Calculus one and several variables 10E Salas solutions manual ch09

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-09 JWDD027-Salas-v1 November 25, 2006 19:21

SECTION 9.1 481

CHAPTER 9

SECTION 9.1

1. y′1(x) = 12 e

x/2; 2y′1 − y1 = 2(

12

)ex/2 − ex/2 = 0; y1 is a solution.

y′2(x) = 2x + ex/2; 2y′2 − y2 = 2(2x + ex/2

)−

(x2 + 2ex/2

)= 4x− x2 �= 0;

y2 is not a solution.

2. y′1 + xy1 = −xe−x2/2 + xe−x2/2 = 0; not a solution

y′2 + xy2 = −Cxe−x2/2 + x + Cxe−x2/2 = x; y2 is a solution.

3. y′1(x) =−ex

(ex + 1)2; y′1 + y1 =

−ex

(ex + 1)2+

1ex + 1

=1

(ex + 1)2= y2

1 ; y1 is a solution.

y′2(x) =−Cex

(Cex + 1)2; y′2 + y2 =

−Cex

(Cex + 1)2+

1Cex + 1

=1

(Cex + 1)2= y2

2 ;

y2 is a solution.

4. y′′1 + 4y1 = −8 sin 2x + 8 sin 2x = 0; y1 is a solution.

y′′2 + 4y2 = −2 cosx + 8 cosx = 6 cosx; not a solution.

5. y′1(x) = 2e2x, y′′1 = 4e2x; y′′1 − 4y1 = 4e2x − 4e2x = 0; y1 is a solution.

y′2(x) = 2C cosh 2x, y′′2 = 4C sinh 2x; y′′2 − 4y2 = 4C sinh 2x− 4C sinh 2x = 0;

y2 is a solution.

6. y′′1 − 2y′1 − 3y1 = e−x + 18e3x − 2(−e−x + 6e3x) − 3(e−x + 2e3x) = 0; not a solution

y′′2 − 2y′2 − 3y2 =74

[(6 + 9x)e3x − 2(1 + 3x)e3x − 3xe3x

]= 7e3x; y2 is a solution.

