 v(t) − 4t−3 v ! (t) + t−2 v !! (t). From which we get t2 y !! = 6t−2 v(t) − 4t−1 v ! (t) + v !! (t) 5ty ! = −10t−2 v(t) + 5t−1 v ! (t) 4y = 4t−2 v(t).

Adding these terms and remembering that we are assuming the y is a solution we get 0 = t−1 v ! (t) + v !! (t). ##

! From this we get vv# = −1 t . Integrating we get ln v (t) = − ln t and hence 1 ! v (t) = t . Integrating again gives v(t) = ln t. It follows that y(t) = t−2 ln t is a second independent solution. The indicial polynomial is q(s) = s(s − 1) + 5s + 4 = (s − 2)2 . Case 3 of the theorem guarantees that one solution is a Frobenius solution and the other has logarithmic term.

14. y1 (t) = =

∞ 7 n + 4 n+2 t n! n=0

∞ ∞ 7 n n+2 7 4 n+2 t + t n! n! n=0 n=0

∞ n 7 tn−1 t 2 = t + 4t (n − 1)! n! n=1 n=0 3

∞ 7

= t3 et + 4t2 et = (t3 + 4t2 )et .

258

1 Solutions

In each case below we let y = tr r is the exponent of singularity.

.∞

n n=0 cn t

where we assume c0 &= 0 and

15. Indicial polynomial: p(s) = s(s − 3); exponents of singularity s = 0 and s = 3. n = 0 c0 (r)(r − 3) = 0 n = 1 c1 (r − 2)(r + 1) = 0 n ≥ 1 cn (n + r)(n + r − 3) = −cn−1 r=3:

n odd cn = 0 m (2m+2) n = 2m c2m = 3c0 (−1) (2m+3)!

y(t) = 3c0

.∞

m=0

(−1)m (2m+2)t2m+3 (2m+3)!

= 3c0 (sin t − t cos t).

r=0: One is lead to the equation 0c3 = 0 and we can take c3 = 0. Thus n odd cn = 0 m+1 (2m−1) n = 2m c2m = c0 (−1) (2m)! y(t) = c0

.∞

m=0

(−1)m+1 (2m−1)t2m (2m)!

= c0 (t sin t + cos t).

General Solution: y = c1 (sin t − t cos t) + c2 (t sin t + cos t). 16. Indicial polynomial: p(s) = (2s − 1)(s − 1); exponents of singularity s = 1/2 and s = 1. n = 0 c0 (2r − 1)(r − 1) = 0 n ≥ 1 cn (2(n + r) − 1)(n + r − 1) = −cn−1 r = 1: cn =

−cn−1 (2n+1)n

and hence cn =

(−1)n 2n c0 , (2n + 1)!

n ≥ 1.

From this we get ∞ 7 (−1)n 2n tn+1 y(t) = c0 (2n + 1)! n=0 √ √ ∞ t 7 (−1)n ( 2t)2n+1 = c0 √ (2n + 1)! 2 n=0 √ c0 √ = √ t sin 2t. 2

r = 1/2: The recursion relation becomes cn (2n − 1)(n) = −cn−1 . The −cn−1 coefficient of cn is never zero for n ≥ 1 hence cn = n(2n−1) and hence

1 Solutions

259

cn =

(−1)n 2n . (2n)!

We now get ∞ 7 (−1)n 2n tn+1/2 (2n)! n=0 √ ∞ √ 7 (−1)n ( 2t)2n = t (2n)! n=0 √ √ = t cos 2t.

y(t) =

√ √ √ √ General Solution: y = c1 t sin 2t + c2 t cos 2t. 17. Indicial polynomial: p(s) = (s − 1)2 ; exponents of singularity s = 1, multiplicity 2. There is one Frobenius solution. .∞ r = 1 : Let y(t) = n=0 cn tn+1 . Then n2 cn − ncn−1 = 0.

n≥1

This is easy to solve. We get cn = y(t) = c0

1 n! c0

and hence

∞ 7 1 n+1 t = c0 tet . n! n=0

Logarithmic Solution: Let y1 (t) = tet . The second independent solution is necessarily of the form y(t) = y1 (t) ln t +

∞ 7

cn tn+1 .

n=0

Substitution into the differential equation leads to t2 et +

∞ 7

n=1

(n2 cn − ncn−1 )tn+1 = 0.

We write out the power series for t2 et and add corresponding coefficients to get 1 n2 cn − ncn−1 + . (n − 1)! The following list is a straightforward verification:

260

1 Solutions

c1 = −1

n=1 n=2

Let sn = 1 +

1 2

c2 =

n=3

c3 =

n=4

c4 =

−1 2 −1 3! −1 4!

% % %

1+ 1+ 1+

1 2 1 2 1 2

&

+ +

1 3 1 3

&

+

1 4

&

.

+ · · · + n1 . Then an easy argument gives that cn =

−sn . n!

We now have a second independent solution y2 (t) = tet ln t − t

∞ 7 s n tn . n! n=1

General Solution: t

y = c1 te + c2

,

∞ 7 s n tn te ln t − t n! n=1 t

-

.

18. Indicial polynomial: p(s) = (2s − 1)(s − 1); exponents of singularity s = 1/2 and s = 1. n = 0 c0 (2r − 1)(r − 1) = 0 n ≥ 1 cn (2(n + r) − 1)(n + r − 1) = cn−1 r = 1: cn =

cn−1 (2n+1)n

and hence cn =

2n c0 , (2n + 1)!

n ≥ 1.

From this we get ∞ 7 2n tn+1 (2n + 1)! n=0 √ ∞ √ 2n+1 t 7 ( 2t) = c0 √ 2 n=0 (2n + 1)! √ √ c0 = √ t sinh 2t. 2

y(t) = c0

r = 1/2: The recursion relation becomes cn (2n − 1)(n) = cn−1 . The cn−1 coefficient of cn is never zero for n ≥ 1 hence cn = n(2n−1) and hence

1 Solutions

261

cn =

2n . (2n)!

We now get ∞ 7 2n tn+1/2 (2n)! n=0 √ ∞ √ 7 ( 2t)2n = t (2n)! n=0 √ √ = t cosh 2t.

y(t) =

√ √ √ √ General Solution: y = c1 t sinh 2t + c2 t cosh 2t. 19. Indicial polynomial: p(s) = (s − 2)(s + 1); exponents of singularity s = 2 and s = −1. n = 0 c0 (r − 2)(r + 1) = 0 n ≥ 1 cn (n + r − 2)(n + r + 1) = −cn−1 (n + r − 1) r=2: cn = 6c0 y(t) = 6c0

.∞

n=0

(−1)n (n+1)tn (n+3)!

(−1)n (n + 1) , n≥1 (n + 3)! " # −t = 6c0 (t+2)e + t−2 . t t

r=-1: The recursion relation becomes cn (n − 3)(n) = cn−1 (n − 2) = 0. Thus n = 1 c1 = − c20 n = 2 c2 = 0 n = 3 0c3 = 0 We can take c3 = %0 and & then cn = 0 for all n ≥ 2. We now have y(t) = c0 t−1 (1 − 2t ) = c20 2−t . t −t

(t+2)e General Solution: y = c1 2−t . t + c2 t

20. Indicial polynomial: p(s) = s(s + 1); exponents of singularity r = 0 and r = −1. n = 0 c0 r(r + 1) = 0 n = 1 c1 (r + 1)(r + 2) = 0 n ≥ 1 cn (n + r)(n + r + 1) = a2 cn−2 r = 0: In the case c1 = 0 and c2n+1 = 0 for all n ≥ 0. We also get a2 c0 c0 a2n c2n = 2n(2n+1) , n ≥ 1 which gives c2n = (2n+1)! . Thus

262

1 Solutions ∞ 7 (at)2n y(t) = c0 (2n + 1)! n=0

∞ c0 7 (at)2n+1 at n=0 (2n + 1)! c0 = sinh at. at

=

r = −1: The n = 1 case gives 0c1 = 0 so c1 may be arbitrary. There is no inconsistency. We choose c1 = 0. The recursion relation becomes cn (n − 1)(n) = a2 cn−2 and it is straightforward to see that c2n+1 = 0 2n c0 while c2n = a(2n)! . Thus ∞ 7 a2n t2n−1 y(t) = c0 (2n)! n=0 ∞ c0 7 (at)2n t n=0 (2n)! c0 = cosh at. t

=

General Solution: y = c1 cosht at + c2 sinht at . 21. Indicial polynomial: p(s) = s(s − 2); exponents of singularity r = 0 and r = 2. n = 0 c0 r(r − 2) = 0 n ≥ 1 cn (n + r)(n + r − 2) = −cn−1 (n + r − 3) n−1 r=2: The recursion relation becomes cn = − n(n+2) cn−1 . For n = 1 we see that c1 = 0 and hence cn = 0 for all n ≥ 1. It follows that y(t) = c0 t2 is a solution.

r=0: The recursion relation becomes cn (n)(n − 2) = −cn−1 (n − 3) = 0. Thus c0 n = 1 −c1 = − −2 n = 2 0c2 = −2c0 ⇒⇐

The n = 2 case implies an inconsistency in the recursion relation since c0 &= 0. Since y1 (t) = t2 is a Frobenius solution a second independent solution can be written in the form y(t) = t2 ln t +

∞ 7

n=0

Substitution leads to

cn tn .

1 Solutions 3

263 2

t + 2t + (−c1 − 2c0 )t +

∞ 7

n=2

(cn (n)(n − 2) + cn−1 (n − 3))tn = 0

and the following relations: −c1 − 2c0 = 0 2 − c1 = 0 1 + 3c3 = 0 n(n − 2)cn + (n − 3)cn−3 = 0.

n=1 n=2 n=3 n≥4

We now have c0 = −1, c1 = 2, c3 = −1/3. c2 can be arbitrary so we choose n−1 c2 = 0, and cn = −(n−3)c , for n ≥ 4. A straightforward calculation n(n−2) gives 2(−1)n cn = . n!(n − 2) A second independent solution is , 2

y2 (t) = t ln t +

∞

t3 7 2(−1)n tn −1 + 2t − + 3 n=4 n!(n − 2)

-

.

General Solution: y = c1 t2 + c2 y2 (t). 22. Write the differential equation in standard form: t2 y !! − 4ty = 0. Clearly, the indicial polynomial is p(s) = s(s − 1); exponents of singularity are r = 0 and r = 1. r=1: Let y(t) =

∞ 7

cn tn+1 ,

n=0

Then n≥1 It follows that cn = get c0 ). Hence y(t)

c0 &= 0.

cn (n + 1)(n) = 4cn−1

n

4 c0 (n+1)!n! , for all n ≥ 1 .∞ 4n tn+1 = c0 n=0 (n+1)!n! = c40

(notice that when n = 0 we .∞ (4t)n+1 n=0 (n+1)!n! .

r = 0: When r = 0 the recursion relations become cn (n)(n − 1) = 4cn−1 ,

n≥1

and hence when n = 1 we get 0c1 = 4c0 , an inconsistency as c0 &= 0. There is no second Frobenius solution. However, there is .a∞second linearly independent solution of the form y(t) = y1 (t) ln t + n=0 cn tn , where .∞ (4t)n+1 y1 (t) = n=0 (n+1)!n! . Substituting this equation into the differential equation and simplifying leads to the recursion relation

264

1 Solutions

n(n − 1)cn = 4cn−1 −

(2n − 1)4n , (n + 1)!n!

n ≥ 1.

When n = 1 we get 0 = 0c1 = 4c0 − 2 and this implies c0 = 12 . Then the recursion relation for n = 1 becomes 0c1 = 0, no inconsistency; we will choose c1 = 0. The above recursion relation is difficult to solve. We give the first few terms: c0 = 12 c1 = 0 c2 = −2 c4 =

c3 =

−46 27

−251 405

It now follows that y2 (t) = y1 (t) ln t +

'

1 46 251 4 − 2t2 − t3 − t + ··· 2 27 405

(

.

23. Indicial polynomial: p(s) = (s2 + 1); exponents of singularity r = ±i. Let r = i (the case r = −i . gives equivalent results). The recursion relation ∞ that arises from y(t) = n=0 cn tn+i is cn ((n + i)2 + 1) − cn−1 ((n − 2 + i)2 + 1) = 0 and hence cn =

(n − 2)(n − 2 + 2i) cn−1 . n(n + 2i)

A straightforward calculation gives the first few terms as follows: n = 1 c1 =

1−2i 1+2i c0

n = 2 c2 = 0 n = 3 c3 = 0 and hence cn = 0 for all n ≥ 2. Therefore y(t) = c0 (ti +

"

1−2i 1+2i

#

t1+i ).

Since t > 0 we can write ti = ei ln t = cos ln t+i sin ln t, by Euler’s formula. Separating the real and imaginary parts we get two independent solutions y1 (t) = −3 cos ln t − 4 sin ln t + 5t cos ln t y2 (t) = −3 sin ln t + 4 cos ln t + 5t sin ln t. 24. Indicial polynomial: p(s) = (s − 1)(s + 1); exponents of singularity r = 1 and r = −1. n = 0 c0 (r − 1)(r + 1) = 0 n ≥ 1 cn (n + r − 1)(n + r + 1) = −cn−1 (n + r − 1)

1 Solutions

265

r = 1 : The recursion relation becomes cn (n(n + 2)) = −ncn−1 and hence 2(−1)n c0 cn = . (n + 2)! We then get y(t) =

∞ 7

cn tn+1

n=0

= 2c0 =

∞ 7 (−1)n tn+1 (n + 2)! n=0

∞ 2c0 7 (−t)n+2 t n=0 (n + 2)!

2c0 −t (e − (1 − t)) t ' ( e−t − (1 − t) = 2c0 . t =

r = −1: Thus

The recursion relation becomes cn (n − 2)(n) = −cn−1 (n − 2). n = 1 c1 = −c0 n = 2 0c2 = 0

Thus c2 can be arbitrary so we can take c2 = 0 and then cn = 0 for all n ≥ 2. We now have y(t) =

∞ 7

n=0

cn tn−1 = c0 (t−1 − 1) = c0

1−t . t

−t

e General Solution: y = c1 1−t t + c2 t .

25. Indicial polynomial: p(s) = (s2 + 1); exponents of singularity r = ±i. Let r = i (the case r = −i . gives equivalent results). The recursion relation ∞ that arises from y(t) = n=0 cn tn+i is n = 1 c1 = c0 n ≥ 2 cn ((n + i)2 + 1) + cn−1 (−2n − 2i + 1) + cn−2 = 0

A straightforward calculation gives the first few terms as follows: n=1 n=2 n=3 n=4

c1 c2 c3 c4

= c0 1 = 2! c0 1 = 3! c0 1 = 4! c0 .

266

1 Solutions

An easy induction argument gives cn =

1 c0 . n!

We now get y(t) =

∞ 7

cn tn+i

n=0 ∞ n+i 7

= c0

n=0 i t

t

n!

= c0 t e .

Since t > 0 we can write ti = ei ln t = cos ln t+i sin ln t, by Euler’s formula. Now separating the real and imaginary parts we get two independent solutions y1 (t) = et cos ln t and y2 (t) = et sin ln t. 26. Indicial polynomial: p(s) = (s − 1)2 ; exponents of singularity s = 1 with multiplicity 2. r=1: Let y(t) =

∞ 7

n=0

Then

n=1 n≥2

cn tn+1 ,

c0 &= 0.

c1 = 0 n2 cn = −cn−1 (n − 2)

It follows that cn = 0 for all n ≥ 1. Hence y(t) = c0 t. Let y1 (t) = t. .∞ logarithmic case: Let y(t) = t ln t + n=0 cn tn+1 . The recursion relations becomes n = 0 −c0 + c0 = 0 n = 1 c1 = 1 n ≥ 2 n2 cn = −(n − 1)(n − 2)cn−1 The n = 0 case allows us to choose c0 arbitrarily. We choose c0 = 0. For n ≥ 2 it is easily to see the cn = 0. Hence y(t) = t ln t + t2 . general solution: y(t) = a1 t + a2 (t ln t + t2 ).

