Student’s Solutions Manual

1

Student's Solutions Manual

for

Differential Equations: Theory, Technique, and Practice with Boundary Value Problems

Second Edition

by Steven G. Krantz (with the assistance of Yao Xie)

Chapter 1

What is a Differential Equation?

1.1 Introductory Remarks

1.2 A Taste of Ordinary Differential Equations

1.3 The Nature of Solutions

1. Verify the function is a solution to the differential equation.

(a) If y = x2 + c, then y = 2x. (b) If y = cx2, then y = 2cx so xy = 2cx2 = 2y. (c) If y2 = e2x + c, then 2yy = 2e2x so yy = e2x. (d) If y = cekx, then y = kcekx so y = ky. (e) If y = c1 sin 2x + c2 cos 2x, then y = 2c1 cos 2x - 2c2 sin 2x and

y = -4c1 sin 2x - 4c2 cos 2x = -4y so y + 4y = 0. (f) If y = c1e2x + c2e-2x, then y = 2c1e2x - 2c2e-2x and y = 4c1e2x +

4c2e-2x = 4y so y - 4y = 0. (g) If y = c1 sinh 2x + c2 cosh 2x, then y = 2c1 cosh 2x + 2c2 sinh 2x

and y = 4c1 sinh 2x + 4c2 cosh 2x = 4y so y - 4y = 0.

1

2

CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?

(h) If y = arcsin xy, then y = xy +y so xy + y = y 1 - x2y2. 1-(xy)2

(i) If y = x tan x, then y = x sec2 x + tan x = x(tan2 x + 1) + tan x. Using tan x = y/x we get y = y2/x + x + y/x or xy = x2 + y2 + y.

(j) If x2 = 2y2 ln y, then 2x = [2y2(1/y) + 4y ln y]y = 2yy (1 + 2 ln y).

Consequently, y

=

x y+2y

ln

y

.

Using

ln y

=

x2 2y2

we get y

=

. xy

x2+y2

(k) If y2 = x2 - cx, then 2yy = 2x - c so 2xyy = 2x2 - cx = x2 + x2 - cx = x2 + y2.

(l) If y = c2 + c/x, then y = -c/x2 so x4(y )2 = c2 = y - c/x. Use the fact that -c/x = xy to obtain x4(y )2 = y + xy .

(m)

If y = cey/x, then y

=

xy -y x2

cey/x

=

. xyy -y2 x2

Solve for y

to obtain

y = y2/(xy - y2).

(n) If y+sin y = x, then y +y cos y = 1 or y = 1/(1+cos y). Multiply the numerator and denominator of the right side by y to obtain y = y/(y + y cos y). Now use the identity y = x - sin y to obtain y = (x - sin y + y cos y).

(o) If x + y = arctan y, then 1 + y = y /(1 + y2). Consequently, (1 + y )(1 + y2) = y . This simplifies to 1 + y2 + y2y = 0.

3. For each of the following differential equations, find the particular solution that satisfies the given initial condition.

(a) If y = xex, then y = xexdx + C = (x - 1)ex + C (integrate by

parts, u = x). When x = 1, y = C so the particular solution is y(x) = (x - 1)ex + 3.

(b) If y = 2 sin x cos x, then y = 2 sin x cos xdx + C = sin2 x + C. When x = 0, y = C so the particular solution is y(x) = sin2 x + 1.

(c) If y = ln x, then y = ln xdx + C = x ln x - x + C (integrate by parts, u = ln x). When x = e, y = C so the particular solution is y(x) = x ln x - x.

(d) If y = 1/(x2 - 1), then y = 1/(x2 - 1)dx + C = 1/2 1/(x -

1) - 1/(x + 1)dx + C

=

1 2

ln

x-1 x+1

+C

(method

of

partial

fractions).

When

x

=

2,

y

=

1 2

ln

1 3

+C

=

C

-

ln 3 2

so

the

particular

solution

is

y(x)

=

1 2

ln

x-1 x+1

+

ln 3 2

.

1.3. THE NATURE OF SOLUTIONS

3

(e)

If y

=

1 x(x2-4)

,

then

y

=

1 x(x2-4)

dx

+

C

=

1/8

1/(x + 2) + 1/(x -

2) - 2/xdx + C

=

1 8

ln

|x2-4| x2

+C

(method

of

partial

fractions).

When

x

=

1,

y

=

1 8

ln

3

+

C

so

the

particular

solution

is

y(x)

=

1 8

ln

|x2-4| x2

-

1 8

ln 3

=

1 8

ln

. |x2-4| 3x2

(f) If y

=

, 2x2+x

(x+1)(x2+1)

then

y

=

2x2+x (x+1)(x2+1)

dx

+

C

=

1 2

1 x+1

+

3x-1 x2+1

dx

+

C

=

1 2

ln(x

+

1)

+

3 4

ln(x2 + 1) -

1 2

arctan x + C

(method

of

partial fractions). When x = 0, y = C so the particular solution

is

y(x)

=

1 2

ln(x

+

1)

+

3 4

ln(x2

+ 1)

-

1 2

arctan x

+ 1.

5. For the differential equation

y - 5y + 4y = 0,

carry out the detailed calculations required to verify these assertions.

(a) If y = ex, then y - 5y + 4y = ex - 5ex + 4ex 0. If y = e4x, then y - 5y + 4y = 16e4x - 20e4x + 4e4x 0.

(b) If y = c1ex + c2e4x, then y - 5y + 4y = c1(ex - 5ex + 4ex) + c2(16e4x - 20e4x + 4e4x 0.

7. For which values of m will the function y = ym = emx be a solution of the differential equation

2y + y - 5y + 2y = 0?

Find three such values m. Use the ideas in Exercise 5 to find a solution containing three arbitrary constants c1, c2, c3. Substitute y = emx into the differential equation to obtain

2m3emx + m2emx - 5memx + 2emx = 0.

Cancel emx in each term (it is never 0) to obtain the equivalent equation

2m3 + m2 - 5m + 2 = 0.

Observing that m = m1 = 1 is a solution (and y1 = ex is a solution to the differential equation). Using this we can factor the polynomial? divide by m - 1?to obtain

2m3 + m2 - 5m + 2 = (m - 1)(2m2 + 3m - 2).

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