7. y′ − 2y = 1; H(x) =∫

(−2) dx = −2x, integrating factor: e−2x

e−2xy′ − 2e−2xy = e−2x

d

dx

[e−2xy

]= e−2x

e−2xy = − 12e−2x + C

y = − 12

+ Ce2x

8. y′ − 2xy = −1; H(x) =

∫− 2xdx, integrating factor: x−2

x−2y′ − 2x3

y = −x−2

d

dx(x−2y) = −x−2

x−2y =1x

+ C

y = x + Cx2



482 SECTION 9.1

9. y′ +52y = 1; H(x) =

∫ (52

)dx =

52x, integrating factor: e5x/2

e5x/2y′ +52e5x/2y = e5x/2

d

dx

[e5x/2y

]= e5x/2

e5x/2y =25e5x/2 + C

y =25

+ Ce−5x/2

10. y′ − y = −2e−x; H(x) =∫

− dx, integrating factor: e−x

e−xy′ − e−xy = −2e−2x

d

dx

(e−xy

)= −2e−2x

e−xy = e−2x + C

y = e−x + Cex

11. y′ − 2y = 1 − 2x; H(x) =∫

(−2) dx = −2x, integrating factor: e−2x

e−2xy′ − 2e−2xy = e−2x − 2xe−2x

d

dx

[e−2xy

]= e−2x − 2xe−2x

e−2xy = − 12e−2x + x e−2x +

12e−2x + C = x e−2x + C

y = x + Ce2x

12. y′ +2xy =

cosxx2

; H(x) =∫

2xdx = 2 ln |x|, integrating factor: x2

x2y′ + 2xy = cosxd

dx[x2y] = cosx

x2y = sinx + C

y =sinx

x2+

C

x2

13. y′ − 4xy = −2n; H(x) =

∫ (− 4

x

)dx = −4 lnx = ln x−4, integrating factor: eln x−4

= x−4

x−4y′ − 4xx−4y = −2nx−4

d

dx

[x−4y

]= −2nx−4

x−4y =23nx−3 + C

y =23nx + Cx4



SECTION 9.1 483

14. y′ + y = 2 + 2x; H(x) =∫

dx, integrating factor: ex

exy′ + exy = (2 + 2x)ex

d

dx(exy) = 2(1 + x)ex

exy = 2xex + C

y = 2x + Ce−x

15. y′ − ex y = 0; H(x) =∫

−ex dx = −ex, integrating factor: e−ex

e−exy′ − ex e−ex y = 0d

dx

[e−ex y

]= 0

e−ex y = C

y = Ceex

16. y′ − y = ex; H(x) =∫


e−xy′ − e−xy = 1d

dx(e−xy) = 1

e−xy = x + C

y = xex + Cex

17. y′ +1

1 + exy =

11 + ex

; H(x) =∫

11 + ex

dx = lnex

1 + ex,

integrating factor: eH(x) =ex

1 + ex

ex

1 + exy′ +

11 + ex

· ex

1 + exy =

11 + ex

· ex

1 + ex

d

dx

[ex

1 + exy

]=

ex

(1 + ex)2

ex

1 + exy = − 1

1 + ex+ C

y = −e−x + C (1 + e−x)

This solution can also be written: y = 1 + K (e−x + 1) , where K is an arbitrary constant.

18. y′ +1xy =

1 + x

xex; H(x) =

∫1xdx, integrating factor: x

xy′ + y = (1 + x)ex

d

dx(xy) = (1 + x)ex

xy = xex + C

y = ex +C

x



484 SECTION 9.1

19. y′ + 2xy = xe−x2; H(x) =

∫2x dx = x2, integrating factor: ex

2

ex2y′ + 2xex

2y = x

d

dx

[ex

2y]

= x

ex2y =

12x2 + C

y = e−x2 (12 x

2 + C)

20. y′ − 1xy = 2 lnx; H(x) =

∫− 1xdx, integrating factor:

1x

1xy′ − 1

x2y =

2x

lnx

d

dx

(1xy

)=

2x

lnx

1xy = (lnx)2 + C

y = x(lnx)2 + Cx

21. y′ +2

x + 1y = 0; H(x) =

∫2

x + 1dx = 2 ln(x + 1) = ln(x + 1)2,

integrating factor: eln(x+1)2 = (x + 1)2

(x + 1)2 y′ + 2(x + 1) y = 0d

dx

[(x + 1)2 y

]= 0

(x + 1)2 y = C

y =C

(x + 1)2

22. y′ +2

x + 1y = (x + 1)5/2; H(x) =

∫2

x + 1dx, integrating factor: (x + 1)2

(x + 1)2y′ + 2(x + 1)y = (x + 1)9/2

d

dx

[(x + 1)2y

]= (x + 1)9/2

(x + 1)2y =211

(x + 1)11/2 + C

y =211

(x + 1)7/2 + C(x + 1)−2

23. y′ + y = x; H(x) =∫

1 dx = x, integrating factor : ex

ex y′ + ex y = xex

d

dx[ex y] = xex

ex y = xex − ex + C

y = (x− 1) + Ce−x



SECTION 9.1 485

y(0) = −1 + C = 1 =⇒ C = 2. Therefore, y = 2e−x + x− 1 is the solution which satisfies

the initial condition.

24. y′ − y = e2x; H(x) =∫


d

dx(e−xy) = ex

e−xy = ex + C

y = e2x + Cex

1 = y(1) = e2 + Ce =⇒ C =1 − e2

eand y = e2x +

1 − e2

eex is the solution which

satisfies the initial condition.

25. y′ + y =1

1 + ex; H(x) =

∫1 dx = x, integrating factor : ex

ex y′ + ex y =ex

1 + ex

d

dx[ex y] =

ex

1 + ex

ex y = ln (1 + ex) + C

y = e−x [ln (1 + ex) + C]

y(0) = ln 2 + C = e =⇒ C = e− ln 2. Therefore, y = e−x [ln (1 + ex) + e− ln 2] is the

solution which satisfies the initial condition.

26. y′ + y =1

1 + 2ex; H(x) =

∫dx, integrating factor: ex

d

dx(exy) =

ex

1 + 2ex

exy =12

ln(1 + 2ex) + C

y = e−x

[12

ln(1 + 2ex) + C

]

e = y(0) = 12 ln 3 + C =⇒ C = e− 1

2ln 3 and y = e−x

[12 ln(1 + 2ex) + e− 1

2 ln 3]

is the


27. y′ − 2xy = x2ex; H(x) =

∫ (− 2

x

)dx = −2 lnx = ln x−2,

integrating factor: elnx−2= x−2

x−2 y′ − 2x−3 y = ex

d

dx

[x−2 y

]= ex

x−2 y = ex + C

y = x2 (ex + C)



486 SECTION 9.1

y(1) = e + C = 0 =⇒ C = − e. Therefore, y = x2 (ex − e) is the solution which satisfies

the initial condition.

