A First Course in Differential Equations with Modeling

A First Course in Differential Equations with Modeling Applications 11th Edition Zill Solutions Manual

Contents

Chapter 1 .......................................................................................................................................... 1 Chapter 2 ........................................................................................................................................ 36 Chapter 3 ...................................................................................................................................... 106 Chapter 4 ...................................................................................................................................... 157 Chapter 5 ...................................................................................................................................... 285 Chapter 6 ...................................................................................................................................... 339 Chapter 7 ...................................................................................................................................... 430 Chapter 8 ...................................................................................................................................... 512 Chapter 9 ...................................................................................................................................... 580 Chapter 10 .................................................................................................................................... 606 Chapter 11 .................................................................................................................................... 639 Chapter 12 .................................................................................................................................... 691 Chapter 13 .................................................................................................................................... 791 Chapter 14 .................................................................................................................................... 849 Chapter 15 .................................................................................................................................... 903 1 / 4

2

Chapter 1

Introduction to Differential Equations

1.1

• Second order; linear

• Third order; nonlinear because of (dy/dx)

4

• Fourth order; linear

• Second order; nonlinear because of cos(r + u)

√

• Second order; nonlinear because of (dy/dx)

2 or 1 + (dy/dx) 2

• Second order; nonlinear because of R

2

• Third order; linear

• Second order; nonlinear because of x˙

2

• Writing the differential equation in the form x(dy/dx) + y

2 = 1, we see that it is nonlinear in y because of y 2 . However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x.

• Writing the differential equation in the form u(dv/du) +(1 + u)v = ue

u we see that it is linear in v. However, writing it in the form (v + uv − ue u )(du/dv) + u = 0, we see that it is nonlinear in u.

• From y = e

−x/2 we obtain y

′ = −

1 e −x/2 . Then 2y ′ + y = −e −x/2

• e

−x/2 = 0.

• From y =

6 − 6 e −20t we obtain dy/dt = 24e −20t , so that

• 5

dy −20t

• 6

−20t

• 20y = 24e

dt

• 20

5 − 5 e = 24.

• From y = e

3x cos 2x we obtain y ′ = 3e 3x cos 2x−2e 3x sin 2x and y ′′ = 5e 3x cos 2x−12e 3x sin 2x, so that y ′′ − 6y ′ + 13y = 0.

1 Definitions and Terminology 2 / 4

• / 4

2 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

cos x ln(sec x + tan x) we obtain y = (x + 2) (y − 1/2 x)(x + 2) x](x + 2)

• 2 2

3/2 1/2 3 3

• From y = −

′ −1 + sin x ln(sec x + tan x) and

′′

y = tan x + cos x ln(sec x + tan x). Then y + y = tan x.

• The domain of the function, found by solving x+2 ≥ 0, is [−2, ∞). From y

′ = 1+2(x+2)

−1/2

we have

′ −

x)y = (y − x)[1 + (2(x + 2) ] = y − x + 2(y −

−1/2

= y − x + 2[x + 4(x + 2) 1/2 −

−1/2

= y − x + 8(x + 2) 1/2

−1/2

= y − x + 8.An interval of definition for the solution of the differential equation is (−2, ∞) because y ′ is not defined at x = −2.

• Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is

.{x .5x /= π/2 + nπ} .or {x .x /= π/10 + nπ/5}. From y ′ = 25 sec 2 5x we have

′ y = 25(1 + tan 5x) = 25 + 25 tan 5x = 25 + y .

An interval of definition for the solution of the differential equation is (−π/10, π/10). Another interval is (π/10, 3π/10), and so on.

.

• The domain of the function is {x

.

• − x

2

y ′ = 2x/(4 − x 2 ) 2 we have .0} or {x .x −2 and x /= 2}. From

1 2

′ 2

y = 2x

• − x

2

= 2xy .

An interval of definition for the solution of the differential equation is (−2, 2). Other intervals are (−∞, −2) and (2, ∞).

• The function is y = 1/

√

• − sin x , whose domain is obtained from 1 − sin x 0 or sin x /= 1. .

′ 1 −3/2

Thus, the domain is {x .x /= π/2 + 2nπ}. From y = − 2 (1 − sin x) (− cos x) we have

′ − −

2y = (1 − sin x) cos x = [(1 − sin x) ] cos x = y cos x.

An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another one is (5π/2, 9π/2), and so on.′′

• / 4