1 Solutions

267

Section 7.4 1. Let y(t) = t2k+1

∞ 7

c n tn =

n=0

Then ty !! (t) = 2ity ! (t) = −2ky ! (t) = −2iky(t) =

∞ 7

n=0 ∞ 7 n=1 ∞ 7 n=0 ∞ 7 n=1

∞ 7

cn tn+2k+1

n=0

(n + 2k)(n + 2k + 1)cn tn+2k 2i(n + 2k)cn−1 tn+2k −2k(n + 2k + 1)cn tn+2k −2ikcn−1 tn+2k

By assumption the sum of the series is zero. The n = 0 terms in the first and third sum give (2k)(2k+1)c0 −2k(2k+1)c0 = 0. Thus we can start all the series at n = 1. For n ≥ 1 we get n(n + 2k + 1)cn + 2i(n + k)cn−1 = 0 which implies −2i(n + k) cn = cn−1 . n(n + 2k + 1) Since c0 &= 0 it follows from this recursion relation that cn &= 0 for all n ≥ 0. Therefore the Frobenius solution y(t) is not a polynomial. 2. Apply Equation 2 to get 2 3 2 3 s b s/b −1 L−1 (t) = L (t) (s2 + b2 )k+1 ((s/b)2 + 1)k+1 b2(k+1) 2 3 1 s = 2k L−1 (bt) b (s2 + 1)k+1 3. Since differentiation respects the real and imaginary parts of complexvalued functions we have Bk (t) = Re(bk (t)eit ) Bk! (t) = Re((bk (t)eit )! ) = Re((b!k (t) + ibk (t))eit ) Bk!! (t) = Re((b!!k (t) + 2ib!k (t) − bk (t))eit ). It follows now from Proposition 4 that

268

1 Solutions

0 = t2 Bk!! (t) − 2ktBk! (t) + (t2 + 2k)Bk (t)

= t2 Re(bk (t)eit )!! − 2kt Re(bk (t)eit )! + (t2 + 2k) Re(bk (t)eit ) = Re((t2 (b!!k (t) + 2ib!k (t) − bk (t)) − 2kt(b!k (t) + ibk (t)) + (t2 + 2k)bk (t))eit ) % & = Re (t2 b!!k (t) + 2t(it − k)b!k (t) − 2k(it − 1)bk (t))eit .

Apply Lemma 6 to get

t2 b!!k (t) + 2t(it − k)b!k (t) − 2k(it − 1)bk (t) = 0. 3 1 . Then the Transform Derivative Principle (s2 − 1)k (Theorem 2.2.20) applies to give ' ( d d 1 L {tf (t)} = − (L {f (t)}) = − ds ds (s2 − 1)k 2ks = . (s2 − 1)k+1

4. Let f (t) = L−1

2

Now divide by 2k and take the inverse Laplace transform of both sides to get 2 3 2 3 s t t −1 1 L−1 = f (t) = L (s2 − 1)k+1 2k 2k (s2 − 1)k which, when written in terms of Ck−1 and Dk , is 1 t 1 Dk (t) = Ck−1 (t). 2k k! 2k 2k−1 (k − 1)! Simplifying and shifting the index gives the required formula. 2 3 s −1 5. Let g(t) = L . Then (s2 − 1)k ' ( d d s L {tg(t)} = − (L {g(t)}) = − ds ds (s2 − 1)k

(s2 − 1)k − 2ks2 (s2 − 1)k−1 (s2 − 1)2k 2 2 2ks − (s − 1) (2k − 1)(s2 − 1) + 2k = = (s2 − 1)k+1 (s2 − 1)k+1 2k − 1 2k = + 2 . (s2 − 1)k (s − 1)k+1 = −

1 Solutions

269

Divide by 2k, solve for the second term in the last line, and apply the inverse Laplace transform to get 2 3 2 3 1 t (2k − 1) −1 1 −1 L = g(t) − L (s2 − 1)k+1 2k 2k (s2 − 1)k 2 3 2 3 t −1 s (2k − 1) −1 1 = L − L . 2k (s2 − 1)k 2k (s2 − 1)k By the definition of Ck and Dk we get 1 t 2k − 1 Ck (t) = k Dk−1 (t) − k Ck−1 (t). 2k k! 2 k! 2 k! Simplifying gives the result. 6. In Exercise 4 replace k by k − 2 to get Dk−1 = tCk−2 and substitute this into the formula verified in Exercise 5 to get Ck (t) = t2 Ck−2 − (2k − 1)Ck−1 (t). Now shift the index k by 2 and the result follows. 7. Multiply the equation in Exercise 5 by t and use the formula in Exercise 4 to get Dk+1 (t) = t2 Dk−1 − (2k − 1)Dk (t). Now shift k and the formula follows. 8. By the Initial Value Theorem, Theorem 5.4.1, we have s =0 (s2 − 1)k+1 s2 Dk (0) = 2k k! lim 2 =0 s→∞ (s − 1)k+1 Ck (0) = 2k k! lim

s→∞

9. By the Input Derivative Principle we have 1 1 L {Ck! (t)} = k (sL {Ck (t)} − Ck (0)) 2k k! 2 k! s = 2 (s − 1)k+1 1 = k L {Dk (t)} . 2 k! Laplace inversion gives the first formula. In a similar way the Input Derivative Principle gives

270

1 Solutions

1 1 L {Dk! (t)} = k (sL {Dk (t)} − Dk (0)) 2k k! 2 k! s2 = 2 (s − 1)k+1 s2 − 1 1 = + 2 2 (s − 1)k+1 (s − 1)k+1 1 1 = + 2 2 k (s − 1) (s − 1)k+1 1 1 = k−1 L {Ck−1 (t)} + k L {Ck (t)} . 2 (k − 1)! 2 k! Simplifying and Laplace inversion gives the result. 10. 1. Differentiate the first equation in Exercise 9 and substitute in the second equation to get Ck!! (t) = Dk! (t) = 2kCk−1 (t) + Ck (t). Now multiply by t and simplify using Exercises 4 and 9 to get tCk!! (t) = 2ktCk−1 (t) + tCk (t) = 2kDk (t) + tCk (t) = 2kCk! (t) + tCk (t). 2. Differentiate the first equation and multiply by t to get t2 Ck!!! (t) + (1 − 2k)tCk!! (t) − t2 Ck! (t) − tCk (t) = 0. Subtract the first equation from this to get t2 Ck!!! (t) − 2kc!!k (t) + (2k − t2 )Ck! (t) = 0. Now use Dk (t) = Ck! (t) to get the desired equation. 1 11. Since s2 − 1 = (s − 1)(s + 1) it follows that (s2 −1) k+1 is an s − 1 -chain and an s + 1 -chain, each of length k + 1. Hence there are constants αn and βn so that k+1 7 αn 1 βn = + . n (s2 − 1)k+1 (s − 1) (s + 1)n n=1

Now replace s by −s. The left-hand side does not change so we get 1 (s2 − 1)k+1

=

=

k+1 7

αn βn + n (−s − 1) (−s + 1)n n=1 k+1 7

αn (−1)n βn (−1)n + . n (s + 1) (s − 1)n n=1

It follows now by the uniqueness of partial fraction decompositions that βn = (−1)n αn . Laplace inversion now gives

1 Solutions

271

L−1 k+1 7

=

2

1 2 (s − 1)k+1

αn

tn−1 t tn−1 −t e + αn (−1)n e (n − 1)! (n − 1)!

αn

tn−1 t (−t)n−1 −t e − αn e (n − 1)! (n − 1)!

n=1 k+1 7

=

n=1

Let f (t) =

.k+1

n=1

3

n−1

t αn (n−1)! . Then

L−1

2

1 2 (s − 1)k+1

3

= f (t)et − f (−t)e−t .

Up to the constant 2k k!, the polynomial f (t) is ck (t). A similar argument gives the second part of the problem. 12. Hint: Follow the procedure detailed in the proof of Proposition 7. 13. 1. It is easy to see from the definition of ck and Exercise 5 that ck satisfies ck+2 (t) = t2 ck (t) − (2k + 3)ck+1 (t)

and therefore ck+2 (0) = −(2k+3)ck+1 (0). An easy check gives c1 (t) = t−1 −1 2 and thus c1 (0) = 2 . Recursively we get c1 (0) =

−1 2

c2 (0) = −3c1 (0) =

3 2

−5·3 2 7·5·3 2

c3 (0) = −5c2 (0) = c4 (0) = 7c3 (0) =

Inductively, we get (−1)k (2k − 1) · (2k − 3) · (2k − 5) · · · 1 2 (−1)k (2k)! = 2 2k k! (−1)k (2k)! = . 2k+1 k!

ck (0) =

2. From Exercise 4 it is easy to see that dk (t) = tck−1 and so d!k (0) = ck−1 (0) =

(−1)k−1 (2(k−1))! . 2k (k−1)!

14. By Exercise 12 ck (t) is the polynomial solution to the differential equak (2k)! tion ty !! +(2t−2k)y ! −2ky = 0 with constant coefficient ck (0) = (−1) . 2k+1 k! It is easy to see that this differential equation has a regular singular point at t = 0. The indicial polynomial is given by q(s) = s(s − 1) − 2ks =

272

1 Solutions

s(s − (2k + 1)). It follows that the exponents of singularity are 0 and 2k + 1. The r = 0 case gives the desired polynomial. We thus let y(t) =

∞ 7

a n tn .

n=0

Then ty !! (t) = 2ty ! (t) = −2ky ! (t) = −2ky(t) =

∞ 7

n=1 ∞ 7 n=1 ∞ 7 n=0 ∞ 7 n=0

(n)(n + 1)an+1 tn 2nan tn −2k(n + 1)an+1 tn −2kan tn

By assumption the sum of the series is zero. We separate the n = 0 and n ≥ 1 cases and simplify to get n=0 n≥1

−2ka1 − 2ka0 = 0 (n − 2k)(n + 1)an+1 + 2(n − k)an = 0

(2)

The n = 0 case tells us that a1 = −a0 . For 1 ≤ n ≤ 2k − 1 we have an+1 = −

2(k − n) an . (2k − n)(n + 1)

Observe that ak+1 = 0 and hence an = 0 for all k + 1 ≤ n ≤ 2k − 1. For n = 2k we get from the recursion relation 0a2k+1 = −2kak = 0. This implies that a2k+1 can be arbitrary. We will choose a2k+1 = 0. Then an = 0 for all n ≥ k + 1 and hence the solution y is a polynomial. From the recursion relation, Equation (2), we get n=0

a1 = −a0

2(k−1) a2 = − (2k−1)2 a1 =

n=2

2(k−2) 2 k(k−1)(k−2) a3 = − (2k−2)3 a2 = − 2k(2k−1)(2k−2)3! a0

n=3

2(k−3) a4 = − (2k−3)4 a3 =

and generally,

2(k−1) (2k−1)2 a0

22 k(k−1) 2k(2k−1)2 a0

n=1

=

3

24 k(k−1)(k−2)(k−3) 2k(2k−1)(2k−2)(2k−3)4! a0

1 Solutions

273

an =

(−2)n k(k − 1) · · · (k − n + 1) a0 2k(2k − 1) · · · (2k − n + 1)n!

n = 1, . . . , k.

We can write this more compactly as an = Since a0 = ck (0) =

(−1)k (2k)! 2k+1 k!

an = Thus

(−2)n k!(2k − n)! a0 . n!(2k)!(k − n)! it follows that

(−1)k (−2)n (2k − n)! . 2k+1 n!(n − k)!

k (−1)k 7 (2k − n)! ck (t) = k+1 (−2t)n . 2 n!(k − n)! n=0

It is easy to check that dk (t) = tck−1 (t) from which the second formula follows. 15. Merely put the previous calculations together.

Section 8.1 

     1 1 1 −3 2 −8 1. B + C = −1 7, B − C =  5 −1, and 2B − 3C =  13 −6 0 3 −2 1 −5 1     8 9 8 9 1 −1 −1 2 0 8 −3 1 6 3 3 7 2. AB = , AC = , BA = 5 −2 18, CA = −2 −3 5 4 6 0 1 5 3 −1 7   8 9 3 −1 7 3 4 1 25 3. A(B + C) = AB + AC = , (B + C)A = 3 1 13 5 0 12   −2 5 4. C = −13 −8 7 0   6 4 −1 −8 2 −8 2 5. AB = 0 2 −1 9 −5

274

1 Solutions

8

2 −2  8  4 7. CA =   8 10  6. BC =

9 3 −8 0 24  0 −5  14 11

6  4 8. B t At =  −1 −8  8 9. ABC =  4 −2

 0 2 2 −1  −8 9 2 −5  9 −48 0 −48 . 3 40 

 1 4 3 1  0 0 0 0  10. AB = −4 and BA =  −1 −4 −3 −1 −2 −8 −6 −2 8 9 1 0 14. 1 −1   0 0 1 15. 3 −5 −1 0 0 5 8 9 ab 0 16. AB − BA = . It is not possible to have ab = 1 and −ab = 1. 0 −ab 8 9 8 9 0 1 0 0 17. (a) Choose, for example, A = and B = . 0 0 1 0 (b) (A + B)2 = A2 + 2AB + B 2 precisely when AB = BA. 8 9 8 9 1 1 1 2 18. A2 = , A3 = 1 2 2 3 8 9 1 n 19. B n = 0 1 8 n 9 a 0 20. An = 0 bn

1 Solutions

275

8 9 8 9 8 9 0 1 v 1 c 21. (a) A = 2 ; the two rows of A are switched. (b) A = 1 0 v1 0 1 8 9 v1 + cv2 ; to the first row is added c times the second row while the v2 second row is unchanged, (c) to the second row is added c times the first row while the first row is unchanged. (d) the first row is multiplied by a while the second row is unchanged, (e) the second row is multiplied by a while the first row is unchanged. 22. E(θ1 )E(θ2 ) 8 98 9 cos θ1 sin θ1 cos θ2 sin θ2 = − sin θ1 cos θ1 − sin θ2 cos θ2 8 9 cos θ1 cos θ2 − sin θ1 sin θ2 cos θ1 sin θ2 + sin θ1 cos θ2 = − sin θ1 cos θ2 − cos θ1 sin θ2 − sin θ1 sin θ2 + cos θ1 cos θ2 8 9 cos(θ1 + θ2 ) sin(θ1 + θ2 ) = − sin(θ1 + θ2 ) cos(θ1 + θ2 ) = E(θ1 + θ2 ).

We used the addition formulas for sin and cos in the second line. 23. F (θ1 )F (θ2 ) 8 98 9 cosh θ1 sinh θ1 cosh θ2 sinh θ2 = sinh θ1 cosh θ1 sinh θ2 cosh θ2 8 9 cosh θ1 cosh θ2 + sinh θ1 sinh θ2 cosh θ1 sinh θ2 + sinh θ1 cosh θ2 = sinh θ1 cosh θ2 + cosh θ1 sinh θ2 sinh θ1 sinh θ2 + cosh θ1 cosh θ2 8 9 cosh(θ1 + θ2 ) sinh(θ1 + θ2 ) = sinh(θ1 + θ2 ) cosh(θ1 + θ2 ) = F (θ1 + θ2 ),

We used the addition formulas for sinh and cosh in the second line. 24.

276

1 Solutions



  0 e1 0 · · · 0   0   0 e2 · · · 0   .. .. . .  . 0 0  . . dn 0 0 · · · en  ··· 0 ··· 0   .. . 0 0 · · · en dn

d1 0 · · ·  0 d2 · · ·  DE =  . . .  .. .. . . 0 0 ···  d1 e1 0  0 d2 c 2  =  .. ..  . . 0

= diag(d1 e1 , . . . , dn en ).

Section 8.2 

       4 3 2 1 4 3 2 x     1 −1 , x =  y , b = 4, and [A|b] = 1 1 −1 4. 1  2 0 0 1 1 1 z 1 −1 6 0 1 −1 6   x1 8 9 8 9 8 9  x2  2 −3 4 1 4 1 0 , b = 0 , and [A|b] = 2 −3 2. A = ,x= .  x3  3 8 −3 −6 1 3 8 −3 −6 1 x4 1 1 1. A =  2 0

− + 3x2 − 2x2 + 2x2  1 0 4. p2,3 (A) = 0 1 0 0 x1 5x1 3. 3x1 −8x1

5. RREF

8

x3 + 4x4 − 3x3 − x4 + 8x3 + 4x4 + 2x4  1 4 0

1 0  0 7. m2 (1/2)(A) = 0 0 6. t2,1 (−2)(A) =

8. RREF

+ 3x5 − 3x5 − 3x5 + x5

9 0 −5 −2 −1 1 3 1 1  1 0 3 0 1 3 0 0 0

= 2 = 1 = 3 = −4

1 Solutions

277



 1 0 1 0 3 9. t1,3 (−3)(A) = 0 1 3 4 1 0 0 0 0 0   1 0 0 2 1 10. 0 1 0 0 0 1 −1   1 0 0 −11 −8 11. 0 1 0 −4 −2 0 0 1 9 6   0 1 0 72 14 0 0 1 3 1  2 12.   0 0 0 0 0 0 0 0 0 0   1 2 0 0 3  0 0 1 0 2  13.   0 0 0 1 0 0 0 0 0 0   1 0 0 1 1 1 14. 0 1 0 −1 3 1 0 0 1 2 1 1   1 0 2 0 1 1    15.  0 0 0 0 0 0 0 0 0   1 0 1 0  0 1 3 0  16.   0 0 0 1 0 0 0 0   1 4 0 0 3 17. 0 0 1 0 1 0 0 0 1 3   1 0 0 0 0 0 1 −1 0 0  18.  0 0 0 1 −1 0 0 0 0 0

278

19.

