Student’s Solutions Manual - Routledge

[Pages:126]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).

4

CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?

The quadratic term factors yield two more roots, m2 = -2, m3 = 1/2, and two more solutions

y2 = e-2x and y3 = ex/2.

These three solutions can be combined, as in Exercise 5, to produce a solution with three arbitrary constants

y = c1ex + c2e-2x + c3ex/2.

1. Use the method of separation of variables to solve each of these ordinary differential equations.

(a)

Write the

equation

x5y

+ y5

=0

in

Leibnitz

form

x5

dy dx

+ y5

=

0

and separate the variables:

dy y5

=

-

dx x5

.

Integrate,

dy y5

=-

dx x5

,

to obtain the solution: y-4/(-4) = x-4/4 + C. This can also be

written

in

the

form

x4

+

y4

=

C x4y 4

or

y

=

(

C

x4 x4 -1

)1/4.

(b)

Write the equation y

= 4xy in Leibnitz form

dy dx

= 4xy and sepa-

rate the variables:

dy y

= 4xdx.

Integrate,

dy y

=

4xdx, to obtain

the solution: ln |y| = 2x2 +C. This can also be written in the form

y = Ce2x2.

(c)

Write the equation y

+y tan x = 0 in Leibnitz form

dy dx

+y

tan

x

=

0

and separate the variables:

dy y

= - tan xdx.

Integrate,

dy y

==

tan xdx, to obtain the solution: ln |y| = ln | cos x| + C. This can

also be written in the form y = C cos x.

(d) The equation (1 + x2)dy + (1 + y2)dx = 0 can be rearranged and

integrated directly,

dy 1+y2

+

dx 1+x2

= C.

Therefore,

the

implicit

solution is arctan y + arctan x = C. This can also be written in

the form y = tan(C - arctan x).

(e) Proceed as in part (d). Rearrange y ln ydx - xdy = 0 to the form

dx x

-

dy y ln y

=

0

and

integrate:

dx x

-

dy y ln y

= C.

This

yields the

implicit solution ln |x| - ln | ln y| = C which can also be written in

the form ln y = Cx or y = eCx.

(f)

From

Leibnitz

form

x

dy dx

=

(1 - 4x2) tan y

we

obtain

cot ydy

=

(1/x - 4x)dx. Integrating, cot ydy = 1/x - 4xdx, gives the

implicit solution ln | sin y| = ln |x|-2x2+C. Consequently, sin y =

Cxe-2x2 so y = arcsin(Cxe-2x2).

1.4. FIRST-ORDER LINEAR EQUATIONS

5

(g)

The

Leibnitz form

dy dx

sin

y

= x2

separates

to sin ydy

= x2dx.

In-

tegrate to obtain - cos y = x3/3 + C or y = arccos(C - x3/3).

(h)

Write the equation y

- y tan x = 0 in Leibnitz form

dy dx

-y

tan

x

=

0 and separate the variables:

dy y

=

tan xdx.

Integrate

dy y

=

tan xdx, to obtain the solution: ln |y| = - ln | cos x| + C. This

can also be written in the form y = C/ cos x or y = C sec x.

(i)

From

Leibnitz

form,

xy

dy dx

=

y

-1

we

obtain

ydy y-1

=

dx x

.

Write

this

in

the

form

y-1+1 y-1

dy

=

dx x

and

integrate

to

obtain

the

implicit

solution y + ln |y - 1| = ln |x| + C.

(j)

Leibnitz

form

xy2-

dy dx

x2

=

0

separates

to

dx x

= y-2dy.

Integrating

yields the implicit solution ln |x| = -1/y + C. The solution can

be

expressed

explicitly

as

y

=

1 C-ln

|x|

.

3.

Substituting y

= p yields

p p

= x2.

