Introductory Differential Equations: with Boundary Value Problems, Student Solutions Manual (e-only)

By Martha L Abell and James P. Braselton

This text is for courses that are typically called (Introductory) Differential Equations, (Introductory) Partial Differential Equations, Applied Mathematics, and Fourier Series. Differential Equations is a text that follows a traditional approach and is appropriate for a first course in ordinary differential equations (including Laplace transforms) and a second course in Fourier series and boundary value problems.

Some schools might prefer to move the Laplace transform material to the second course, which is why we have placed the chapter on Laplace transforms in its location in the text. Ancillaries like Differential Equations with Mathematica and/or Differential Equations with Maple would be recommended and/or required ancillaries.

Because many students need a lot of pencil-and-paper practice to master the essential concepts, the exercise sets are particularly comprehensive with a wide range of exercises ranging from straightforward to challenging. Many different majors will require differential equations and applied mathematics, so there should be a lot of interest in an intro-level text like this. The accessible writing style will be good for non-math students, as well as for undergrad classes.

• Language: English
• Publisher: Academic Press
• Release date: Apr 20, 2010
• ISBN: 9780123846655

Martha L Abell

Martha L. Abell and James P. Braselton are graduates of the Georgia Institute of Technology and the Ohio State University, respectively, and teach at Georgia Southern University, Statesboro where they have extensive experience instructing students at both the undergraduate and graduate levels. Other books by the authors include Differential Equations with Mathematica and Mathematica by Example.

Book preview

CHAPTER 1

Introduction to Differential Equations

Exercises 1.1

1. Second-order linear ordinary differential equation. The forcing function is f(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 the deriviatve dy/dx is squared.

7. Second-order linear partial differential equation

9. Second-order nonlinear ordinary differential equation

11. This is a second-order partial differential equation. It is nonlinear because of the product, , 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 is nonlinear. 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 (because of the 2x dx term) and y (because of the −y dy term).

31. Differentiate and collect dy and dx terms:

If x = 1, y = 99.

35. − cos(x²) + C

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

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

41. tan−1(x) − ln (x + 1) + C

45. y(x) = 2e−2x

53. y(t) = −t⁷ + t⁶

55. y(x) = x⁴ − x² + 2x + 1

57. y(x) = −sin(x−1) + 2

67. y(x) = c1x + c2x²

69. y(x) = (ex + C)e−2x

79.

80. y = (12x − 8 sin 2x + sin 4x)

81. (74ex cos 2x − 74 cos 3x − 111ex sin 2x − 20 sin x)

Exercises 1.2

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. , where

Chapter 1 Review Exercises

1. First-order ordinary linear homogeneous differential equation

3. Second-order linear homogeneous differential equation

5. Second-order nonlinear partial differential equation

17. y = (2 − x²) cos x + 2x sin x

CHAPTER 2

First-Order Equations

Exercises 2.1

1. (a) yes; (b) no; (c) no

3. f (x, y) = y¹/⁵, so is not continuous at (0, 0) and uniqueness is not guaranteed. Solutions:

5.

, so fy(t, y) = is not continuous at (0, 0). Therefore, the hypotheses of the Existence and Uniqueness Theorem are not satisfied.

7. Yes.

. Application of the initial conditions yields .

9. Yes. f (t, y) = sin y − cos t and fy(t, y) = cos y are continuous on a region containing (π, 0).

11. y = sec t so y′ = sec t tan t = y tan t and y(0) = 1. fy(t, y) = t is continuous on −π/2 < t < π/2 and f (t, y) = sec t is continuous on −π/2 < t < π/2 so the largest interval on which the solution is valid is −π/2 < t < π/2.

13.

unique solution guaranteed for (a) only.

15. (0, ∞). The solution is .

17. (0, ∞), because y = ln t has domain t > 0.

19. (−∞, 1), because y = 1/(t − 1) has domain (−∞, 1) u1 (1, ∞) and y = 1/(t − 3) has domain (−∞, 3) (3, ∞).

21. (−2, 2)

23. t > 0, y = t−1 sin t − cos t, −∞ < t < ∞

25. First, we solve the equation (see next section):

Applying the initial condition indicates that −1/C = a ⇒ C = −1/a so y = a/(1 − at). This solution is defined for t > 1/a or t < 1/a.

27. Separating variables (see next section) gives us Applying the initial conditions indicates that C = a² so (because y(0) = a is positive). Thus, the interval of definition of the solution is |t| < a.

29. General solution: solution is not unique if x0 = π/2 or 3π/2.

Exercises 2.2

1. Separate variables and integrate:

3. Separate variables and integrate:

9. y−3/2 − x³ − 4x³/² = C

13. Separate variables and integrate:

In the calculation above, remember that e−kt + C = e−kteC. C is arbitrary so eC is positive and arbitrary.

21. y sin 2x + 2xy + 2y² + 5 + 4Cy = 0

31. Factor first, then separate, use partial fractions, and simplify:

FIGURE 1

Direction fields for dy/dt = y³ + y², dy/dt = y³ − y², and dy/dt = y² − y³.

43. Integrating gives us x = sin y + C and applying the initial condition gives us C = 2 so x(y) = sin(y) + 2.

51. y(x) = arctan(x) + 1

53. The solution for (a) y = esin t, for (b) it is , and for (c) it is .

55. y = 2x(x − 2)−1

57. (a) y = −x + 4 − tan(C − x); (b)