20.

21.

22.

23.

1 Solutions

      x −1 −3  y  =  1 + α  1, α ∈ R z 0 5         x1 4 −1 −2 x2  −1 −3 −1   =   + α   + β  , α, β ∈ R x3   0  1  0 x4 0 0 1 8 9 8 9 x −2 =α ,α∈R y 1       x1 3 −4 x2   0  1   =   + α  , α ∈ R x3  −2  0 x4 5 0     x 14/3  y  =  1/3 z −2/3

24. no solution     0 1 25. 3 + α 0, α ∈ R 4 0 28 93 2 26. −1 27. ∅ 28 93 −3 28. 1     −1  29.  0   1     −11 3       −1 2     30.  , 3  0      0 1     −34    −40  31.      39     1

1 Solutions

279

       −3 −5 3               0  3 −8       32.  0 ,  1 ,  0     0  0  2       2 0 0       5 1 1 33. The equation −1 = a 1 + b −1 has solution a = 2 and b = 3. By 2 0  4 5 Proposition 7 −1 is a solution. 4 34. k = 2



   −7/2 −3/2 35. If xi is the solution set for Ax = bi then x1 =  7/2, x2 =  3/2, −3/2 −1/2   7 and x3 = −6. 3 36. The augmented matrix [A|b1 |b2 |b3 ] reduces to   1 0 0 −7/2 −3/2 7 0 1 0 7/2 3/2 −6 . 0 0 1 −3/2 −1/2 3

The last three columns correspond in order to the solutions.

Section 8.3 8

9 4 −1 1. −3 1 8 9 3 −2 2. −4 3 3. not invertible 8 9 −2 1 4. −3/2 1/2 5. not invertible

280

1 Solutions



 1 −1 1 1 −2 6. 0 0 0 1   −6 5 13 7.  5 −4 −11 −1 1 3   −1/5 2/5 2/5 8. −1/5 −1/10 2/5 −3/5 1/5 1/5   −29 39/2 −22 13  7 −9/2 5 −3  9.  −22 29/2 −17 10 9 −6 7 −4   −1 0 0 −1  0 −1 0 −1  10. 12   0 0 −1 −1 −1 −1 −1 −1   0 0 −1 1  1 0 0 0  11.   0 1 1 −1 −1 −1 0 1

12. not invertible 8 98 9 8 9 4 −1 2 5 13. x = A−1 b = = −3 1 3 −3      1 −1 1 1 −2 1 −2  0 =  6 14. x = A−1 b = 0 0 0 1 −3 −3      −2 4 4 −2 16 1  1   −2 −1 4  1 = 10 11 15. x = A−1 b = 10 −6 2 2 2 18      1 −1 3 1 1 1 0 1 = 1 16. x = A−1 b = 13 2 4 2 −3 1 1      −58 39 −44 26 1 19       14 −9 10 −6 0 −4   =   17. x = A−1 b = 12  −44 29 −34 20 −1  15 18 −12 14 −8 2 −6

1 Solutions

281



    0 0 −1 1 1 3  1 −1  1 0 0 0 −1      . 18. x = A b =  = 0 1 1 −1 −2 −4 −1 −1 0 1 1 1 19. (At )−1 = (A−1 )t

20. (E(θ))−1 = E(−θ) 21. F (θ)−1 = F (−θ) 8 9 8 9 8 9 1 0 1 2 1 2 22. Example. A = ,B= ,C= . 0 0 3 4 5 6

Section 8.4 1. 1 2. 0 3. 10 4. 8 5. −21 6. 6 7. 2 8. 15 9. 0 10. 11. 12.

13.

9 −2 + s 2 s = 0, 3 1 −1 + s 8 9 s−3 1 1 s = 2, 4 s2 −6s+8 1 s−3 8 9 s−1 1 1 s=1±i s2 −2s+s −1 s − 1   (s − 1)2 3 s−1 1  0 (s − 1)2 0 s = 1 (s−1)3 0 3(s − 1) (s − 1)2 1 s2 −3s

8

282

1 Solutions



 s2 − 2s + 10 −3s − 6 3s − 12 3s − 12 s = −2, 1, 4 14. s3 −3s21−6s+8  −3s + 12 s2 − 2s − 8 2 3s + 6 −3s − 6 s − 2s − 8  2  s + s 4s + 4 0 0 s = −1, ±2i 15. s3 +s21+4s+4 −s − 1 s2 + s 2 s − 4 4s + 4 s + 4 8 9 9 −4 16. −2 1 17. no inverse 8 9 6 −4 1 18. 10 −2 3   4 −4 4 3 −1 19. 18 −1 −5 −1 3   27 −12 3 1  −13 5 4 20. 21 −29 16 −4   2 −98 9502 3 −297 21. 16 0 0 0 6   −13 76 −80 35 −14 76 −80 36  22. 12   6 −34 36 −16 7 −36 38 −17   55 −95 44 −171  −85 40 −150 1 50  23. 15 70 −125 59 −216 65 −115 52 −198 24. no inverse

9 8 9 2 1 1 2 = 5, and det A(2, b) = det = 3 4 3 3 −3. It follows that x1 = 5/1 = 5 and x2 = −3/1 = −3

25. det A = 1, det A(1, b) = det

8

1 Solutions

283



   1 1 1 1 1 1 0 2 = 26. det A = 1, det A(1, b) = det  0 1 2 = −2, det A(2, b) = det 0 −3 0 1 0 −3 1  1 1 1 0 = −3. It follows that x1 = −2/1 = 6, and det A(3, b) = det 0 1 0 0 −3 −2, x2 = 6/1 = 6, and x3 = −3/1 = −3.   −2 0 −2 0 = −16, det A(2, b) = 27. det A = −10, det A(1, b) = det  1 −2 2 2 −1     1 −2 −2 1 0 −2 1 0 = −11, and det A(3, b) = det 2 −2 1 = −18. It det 2 1 2 −1 1 2 2 follows that x1 = 16/10, x2 = 11/10, and x3 = 18/10.   1 3 0 1  0 2 −2 0  = 38, det A(2, b) = 28. det A = 2, det A(1, b) = det  −1 −1 0 4 2 2 3 9     1 1 0 1 1 3 1 1 2  0 −2 0 2 0 0  = −8, det A(3, b) = det 2  det  1 −1 1 −1 −1 4 = 30, and 0 4 1 2 3  9 1 2 2 9  1 3 0 1 2 2 −2 0   = −12. It follows that x1 = 38/2 = det A(4, b) = det  1 −1 0 −1 1 2 3 2 19, x2 = −8/2 = −4, x3 = 30/2 = 15, and x4 = −12/2 = −6.

Section 8.5 1. The characteristic polynomial is cA (s) = (s − 1)(s − 2). 28The 93 eigenvalues 0 are thus s = 1, 2. The eigenspaces are E1 = Span and E2 = 1 28 93 1 Span . −1 2. The characteristic polynomial is cA (s) = s2 − 1 = (s + 1)(s −28 1). The 93 1 eigenvalues are thus s = −1, 1. The eigenspaces are E−1 = Span −1 28 93 5 and E1 = Span . −7

284

1 Solutions

2 3. The characteristic polynomial is cA (s) = s2 − 2s +28 1 = (s 93− 1) . The only 1 eigenvalue is s = 1. The eigenspace is E1 = Span . −1

4. The characteristic polynomial is cA (s) = s2 + 2s + 1 = 1)2 . The only 28(s +93 1 eigenvalue is s = −1. The eigenspace is E−1 = Span . −2 5. The characteristic polynomial is cA (s) = s2 + 2s − 3 = (s + 3)(s − 1). The 28 93 1 eigenvalues are thus s = −3, 1. The eigenspaces are E−3 = Span −1 28 93 −3 and E1 = Span . 2 2 6. The characteristic polynomial is cA (s) = s2 − 2s + 5 = (s − 1)2 + 228 . The 93 i eigenvalues are thus s = 1 ± 2i. The eigenspaces are E1+2i = Span 1 28 93 −i and E1−2i = Span . 1

7. The characteristic polynomial is cA (s) = s2 + 2s + 10 = (s + 1)2 + 32 . The eigenvalues 3i. The93 eigenspaces are E−1+3i = 28 93 are thus s = −1 ±28 7+i 7−i Span and E−1−3i = Span . 10 10        1   1  8. The eigenvalues are s = −1, 1, 2. E−1 = Span 3 , E1 = Span  0 ,     1 −1    0  E2 = Span 1   1      0   1 9. The eigenvalues are s = −2, 3. E−2 = Span 2 , 1 , E3 =   0 1     1  Span  0 ,   −1        1   0  10. The eigenvalues are s = 1, 3. E1 = Span 0 , E3 = Span 1     0 1

1 Solutions

285

    0  11. The eigenvalues are s = 0, 2, 3. E0 = NS(A) = Span 2 , E2 =   1        2   0  Span 2 , E3 = Span 1     1 1     2  12. The only eigenvalue is s = 2. E2 = Span  0   −1

13. We write cA (s) = (s − 2)((s −2)2 + 1) to see that are    the eigenvalues  2 −4 + 3i     s = 2, 2 ± i. E2 = Span 3 , E2+i = Span  4 + 2i  , E2−i =     1 5    −4 − 3i  Span  4 − 2i    5

Section 9.2 1. nonlinear, because of the presence of the product y1 y2 . 2. We may write the system in the form 8 9 8 29 1 1 t y! = y+ , −1 1 1 8 9 y1 where y = . It follows that it is linear, constant coefficient, and y2 nonhomogeneous. 3. We may write the system in the form 8 9 sin t 0 y! = y. 1 cos t It is linear and homogeneous, but not constant coefficient. 4. This is not a linear system because of the presence of sin y1 or cos y2 . 5. We write the system in the form

286

1 Solutions



1 2 !  y = 0 0

0 0 0 1

0 0 0 2

 0 1  y. 1 0

It is linear, constant coefficient, and homogeneous. 6. We write the system in the form 8 1 9 8 9 −1 5 ! 2 y = . 1 y+ −5 −1 2 It is linear, constant coefficient, and nonhomogeneous. 7. First note that y1 (0) = 0 and y2 (0) = 1, so the initial condition is satisfied. Then 8 ! 9 8 t 9 y1 (t) e − 3e3t ! y (t) = ! = y2 (t) 2et − 3e3t while

8

9 8 t 9 5 −2 5(e − e3t ) − 2(2et − e3t ) y(t) = 4 −1 4(et − e3t ) − (2et − e3t ) 8 t 9 e − 3e3t = . 2et − 3e3t

9 5 −2 Thus y (t) = y(t), as required. 4 −1 !

8

8. First note that y1 (0) = 1 and y2 (0) = 0, so the initial condition is satisfied. Then 8 ! 9 8 t 9 8 t 9 y1 (t) e + 2et + 2tet 3e + 2tet ! y (t) = ! = = y2 (t) 4et + 4tet 4et + 4tet while 8

98 9 8 t 9 3 −1 et + 2tet 3e + 6tet − 4tet = 4 −1 4tet 4et + 8tet − 4tet 8 t 9 3e + 2tet = . 4et + 4tet 8 9 3 −1 Thus y ! (t) = y(t), as required. 4 −1 9. First note that y1 (0) = 1 and y2 (0) = 3, so the initial condition is satisfied. Then 8 ! 9 8 −t 9 y1 (t) −e + et + tet ! y (t) = ! = y2 (t) −3e−t + et + tet

1 Solutions

287

while 8 98 9 8 t9 8 −t 9 2 −1 e−t + tet e 2e + 2tet − 3e−t − tet + et + = 3 −2 3e−t + tet et 3e−t + 3tet − 6e−t − 2tet + et 8 −t 9 −e + tet + et = . −3e−t + tet + et Thus y ! (t) =

8

9 8 t9 2 −1 e y(t) + t , as required. 3 −2 e

10. First note that y1 (0) = 1 and y2 (0) = 1, so the initial condition is satisfied. Then 8 ! 9 8 9 y (t) −1 + 2 cos t y ! (t) = 1! = y2 (t) −1 − 2 sin t while

8

98 9 8 9 8 9 0 1 1 − t + 2 sin t t −1 − t + 2 cos t + t + = −1 0 −1 − t + 2 cos t −t −1 + t − 2 sin t − t 8 9 −1 + 2 cos t = . −1 − 2 sin t 8 9 8 9 0 1 t Thus y ! (t) = y(t) + , as required. −1 0 −t 8 9 8 9 y y In solutions, 11–15, y = 1 = ! . y2 y ! !! ! 11. Let y1 = y and y2 = y ! . Then y1! =8y ! = 9 y2 and y2 = y = −5y − 6y + y e2t = −6y1 − 5y2 + e2t . Letting y = 1 , this can be expressed in vector y2 form as 8 9 8 9 8 9 0 1 0 1 ! y = y + 2t , y(0) = . −6 −5 e −2 ! ! ! ! !! 2 2 12. Let y1 = y and 8 y92 = y . Then y1 = y = y2 and y2 = y = −k y = −k y1 . y Letting y = 1 , this can be expressed in vector form as y2

y! =

8

9 0 1 y, −k 2 0

y(0) =

8

9 −1 . 0

13. Let y1 = y and y2 = y ! . Then y1! = y !8 = 9y2 and y2! = y !! = k 2 y + y A cos ωt = k 2 y1 + A cos ωt. Letting y = 1 , this can be expressed in y2 vector form as

288

1 Solutions

y! =

8

9 8 9 0 1 0 y + , k2 0 A cos ωt

y(0) =

8 9 0 . 0

b ! c ! ! ! !! 14. Let y1 = y and y2 = y ! . Then 8 9 y1 = y = y2 and y2 = y = − a y − a y = y − ac y1 − ab y2 . Letting y = 1 , this can be expressed in vector form as y2

y! =

8

0

1

− ac − ab

9

y,

y(0) =

8 9 α . β

2 ! 1 ! ! ! !! 15. Let y1 = y and y2 = y ! . Then 8 9y1 = y = y2 and y2 = y = − t y − t2 y = y − t12 y1 − 2t y2 . Letting y = 1 , this can be expressed in vector form as y2 !

y =

16. 17.

18.

19.

20. 21. 22.

0

1

− t12 − 2t

9

y,

9 −2 y(1) = . 3

9 −2 sin 2t 2 cos 2t A (t) = −2 cos 2t −2 sin 2t 8 9 −3e−3t 1 A! (t) = 2t 2e2t  −t  −e (1 − t)e−t (2t − t2 )e−t −e−t (1 − t)e−t  A! (t) =  0 0 0 −e−t   1 y ! (t) =  2t  t−1 8 9 0 0 A! (t) = 0 0 L M v ! (t) = −2e−2t t22t +1 −3 sin 3t !

8

8

8

1 Solutions

289

*

π 2

0

9 e2 − e−2 e2 + e−2 − 2 2 − e2 − e−2 e2 − e−2   3/2 24.  7/3  ln 4 − 1 8 9 4 8 25. 12 16 23.

8

1 4

26. Continuous on I1 , I4 , and I5 N1 O 1 28.

N

29.

N

30.