28. y′ +2xy = e−x; H(x) =

∫2xdx, integrating factor: x2

d

dx(x2y) = x2e−x

x2y = −e−x(x2 + 2x + 2) + C

y = −e−x

x2(x2 + 2x + 2) +

C

x2

−1 = y(1) = −5e−1 + C =⇒ C = 5e−1 − 1 and y = −e−x

x2(x2 + 2x + 2) +

5e−1 − 1x2

is the


29. Set z = y′ − y. Then z′ = y′′ − y′.

y′ − y = y′′ − y′ =⇒ z = z′ =⇒ z = C1ex

Now,

z = y′ − y = C1ex =⇒ e−xy′ − e−xy = C1 =⇒ (e−xy)′ = C1

=⇒ e−xy = C1x + C2 =⇒ y = C1xex + C2e

x

30. General solution: y = Ce−rx. (a) y(a) = 0 = Ce−ra =⇒ C = 0 =⇒ y(x) = 0 for all x.

(b) r < 0 and y �= 0 =⇒ y → ∞ as x → ∞(c) r > 0 and y �= 0 =⇒ y → 0 as x → ∞(d) If r = 0, then y(x) = C, constant.

31. (a) Let y1 and y2 be solutions of y′ + p(x)y = 0, and let u = y1 + y2. Then

u′ + pu = (y1 + y2)′ + p (y1 + y2)

= y′1 + y′2 + py1 + py2

= y′1 + py1 + y′2 + py2 = 0 + 0 = 0

Therefore u is a solution.

(b) Let u = Cy where y is a solution of y′ + p(x)y = 0. Then

u′ + pu = (Cy)′ + p(Cy) = Cy′ + Cpy = C(y′ + py) = C × 0 = 0

Therefore u is a solution.



SECTION 9.1 487

32. (a) y′ + p(x)y = 0

e∫ x

ap(t)dt

y′ + p(x)e∫ x

ap(t)dt

y = 0[e∫ x

ap(t)dt

y

]′= 0

e∫ x

ap(t)dt

y = C

y(x) = Ce−∫ x

ap(t)dt

(b) y(b) = 0 =⇒ Ce−∫ b

ap(t)dt = 0 =⇒ C = 0 =⇒ y(x) = 0 for all x.

(c) Let z = y1 − y2. Then z is a solution of y′ + p(x)y = 0. If y1(b) = y2(b), then z(b) = 0 =⇒z(x) = 0 for all x.

33. Let y(x) = e−H(x)

∫ x

a

q(t) eH(t) dt.

Note first that y(a) = e−H(a)

∫ a

a

q(t) eH(t) dt = 0 so y satisfies the initial condition.

Now,

y′ + p(x)y =[e−H(x)

∫ x

a

q(t) eH(t) dt

]′+ p(x) e−H(x)

∫ x

a

q(t) eH(t) dt

= e−H(x)q(x) eH(x) + e−H(x)[−p(x)]∫ x

a

q(t) eH(t) dt + p(x) e−H(x)

∫ x

a

q(t) eH(t) dt

= q(x)

Thus, y(x) = e−H(x)

∫ x

a

q(t) eH(t) dt is the solution of the initial value problem.

34. Let z = y1 − y2. Then

z′ = y′1 − y′2 = q − py1 − (q − py2) = −p (y1 − y2) = −pz =⇒ z′ + pz = 0

35. According to Newton’s Law of Cooling, the temperature T at any time t is given by

T (t) = 32 + [72 − 32]e−kt

We can determine k by applying the condition T (1/2) = 50◦:

50 = 32 + 40 e−k/2

e−k/2 =1840

=920

− 12k = ln(9/20)

k = −2 ln(9/20) ∼= 1.5970

Therefore, T (t) ∼= 32 + 40 e−1.5970t.



488 SECTION 9.1

Now, T (1) ∼= 32 + 40 e−1.5970 ∼= 40.100; the temperature after 1 minute is (approx.) 40.10◦.

To find how long it will take for the temperature to reach 35◦, we solve

32 + 40 e−1.5970t = 35

for t:

32 + 40 e−1.5970t = 35

40 e−1.5970t = 3

−1.5970t = ln(3/40)

t =ln(3/40)−1.5970

∼= 1.62

It will take approximately 1.62 minutes for the thermometer to read 35◦.

