Answers, Hints, andSolutions to SelectedExercises for IntroductoryDifferential Equations,Fourth Edition, Abell &Braselton

Martha L. Abell and James P. Braseltonc©2013

Exercises 1.1

1. Second order linear ordinary differential equation. The forcing function isf(x) = x3 so the equation is nonhomogeneous.3. Second order linear partial differential equation.5. This is a first order ordinary differential equation. It is nonlinear because thederiviatve dy/dx is squared.7. Second order linear partial differential equation.9. Nonlinear second order ordinary differential equation.11. This is a second order partial differential equation. It is nonlinear because

of the product, uux = u∂u

∂x, of functions involving the dependent variable, u =

u(x, t).13. This is a first order ordinary differential equation. If we write it as (2t −y)dt/dy − 1 = 0, y is independent, t = t(y) is dependent, and the equation isnonlinear. If we write the equation as dy/dt + y = 2t, t is independent, y = y(t)is dependent and the equation is linear.15. This is a first order ordinary differential equation. It is nonlinear in both x

1

2

(because of the 2x dx term) and y (because of the −y dy term.31. Differentiate and collect dy and dx terms:

3x2 dx + 2xy dx + x2 dy = 0

3x dx + 2y dx + x dy = 0

x dy = (−2y − 3x) dx

dy/dx = −2y/x− 3.

If x = 1, y = 99.

35. −1

2cos(x2)

+ C

37. Use a u-substitution with u = lnx⇒ du = 1/x dx. Then,

y =

∫1

x lnxdx =

∫1

udu = ln |u|+ C = ln(lnx) + C.

39. Use integration by parts with u = x ⇒ du = dx and dv = e−x dx ⇒ v =−e−x.

y =

∫xe−x dx = −xe−x +

∫e−x dx = −xe−x − e−x + C.

41. tan−1(x)− ln(x + 1) + C43. 6 sin−1

(x2

)− 1

4x√

4− x2(x2 − 10

)45. Start by rewriting cotx =

cosx

sinx. Then,

y =

∫cos2 x

sinxdx

=

∫1− sin2 x

sinxdx

=

∫(cscx− sinx) dx

so y = ln(cscx + cotx) + cosx + C.47. y (x) = 2e−2x

49. y (x) = 57e−3x + 2

7e4x

51. y (x) = − 14

(−3 + 3e2x − 2x

)53. y (t) = −t7 + t6

55. y (x) = x4 − 1/2x2 + 2x + 157. y (x) = − sin

(x−1

)+ 2

67. y (x) = c1 x + c2 x2

69. y (x) = (ex + C) e−2 x

80. y = 132 (12x− 8 sin 2x + sin 4x

3

-2 -1 1 2x

-2

-1

1

2

y79.

2 4 6 8 10 12x

1

2

3

4

y80.

81. y = − 1208e

−x/2(74ex cos 2x− 74 cos 3x− 111ex sin 2x− 20 sinx)

1 2 3 4 5 6x

-10

-5

5

y81.

83. x = 49e−6t (e9t + 8

), y = 8

9e−6t (e9t − 1

)

0.2 0.4 0.6 0.8 1.0t

5

10

15

x,y83.

3 4 5 6 7 8 9x

5

10

15

y83.

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y82.

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y83.

4

Exercises 1.2

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y1. No

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y2. Yes

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y3. Yes

-2 -1 1 2x

-2

-1

1

2

y4. No

-2 -1 1 2x

-2

-1

1

2

y7.

-2 -1 1 2x

-2

-1

1

2

y8.

5

-6 -4 -2 2 4 6t

-6

-4

-2

2

4

6

y9.

-6 -4 -2 2 4 6x

-6

-4

-2

2

4

6

y10.

-4 -2 2 4x

-4

-2

2

4

y11.

-4 -2 2 4x

-4

-2

2

4

y12.

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y13.

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y14

15. x′ = y so y′ = x′′ = −4x: {x′ = y, y′ = −4x}17. {x′ = y, y′ = −13x− 4y}19. {x′ = y, y′ = −16x + sin t}21. y = C +

√π2 erf (x), where erf (x) = 2√

π

∫ x0e−t

2

dt.

1 2 3 4 5x

-2

-1

1

2

y21.

5 10 15 20 25x

-10

-5

5

10

y22.

6

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y23 HaL.

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

y23 HbL.

0.2 0.4 0.6 0.8 1.0x

0.2

0.4

0.6

0.8

1.0

y25 HaL.

0.2 0.4 0.6 0.8 1.0x

0.2

0.4

0.6

0.8

1.0

y25 HbL.

Chapter 1 Review Exercises

1. First-order ordinary linear homogeneous differential equation3. Second-order linear homogeneous differential equation5. Second-order non-linear partial differential equation17. y = (2− x2) cosx + 2x sinx19. y = 1

2

(x√x2 − 1 + ln

(2x + 2

√x2 − 1

))21. y = 1

4 (x + 4 cos 2x)23. y = 1

3 (1− cos3 x)