31.

s2 1 s−2

s 2 s3

27.

s s2 +1 − s21+1 3! s4 2−s s3



1  s2 2  s3  6 s4 2 s2 −1

1 s

 32.  0 0

8

1 s2 +1 s s2 +1

O

2s (s2 +1)2 s−3 s2 −6s+13

9 1 −1 −1 1

1 s2 +1 s s2 +1 −1 s2 +1

1 (s+1)2 3 s

 1 s(s2 +1) 1  s2 +1  s s2 +1

L M 33. 1 2t 3t2

*

π 2

8

cos 2t sin 2t − sin 2t cos 2t 0 N O$ π sin 2t cos 2t $ 2 − $ 2 2 = cos 2t sin 2t $$ 2 2 0 8 9 1 0 2 = 2 −2 0 8 9 0 1 = −1 0

A(t) dt =

O

9

290

1 Solutions



1  s 34. We have  s 2 s −N 1 inversion gives

 1 1   2 s s =  s  1 1 − 2 s + 1 O 2(s − 1) 2(s + 1) 1 t

et −e−t 2

 1 s2  . Laplace s  s2 + 1

cos t

 2s 2   1 1 −1 1  + + s2 − 1   s + 1 s − 1 s + 1 s − 1  s2 − 1 35. We have  =  . Laplace 2 2s −1 1 1 1 + + s2 −N1 s2 − 1 s +O1 s − 1 s+1 s−1 t e + e−t et − e−t inversion gives t e − e−t et + e−t     1 1 1 1   s−1 (s − 1)2  s−1 s2 − 2s + 1         −4/3  4 1 1/3 1 1    . 36. We have  3 = + +  2 2 2 s +1   s s+3 s−1 s +1   s + 2s − 3s      3s 1 3s 1 s2 + 9

s−3 et



 Laplace inversion gives − 43 +

e−3t 3

t

te

+ et

3 cos 3t

 sin t e3t

Section 9.3 1. 9 1 0 0 −2 8 9 1 0 = 0 4 8 9 1 0 = 0 −8 .. . 8 9 1 0 = . 0 (−2)n

A = A2 A3

An It follows now that

8



s2 + 9

s−3

1 Solutions

291

A2 2 A3 3 eAt = I + At + t + t + ··· 8 9 82! 93! 8 9 8 9 1 t2 1 tn 1 0 t 0 0 0 = + + + ··· + + ··· 0 1 0 −2t 2! 0 (−2t)2 n! 0 (−2t)n 8 t 9 e 0 = . 0 e−2t 2. 8

9 3 −1 A = 9 −3 8 9 0 0 2 A = 0 0 .. . 8 9 0 0 n A = n ≥ 2. 0 0 It follows now that A2 2 A3 3 eAt = I + At + t + t + ··· 3! 8 9 82! 9 1 0 3t −t = + 0 1 9t −3t 8 9 1 + 3t −t = . 9t 1 − 3t 3. 9 0 1 1 0 8 9 1 0 = =I 0 1 8 9 0 1 = =A 1 0

A = A2 A3

8

It follows now that An = I if n is even and An = A if n is odd. Thus

292

1 Solutions

A2 2 A3 3 t + t + ··· 2! 3! t2 t3 t4 t5 I + At + I + A + I + A + · · · 2! 3! ( 4! ' 5! ' ( t2 t4 t3 t5 I 1 + + + ··· + A t + + + ··· 2! 4! 3! 5! I cosh t + A sinh t 8 9 cosh t sinh t . sinh t cosh t

eAt = I + At + = = = = 4.

9 1 2 0 1 8 9 1 4 = 0 1 8 9 1 6 = 0 1 .. . 8 9 1 2n = . 0 1

A = A2 A3

An

8

It follows now that A2 2 A3 3 eAt = I + At + t + t + ··· 8 9 82! 9 3! 8 2 9 8 9 1 t 4t2 1 tn 2ntn 1 0 t 2t = + + + · · · + + ··· 0 1 0 t tn 2! 0 t2 n! 0 N O 2 3 2 3 n 1 + t + t2! + t3! + · · · 2t + 4t2! + 6t3! + · · · + 2nt + ··· n! = 2 3 0 1 + t + t2! + t3! + · · · 8 t 9 e 2tet = . 0 et 5.

1 Solutions

293

9 1 1 1 1 8 9 2 2 = 2 2 8 9 4 4 = 4 4 .. . 8 9 2n−1 2n−1 = . 2n−1 2n−1

A = A2 A3

An

8

It follows now that A2 2 A3 3 eAt = I + At + t + t + ··· 8 9 82! 9 3! 8 2 9 8 9 1 2t 2t2 1 2n−1 tn 2n−1 tn 1 0 t t = + + + · · · + + ··· 0 1 t t 2! 2t2 2t2 n! 2n−1 tn 2n−1 tn The (1, 1) entry is 1+

∞ ∞ 7 2n−1 tn 1 7 (2t)n = 1+ n! 2 n=1 n! n=1

= =

∞ 1 1 7 (2t)n + 2 2 n=0 n!

1 1 2t + e 2 2

The (1, 2) entry is ∞ ∞ 7 2n−1 tn 1 7 (2t)n 0+ = n! 2 n=1 n! n=1

∞ 1 1 7 (2t)n = − + 2 2 n=0 n!

1 1 = − + e2t 2 2

Since the (1, 1) entry and the (2, 2) entry are equal and the (1, 2) entry and the (2, 1) entry are equal we have N 1 1 O 2t − 12 + 12 e2t 2 + 2e At e = 1 1 2t − 12 + 12 e2t 2 + 2e

294

1 Solutions

6. 

Thus

0 A = 1 0  2 A2 = 0 2  0 A3 = 0 0

 2 0 0 −1 2 0  0 −2 0 0 0 −2  0 0 0 0 0 0

A2 2 A3 3 eAt = I + At + t + t + ··· 3!  2!   2  1 0 0 0 2t 0 t 0 −t2 0 = 0 1 0 +  t 0 −t +  0 0 0 0 1 0 2t 0 t2 0 −t2   1 + t2 2t −t2 1 −t  =  t t2 2t 1 − t2 7. A =

A2 =

A3 =

A4 =

A5 =



 0 1 0 −1 0 0 0 0 2   −1 0 0  0 −1 0 0 0 4   0 −1 0 1 0 0 0 0 8   1 0 0  0 1 0 0 0 16   0 1 0 −1 0 0 0 0 32

The (1, 1) entry and the (2, 2) entry of eAt are equal and are

1 Solutions

295

cos t = 1 −

t2 t4 + − ··· . 2! 4!

The (1, 2) entry and the (2, 1) entry of eAt have opposite signs. The (2, 1) entry is t3 t5 sin t = t − + − · · · . 3! 5! The (3, 3) entry is (2t)2 e2t = 1 + 2t + + ··· . 2! All other entries are zero thus   cos t sin t 0 eAt = − sin t cos t 0  0 0 e2t 8. The characteristic polynomial is cAN(s) = (s −O 1)(s − 2) and sI − A = 8 9 1 0 s−1 0 s−1 . Thus (sI − A)−1 = and 1 0 s−2 0 s−2

eAt = L−1 (sI − A)−1 =

8

et 0 0 e2t

9

9. The characteristic polynomial is cN A (s) = s(s −O3) and sI − A = 8 9 s−2 −1 s−1 1 s(s−3) s(s−3) . Thus (sI − A)−1 = . A partial fraction −2 s−1 2 s−2 s(s−3) s(s−3) decomposition of each entry gives N2 1O N 1 O 1 1 3 3 1 3 −3 −1 (sI − A) = + . 2 s 23 13 s − 3 − 23 3 Thus eAt = L−1 (sI − A)−1 =

N2 3 2 3

+ 13 e3t − 23 e3t

1 3 1 3

− 13 e3t + 23 e3t

O

10. The characteristic polynomial is cA (s) = (s − 1)2 − 1 = s(s − 2) and 8 9 s − 1 −1 sI − A = . Thus the resolvent matrix is (sI − A)−1 = −1 N O s−1 s−1 s(s−2) 1 s(s−2)

1 s(s−2) s−1 s(s−2)

A partial fraction decomposition gives

(sI − A)

−1

N 1 O N1 1 1 1 2 −2 2 = + 1 1 s − 12 s − 2 2 2

1 2 1 2

O

296

1 Solutions

and therefore N e

At

=

1 2 − 12

− 12 1 2

O

+e

2t

N1 2 1 2

1 2 1 2

O

=

N

1 2 − 12

+ 12 e2t − 12 + 12 e2t + 12 e2t

1 2

+ 12 e2t

O

2 11. The characteristic polynomial 8 is cA (s) = 9(s−3)(s+1)+5 = s −2s+2 = s − 3 −5 (s − 1)2 + 1 and sI − A = . Thus the resolvent matrix is 1 Os+1 N

(sI − A)

−1

s+1 (s−1)2 +1 −1 (s−1)2 +1

=

(sI − A)−1 =

N

5 (s−1)2 +1 s−3 (s−1)2 +1

s−1 (s−1)2 +1 −1 (s−1)2 +1

which we write as

5 (s−1)2 +1 s−1 (s−1)2 +1

O

+

N

2 (s−1)2 +1

0

0

−2 (s−1)2 +1

O

.

Therefore e

At

8

9 et cos t + 2et sin t 5et sin t = −et sin t et cos t − 2et sin t

12. The characteristic polynomial is cA (s) = s2 − 2s + 2 = (s − 1)2 + 1 8 9 s − 4 10 and sI − A = . Thus the resolvent matrix is (sI − A)−1 = −1 s + 2 N O s+2 (s−1)2 +1 1 (s−1)2 +1

−10 (s−1)2 +1 s−4 (s−1)2 +1

(sI − A)−1 =

N

which we write as

s−1 (s−1)2 +1 1 (s−1)2 +1

10 − (s−1) 2 +1 s−1 (s−1)2 +1

O

+

N

3 (s−1)2 +1

0

0

3 − (s−1) 2 +1

O

.

Therefore e

At

8

9 et cos t + 3et sin t −10et sin t = et sin t et cos t − 3et sin t 

 s −1 −1 s −1. 13. The characteristic polynomial is cA (s) = s3 and sI − A = 0 0 0 s  1 1 s+1    t2 2 3 1 t t+ 2 s s s    1  At Thus (sI − A)−1 =  0 1s and e =   0 1 t  s2 1 0 0 1 0 0 s

1 Solutions

297

8 2 9 M 0 14. It is easy to see that A2 = and more generally An = 0 N2 8 n 9 M 0 . Thus the power series expansion for eAt is 0 Nn A2 t2 eAt = I + At + + ··· 8 9 2! 8 9 8 2 9 8 n 9 Ir 0 M 0 M 0 t2 M 0 tn = + t+ + ··· + + ··· 0 Is 0 N 0 N 2 2! 0 N n n! N O 2 2 I + M t + M2!t + · · · 0 = 2 2 0 I + N t + N2!t + · · · 8 Mt 9 e 0 = 0 eN t 8

9 8 9 0 1 cos t sin t Mt 15. Let M = and N = 2. Then by Example 6 e = . −1 0 − sin t cos t Thus 8 Mt 9 e 0 At e = 0 eN t   cos t sin t 0 = − sin t cos t 0  . 0 0 e2t 9 8 9 1 1 1 −1 and N = . By Exercises 9 and 10 1 1 −2 2 N 1 1 O N2 1 O 1 3t 2t 3t 1 − 12 + 12 e2t 2 + 2e 3 + 3e 3 − 3e Nt = and e = 2 2 3t 1 2 3t 1 1 2t − 12 + 12 e2t 2 + 2e 3 − 3e 3 + 3e

16. Let M =

eM t

8

It follows now that  1

1 2t 2 + 2e  1 1 2t − 2 + 2 e 

eAt =  

− 12 + 12 e2t 1 2

0

+ 12 e2t

0

0

0

0

0

0 2 3 2 3

+ 13 e3t − 23 e3t

   1 3t  − 3e  + 23 e3t 0

1 3 1 3



298

1 Solutions

Section 9.4 1. The characteristic matrix and characteristic polynomial are 8 9 s−2 1 sI − A = and cA (s) = s2 − 2s + 1 = (s − 1)2 . −1 s The characteristic polynomial has root 1 with multiplicity 2. The standard basis of EcA is BcA = {et , tet }. It follows that eAt = M et + N tet . Differentiating we obtain AeAt = M et + N (et + tet ) = (M + N )et + N tet . Now, evaluate each equation at t = 0 to obtain: I = M A = M + N. from which we get M = I N = A − I. Thus, eAt = = =

8

Iet + (A − I)tet 9 8 9 1 0 t 1 −1 e + tet 0 1 1 −1 8 t 9 e + tet −tet tet et − tet

2 2. The characteristic polynomial of A is cA (s) = (s 0 + 1)(s − 1 4) + 6 = s − t 2t 3s + 2 = (s − 1)(s − 2). It follows that BcA = e , e and Fulmer’s method gives eAt = M1 et + M2 e2t .

Differentiating and evaluating at t = 0 gives I = M1 + M2 A = M1 + 2M2 .

1 Solutions

299

Hence, 8

9 3 −2 = −(A − 2I) = . 3 −2 8 9 −2 2 = A−I = −3 3

M1 M2 We now have that e

At

=e

t

8

9 8 9 3 −2 2t −2 2 +e . 3 −2 −3 3

3. cA (s) = (s − 2)(s + 2) + 4 = s2 . Thus BcA = {1, t} and Fulmer’s method gives eAt = M1 + M2 t. Differentiating and evaluating at t = 0 gives M1 = I M2 = A. Thus e

At

=

N

1 + 2t

t

−4t

1 − 2t

O

.

4. The characteristic polynomial is s2 − 5s +06 = (s −12)(s − 3) and has roots 2, 3. The standard basis of EcA is BcA = e2t , e3t . It follows that eAt = M e2t + N e3t .

Differentiating and evaluating at t = 0 gives I=

M +N

A = 2M + 3N. from which we get M = 3I − A

N = A − 2I.

Thus,

300

1 Solutions

eAt = (3I − A)e2t + (A − 2I)e3t 8 9 8 9 2 −1 2t −1 1 3t = e + e 2 −1 −2 2 8 2t 9 2e − e3t −e2t + e3t = 2e2t − 2e3t −e2t + 2e3t 5. The characteristic polynomial is cA (s) = s2 − 2s + 2 = (s − 1)2 + 1. The standard basis of EcA is BcA = {et cos t, et sin t}. It follows that eAt = M et cos t + N et sin t. Differentiating and evaluating at t = 0 gives I = M A = M + N. from which we get M = I N = A − I. Thus, eAt = Iet cos t + (A − I)et sin t =

=

8 8

9 8 t 9 et cos t 0 3e sin t −10et sin t + t 0 et cos t e sin t −3et sin t et cos t + 3et sin t −10et sin t t t e sin t e cos t − 3et sin t

9

2 6. The characteristic 0 3t 3t 1 polynomial is cA (s) = (s − 3) . The standard basis is BcA = e , te . It follows that

eAt = M e3t + N te3t .

Differentiating and evaluating at t = 0 gives I = M A = 3M + N. from which we get

1 Solutions

301

M = I N = A − 3I. Thus, eAt = Ie3t + (A − 3I)te3t 8 3t 9 8 3t 9 e 0 te −te3t = + 0 e3t te3t −te3t 8 3t 9 e + te3t −te3t = te3t e3t − te3t 7. The characteristic polynomial is0 cA (s) = 1s2 − 4 and has roots −2, 2. The standard basis of EcA is BcA = e2t , e−2t . It follows that eAt = M e2t + N e−2t .

Differentiating and evaluating at t = 0 gives I = M +N A = 2M − 2N. from which we get 1 (A + 2I) 4 1 N = − (A − 2I). 4

M =

Thus, 1 1 (A + 2I)e2t − (A − 2I)e−2t 4 4 8 9 8 9 1 −7 11 2t 1 −11 11 −2t = e − e 4 −7 11 4 −7 7 8 9 1 −7e2t + 11e−2t 11e2t − 11e−2t = 4 −7e2t + 7e−2t 11e2t − 7e−2t

eAt =

8. The characteristic polynomial is cA (s) = s2 + 2s + 17 = ((s + 1)2 + 42 ). The standard basis of EcA is BcA = {e−t cos 4t, e−t sin 4t}. It follows that eAt = M e−t cos 4t + N e−t sin 4t. Differentiating and evaluating at t = 0 gives

302

1 Solutions

I = M A = −M + 4N. from which we get M = I 8 9 1 −1 −2 N = (A + I) = . 1 1 4 Thus, 1 eAt = Ie−t cos 4t + (A + I)e−t sin 4t 4 8 −t 9 8 −t 9 e cos 4t 0 −e sin 4t −2e−t sin 4t = + 0 e−t cos 4t e−t sin 4t e−t sin 4t 8 −t 9 e cos 4t − e−t sin 4t −2e−t sin 4t = e−t sin 4t e−t cos 4t + e−t sin 4t 2 2 2 9. The characteristic polynomial is cA (s) 0 2t= s − 4s2t+ 13 =1 ((s − 2) + 3 ). The standard basis of EcA is BcA = e cos 3t, e sin 3t . It follows that

eAt = M e2t cos 3t + N e2t sin 3t.

Differentiating and evaluating at t = 0 gives I = M A = 2M + 3N. from which we get M = I 8 9 1 8 13 N = (A − 2I) = . −5 −8 3 Thus, 1 eAt = Ie2t cos 3t + (A − I)e2t sin 3t 3 8 2t 9 8 9 e cos 3t 0 8e2t sin 3t 13e2t sin 3t = + 0 e2t cos 3t −5e2t sin 3t −8e2t sin 3t 8 2t 9 e cos 3t + 8e2t sin 3t 13e2t sin 3t = −5e2t sin 3t e2t cos 3t − 8e2t sin 3t

1 Solutions

303

10. The characteristic polynomial is cA (s) = s2 − s = s(s − 1) and has roots 0, 1. The standard basis of EcA is BcA = {1, et }. It follows that eAt = M + N et . Differentiating and evaluating at t = 0 gives I = M +N A = N. from which we get M = I −A= N = A.

