Answers 1

Chapter 2

Section 2.1

1. y ′ = − (1 + t)y

t2.

3.

–3–2–10

123

–2 –1 1 2

5.

–30–20–10

0

102030

–4 –2 2 4

7. For y(t) = 0, y ′(t) = 0 and y(t)(4 − y(t)) = 0(4 − 0) = 0.

9. (a) [t2 − 4y2]′ = 2(t − 4yy ′) = 0

9. (c) Either −∞ < t < −C or C < t < ∞.

9. (d)

2 Answers

−10 −5 0 5 10

−10

−5

0

5

10

11. The interval of existence is (−∞, ln(5)/4).

(0, −1)−2 −1

13. y(t) = 1

3t2 + 5

3t. The interval of existence is (0,∞).

(1,2)

1 2 3

5

10

15. y(t) = 2/(−1 + e−2t/3). The interval of existence is (− ln(3)/2,∞).

Answers 3

(0,−3)

−1 1 2

17.

−2 −1 0 1 2

−1

0

1

t

yy ’ = y + t

19.

−2 −1 0 1 2

−1

0

1

t

y

y ’ = t tan(y/2)

21.

4 Answers

−2 0 2

−5

0

5

t

y

y ’ = − t y

23.

−5 0 5

−6

−4

−2

0

2

4

6

t

y

y ’ = t − y + 1

25.

–12–10

–8–6–4–2

0

2–2 2 4 6 8 10

27.

Answers 5

−15 −10 −5 0

−30

−20

−10

0t

yy ’ = 3 y + sin(t)

29.

−4 −2 0 2 4

−4

−3

−2

−1

0

1

x

r

31.

−2 −1 0 1 2

−2

0

2

4

t

y

33.

6 Answers

1

1.5

2

2.5

3

–4 –2 0 2 4

35.

0

40

80

120

160

–4 –2 2 4

37.

1

2

3

4

5

–4 –2 2 4

Section 2.21. y(x) = Aex

2/2

3. y(x) = ln(ex + C)

5. y(x) = De(1/2)x2+x

7. y(x) = −2 ± √x2 + E

9. y(x) = eDetan−1 x

11. The solution is given implicitly by the equation y4 − 8y − 2x2 = A.

13. y(x) = −2x, (0,∞)

Answers 7

15. y(x) = √1 − 2 cos x, (π/3, 5π/3)

17. y(t) = tan(t + π/4), (−3π/4, pi/4)

19. For y(0) = 1, y(x) = √1 + x2, (−∞,∞), shown with a solid curve in the next

figure. For y(0) = −1, y(x) = −√1 + x2, (−∞,∞), shown with a dashed curve

in the next figure.

−2 0 2

−4

−2

0

2

4

x

y

21. y(x) = 2 + e−x , and y(x) = 2 − e−x + 2 on (−∞,∞),

−2 0 2 4

−5

0

5

10

t

y

23.dN

dt= lim�t→0

N(t +�t)−N(t)

�t= λN(t).

27. 14.26 hours

29. 3.54g

31. The slope is −λ = −0.0869. The half-life is 7.927 days.

8 Answers

2 4 6 8 106

6.2

6.4

6.6

6.8

7

ln R

Day

33. 11.5 hours35. (b) 10:41 PM37. 5.85 minutes39. y(x) = C/x

41. r(θ) = C sin2(θ/2)43. (x, y, z) = (Cy4, y,

√4 − C2y8 − y2/4), where y is the independent variable.

45. y(x) = Kx4

Section 2.31. t = c/5g =612,240 seconds, and x = gt2/2 = 1.84 × 1014 meters.3. t = 7.2438sec and d = 340(8 − t) = 257.1m5. 57.12m and 3.4142s.7. a = −5.8ft/s2.9. vterm = −mg/r = −0.196m/s11. v0 = 16.2665m/s without air resistance and v0 = 18.1142m/s with air resistance.13. 7.9010m15. 4.93m/s17. 0.405 seconds19. mx ′′ = −mg − ke−axx ′