36. By (9.1.4) T (t) = 100 − 80e−kt

T (2) = 22 =⇒ 100 − 80e−2k = 22 =⇒ k =ln(39/40)

−2∼= 0.01266

T (6) ∼= 100 − 80e−0.01266(6) ∼= 25.85◦ C; T (t) = 90 =⇒ −80e−0.01266t = −10 =⇒ t ∼= 164.25 secs.

37. (a) The solution of the initial value problem v′ = 32 − kv, (k > 0) v(0) = 0 is:

v(t) =32k

(1 − e−kt

).

(b) At each time t, 1 − e−kt < 1. Therefore

v(t) =32k

(1 − e−kt

)<

32k

and limt→∞

v(t) =32k

(c)

38. (a)dP

dt+ (b− a)P = 0; H(t) =

∫(b− a) dt = (b− a)t, integrating factor : e(b−a)t

e(b−a)t dP

dt+ (b− a)e(b−a)t P = 0

d

dt

[e(b−a)t P

]= 0

e(b−a)tP = C

P = Ce(a−b)t

P (0) = P0 =⇒ P (t) = P0e(a−b)t.



SECTION 9.1 489

(b) (i) a > b =⇒ P0e(a−b)t is increasing.

P (t) → ∞ as t → ∞.

(ii) a = b =⇒ P (t) = P0 is a constant.

(iii) a < b =⇒ P0e(a−b)t is decreasing.

P (t) → 0 as t → ∞.

39. (a)di

dt+

R

Li =

E

L; H(t) =

∫R

Ldt =

R

Lt, integrating factor : e

RL t

eRL t di

dt+

R

Le

RL ti =

E

Le

RL t

d

dt

[e

RL ti

]=

E

Le

RL t

eRL ti =

E

Re

RL t + C

i(t) =E

R+ Ce−

RL t

i(0) = 0 =⇒ C = −ER , so i(t) =

E

R

[1 − e−(R/L) t

].

(b) limt→∞

i(t) = limt→∞

ER

(1 − e−(R/L) t

)= E

R amps

(c) i(t) = 0.9ER =⇒ e−(R/L) t = 1

10 =⇒ −RL t = − ln 10 =⇒ t = L

R ln 10 seconds.

40. (a)di

dt+

R

Li =

E

Lsinωt; H(t) =

∫R

Ldt =

R

Lt, integrating factor : e

RL t

eRL t di

dt+

R

Le

RL ti =

E

Le

RL t sinωt

d

dt

[e

RL ti

]=

E

Le

RL t sinωt

eRL ti =

E

Le

RL t L2

R2 + ω2L2

[R

Lsinωt− ω cosωt

]+ C

i(t) =EL

R2 + ω2L2

[R


]+ Ce−

RL t

i(0) = i0 =⇒ i(t)=EL

R2 + ω2L2

[R


]+

[i0 + ω

EL

R2 + ω2L2

]e−

RL t.

(b) limt→∞

does not exist because the trigonometric functions continue to oscillate.



490 SECTION 9.1

(c) A sample graph is:

41. (a) V ′(t) = kV (t) =⇒ V (t) = V0ekt

Loses 20% in 5 minutes, so V (5) = V0e5k = 0.8V0 =⇒ k = 1

5 ln 0.8

=⇒ V (t) = V0e15 (ln 0.8)t = V0

(eln 0.8

)t/5= V0(0.8)t/5 = V0

(45

)t/5.

Since V0 = 200 liters, we get V (t) = 200(

45

)t/5(b)

V ′(t) = ktV (t)

V ′(t) − ktV (t) = 0

e−kt2/2V ′(t) − kte−kt2/2 V (t) = 0d

dt

[e−kt2/2 V (t)

]= 0

e−kt2/2 V (t) = C

V (t) = Cekt2/2.

V (0) = C = 200 =⇒ V (t) = 200ekt2/2.

V (5) = 160 =⇒ 200ek(25/2) = 160, ek(25/2) = 45 , ek =

(45

)2/25.

Therefore V (t) = 200(

45

)t2/25 liters.

42. Let s(t) be the number of pounds of salt present after t minutes. Since

s′(t) = rate in − rate out = 3 (0.2) − 3(s(t)100

),

we have

s′(t) + 0.03s(t) = 0.6.