8

−2 3 −2 3

9

Thus, eAt = (I − A) + Aet 8 9 8 t 9 −2 3 3e −3et = + −2 3 2et −2et 8 9 −2 + 3et 3 − 3et = −2 + 2et 3 − 2et 2 11. The characteristic 0 −2t −2t 1polynomial is cA (s) = (s + 2) . The standard basis is BcA = e , te . It follows that

eAt = M e−2t + N te−2t .

Differentiating and evaluating at t = 0 gives I = M A = −2M + N. from which we get M = I N = A + 2I = Thus,

8

9 −1 1 . −1 1

304

1 Solutions

eAt = Ie−2t + (A + 2I)te−2t 8 −2t 9 8 9 e 0 −te−2t te−2t = + 0 e−2t −te−2t te−2t 8 −2t 9 e − te−2t te−2t = −te−2t e−2t + te−2t 12. The characteristic polynomial is cA (s)0= (s − 1 2)(s − 4) and has roots 2, 4. The standard basis of EcA is BcA = e2t , e4t . It follows that eAt = M e2t + N e4t .

Differentiating and evaluating at t = 0 gives I = M +N A = 2M + 4N. from which we get 8 9 1 −1 2 M = − (A − 4I) = −1 2 2 8 9 1 2 −2 N = (A − 2I) = . 1 −1 2 Thus, e

At

9 8 9 −1 2 4t 2 −2 = e +e −1 2 1 −1 8 2t 9 −e + 2e4t 2e2t − 2e4t = −e2t + e4t 2e2t − e4t 2t

8

13. In this case BcA = {1, et , e−t }. It follows that eAt = M + N et + P e−t . Differentiating and evaluating at t = 0 gives M +N +P = I N −P = A

N + P = A2

from which we get

1 Solutions

305

M = I − A2 A2 + A N = 2 A2 − A P = . 2 Thus, eAt = M + N et + P e−t       2 0 −1 0 0 0 −1 0 1 0 + 0 1 0 et +  0 0 0 e−t = 0 0 2 0 −1 0 0 0 −2 0 2   −t −t 2−e 0 −1 + e  0 et 0 =  2 − 2e−t 0 −1 + 2e−t 0 1 14. In this case BcA = et , tet , e2t . It follows that

eAt = M et + N tet + P e2t .

Differentiating and evaluating at t = 0 gives M +P = I M + N + 2P = A M + 2N + 4P = A2 from which we get M = 2A − A2 N = −A2 + 3A − 2I P = I − 2A + A2 .

Thus, eAt = M et + N tet + P e2t       −2 1 2 −1 0 1 3 −1 −2 0 0 e2t =  0 1 0 et + −1 0 1 tet + 0 −3 1 3 −1 0 1 3 −1 −2   t t 2t t 2t t −2e − te + 3e e −e 2e + tet − 2e2t  −tet et tet =  t t 2t t 2t t t 2t −3e − te + 3e e −e 3e + te − 2e 15. The standard basis of EcA is

306

1 Solutions

0 1 BcA = et , et cos t, et sin t .

Therefore

eAt = M et + N et sin t + P et cos t.

Differentiating twice and simplifying we get the system: eAt = M et + N et sin t + P et cos t AeAt = M et + (N − P )et sin t + (N + P )et cos t A2 eAt = M et − 2P et sin t + 2N et cos t. Now evaluating at t = 0 gives I = M +P A = M +N +P A2 = M + 2N and solving gives N = A−I

M = A2 − 2A + 2I P = −A2 + 2A − I. 1  1 2 −1 2 0 −2 and A straightforward calculation gives A2 =  2 1 1 1 2 2 

 1   1  1 1 0 − 12 0 2 0 2 2 0 −2 0 −1 , M =  0 0 0 , and P =  0 1 0 . N = 1 1 1 1 1 1 0 0 −2 0 2 2 0 2 2

Hence, eAt

1

    1  1 0 12 0 − 12 0 2 0 −2 0 −1 et sin t +  0 1 0 et cos t =  0 0 0  et +  1 1 1 1 1 1 0 0 −2 0 2 0 2 2 2 2



 et + et cos t −et sin t et − et cos t 1 2et sin t 2et cos t −2et sin t  . = 2 t t e − e cos t et sin t et + et cos t

0 1 16. In this case BcA = e2t , te2t , t2 e2t . It follows that eAt = M e2t + N te2t + P t2 e2t .

1 Solutions

307

Differentiating twice and evaluating at t = 0 gives M = I 2M + N = A 4M + 4N + 2P = A2 from which we get M =

I

N =

A − 2I

P =

(A − 2I)2 2

Thus,

 1 0 0 0 1 0 = 0 0 1   0 1 0 = −1 0 1 0 1 0   −1 0 1 1 0 0 0 . = 2 −1 0 1

eAt = M e2t + N te2t + P t2 e2t    1 0 0 0 1 = 0 1 0 e2t + −1 0 0 0 1 0 1  2t e − (1/2)t2 e2t te2t  −te2t e2t = −(1/2)t2 e2t te2t



   0 −1 0 1 2 t 1 te2t +  0 0 0 e2t 2 0 −1 0 1  (1/2)t2 e2t  te2t e2t + (1/2)t2 e2t

17. In this case BcA = {et , cos 2t, sin 2t}. It follows that eAt = M et + N cos 2t + P sin 2t. Differentiating twice and evaluating at t = 0 gives M +N = I M + 2P = A M − 4N = A2 from which we get

308

1 Solutions

 −1 0 1  0 0 0 M= −2 0 2   2 0 −1 I − A2 0 1 0 N= = 5 2 0 −1   0 −1 0 −A2 + 5A − 4I 0 −1 . P = = 2 10 0 −1 0 A2 + 4I = 5

Thus,



eAt = M et + N cos 2t + P sin 2t       −1 0 1 2 0 −1 0 −1 0 0 cos 2t + 2 0 −1 sin 2t =  0 0 0 et + 0 1 −2 0 2 2 0 −1 0 −1 0  t  t −e + 2 cos 2t − sin 2t e − cos 2t 2 sin 2t cos 2t − sin 2t  =  t −2e + 2 cos 2t − sin 2t 2et − cos 2t 0 1 18. In this case BcA = 1, t, e2t , te2t . It follows that

eAt = M + N t + P e2t + Qte2t .

Differentiating three times and evaluating at t = 0 gives M +P = I N + 2P + Q = A 4P + 4Q = A2 8P + 12Q = A3 from which we get

1 Solutions

309



 −1 0 1 0  0 −1 0 1 (A − 2I) (A + I)  M= =  −2 0 2 0 4 0 −2 0 2 2

N=

P =

Q=

Thus,

A(A − 2I)2 = 4

−A2 (A − 3I) = 4

A2 (A − 2I) = 4



 0 −1 0 1 0 0 0 0   0 −2 0 2 0 0 0 0 

2 0  2 0 

0 0  0 0

 0 −1 0 2 0 −1  0 −1 0 2 0 −1 2 0 2 0

 0 −1 0 0 . 0 −1 0 0

eAt = M + N t + P e2t + Qte2t   −1 + 2e2t −t + 2te2t 1 − e2t t − te2t  0 −1 + 2e2t 0 1 − e2t   =  2t 2t 2t −2 + 2e −2t + 2te 2−e 2t − te2t  0 −2 + 2e2t 0 2 − e2t 19. In this case BcA = {cos t, sin t, t cos t, t sin t}. It follows that eAt = M cos t + N sin t + P t cos t + Qt sin t. Differentiating three times and evaluating at t = 0 gives M = I N +P = A −M + 2Q = A2 −N − 3P = A3 from which we get

310

1 Solutions

M=

N=



1 0  0 0

I=

0 1 0 0



0 −1 A(A2 + 3I) =   0 2 0 

0 0 1 0

 0 0  0 1

1 0 0 0 0 0 0 −1

 0 0  1 0

 −1 0 1 0  0 −1 0 1 −A(A2 + I)  P = =  −1 0 1 0 2 0 −1 0 1 A2 + I = 2

Q=

Thus,



 0 −1 0 1 1 0 −1 0  . 0 −1 0 1 1 0 −1 0

eAt = M cos t + N sin t + P t cos t + Qt sin t   cos t − t cos t sin t − t sin t t cos t t sin t − sin t + t sin t cos t − t cos t  −t sin t t cos t . =   −t cos t −t sin t cos t + t cos t sin t + t sin t  t sin t −t cos t − sin t − t sin t cos t + t cos t 20. The standard basis is BcA = {er1 t , er2 t } so that eAt = M er1 t + N er2 t . Fulmer’s method gives I = M +N A = r1 M + r2 N which are easily solved to give M=

A − r2 I r1 − r2

Hence, eAt =

and

N=

A − r1 I . r2 − r1

A − r2 I r1 t A − r1 I r2 t e + e . r1 − r2 r2 − r1

(3)

1 Solutions

311

21. The standard basis is BcA = {ert , tert } so that eAt = M ert + N tert . Fulmer’s method gives I = M A = rM + N which are easily solved to give and

M =I Hence,

N = (A − rI).

eAt = Iert + (A − rI)tert = (I + (A − rI)t) ert .

(4)

22. In this case the standard basis is BcA = {eαt cos βt, eαt sin βt} so that eAt = M eαt cos βt + M eαt sin βt. Fulmer’s method now gives I = M A = αM + βN. and therefore M = I and N =

A−αI β .

eAt = Ieαt cos βt +

Hence,

(A − αI) αt e sin βt. β

(5)

Section 9.5 1. It is easy to see that e

At

y(t) =

9 e−t 0 = . Thus 0 e3t 8

8

98 9 8 9 e−t 0 1 e−t = . 0 e3t −2 −2e3t

2. The characteristic polynomial is cA (s) = s2 +4. Thus BcA = {cos 2t, sin 2t} and eAt = M1 cos 2t + M2 sin 2t. Differentiating and evaluating at t = 0 gives M1 = I 2M2 = A.

312

1 Solutions

8 9 8 98 9 cos 2t sin 2t cos 2t sin 2t 1 Thus eAt = and y(t) = = − sin 2t cos 2t − sin 2t cos 2t −1 8 9 cos 2t − sin 2t sin 2t − cos 2t 0 1 3. The characteristic polynomial is cA (s) = (s − 2)2 . Thus BcA = e2t , te2t and eAt = M1 e2t + M2 te2t . Differentiating and evaluating at t = 0 gives I = M1 A = 2M1 + M2 . and hence M1 = I and M2 = A − 2I. We thus get e 8 2t 9 8 9 8 2t 9 e te2t −1 −e + 2te2t y(t) = = 0 e2t 2 2e2t

At

8

9 e2t te2t = and 0 e2t

4. The characteristic polynomial is cA (s) = (s + 1)2 + 4. Thus BcA = {e−t cos 2t, e−t sin 2t} and eAt = M1 e−t cos 2t + M2 e−t sin 2t. Differentiating and evaluating at t = 0 gives I = M1 A = −M1 + 2M2 .

8 −t 9 e cos 2t e−t sin 2t and hence M1 = I and M2 = 12 (A+I). We thus get eAt = −t −t 8 −t 9 8 9 8 −t 9 −e sin 2t e cos 2t −t e cos 2t e sin 2t 1 e cos 2t and y(t) = = −e−t sin 2t e−t cos 2t 0 −e−t sin 2t 5. The characteristic polynomial is cA (s) = s2 − 1 = (s + 1)(s − 1). Thus BcA = {e−t , et } and eAt = M1 e−t + M2 et . Differentiating and evaluating at t = 0 gives I = M1 + M2 A = −M1 + M2 . and M1 = 12 (−A +9I) and M2 = 8 hence t −t 3e − e −et + e−t 1 2 3et − 3e−t −et + 3e−t and

1 2 (A

+ I). We thus get eAt =

1 Solutions

313

y(t) =

8 9 8 9 8 −t 9 1 3et − e−t −et + e−t 1 e = . 3e−t 2 3et − 3e−t −et + 3e−t 3

6. The characteristic polynomial is cA (s) = s2 + 1. Thus BcA = {cos t, sin t} and eAt = M1 cos t + M2 sin t. Differentiating and evaluating at t = 0 gives I = M1 A = M2 . We thus get e and hence

At

8

9 8 9 1 0 2 −5 = cos t + sin t 0 1 1 −2

y h (t) = eAt y 0 ' 8 9 8 9( 8 9 1 0 2 −5 1 = cos t + sin t 0 1 1 −2 −1 8 9 cos t + 2 sin t −5 sin t = sin t cos t − 2 sin t 7. The characteristic polynomial is cA (s) = (s − 1)2 . Thus BcA = {et , tet } and eAt = M1 et + M2 tet . Differentiating and evaluating at t = 0 gives I = M1 A = M1 + M2 . and hence M1 = I and M2 = A−I. We thus get eAt = 8 t 98 9 8 t 9 e + 2tet −4tet 1 e − 2tet and y(t) = = tet et − 2tet 1 et − tet

8

et + 2tet −4tet t t te e − 2tet

8. The characteristic 0 1polynomial is cA (s) = (s + 1)(s − 2)(s − 1). Thus BcA = e−t , e2t , et and eAt = M1 e−t + M2 e2t + M3 et .

Differentiating twice and evaluating at t = 0 gives

9

314

1 Solutions

I = M1 + M2 + M3 A = −M1 + 2M2 + M3 A2 = M1 + 4M2 + M3 . Solving this system gives M1

M2

M3

We thus get eAt



1 1 2 = (A − 3A + 2I) = 0 6 0   0 0 0 1 2 = (A − I) = 0 1 0 3 0 0 0  0 1 2 = − (A − A − 2I) = 0 2 0



e−t 0  0 e2t = 0 0



e−t 0  0 e2t y(t) = 0 0

3 t 2 (e

3 t 2 (e

 − e−t ) 0 and et

 0 −3/2 0 0 0 0  0 3/2 0 0 . 0 1

    1 −2e−t + 3et − e−t ) e2t  . 0 1 =  t 2 2et e

9. The characteristic polynomial is cA (s) = (s + 1)(s2 + 4). Thus BcA = {e−t , cos 2t, sin 2t} and eAt = M1 e−t + M2 cos 2t + M3 sin 2t. Differentiating and evaluating at t = 0 gives I = M1 + M2 A = −M1 + 2M3 A2 = M1 − 4M2 . and hence

1 Solutions

315

M1

M2

M3

We thus get eAt 

y(t) = 

  0 0 0 1 2 = (A + 4I) =  0 0 0 5 −1 0 1   1 0 0 1 2 = − (A − I) = 0 1 0 5 1 0 0   0 2 0 1 2 = (A + 5A + 4I) = −1/2 0 0 . 10 0 2 0 

 cos 2t 2 sin 2t 0 − 12 sin 2t cos 2t 0 and hence = −t −t −e + cos 2t 2 sin 2t e

    cos 2t 2 sin 2t 0 2 2 cos 2t + 2 sin 2t sin 2t cos 2t 0 1 =  cos 2t − sin 2t −t 2 2 cos 2t + 2 sin 2t + cos 2t 2 sin 2t e

− 12 −t

−e

10. A straightforward calculation gives N O 8 9 et 3 −1 e−t −1 1 At e = + . 2 3 −1 2 −3 3 It follows that y h (t) = eAt y 0 8 98 9 8 98 9 e−t 3 −1 1 e−t −1 1 1 = + 2 3 −1 3 2 −3 3 3 8 9 1 = e−t 3 and y p = eAt ∗ f (t) , N O 8 9- 8 t 9 et 3 −1 e−t −1 1 e = + ∗ t −3 3 e 2 3 −1 2 N O8 9 8 98 9 et ∗ et 3 −1 1 e−t ∗ et −1 1 1 = + −3 3 1 2 2 3 −1 1 8 9 1 = tet . 1 It now follows that