Section 2.41. y(t) = 2 + Ce−t

3. y(x) = sin x + C

x2

5. x(t) = t (t + 1)2 + C(t + 1)2

7. y(x) = sin x + C

1 + x

9. i(t) = E

R+ Ce

−R

Lt

11. y(x) = x − cos x + C

sec x + tan x13. (a) y(x) = 1 + Ce− sin x , C any real number13. (b) We get the same solutions.15. y(x) = 2 − 3(x2 + 1)−3/2

17. x(t) = sin t − 1 + 2e− sin t

Answers 9

19. y(x) = (1/2)(2x + 3)1/2 ln(2x + 3)

−2 6−1

6

(−1,0)

x

y21. x(t) = 1 + sin t

1 + t

−4 −1−1

0 (−π/2,0)

t

x

23. y(x) = 1

x(C − x).

25. y(x) = ±1/√Cx2 − x4

27. (b) y(t) = 1

t− 2Bt

1 + Bt2.

29. At approximately 10:57 PM.

31. y(t) = 5

2+ Ce−2t .

33. y(t) = 4

3t2 + C/t.

35. y(t) = 2 + Cex2.

37. y(t) = (2t − 4)+ 5e−t/2.39. y(x) = x2 − 1.

41. y(t) = 4 + 2t2 + t4

4(1 + t2)2.

Section 2.51. (a) 9.038lb

10 Answers

1. (b) 46.2098m

1. (c) 20 lb

3. (a) 0.25 lb/gal or 25 lb.

3. (b)

0 50 1000

0.1

0.2

0.3

t

c

c ’ = 1/80 − 1/20 c

5. 21 lb.

7. (a) 21.5342 lb.

7. (b) t = 8.7868 min.

9. (b) Approximately 1.41 years.

11. (a) The concentration varies periodically.

30 40 50 60

0.035

0.04

t

c

11. (b) Early in February.

Answers 11

Feb Jul Jant

c

13. (a) 0.0298km3.13. (b) 10.18 yrs

0 5 100

0.05

0.1

0.15

0.2y

Time in years

Section 2.61. dF = 2ydx + (2x + 2y)dy

3. dF = xdx + ydy√x2 + y2

5. dF = x2ydx + y3dx − ydx + x3dy + xy2dy + xdy

x2 + y2

7. dF =(

2x

x2 + y2+ 1

y

)dx +

(2y

x2 + y2− x

y2

)dy

9. Exact. F(x, y) = x2 + xy − 3y2 = C

10. Exact. F(x, y) = x + y cos x = C

11. Not exact.13. Exact. F(x, y) = x3 + xy − y3 = C

15. Exact. F(u, v) = u2 + 2vu− v2

2= C

17. Not exact.19. Exact. F(x, t) = x sin 2t − t2 = C

21. Not exact.

23. F(x, y) = x2y2

2− ln x + ln y = C

12 Answers

25. F(x, y) = x − 12 ln(x2 + y2) = C

27. µ(x) = 1

x. F(x, y) = xy − ln x − y2

2= C

29. µ(y) = 1

y2. F(x, y) = yx + x2

y= C

31. Degree one

33. Degree one.

35. x2 − Cx = y2

37. F(x, y) = xy + 32x

2 = C.

39. y(x) = x + Cx4

1 − 2Cx3

41. (c) The following three graphs show the cases where a = 1, and k = 1/2, 1, 3/2.

0 0.5 10

0.1

0.2

x

y

0 0.5 1x

y

Answers 13

0 0.5 10

1

2

3

x

y

43. (b) The orthogonal family is defined implicitly by

G(x, y) = y

x2 + y2= C.

The original curves are the solid curves in the following figure, and the orthogonalfamily is dashed.

−2 −1 0 1 2−2

−1

0

1

2

y

x

45. arctan(yx

)− y4

4= C

47. arctan(yx

)− (y2 + x2)2

4= C

49. Cxy = x + y

x − y

Section 2.71. Yes.3. Yes.5. Yes.7. The general solution is y(t) = t sin t + Ct

14 Answers

−2 0 2

−4

−2

0

2

4

t

x

9. The y-derivative of the right hand side f (t, y) = 3y2/3 is 2y−1/3, which is notcontinuous at y = 0. Hence the hypotheses of Theorem 7.16 are not satisfied.11. The exact solution is y(t) = −1 + √

t2 − 3. The interval of solution is (√

3,∞).The solver has trouble near

√3. The point where the difficulty arises is circled in the

following figure.