Multiply by e∫

0.03dt = e0.03t:

e0.03ts′(t) + 0.03e0.03ts(t) = 0.6e0.03t

d

dt

[e0.03ts(t)

]= 0.6e0.03t

e0.03ts(t) = 20e0.03t + C

s(t) = 20 + Ce−0.03t.

Use the initial condition s(0) = 100(0.25) = 25 to determine C: 25 = 20 + Ce0 so C = 5.

Thus, s(t) = 20 + 5e−0.03t lb.



SECTION 9.1 491

43. (a)dP

dt= k(M − P )

(b)dP

dt+ kP = kM ; H(t) =

∫k dt = kt, integrating factor : ekt

ektdP

dt+ kekt P = kM ekt

d

dt

[ekt P

]= kM ekt

ektP = M ekt + C

P = M + Ce−kt

P (0) = M + C = 0 =⇒ C = −M and P (t) = M(1 − e−kt

)P (10) = M

(1 − e−10k

)= 0.3M =⇒ k ∼= 0.0357 and P (t) = M

(1 − e−0.0357t

)(c) P (t) = M

(1 − e−0.0357t

)= 0.9M =⇒ e−0.0357t = 0.1 =⇒ t ∼= 65

Therefore, it will take approximately 65 days for 90 % of the population to be aware of the product.

44. (a)dQ

dt= rate in − rate out = r − kQ, k > 0

(b)dQ

dt+ kQ = r, Q(0) = 0 =⇒ Q(t) =

r

k

(1 − e−kt

)(c) lim

t→∞Q(t) =

r

k.

45. (a)dP

dt− 2 cos(2πt)P = 0 =⇒ P = Ce

1π sin 2πt.

P (0) = C = 1000 =⇒ P = 1000e1π sin 2πt.

(b) dP

dt− 2 cos(2πt)P = 2000 cos 2πt =⇒ P = Ce

1π sin 2πt − 1000.

P (0) = 1000 =⇒ C = 2000 =⇒ P = 2000e1π sin 2πt − 1000.

46. (a) Let Q = lnP. ThendQ

dt=

1P

dP

dt= a− bQ.

Solving the differential equationdQ

dt+ bQ = a =⇒ Q =

a

b+ Ce−bt, so P = e

ab +Ce−bt

.

P (0) = P0 =⇒ ec = P0e− a

b . Thus P = eab

[P0e

− ab

]e−bt

.



492 SECTION 9.2

(b) e−bt → 0 as t → ∞, so P → eab .

(c) P ′ = P (a− b lnP ) =⇒ P ′′ = P

(−b

P

)P ′ + P ′(a− b lnP ) = P (a− b lnP )(a− b− b lnP ).

If 0 < P < ea/b−1, then P is increasing and the graph is concave up; if ea/b−1 < P < ea/b, then P

is increasing and the graph is concave down; if ea/b < P , then P is decreasing and the graph is

concave down.

(d)

SECTION 9.2

1. y′ = y sin(2x + 3)

1ydy = sin(2x + 3) dx

∫1ydy =

∫sin(2x + 3) dx

ln | y | = − 12

cos(2x + 3) + C

This solution can also be written: y = Ce−(1/2) cos(2x+3).

2. y′ = (x2 + 1)(y2 + y)∫dy

y2 + y=

∫(x2 + 1) dx

ln∣∣∣∣ y

y + 1

∣∣∣∣ =x3

3+ x + C

This solution can also be written: y =1

Ke−x−x3/3 − 1(K = eC).

3. y′ = (xy)3

1y3

dy = x3 dx, y �= 0∫1y3

dy =∫

x3 dx

− 12y−2 =

14x4 + C

This solution can also be written: x4 +2y2

= C, or y2 =2

C − x4;



SECTION 9.2 493

4. y′ = 3x2(1 + y2)∫dy

1 + y2=

∫3x2 dx

tan−1 y = x3 + C

y = tan(x3 + C)

5. y′ =sin(1/x)x2y cos y

y cos y dy =1x2

sin(1/x) dx∫y cos y dy =

∫1x2

sin(1/x) dx

y sin y + cos y = cos(1/x) + C

6.y′ =

y2 + 1y + yx∫

y

1 + y2dy =

∫1

1 + xdx

ln√

1 + y2 = ln |1 + x| + C

1 + y2 = K(1 + x)2 (K = lnC)

7.y′ = x ey+x

e−y dy = x ex dx∫e−y dy =

∫x ex dx

−e−y = x ex − ex + C

e−y = ex − x ex + C

This solution can also be written: y = − ln(ex − xex + C).