316

1 Solutions

y(t) = y h + y p 8 −t 9 e + tet = 3e−t + tet 11. A straightforward calculation gives 8 9 8 9 1 0 0 1 eAt = e−t cos 2t + e−t sin 2t 0 1 −1 0 It follows that y h (t) = eAt y 0 ' 8 9 8 9( 8 9 1 0 0 1 1 −t −t = e cos 2t + e sin 2t 0 1 −1 0 0 8 −t 9 e cos 2t = . −e−t sin 2t and y p = eAt ∗ f (t) ' 8 9 8 9( 8 9 1 0 0 1 5 −t −t = e cos 2t + e sin 2t ∗ 0 1 −1 0 0 8 9 8 9 5 0 = e−t cos 2t ∗ 1 + e−t sin 2t ∗ 1 0 −5 8 9 8 9 1 0 −t −t −t −t = (1 − e cos 2t + 2e sin 2t) + (2 − 2e cos 2t − e sin 2t) 0 −1 8 9 1 − e−t cos 2t + 2e−t sin 2t = . −2 + 2e−t cos 2t + e−t sin 2t It now follows that y(t) = y h + y p 8 9 1 + 2e−t sin 2t = −2 + 2e−t cos 2t 12. A straightforward calculation gives 8 9 8 9 1 0 2 −5 eAt = cos t + sin t . 0 1 1 −2 It follows that

1 Solutions

317

y h (t) = eAt y 0 ' 8 9 8 9( 8 9 1 0 2 −5 1 = cos t + sin t 0 1 1 −2 −1 8 9 8 9 1 7 = cos t + sin t −1 3 and y p = eAt ∗ f (t) ' 8 9 8 9( 8 9 1 0 2 −5 2 = cos t + sin t ∗ cos t 0 1 1 −2 1 8 9 8 9 2 −1 = cos t ∗ cos t + sin t ∗ cos t 1 0 8 9 8 9 1 1 2 −1 = (t cos t + sin t) + (t sin t) 1 0 2 2 8 9 1 2t cos t − t sin t + 2 sin t = . t cos t + sin t 2 It now follows that y(t) = y h + y p 8 9 8 9 8 9 1 2t cos t − t sin t + 2 sin t 1 7 = cos t + sin t + −1 3 t cos t + sin t 2 8 9 1 2 cos t + 16 sin t + 2t cos t − t sin t = −2 cos t + 7 sin t + t cos t 2 13. A straightforward calculation gives 8 9 8 9 1 0 0 −2 At −t −t e = e cos 2t + e sin 2t 0 1 1/2 0 It follows that y h (t) = eAt y 0 ' 8 9 8 9( 8 9 1 0 0 −2 2 = e−t cos 2t + e−t sin 2t 0 1 1/2 0 −1 8 9 8 9 2 2 = e−t cos 2t + e−t sin 2t −1 1 and

318

1 Solutions

y p = eAt ∗ f (t) ' 8 9 8 9( 8 9 1 0 0 −2 4 −t −t = e cos 2t + e sin 2t ∗ 0 1 1/2 0 1 8 9 8 9 4 −2 = ((e−t cos 2t) ∗ 1) + ((e−t sin 2t) ∗ 1) 1 2 8 9 8 9 % & % & −2 1 1 4 = 1 − e−t cos 2t + 2e−t sin 2t + 2 − 2e−t cos 2t − e−t sin 2t 1 2 5 5 8 9 −t 2e sin 2t = . 1 − e−t cos 2t It now follows that y(t) = y h + y p

8 9 8 9 8 9 2 2 2e−t sin 2t = e−t cos 2t + e−t sin 2t + −1 1 1 − e−t cos 2t 8 9 2e−t cos 2t + 4e−t sin 2t = . 1 + e−t sin 2t − 2e−t cos 2t

14. A straightforward calculation gives 8 9 8 9 et e3t 1 1 1 −1 eAt = + 1 2 −1 2 1 1 It follows that y h (t) = eAt y 0 ' t8 9 8 9( 8 9 e e3t 1 1 1 −1 1 = + −1 1 1 1 1 2 2 8 9 1 = e3t 1 and y p = eAt ∗ f (t) ' t8 9 8 9( 8 9 e e3t 1 1 1 −1 1 t = + ∗e 1 −1 2 −1 2 1 1 8 9 et ∗ et 2 = −2 2 8 9 1 t = te . −1 It now follows that

1 Solutions

319

y(t) = y h + y p 8 9 8 9 1 3t 1 t = e + te 1 −1 8 9 tet + e3t = . −tet + e3t 15. A straightforward calculation gives 8 9 8 9 1 0 4 2 eAt = et + tet 0 1 −8 −4 It follows that y h (t) = eAt y 0 ' 8 9 8 9( 8 9 4 2 0 t 1 0 t = e + te 0 1 −8 −4 1 8 9 t 2te = et − 4tet and y p = eAt ∗ f (t) ' 8 9 8 9( 8 9 4 2 1 t 1 0 t = e + te ∗t 0 1 −8 −4 −2 8 9 1 t = e ∗t −2 8 9 1 t = (e − t − 1) . −2 It now follows that y(t) = y h + y p 8 9 8 9 2tet 1 t = + (e − t − 1) et − 4tet −2 8 9 2tet + et − t − 1 = −4tet − et + 2t + 2 16. A straightforward calculation gives       1 0 1 0 1 0 2 −3 −1 −3t 1 e  0 1 0 + 0 0 eAt = 0 0 0 + e−t 0 3 3 2 0 2 0 −1 0 −2 3 1

320

1 Solutions

It follows that y h (t) = eAt y 0          1 0 1 0 1 0 2 −3 −1 2 −3t 1 e  0 1 0 + 0 0 1 =  0 0 0 + e−t 0 3 3 2 0 2 0 −1 0 −2 3 1 1     1 1 = 0 + e−t  1 2 −1 and

y p = eAt ∗ f (t)       1 0 1 0 1 0 2 −3t 1 e  0 1 0 + =  0 0 0 + e−t 0 3 3 2 0 2 0 −1 0 −2     1 0 1 1 0 1 1 ∗ e−2t  0 0 0  0 + e−t ∗ e−2t 0 1 = 3 2 0 2 −1 0 −1  2 −3 e−3t ∗ e−2t  0 0 + 3 −2 3   1 = e−2t − e−3t  0 −1

   −3 −1 1 0 0 ∗ e−2t  0 3 1 −1   0 1 0  0  0 −1   −1 1 0  0 1 −1

It now follows that

y(t) = y h + y p       1 1 1 = 0 + e−t  1 + (e−2t − e−3t )  0 2 −1 −1   −t −2t −3t 1+e +e −e e−t  =  −t −2t 2−e −e + e−3t 17. A straightforward calculation gives       1 0 0 −1 1 1 0 0 0 eAt = et −1 1 1 + tet  0 0 0 + e2t  1 0 −1 1 0 0 −1 1 1 −1 0 1 Clearly

1 Solutions

321

  0 y h (t) = 0 0

while y p (t) = eAt ∗ f (t)   1 0 = et −1 1 1 0 

  0 −1 1 + tet  0 0 −1 

     1 1 0 0 0 1 0 0 + e2t  1 0 −1 ∗ e2t  1 1 1 −1 0 1 −1     1 −1 0 = (et ∗ e2t ) −1 + (tet ∗ e2t )  0 + (e2t ∗ e2t )  2 1 −1 −2       1 −1 0 = (e2t − et ) −1 + (e2t − tet − et )  0 + te2t  2 1 −1 −2   t te = 2te2t − e2t + et  −2te2t + tet

It now follows that



 tet y(t) = y h + y p = 2te2t − e2t + et  −2te2t + tet 18. y1 and y2 are related to each other as follows: y1! = −2y1 + 2y2 y2! = 2y1 − 2y2 with initial conditions y1 (0) = 4 and y2 (0) = 0. Let A = 8 9 4 y(0) = . Then it is easy to check that 0 8 9 8 9 1 1 1 1 1 −1 −4t eAt = + e . 1 2 1 1 2 −1

8

9 −2 2 and 2 −2

It follows that y(t) = eAt y 0 =

' 8 9 8 9 (8 9 8 9 1 1 1 1 1 −1 −4t 4 2 + 2e−4t + e = . 1 0 2 − 2e−4t 2 1 1 2 −1

After 30 seconds (1/2 minute) we have

322

1 Solutions

y(1/2) =

8

9 8 9 2 + 2e−2 2.27 = . 2 − 2e−2 1.73

After 30 seconds Tank 1 has 2.27 pounds of dissolved salt and Tank 2 has 1.73 pounds of dissolved salt. 19. y1 and y2 are related to each other as follows: y1! = 2 − 2y1 y2! = 2y1 − y2

8 9 −2 0 with initial conditions y1 (0) = 4 and y2 (0) = 0. Let A = , 2 −1 8 9 8 9 2 4 f (t) = , and y(0) = . We need to solve the system y ! = Ay + f , 0 0 8 9 4 y(0) = . It is easy to check that 0 8 9 8 9 0 0 −t 1 0 −2t eAt = e + e . 2 1 −2 0 It follows that '8 9 8 9 (8 9 8 9 8 9 0 0 −t 1 0 −2t 4 0 −t 4 −2t y h (t) = e + e = e + e 2 1 −2 0 0 8 −8 and yp

8 9 8 98 9 8 98 9 2 0 0 2 1 0 2 −t −2t = e ∗ = (e ∗ 1) + (e ∗ 1) 0 2 1 0 −2 0 0 8 9 8 9 0 1 = (1 − e−t ) + (1 − e−2t ) 4 −2 8 9 8 9 8 9 1 0 −t −1 −2t = + e + e . 2 −4 2 At

It now follows that y(t) = y h (t) + y p (t) 8 9 8 9 8 9 1 0 −t 3 −2t = + e + e . 2 4 −6 The concentration of salt in Tank 2 is 1/2 if y2 (t) = 1. We thus solve y2 (t) = 1, i.e. 2 + 4e−t − 6e−2t = 1 for t. Let x = e−t . Then −6x2 + √ 2± 10 4x + 1 = 0. The quadratic formula gives x = . Since x > 0 we 6

1 Solutions

323 √

have e−t = x = 2+6 10 . Solving for t we get t = − ln minutes or 9.02 seconds.

"

√ # 2+ 10 6

= 0.1504

20. y1 and y2 are related to each other as follows: y1! = 8 − 2y1 y2! = y1 − 2y2

8 9 −2 0 with initial conditions y1 (0) = 0 and y2 (0) = 4. Let A = , 1 −2 8 9 8 9 8 0 f (t) = , and y(0) = . We need to solve the system y ! = Ay + f , 0 4 8 9 0 y(0) = . It is easy to check that 4 eAt =

8 9 8 9 1 0 −2t 0 0 e + te−2t . 0 1 1 0

It follows that y h (t) =

'8

9 8 9 (8 9 8 9 1 0 −2t 0 0 0 0 −2t −2t e + te = e 0 1 1 0 4 4

and yp

8 9 8 98 9 8 98 9 8 1 0 8 0 0 8 −2t −2t = e ∗ = (e ∗ 1) + (te ∗ 1) 0 0 1 0 1 0 0 8 9 8 9 4 0 = (1 − e−2t ) + (1 − e−2t − 2te−2t ) 0 2 8 9 8 9 8 9 4 4 −2t 0 = − e − te−2t . 2 2 4 At

It now follows that y(t) = y h (t) + y p (t) 8 9 8 9 8 9 4 −4 −2t 0 = + e − te−2t . 2 2 4 Thus y1 (t) = 4 − 4e−2t

y2 (t) = 2 + 2e−2t − 4te−2t . The amount of salt in Tank 2 is minimal when y2! (t) = 0. Since y2! (t) = 8(t − 1)e−2t we have t = 1 as a critical point. Since y2 (0) = 4, y2 (1) =

324

1 Solutions

2 − 2e−2 and limt→∞ y2 (t) = 2, it follows the y2 is a minimum at t = 1 and the minimum is y2 (1) = 2 − 2e−2 ≈ 1.73 lbs salt. 21. y1 and y2 are related to each other as follows: y1! = 2 + 3y2 − 5y1 y2! = 2 + 5y1 − 7y2

8 9 −5 3 with initial conditions y1 (0) = 0 and y2 (0) = 0. Let A = , 5 −7 8 9 8 9 2 0 f (t) = , and y(0) = . We need to solve the system y ! = Ay + f , 2 0 8 9 0 y(0) = . It is easy to check that 0 e

At

8 9 8 9 e−2t 5 3 e−10t 3 −3 = + . −5 5 8 5 3 8

Clearly y h = 0 while y(t) = y p (t) = eAt ∗ f (t) ' −2t 8 9 8 9( 8 9 e e−10t 5 3 3 −3 2 = + ∗1 −5 5 2 8 5 3 8 8 9 −2t e ∗ 1 16 = 16 8 8 9 1 −2t = (1 − e ) 1 Thus y1 (t) = 1 − e−2t

y2 (t) = 1 − e−2t .

Section 9.6 1. The characteristic polynomial is cA (s) = s2 −1 = (s+1)(s−1). There are two distinct λ1 = −1 and λ2 = 1. An easy calculation 8 eigenvalues, 9 8 9 give 1 3 that v1 = is an eigenvector with eigenvalue −1 and v2 = is an −1 −1 8 9 1 3 eigenvector with eigenvalue 1. Let P = . Then J = P −1 AP = −1 −1

1 Solutions

325

8

9 −1 0 . Since there is a distinct positive and negative real eigenvalue 0 1 the critical point is a saddle. 2. The characteristic polynomial is cA (s) = (s + 1)28. 9So λ = −1 is 8 the 9 only 1 1 eigenvalue and all eigenvectors are multiples of Let v1 = . Then 18 9 8 90 8 9 −2 2 1 −2 v1 is not an eigenvector. Let v2 = (A − (−1)I)v1 = = . −2 2 0 −2 8 9 8 9 1 −2 −1 0 Let P = . Then J = P −1 AP = . Since the eigenvalue 0 −2 1 −1 is negative it follows that the origin is a stable improper node. 3. The characteristic polynomial is cA (s) = s2 +4s+5 = (s+2)28+1 and9has −3 − i complex roots −2 ± i. A calculation gives an eigenvector v = for 5 9 8 9 8 9 8 L M −3 −1 −3 −1 −2 − i. Let v1 = and v2 = . Let P = v1 v2 = , 58 0 5 0 9 −2 −1 Then J = P −1 AP = and the origin is a stable spiral node. 1 −2 2 4. The characteristic polynomial is cA (s) = s2 − 8 4 + 98 = s + 4. It has 1−i complex roots ±2i. An eigenvector for −2i is . From this we get 2 8 9 8 9 1 −1 0 −2 P = , J = P −1 AP = , and the origin is a center. 2 0 2 0

5. In this case A is of type J3 with positive eigenvalue 2. The origin is an unstable star node. 6. The characteristic polynomial is cA (s) = s2 − 6s + 8 + 1 = (s −8 3)29. So 3 1 is the only eigenvalue. The eigenspace for 3 is all multiples of . Let −1 8 9 8 9 1 1 v1 = , a vector not an eigenvector. Let v2 = (A − 3I)v1 = . Let 0 −1 8 9 8 9 1 1 3 0 P = . Then J = P −1 AP = and the origin is an unstable 0 −1 1 3 improper node. 7. The characteristic polynomial is cA (s) = s2 −6s+8 = (s−2)(s−4). There are two distinct 8 9 eigenvalues, λ1 = 2 and λ2 = 4. An easy calculation 8 9 gives 1 3 that v1 = is an eigenvector with eigenvalue 2 and v2 = is an −1 −1

326

1 Solutions

8 9 1 3 eigenvector with eigenvalue 4. Let P = . Then J = P −1 AP = −1 −1 8 9 2 0 . Since both eigenvalues are positive the origin is an unstable node. 0 4 8. The characteristic polynomial is cA (s) = s2 + 4s + 3 = (s + 1)(s + 3). There are two distinct eigenvalues, λ1 = −1 and λ2 = −3. An easy 8 9 1 calculation gives that v1 = is an eigenvector with eigenvalue 1 and 1 8 9 8 9 1 1 1 v2 = is an eigenvector with eigenvalue 3. Let P = . Then 2 1 2 8 9 −1 0 J = P −1 AP = . Since both eigenvalues are negative the origin 0 −3 is a stable node. 2 2 9. The characteristic polynomial is cA (s) = s2 − 2s + 5 = (s 8 − 1) 9+ 2 . −1 + i So 1 ± 2i are the eigenvalues. An eigenvector for 1 − 2i is . Let 4 8 9 8 9 −1 1 1 −2 P = . Then J = P −1 AP = . The origin is an unstable 4 0 2 1 star node.