0 1 2 3 4

−4

−2

0

2

4

t

y

13. The exact solution is y(t) = −1 + √4 + 2 ln(1 − t). The interval of existence is

(−∞, 1 − e−2). The solver has trouble near 1 − e−2. The point where the difficultyarises is circled in the following figure.

−2 −1 0 1 2

−4

−2

0

2

4

t

y

Answers 15

15. The solution is defined implicitly by the equation y3/3 + y2 − 3y = 2t3/3.The solver has trouble near (t1, 1), where t1 = −(5/2)1/3 ≈ −1.3572, and also near(t2,−3), where t2 = (22/2)1/3 ≈ 2.3811.. The points where the difficulty arises arecircled in the following figure.

−2 −1 0 1 2 3

−4

−3

−2

−1

01

2

t

y

17. The computed solution is shown in the following figure.

0 1 2 3 40

1

2

3

4

5

t

q

q(4) = 0.5851.19. The computed solution is shown in the following figure.

0 1 2 3 40

1

2

3

t

q

16 Answers

q(4) = 2(1 + e−2)e−2 ≈ 0.3073.

21. (b) The right hand side of the equation, f (t, y = 3y2/3, is continuous, but∂f/∂y = 2y−1/3 is not continuous where y = 0. Hence the hypotheses of Theo-rem 7.16 are not satisfied.

25. No.

27. This is true because of the uniqueness theorem.

29. This is true because of the uniqueness theorem.

31. This is true because of the uniqueness theorem.

Section 2.8

1. x(0) = 0.8009

3. x(0) = 0.9596

5. x(0) = 0.7275

7. x(0) = 0.7290106

9. x(0) = −3.2314

11. x(0) = −3.2320923

17. (a) |y(t)− x(t)| ≤ |y(0)− x(0)|e2t .

17. (b) x(t) = [2 sin t − cos t]/5, and y(t) = [2 sin t − cos t]/5 − e−2t /10.Yes.

17. (c) The maximum predicted error is achieved for all t < 0.

Section 2.9

1. Autonomous. P = 20000 is an unstable equilibrium point.

3. Not autonomous.

5. q = 2 and kπ , where k is any integer, positive of negative are equilibrium points.q = 2 is unstable and the stability alternates.

0 2 4 6 8 10

−2π

−π

0

2π

2π

t

q

7. y = 3 is an unstable equilibrium point.

Answers 17

3

t

y

9. y = −1 and y = 1 are unstable equilibrium points.

0 2 4 6 8 10

−1

1

t

y

11.

t

y

13.

18 Answers

t

y

15. The graph of f .

y

f (y)=2−y

(2,0)

The phase line.

y2

The solutions.

Answers 19

t

y

(0,2)

17. The graph of f .

y

f (y)=(y+1)(y−4)

(−1,0) (4,0)

The phase line.

y−1 4

The solutions.

20 Answers

t

y

(0,−1)

(0,4)

19. The graph of f .

y

f (y)=9y−y3

(−3,0)

(0,0)

(3,0)

The phase line.

y−3 0 3

The solutions.

Answers 21

t

y

(0,−3)

(0,0)

(0,3)

21. The graph of f .

y

f (y)=sin y

(0,0) (π,0)

(2π,0)

The phase line.

y0 π 2π

The solutions.

22 Answers

t

y

(0,0)

(0,π)

(0,2π)

23. y(t) = 6 + Ce−t . limt→∞ y(t) = 6. The graph of f (y) = 6 − y and the phaseline on the y-axis.

y

f (y)=6−y

(6,0)

25. The graph of the right hand side, with the phase line information on the y-axis.

−5

0

5

y

f(y) = 5−2y

Answers 23

The solution is y(t) = 5 − e−6t

1 + e−6t. limt→∞ y(t) = 5.

27. x = −2 is unstable. x = 2 is asymptotically stable.29. (a) f (x) = x2, f (x) = x3, or f (x) = x4.29. (b) f (x) = −x3, f (x) = −x5, or f (x) = −x7.31. c = 3 lb/gal.

Section 2.10