8.y′ = xy2 − x− y2 + 1 = (x− 1)(y2 − 1)∫

dy

y2 − 1=

∫dx

x− 112

ln∣∣∣∣y − 1y + 1

∣∣∣∣ = ln |x− 1| + C

This solution can also be written: y =1 + Kex

2−2x

1 −Kex2−2x(K = eC).

9. (y lnx)y′ =(y + 1)2

x

y

(y + 1)2dy =

1x lnx

dx

∫y

(y + 1)2dy =

∫1

x lnxdx

ln | y + 1 | + 1y + 1

= ln | lnx | + C



494 SECTION 9.2

10. ey sin 2x dx + cosx(e2y − y) dy = 0∫sin 2xcosx

dx +∫

(ey − ye−y) dy = C

−2 cosx + ey + e−y(1 + y) = C

11.(y lnx)y′ =

y2 + 1x

y

y2 + 1dy =

1x lnx

dx

∫y

y2 + 1dy =

∫1

x lnxdx

12 ln (y2 + 1) = ln | lnx| + K = ln |C lnx| (K = ln |C|)

ln (y2 + 1) = 2 ln |C lnx| = ln (C lnx)2

y2 = C(lnx)2 − 1

12. y′ =1 + 2y2

y sinxy

1 + 2y2dy = cscx dx

14 ln(1 + 2y2) = ln | cscx− cotx| + C

the integral curves can be written as: ln(1 + 2y2) = ln[C(cscx− cotx)4

], or as

y2 = K(cscx− cotx)4 − 12 .

13.y′ = x

√1 − y2

1 − x2, y(0) = 0

1√1 − y2

dy =x√

1 − x2dx

∫1√

1 − y2dy =

∫x√

1 − x2dx

sin−1 y = −√

1 − x2 + C

y(0) = 0 =⇒ arcsin 0 = −1 + C =⇒ C = 1

Thus, arcsin y = 1 −√

1 − x2.

14. y′ =ex−y

1 + ex∫ey dy =

∫ex

1 + exdx

ey = ln(1 + ex) + C

y(1) = 0 =⇒ 1 = ln(1 + e) + C =⇒ C = 1 − ln(1 + e) and ey = ln(1 + ex) + 1 − ln(1 + e)



SECTION 9.2 495

15. y′ =x2y − y

y + 1, y(3) = 1

y + 1y

dy = (x2 − 1) dx, y �= 0∫y + 1y

dy =∫

(x2 − 1) dx

y + ln | y | =13x3 − x + C

y(3) = 1 =⇒ 1 + ln 1 =13

(3)3 − 3 + C =⇒ C = −5.

Thus, y + ln | y | = 13 x

3 − x− 5.

16. x2y′ = y − xy∫1ydy =

∫(1 − x)x−2 dx

ln |y| = − 1x− ln |x| + C

−1 = y(−1) =⇒ C = −1 and ln |xy| + 1x

= −1

17. (xy2 + y2 + x + 1) dx + (y − 1) dy = 0, y(2) = 0

(x + 1)(y2 + 1) dx + (y − 1) dy = 0

(x + 1) dx +y − 1y2 + 1

dy = 0∫(x + 1) dx +

∫y − 1y2 + 1

dy = C

x2

2+ x +

12

ln (y2 + 1) − tan−1 y = C

y(2) = 0 =⇒ C = 4. Thus, 12 x

2 + x + 12 ln (y2 + 1) − tan−1 y = 4

18. cos y dx + (1 + e−x) sin y dy = 0∫dx

1 + e−x+

∫sin y

cos ydy = C

ln(ex + 1) + ln | sec y| = C;π

4= y(0) =⇒ ln 2 + ln

√2 = C

ln(ex + 1) + ln | sec y| = 32 ln 2

19. y′ = 6 e2x−y, y(0) = 0

y′ = 6 e2x−y

ey dy = 6 e2x dx

ey = 3 e2x + C

y(0) = 0 =⇒ 1 = 3 + C =⇒ C = −2

Thus, ey = 3 e2x − 2 =⇒ y = ln[3 e2x − 2

]



496 SECTION 9.2

20. xy′ − y = 2x2y, y(1) = 1

xy′ = y(1 + 2x2

)1ydy =

1 + 2x2

xdx =

(1x

+ 2x)

dx

ln |y| = ln |x| + x2 + C

y(1) = 1 =⇒ 0 = 1 + C =⇒ C = −1

Thus, ln |y| = ln |x| + x2 − 1 or y = xex2−1

21. We assume that C = 0 at time t = 0. (a) Let A0 = B0. Then

dC

dt= k(A0 − C)2 and

dC

(A0 − C)2= k dt.