10. In this case A is of type J3 with negative eigenvalue −3. stable star node. 8 a −1 11. Let (x, y) be a point on the P (L). Suppose P = c 8 9 8 9 x ax + by P −1 = . Then (u, v) is on L and so y cx + dy

The origin is a 9 8 9 b u . Let = d v

0 = Du + Ev + F = D(ax + by) + E(cx + dy) + F = (Da + Ec)x + (Db + Ed)y + F = D! x + E ! y + F, where (D! , E ! ) = (Da + Ec, Db + Ed) = (D, E)P −1 . It follows that (x, y) satisfies the equation of a line. A line goes through the origin if and only if F = 0. If the equation for L has F = 0 then the above calculation shows the equation for P (L) does too. 12. Let C be the graph of the general equation in the (u, v) plane and P (C) the transform of C. 8 Let 9 (x, 8 y) 9 be a point8 of P9(C) and 8 (u, 9 v)8the point 9 u x a b u ax + by −1 on C such that P = . If P = then = . v y c d v cx + dy Replace u and v in the quadratic equation above by ax + by and cx + dy, respectively. The only quadratic terms are

1 Solutions

327

A(ax + by)2 + B(ax + by)(cx + dy) + C(cx + dy)2 . Collect together the coefficients of x2 , xy, and y 2 and call them A! , B ! , and C ! . It is a straightforward although tedious calculation to see that he discriminant in the (x, y) variable, ∆! , becomes ∆! = ∆(det P −1 )2 . It follows that the discriminants have the same sign, or are both zero. Hence P (C) is an ellipse, parabola, or hyperbola according as C is an ellipse, parabola, or hyperbola. 13. Let C be the graph of a power curve in the (u, v) plane and P (C) the transform of C.8Let 9 (x,8y)9be a point of8 P (C) 9 and (u, 8 9v) the8 point on 9 u x a b u ax + by −1 C such that P = . If P = then = . v cx + dy v y c d Replace u and v in the equation Au + Bv = (Cu + Dv)p by ax + by and cx + dy, respectively. We then get (Aa + Bc)x + (Ab + Bd)y = ((Ca + Dc)x + (Cb + Dd)y)p . Thus P (C) is the graph of a power curve. 14. Suppose C is the graph of the differentiable function G(u, v) = 0 and L is given by the graph of Du + Ev + F = 0. Then L is tangent to C at (u0 , v0 ) if ∇G(u0 , v0 ) = (D, E). The P 8 curve, 9 8 (C), 9 is given as the graph x0 u0 −1 of GP = G ◦ P (x, y) = 0. Let = P . By the chain rule we y0 v0 have 8 9 a b ∇GP (x0 , y0 ) = ∇G(u0 , v0 ) = (Da + Ec, Db + Ed). c d It follows that the tangent line to P (C) at (x0 , y0 ) is given by (Da + Ec)x + (Db + Ed)y + F = 0. The graph of this line is P (L) (see the solution to Problem 11 above). 15. The characteristic polynomial takes the form cA (s) = s2 −(tr A)s+det A. Let λ = tr A. Since det A = 0 we have cA (s) = s2 − λs = s(s − λ). Now consider two cases: λ &= 0: In this case A has two distinct eigenvalues, 0 and λ. Let v1 be an eigenvector with eigenvalue 0 and v2 an eigenvectorL with eigenvalue M λ. Then v1 and v2 are linearly independent. If P = v1 v2 then P is invertible and 8 9 L M L M L M 0 0 AP = Av1 Av2 = 0 λv2 = v1 v2 = P J1 . 0 λ

Now multiply both sides on the left by P −1 to get that P −1 AP = J1 .

328

1 Solutions

λ = 0: In this case cA (s) = s2 . Since A is not zero there must be a vector v1 that is not an eigenvector. Let v2 = Av1 . Then v2 is an eigenvector with eigenvalue 0 since, by the Cayley-Hamilton theorem, Av2 = A2 v1 = 0. L M Now let P = v1 v2 . Then

8 9 8 9 L M L M L M L M 0 0 0 0 AP = A v1 v2 = Av1 Av2 = v2 0 = v1 v2 =P . 1 0 1 0 Now multiply both sides on the left by P −1 to get that P −1 AP = J2 .

16. First notice that J1 c = 0 if and only if c2 = 0. Therefore every point 8 9 in c the u-axis is an equilibrium point. Now assume c2 &= 0 and w(0) = 1 is c2 8 9 c 1 an initial point. Then the solution to w! = J1 w, w(0) = is u(t) = c1 c2 and v(t) = c2 eλt . The path (u(t), v(t)) = (c1 , c2 eλt ), t ∈ R, is a vertical ray going through (c1 , c2 ) and pointing away from the u-axis if λ > 0 and toward the u-axis if λ < 0. See the diagrams below.

λ0

8 9 c1 then the equation J2 c = 0 implies c1 = 0. It follows that c2 each point on the v-axis is an equilibrium point. Now assume c1 &= 0. 8 9 c1 ! The solution to w = J2 w, w(0) = is u(t) = c1 and v(t) = tc1 + c2 . c2 The path (u(t), v(t)) = (c1 , c2 ) + t(0, c1 ), t ∈ R, is a vertical line that passes through the initial condition (c1 , c2 ) and points upward if c1 > 0 and downward it c1 < 0. The phase portrait is given below:

17. If c =

1 Solutions

329

18. We present two solutions. Solution 1: Let v1 be an eigenvector L withMeigenvalue 0 and v2 an eigenvector with eigenvalue λ. Let P = v1 v2 . Then J1 = P −1 AP and P maps the trajectories in the (u, v) phase plane for J1 to the trajectories in the (x, y) phase plane 8for9 A. The equilibrium 8 9points in the (u, v) a a phase plane are of the form . Observe that P = av1 . So av1 is 0 0 an equilibrium point. Since a is arbitrary it follows that each point in the zero eigenspace is an equilibrium point. 8 The 9 other trajectories in the a (u, v) phase plane for J1 take the form , b &= 0. Now observe that beλt 8 9 a P = av1 + beλt v2 . It follows that the corresponding trajectories in beλt the (x, y) phase plane are half lines with one end at av1 and parallel to v2 . If λ > 0 they point away from the equilibrium point and if λ < 0 they point toward the equilibrium point. Solution 2: Let v1 be an eigenvector with eigenvalue 0 and v2 an eigenvector with eigenvalue λ. By Lemma 9.5.9 eAt v1 = v1 and eAt v2 = eλt v2 . it follows that each eigenvector with eigenvalue 0 is an equilibrium point. Let c be any initial value. Then we can write c = av1 + bv2 , for some a and b, and it follows that eAt (av1 + bv2 ) = av1 + beλt v2 . If b &= 0 then c is not an equilibrium point. The path eAt c is a half line starting at av1 and parallel to v2 . If λ > 0 then the trajectory points away from the equilibrium point and if λ < 0 it points toward the equilibrium point. 19. It is not difficult to see that eAt = I + tA. Let v1 be an eigenvector. By Lemma 9.5.9, eAt v1 = v1 . So each eigenvector is an equilibrium point. Let v be a vector that is not an eigenvector. By the Cayley-Hamilton theorem A2 v = 0 so Av is an eigenvector. Furthermore eAt v = v + tAv. The trajectory is a line parallel to Av going through v. 20. It is enough to show that lim x ln |x| = 0. But this follows from L’Hospitals rule:

x→0

330

1 Solutions

ln |x| 1/x = lim = lim −x = 0. x→0 1/x x→0 −1/x2 x→0

lim x ln |x| = lim

x→0

21. Assume c1 > 0 and thus x > 0 (the case where c1 < 0 is similar). We ln x/c1 1 c2 have y ! = + + . It follows that lim+ y ! = −∞ if λ > 0 and λ λ c1 x→0 lim y ! = ∞ if λ < 0. x→0+

22. Assume c1 > 0 and thus x > 0 (the case where c1 < 0 is similar). Assume x 1 c2 c1 1 y ! = 0. Solving ln + + = 0 gives x = c1 +λc2 . λ c1 λ c1 e c1 23. y !! =

1 and the result follows. λx

Section 9.7 8 −t 9 8 98 9 −e 2e2t −2 1 e−t e2t 1. Observe that Φ (t) = = = A(t)Φ(t). −e−t 8e2t −4 3 e−t 4e2t Also, det Φ(t) = 4et − et = 3et &= 0. Thus Φ(t) is a fundamental matrix. The 8general solution can be written in the form y(t) = Φ(t)c, where 9 c1 c= is a constant vector. The initial condition implies y(0) = Φ(0)c c2 or 8 9 8 98 9 8 9 1 1 1 c1 c1 + c 2 = = . −2 1 4 c2 c1 + 4c2 !

Solving for c we get c1 = 2 and c2 = −1. It follows that 8 −t 9 8 2t 9 8 −t 9 e e 2e − e2t y(t) = Φ(t)c = 2 −t − = . e 4e2t 2e−t − 4e2t

The standard fundamental matrix at t = 0 is 8 −t 98 9−1 e e2t 1 1 −1 Ψ (t) = Φ(t)Φ(0) = e−t 4e2t 1 4 8 98 9 1 e−t e2t 4 −1 = 1 3 e−t 4e2t −1 8 −t 9 2t −t 1 4e − e −e + e2t = . 3 4e−t − 4e2t −e−t + 4e2t 9 8 98 9 2e2t 9e3t 5 −3 e2t 3e3t 2. Observe that Φ (t) = = = A(t)Φ(t). 2e2t 6e3t 2 0 e2t 2e3t 5t 5t 5t Also, det Φ(t) = 2e − 3e = −e &= 0. Thus Φ(t) is a fundamental !

8

1 Solutions

331

matrix. The solution can be written in the form y(t) = Φ(t)c, 8 general 9 c1 where c = is a constant vector. The initial condition implies y(0) = c2 Φ(0)c or 8 9 8 98 9 8 9 2 1 3 c1 c + 3c2 = = 1 . 1 1 2 c2 c1 + 2c2 Solving for c we get c1 = −1 and c2 = 1. It follows that 8 2t 9 8 3t 9 8 2t 9 e 3e −e + 3e3t y(t) = Φ(t)c = − 2t + = . e 2e3t −e2t + 2e3t

The standard fundamental matrix at t = 0 is 8 2t 98 9−1 e 3e3t 1 3 −1 Ψ (t) = Φ(t)Φ(0) = e2t 2e3t 1 2 8 2t 98 9 e 3e3t −2 3 = e2t 2e3t 1 −1 8 9 2t 3t −2e + 3e 3e2t − 3e3t = . −2e2t + 2e3t 3e2t − 2e3t 3. Observe that 9 t cos(t2 /2) −t sin(t2 /2) −t sin(t2 /2) − cos(t2 /2) 8 98 9 0 t sin(t2 /2) cos(t2 /2) = −t 0 cos(t2 /2) − sin(t2 /2)

Φ! (t) =

8

= A(t)Φ(t).

Also, det Φ(t) = − sin2 (t2 /2) − cos2 (t2 /2) = −1 &= 0. Thus Φ(t) is a fundamental matrix. The8 general solution can be written in the form 9 c1 y(t) = Φ(t)c, where c = is a constant vector. The initial condition c2 implies y(0) = Φ(0)c or 8 9 8 98 9 8 9 1 0 1 c1 c = = 2 . 0 1 0 c2 c1 Thus c1 = 0 and c2 = 1. It follows that 8 9 8 9 8 9 sin(t2 /2) cos(t2 /2) cos(t2 /2) y(t) = Φ(t)c = 0 + = . cos(t2 /2) − sin(t2 /2) − sin(t2 /2) The standard fundamental matrix at t = 0 is

332

1 Solutions

Ψ (t) = Φ(t)Φ(0)

−1

8

98 sin(t2 /2) cos(t2 /2) 0 = cos(t2 /2) − sin(t2 /2) 1 8 98 sin(t2 /2) cos(t2 /2) 0 = cos(t2 /2) − sin(t2 /2) 1 8 9 cos(t2 /2) sin(t2 /2) = . − sin(t2 /2) cos(t2 /2)

1 0 1 0

9−1 9

9 8 98 9 2t 2t t t 1 + t2 3 + t2 = = −2t −2t −t −t 1 − t2 −1 − t2 A(t)Φ(t). Also, det Φ(t) = −4 &= 0. Thus Φ(t) is a fundamental matrix.8The 9 general solution can be written in the form y(t) = Φ(t)c, where c1 c= is a constant vector. The initial condition implies y(0) = Φ(0)c c2 or 8 9 8 98 9 8 9 4 1 3 c1 c1 + 3c2 = = . 0 1 −1 c2 c1 − c 2

4. Observe that Φ! (t) =

8

Solving for c we get c1 = 1 and c2 = 1. It follows that 8 9 8 9 8 9 1 + t2 3 + t2 4 + 2t2 y(t) = Φ(t)c = + = . 1 − t2 −1 − t2 −2t2

The standard fundamental matrix at t = 0 is 8 98 9−1 1 + t2 3 + t2 1 3 Ψ (t) = Φ(t)Φ(0)−1 = 1 − t2 −1 − t2 1 −1 8 98 9 1 1 + t2 3 + t2 1 3 = 4 1 − t2 −1 − t2 1 −1 8 9 1 4 + t2 2t2 = . 4 −2t2 4 − t2 5. Observe that 8

− cos t + t sin t − sin t − t cos t Φ (t) = sin t + t cos t − cos t + t sin t 8 98 9 1/t 1 −t cos t −t sin t = −1 1/t t sin t −t cos t !

9

= A(t)Φ(t).

Also, det Φ(t) = t2 cos2 t + t2 sin2 t = t2 &= 0. Thus Φ(t) is a fundamental matrix. The solution can be written in the form y(t) = Φ(t)c, 8 general 9 c1 where c = is a constant vector. The initial condition implies y(π) = c2 Φ(π)c or

1 Solutions

333

8

9 8 98 9 8 9 1 π 0 c1 πc1 = = . −1 0 π c2 πc2

Thus c1 = 1/π and c2 = −1/π. It follows that 8 9 8 9 8 9 1 −t cos t 1 −t sin t t − cos t + sin t y(t) = Φ(t)c = − = . t sin t cos t + sin t π π −t cos t π The standard fundamental matrix at t = π is 8 98 9−1 −t cos t −t sin t π 0 −1 Ψ (t) = Φ(t)Φ(π) = t sin t −t cos t 0 π 8 9 1 −t cos t −t sin t = . t sin t −t cos t π 6. By the product rule we get 8 9 8 9 8 9 1 t2 2t 1 t2 + 2t t 0 t Φ! (t) = et + e = e −1 1 − t2 0 −2t −1 1 − t2 − 2t 8 9 8 9 1 + 2t 2t t 1 t2 = e −2t 1 − 2t −1 1 − t2 = A(t)Φ(t).

Also, det Φ(t) = e2t &= 0. Thus Φ(t) is a fundamental matrix. The 8 general 9 c1 solution can be written in the form y(t) = Φ(t)c, where c = is a c2 constant vector. The initial condition implies y(0) = Φ(0)c or 8 9 8 98 9 8 9 1 1 0 c1 c1 = = . 0 −1 1 c2 −c1 + c2 Thus c1 = 1 and c2 = 1. It follows that 8 9 8 9 8 9 2 1 t2 t t t 1+t y(t) = Φ(t)c = e +e =e . −1 1 − t2 −t2 The standard fundamental matrix at t = 0 is 8 98 9−1 1 t2 1 0 Ψ (t) = Φ(t)Φ(0)−1 = et −1 1 − t2 −1 1 8 98 9 1 t2 1 0 = et −1 1 − t2 1 1 8 9 1 + t2 t2 = et . −t2 1 − t2

334

1 Solutions

7. Observe that 8 9 0 tet 0 et 8 98 9 1 1 1 (t − 1)et = 1/t 1/t −1 et

Φ! (t) =

= A(t)Φ(t).