Separation of variables (or direct

integration) gives ln |p| = x3/3 + C. This implies that p = Cex3/3 and

so y = Cex3/3. Consequently, y = C ex3/3dx + D. As we expect,

the solution contains two arbitrary constants. The integral cannot be

evaluated in terms of elementary functions.

1.4 First-Order Linear Equations

1. Find the general solution of the following equations.

(a) The equation is linear and separable. The integrating factor is e-x2/2 so it simplifies to (e-x2/2y) = 0 and e-x2/2y = C. Therefore, y = Cex2/2.

(b) This equation is also linear and separable. The integrating factor is ex2/2 so it simplifies to (ex2/2yy) = xex2/2. Integrate to obtain ex2/2y = ex2/2 + C and y = 1 + Ce-x2/2.

(c) The integrating factor is ex so the equation simplifies to (exy) =

. ex

1+e2x

Integrating we obtain exy = arctan ex + C.

The general

solution is y = e-x arctan(ex) + Ce-x.

(d) The integrating factor is ex so the equation simplifies to (exy) = 2x+x2ex. Integrate to get exy = x2 +(2-2x+x2)ex +C (integrate

6

CHAPTER 1. WHAT IS A DIFFERENTIAL EQUATION?

x2ex by parts, twice, or use an integral table). The general solution is y = x2e-x + 2 - 2x + x2 + Ce-x.

(e) Write as xy = 2y-x3 and then y -(2/x)y = -x2. The integrating factor is x-2 so the equation simplifies to (x-2y) = -1. Integrate to obtain x-2y = -x + C. The general solution is y = -x3 + Cx2.

(f) The integrating factor is ex2 so the equation simplifies to (ex2y) = 0. Consequently, ex2y = C and y = Ce-x2.

(g) Write as y - (3/x)y = x3. The integrating factor is x-3 so the equation simplifies to (x-3y) = 1. Integrating we obtain x-3y = x + C so y = x4 + Cx3.

(h)

Express the equation in the form y

+

2x 1+x2

y

=

cot x 1+x2

.

The integrat-

ing factor is 1 + x2 and the equation simplifies to ((1 + x2)y) =

cot x.

Consequently,

(1 + x2)y

=

ln | sin x| + C

and

y

=

ln

|

sin x|+C 1+x2

.

(i) The integrating factor is sin x and the equation simplifies to (y sin x) =

2x.

Therefore,

y sin x

=

x2

+C

and

y

=

x2+C sin x

.

(j) Express the equation in the form y + (1/x + cot x)y = 1. The in-

tegrating factor is x sin x so the equation simplifies to (xy sin x) =

x sin x. Integrate to obtain xy sin x = sin x - x cos x + C (use a

table of integrals or integrate x sin x by parts, u = x). Therefore,

y

=

sin

x-x cos x sin x

x+C

.

3. Bernoulli Equations To verify the technique write the Bernoulli

equation in the form y-ny + P y1-n = Q. The substitution z = y1-n

and z

= (1 - n)y-ny

yield

1 1-n

z

+ P z = Q.

(a) Bernoulli, n = 3. Substitute z = y-2, z = -2y-3y into xy-3y +

y-2 = x4 to obtain (-1/2)xz +z = x4. This is linear, z -(2/x)z =

-2x3, with integrating factor x-2. It simplifies to (x-2z) = -2x.

COnsequently, x-2z = -x2 + C and z = -x4 + Cx2. This means

that

y-2

=

C x2

-

x4

so

y2

=

. 1

C x2 -x4

(b) Write the equation in the form y + (1/x)y = y-2 cos x to see

that it is Bernoulli, n = -2. Substitute z = y3, z = 3y2y into

the equation y2y + (1/x)y3 = cos x to obtain the linear equation

(1/3)z + (1/x)z = cos x. This is z + (3/x)z = 3 cos x, with inte-

grating factor x3. It simplifies to (x3z) = 3x3 cos x. Consequently,

x3z = 3F (x) + C where F (x) is and antiderivative for x3 cos x.

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