Integrating, we get ∫1

(A0 − C)2dC =

∫k dt

1A0 − C

= kt + M M a constant.

Since C(0) = 0, M =1A0

and

1A0 − C

= kt +1A0

.

Solving this equation for C gives

C(t) =kA2

0t

1 + kA0t.

(b) Suppose that A0 �= B0. Then

dC

dt= k(A0 − C)(B0 − C) and

dC

(A0 − C)(B0 − C)= k dt.

Integrating, we get ∫1

(A0 − C)(B0 − C)dC =

∫k dt

1B0 −A0

∫ (1

A0 − C− 1

B0 − C

)dC =

∫k dt

1B0 −A0

[− ln (A0 − C) + ln (B0 − C)] = kt + M

1B0 −A0

ln(B0 − C

A0 − C

)= kt + M M an arbitrary constant

Since C(0) = 0, M = 1B0−A0

ln(

B0A0

)and

1B0 −A0

ln(B0 − C

A0 − C

)= kt +

ln(B0/A0)B0 −A0

.

Solving this equation for C, gives

C(t) =A0B0

(ekA0t − ekB0t

)A0ekA0t −B0ekB0t

.



SECTION 9.2 497

22. (a) From (9.2.4) with K = 0.0020, M = 800, R = P (0) = 100, we have

P (t) =80, 000

100 + 700e−1.6 t

(b) (c)dP

dtis maximal at t ∼= 1.2162

Maximum value = 320

23. (a) mdv

dt= −αv − βv2

dv

v(α + βv)= − 1

mdt

∫1

v(α + βv)dv = −

∫1m

dt

1α

∫1vdv − β

α

∫1

α + βvdv = −

∫1m

dt

1α

ln v − 1α

ln(α + βv) = − 1mt + M, M a constant

ln(

v

α + βv

)= − α

mt + αM

v

α + βv= Ke−αt/m

[K = eαM

]Solving this equation for v we get v(t) =

αK

eαt/m − βK=

α

Ceαt/m − β[C = 1/K].

(b) Setting v(0) = v0, we get

C =α + βv0

v0and

v(t) =αv0

(α + βv0)eαt/m − βv0

(c) limt→∞

v(t) = 0

24. F = ma = mdv

dt

(a) mdv

dt= mg − βv2

dt =mdv

mg − βv2=

m

β

(dv

vc2 − v2

)

t =m

β

∫1

vc2 − v2dv =

m

2vcβ

∫ (1

vc + v+

1vc − v

)dv

=m

2vcβ

[ln

(vc + v

vc − v

)]+ C



498 SECTION 9.2

At t = 0, v(0) = v0. Therefore

C = − m

2vcβ

[ln

(vc + v0

vc − v0

)]=

m

2vcβ

[ln

(vc − v0

vc + v0

)].

Thus

t =m

2vcβ

[ln

(vc + v

vc + v0· vc − v0

vc − v

)]=

vc2g

[ln

(vc + v

vc + v0· vc − v0

vc − v

)].

vc =√mg/β

(b)vc + v

vc + v0· vc − v0

vc − v= e2tg/vc

vc + v =vc + v0

vc − v0e2tg/vc(vc − v)

v

[1 +

(vc + v0

vc − v0

)e2tg/vc

]= vc

[(vc + v0

vc − v0

)e2tg/vc − 1

]

v = vc

[(vc + v0)e2tg/vc − (vc − v0)(vc + v0)e2tg/vc + (vc − v0)

].

We can bring the hyperbolic functions into play by writing

v = vc

[(vc + v0)egt/vc − (vc − v0)e−gt/vc

(vc + v0)egt/vc − (vc − v0)e−gt/vc

]

= vc

[v0 cosh(gt/vc) + vc sinh(gt/vc)v0 sinh(gt/vc) + vc cosh(gt/vc)

]

(c) a = g

{vc

2 − v02

[v0 sinh(gt/vc) + vc cosh(gt/vc)]2

}

The acceleration can not change sign since the denominator is always positive and the numerator

is constant. As t → ∞, the denominator → ∞, and the fraction → 0.