Also, det Φ(t) = et +tet −et = tet &= 0. Thus Φ(t) is a fundamental matrix. The 8general solution can be written in the form y(t) = Φ(t)c, where 9 c1 c= is a constant vector. The initial condition implies y(0) = Φ(0)c c2 or 8 9 8 98 9 8 9 −3 1 0 c1 c1 = = . 4 −1 e c2 −c1 + ec2 Thus c1 = −3 and c2 = 1/e. It follows that 8 9 8 9 8 9 1 (t − 1)et 1 (t − 1)et−1 − 3 y(t) = Φ(t)c = −3 + = . −1 et et−1 + 3 e The standard fundamental matrix at t = 0 is 8 98 9−1 1 (t − 1)et 1 0 −1 Ψ (t) = Φ(t)Φ(0) = −1 et −1 e 8 98 9 1 1 (t − 1)et e 0 = et 1 1 e −1 8 9 1 e + (t − 1)et (t − 1)et = −e + et et e 8 9 1 + (t − 1)et−1 (t − 1)et−1 = . −1 + et−1 et−1 9 8 u 9 1 1 0 e 0 . Then A(t) = A. It is easy to see that eAu = . 0 1 0 eu t )t 1 Let ln t = 1 du. Then the standard fundamental matrix at t = 1 is u

8. Let A =

8

Ψ (t) = eAu |u=ln t 8 u 9$ 8 9 e 0 $$ t 0 = = . 0 eu $u=ln t 0 t

The homogeneous solution is given by

1 Solutions

335

y h (t) = Ψ (t)y(1) 8 98 9 8 9 t 0 1 t = = . 0 t 2 2t The particular solution is given by * t y p (t) = Ψ (t) Ψ (u)f (u) du 1 8 9* t8 98 9 t 0 1/u 0 1 = du 0 t 1 0 1/u u 8 9* t8 9 t 0 1/u = du 0 t 1 1 8 98 9 8 9 t 0 ln t t ln t = = 2 . 0 t t−1 t −t It follows that 8 9 8 9 8 9 t t ln t t + t ln t y(t) = y h (t) + y p (t) = + 2 = . 2t t −t t + t2 9 1 0 −1 . Then A(t) = A. We first compute eAu . The charac1 2 t teristic polynomial, cA (s) is 8 9 s 1 cA (s) = det (sI − A) = det = s2 − 2s + 1 = (s − 1)2 . −1 s − 2

9. Let A =

8

It follows that BcA = {eu , ueu }. Using Fulmer’s method we have eAu = eu M1 + ueu M2 . Differentiating and evaluating at u = 0 gives I = M1 A = M1 + M2 . 8 9 8 9 1 0 −1 −1 It follows that M1 = I = and M2 = A − I = . Thus 0 1 1 1 8 9 8 9 Au u 1 0 u −1 −1 e =e + ue . 0 1 1 1 Since ln t is an antiderivative of 12

1 t

and ln 1 = 0 we have by Proposition

336

1 Solutions

8 9 8 9 1 0 −1 −1 Ψ (t) = eln t + (ln t)eln t 0 1 1 1 8 9 t − t ln t −t ln t = , t ln t t + t ln t is the standard fundamental matrix for y ! (t) = A(t)y(t) at t = 1. The homogeneous solution is given by y h (t) = Ψ (t)y(1) 8 98 9 8 9 t − t ln t −t ln t 2 2t − 2t ln t = = . t ln t t + t ln t 0 2t ln t The particular solution is given by * t y p (t) = Ψ (t) Ψ (u)−1 f (u) du 1 8 9* t 8 98 9 1 1 + ln u t − t ln t −t ln t ln u 1 = du t ln t t + t ln t 1 u − ln u 1 − ln u −1 8 9 8 9 t − t ln t −t ln t 1 = ln t du t ln t t + t ln t −1 8 9 1 = t ln t −1 It follows that 8

9 8 9 8 9 2t − 2t ln t t ln t 2t − t ln t y(t) = y h (t) + y p (t) = + = . 2t ln t −t ln t t ln t 9 2t 2 3 . Then A(t) = 2 A. We first compute eAu . The −1 −2 t +1 characteristic polynomial is 8 9 s−2 −3 cA (s) = det (sI − A) = det = s2 − 1 = (s − 1)(s + 1). 1 s+2

10. Let A =

8

The resolvent matrix is (sI − A)−1



s+2  (s − 1)(s + 1) =   −1 (s − 1)(s + 1)

 3 (s − 1)(s + 1)    s−2 (s − 1)(s + 1)

8 9 8 9 1 1 3 3 −1 −3 = + . 1 3 2(s − 1) −1 −1 2(s + 1)

1 Solutions

337

It follows that eAu =

8 9 8 9 eu e−u −1 −3 3 3 + . 1 3 2 −1 −1 2

)t

2u du = ln(t2 + 1) it follows that u2 + 1 8 9 8 9 t2 + 1 1 3 3 −1 −3 Ψ (t) = + −1 −1 1 3 2 2(t2 + 1)

Since u =

0

N O 3(t2 + 1)2 − 1 3(t2 + 1)2 − 3 1 = . 2(t2 + 1) 1 − (t2 + 1)2 3 − (t2 + 1)2

is the standard fundamental matrix at t = 0. The homogeneous solution is given by y h (t) = Ψ (t)y(0)

N O8 9 3(t2 + 1)2 − 1 3(t2 + 1)2 − 3 1 1 = 2(t2 + 1) 1 − (t2 + 1)2 3 − (t2 + 1)2 −1 8 9 8 9 1 1 2 1 − = 2 . 2(t2 + 1) −2 (t + 1) −1

We compute the particular solution in pieces: First, N O8 9 3 − (u2 + 1)2 3 − 3(u2 + 1)2 −3u 1 −1 Ψ (u) f (u) = u 2(u2 + 1)2 (u2 + 1)2 − 1 3(u2 + 1)2 − 1 8 9 u −6 = 2 2 2 2(u + 1) 8 9 u −3 = . 2 2 1 (u + 1) Second, *

t 0

$u=t 8 9 ' (8 9 −1 $$ 1 1 −3 −3 Ψ (u)f (u) du = = 1− 2 $ 2 1 1 2(u + 1) u=0 2 t +1 8 9 1 t2 −3 = . 2 t2 + 1 1

Finally,

338

1 Solutions

y p (t) = Ψ (t)

*

t 0

Ψ (u)−1 f (u) du N 3(t2 + 1)2 − 1

t2 = 4(t2 + 1)2 t2 4(t2 + 1)2 8 9 t2 −3 = . 1 2 =

8

1 − (t2 + 1)2 9 −6(t2 + 1)2 2(t2 + 1)2

O8 9 3(t2 + 1)2 − 3 −3 1 3 − (t2 + 1)2

It now follows that y(t) =

8 9 8 9 8 9 1 t2 −3 1 1 −3t4 − 3t2 + 2 + = . 1 t4 + t2 − 2 t2 + 1 −1 2 2(t2 + 1)

8

9 3 5 11. Let A = . Then A(t) = sec(t)A. The characteristic polynomial −1 −3 0 1 2 of A is cA (s) = s − 4 = (s − 2)(s + 2). Hence BcA = e2t , e−2t and eAu = M1 e2u + M2 e−2u .

Differentiating and evaluating at u = 0 give the equations I = M1 + M2 and A = 2M1 − 2M2 . It follows that 8 9 8 9 1 1 −1 −5 5 5 M1 = and M2 = 1 5 4 −1 −1 4 and eAu =

8 9 1 5e2u − e−2u 5e2u − 5e−2u . 4 −e2u + e−2u −e2u + 5e−2u

)t If b(t) = 0 sec u du = ln |sec t + tan t| then Ψ (t) = eAb(t) . If X = (sec t + tan t)2 then X −1 = (sec t − tan t)2 and 

1 1 5X − 5 1 X X Ψ (t) =   1 1 4 −X + −X + 5 X X 8 2 9 sec t + 3 sec t tan t + tan2 t 5 sec t tan t = . − sec t tan t sec2 t − 3 sec t tan t + tan2 t 5X −

From this it follows that the homogeneous solution is

1 Solutions

339

8

sec2 t + 3 sec t tan t + tan2 t 5 sec t tan t − sec t tan t sec2 t − 3 sec t tan t + tan2 t 8 9 2 sec2 t + 11 sec t tan t + 2 tan2 t = sec2 t − 5 sec t tan t + tan2 t

y h (t) +

98 9 2 1

Since the forcing function f is identically zero the particular solution is zero. Hence y = y h . 8 9 1 1 12. Let A = . Then A(t) = tA. The characteristic polynomial of −1 −1 2 A is cA (s) = s . Therefore BcA = {1, t}. We can write eAu = M1 + uM2 . Differentiating and evaluating at u = 0 gives I = M1 and A = M2 . Thus 8 9 1+u u eAu = . −u 1 − u Since b(t) =

)t 0

u du =

t2 we have 2  t2 1 +  2 Ψ (t) =  t2 − 1− 2

 t2 2 t2  2

is the standard fundamental matrix for y ! (t) = A(t)y(t) at t = 0. The homogeneous solution is given by y h (t) = Ψ (t)y(0)  t2 1 +  2 =  t2 − 1− 29 8 4 + 2t2 = . −2t2 The particular solution is given by

 t2 8 9 2 4 t2  0 2

340

1 Solutions

* t y p (t) = Ψ (t) Ψ (u)−1 f (u) du 0     t2 t2 * t u2 u2 8 9 1 + 1 − −   2 2 2 2  4u du =  t2 t2  0  u2 u2  4u − 1− 1+ 2 2 2 2   t2 t2 * t 8 9 4u − 4u3 1 + 2  2 =  du t2 t2  0 4u + 4u3 − 1− 2 2   2 t t2 8 9 1 + 2t2 − t4  22 22  =  t t  2t2 + t4 − 1− 2 8 2 2 49 2t + t = . 2t2 − t4 It follows that 8

9 8 2 9 8 9 4 + 2t2 2t + t4 4 + 4t2 + t4 y(t) = y h (t) + y p (t) = + = . −2t2 2t2 − t4 −t4 13. Let v1 (t) and v2 (t) denote the volume of brine in Tank 1 and Tank 2, respectively. Then v1 (t) = v2 (t) = 2 − t. The following differential equations describe the system 3 1 y1 (t) + y2 (t) + 6 2−t 2−t 1 3 y2! (t) = y1 (t) − y2 (t) + 0, 2−t 2−t y1! (t) = −

with initial conditions y1 (0) = 0 and y2 (0) = 20. In matrix form, y ! (t) = A(t)y(t) + f (t), we have 8 9 8 9 8 9 −3/(2 − t) 1/(2 − t) 6 0 A(t) = , f (t) = , y(0) = . 1/(2 − t) −3/(2 − t) 0 20 Let a(t) =

−1 . Then we can write A(t) = a(t)A where 2−t 8 9 3 −1 A= . −1 3

2 The characteristic polynomial + 8 = (s − 2)(s − 4). 0 2t 4t 1is cA (s) = s − 6sAu We now have BcA = e , e It follows that e = M1 e2u + M2 e4u . Differentiating and setting u = 0 we get

1 Solutions

341

I = M1 + M2 A = 2M1 + 4M2 . An easy calculation gives 1 (4I − A) = 2 1 = (A − 2I) = 2

M1 = M2

8 9 1 1 1 2 1 1 8 9 1 1 −1 1 2 −1

and

8 9 8 9 e2u 1 1 e4u 1 −1 e = + . 1 2 1 1 2 −1 )t ) t −1 Let b(t) = 0 a(u) du = 0 du = ln 2−t 2 . Then the standard funda2−u mental matrix is 8 9 8 9 (2 − t)2 1 1 (2 − t)4 1 −1 Au Ψ (t) = e |u=b(t) = + . 1 1 −1 1 8 32 Au

For the homogeneous solution we have y h (t) = Ψ (t)y(0) 8 98 9 8 98 9 (2 − t)2 1 1 (2 − t)4 0 1 −1 0 = + 1 1 20 −1 1 20 8 32 8 9 8 9 (2 − t)2 5 (2 − t)4 −5 = + . 5 5 2 8 For the particular solution straightforward calculations give 8 9 2 (2 − u)2 + 4 (2 − u)2 − 4 −1 Ψ (u) = , (2 − u)4 (2 − u)2 − 4 (2 − u)2 + 4 8 98 9 2 (2 − u)2 + 4 (2 − u)2 − 4 6 (2 − u)4 (2 − u)2 − 4 (2 − u)2 + 4 0 8 9 12 (2 − u)2 + 4 = (2 − u)4 (2 − u)2 − 4 8 9 (2 − u)−2 + 4(2 − u)−4 = 12 , (2 − u)−2 − 4(2 − u)−4

Ψ −1 f (u) =

and *

t 0

Ψ −1 (u)f (u) du =

8 9 8 9 4 3(2 − t)2 + 4 8 − . 4 (2 − t)3 3(2 − t)2 − 4

342

1 Solutions

Finally, we get y p (t) = Ψ (t)

=

'

*

t

Ψ −1 (u)f (u) du

0

8 9 8 9( * t (2 − t)2 1 1 (2 − t)4 1 −1 + Ψ −1 (u)f (u) du 1 1 −1 1 8 32 0

8 9 8 9 8 9 (2 − t)2 3 (2 − t)4 4 1 = (2 − t) − − . 2 3 −1 2 8

We now add the homogeneous and particular solutions together and simplify to get y(t) = y h + y p 8 9 8 9 8 9 (2 − t)4 −3 4 2 1 = (2 − t) + (2 − t) + . 2 1 3 4 We now get 3 y1 (t) = 4(2 − t) + (2 − t)2 − (2 − t)4 4 3 y2 (t) = 2(2 − t) + (2 − t)2 + (2 − t)3 . 4 The amount of fluid in each tank after 1 minute is v1 (1) = v2 (1) = 1. Thus the concentrations (grams/L) of salt in Tank 1 is y1 (1)/1 and in Tank 2 is y2 (1)/1, i.e. y1 (1) 17 y2 (1) 15 = and = . 1 4 1 4 14. Let v1 (t) and v2 (t) denote the volume of brine in Tank 1 and Tank 2, respectively. Then v1 (t) = 2 + 2t and v2 (t) = 1 + t. The following differential equations describe the system 3 1 y1 (t) + y2 (t) + 28 2 + 2t 1+t 3 4 y2! (t) = y1 (t) − y2 (t) + 28, 2 + 2t 1+t y1! (t) = −

with initial conditions y1 (0) = 0 and y2 (0) = 0. In matrix form, y ! (t) = A(t)y(t) + f (t), we have

1 Solutions

343

A(t) = Let a(t) =

8

9 8 9 8 9 −3/(2 + 2t) 1/(1 + t) 28 0 , f (t) = , y(0) = . 3/(2 + 2t) −4/(1 + t) 28 0

1 . Then we can write A(t) = a(t)A where 2 + 2t 8 9 −3 2 A= . 3 −8

2 The characteristic0polynomial = (s+2)(s+9). We 1 is cA (s) = s +11s+18 −2t −9t now have BcA = e , e It follows that eAu = M1 e−2u + M2 e−9u . Differentiating and setting u = 0 we get

I = M1 + M2 A = −2M1 − 9M2 . An easy calculation gives 1 (A + 9I) = 7 1 = (A + 2I) = 7

M1 = M2

8 9 1 6 2 7 3 1 8 9 1 1 −2 6 7 −3

and

8 9 8 9 e−2u 6 2 e−9u 1 −2 + . 3 1 −3 6 7 7 )t )t 1 √ Let b(t) = 0 a(u) du = 0 du = ln 1 + t. Then the standard 2 + 2u fundamental matrix is 8 9 8 9 1 1 6 2 1 −2 Au Ψ (t) = e |u=b(t) = + . 9 6 7(1 + t) 3 1 7(1 + t) 2 −3 eAu =

Since the initial value is trivial we have y h = 0. For the particular solution straightforward calculations give 8 9 99 1 (1 + u) 2 + 6(1 + u) 2(1 + u) − 2(1 + u) 2 Ψ −1 (u) = , 9 9 7 3(1 + u) − 3(1 + u) 2 6(1 + u) 2 + (1 + u) Ψ

−1

(u)f (u) = Ψ

−1

8

8

9 28 (u) 28

9 9 −(1 + u) 2 + 8(1 + u) = 4 , 9 3(1 + u) 2 + 4(1 + u) and

344

1 Solutions

*

t

Ψ

−1

(u)f (u) du = 4

0

N

11

2 − 11 (1 + t) 2 + 4(1 + t)2 6 11 (1

11

=

8(1 + t) 2 11

11

+ t) 2 + 2(1 + t)2 8

O

−4

N 42 O 11 28 11

9 8 9 8 9 56 3 −1 2 + 8(1 + t)2 − . 3 1 11 2

Finally, we get y p (t) = Ψ (t) =

*

t

Ψ −1 (u)f (u) du

0

8 9 8 9 8 9 8 9 8(1 + t) −1 8 8 2 2 −1 + 8(1 + t) − − 9 3 1 3 11 1+t 1 11(1 + t) 2

8 9 8 9 8 9 56(1 + t) 3 8 8 2 1 = − + . 9 2 −3 11 1+t 1 2 (1 + t) It follows that y1 (t) =

168 16 8 (1 + t) − + 11 1 + t 11(1 + t) 92

y2 (t) =

112 8 24 (1 + t) − − 11 1 + t 11(1 + t) 92

The concentration (grams/L) of salt in Tank 1 is y1 (t)?(1 + t) and in Tank 2 is y2 (t)/(1 + t). At t = 3 we get y1 (3) y2 (3) = 14.27 and = 9.68. 4 4 The long term concentrations are given by 8 9 y(t) 168/11 lim = . 112/11 t→∞ 1 + t

http://www.springer.com/978-1-4614-3617-1