(d) We can write

v = vc

[(vc + v0) − (vc − v0)e−2tg/vc

(vc + v0) + (vc − v0)e−2tg/vc

].

As t → ∞, −2gt/vc → −∞ and e−2tg/vc → 0. Thus v → vc

25. (a) Let P = P (t) denote the number of people who have the disease at time t. Then, substituting

into (9.2.4) with M = 25, 000 and R = 100, we get

P (t) =25, 000(100)

100 + (249, 00)e−25,000kt=

25, 0001 + 249e−25,000kt

.

P (10)25, 000

1 + 249e−25,000(10k)= 400 =⇒ −25, 000k ∼= −0.1398.

Therefore, P (t) =25, 000

1 + 249e−0.1398t.



SECTION 9.2 499

(b)25, 000

1 + 249e−0.1398 t= 12, 500 =⇒ t ∼= 40;

It will take 40 days for half the

population to have the disease.

(c)

26.dy

dt= ky(M − y) = kMy − ky2 =⇒ d2y

dt2= (kM − 2ky)

dy

dt= k2(M − 2y)(M − y)y

d2y

dt2> 0 for 0 < y <

M

2, so

dy

dtis increasing

d2y

dt2< 0 for

M

2< y < M, so

dy

dtis decreasing

Thereforedy

dtis maximal at y =

M

2. The disease is spreading fastest when half the population is

infected.

27. Assume that the package is dropped from rest.

(a) Let v = v(t) be the velocity at time t, 0 ≤ t ≤ 10. Then

100dv

dt= 100g − 2v or

dv

dt+

150

v = g (g = 9.8 m/sec2)

This is a linear differential equation; et/50 is an integrating factor.

et/50dv

dt+

150

et/50v = g et/50

d

dt

[et/50v

]= g et/50

et/50v = 50g et/50 + C

v = 50g + Ce−t/50

Now, v(0) = 0 =⇒ C = −50g and v(t) = 50g(1 − e−t/50

).

At the instant the parachute opens, v(10) = 50g(1 − e−1/5

) ∼= 50g(0.1813) ∼= 88.82 m/sec.

(b) Now let v = v(t) denote the velocity of the package t seconds after the parachute opens. Then

100dv

dt= 100g − 4v2 or

dv

dt= g − 1

25v2



500 SECTION 9.2

This is a separable differential equation:

dv

dt= g − 1

25v2 set u = v/5, du = (1/5)dv

du

g − u2=

15dt

12√g

ln∣∣∣∣u +

√g

u−√g

∣∣∣∣ =t

5+ K

ln∣∣∣∣u +

√g

u−√g

∣∣∣∣ =2√g

5t + M

u +√g

u−√g

= Ce2√gt/5 ∼= Ce1.25 t

u =√gCe1.25 t + 1Ce1.25t − 1

v = 5√gCe1.25 t + 1Ce1.25t − 1

Now, v(0) = 88.82 =⇒ 5√gC + 1C − 1

= 88.82 =⇒ C ∼= 1.43.

Therefore, v(t) = 5√g

1.43e1.25 t + 11.43e1.25t − 1

=15.65

(1 + 0.70e−1.25 t

)1 − 0.70e−1.25 t

(c) From part (b), limt→∞

v(t) = 15.65 m/sec.

28. (a) By the hint ∫dC(

A0 −12C

)2 =∫

k dt

2

A0 −12C

= kt + K.

First, C(0) = 0 =⇒ K = 2/A0. Then, C(1) = A0 =⇒ k = 2/A0. Thus,

2A0 − 1

2C=

2A0

(t + 1), which gives C(t) = 2A0

(t

t + 1

).

(b) By the hint ∫dC(

A0 −12C

) (2A0 −

12C

) =∫

k dt

1A0

∫ ⎡⎢⎣ 1

A0 −12C

− 1

2A0 −12C

⎤⎥⎦ dC =

∫k dt



SECTION 9.2 501

1A0

[−2 ln |A0 −

12C| + 2 ln |2A0 −

12C|

]= kt + K

2A0

ln

∣∣∣∣∣∣∣2A0 −

12C

A0 −12C

∣∣∣∣∣∣∣ = kt + K.

First, C(0) = 0 =⇒ K =2A0

ln 2. Then,

C(1) = A0 =⇒ 2A0

ln 3 = k +2A0

ln 2 =⇒ k =2A0

ln32.

Thus,

2A0

ln∣∣∣∣2A0 − 1

2C

A0 − 12C

∣∣∣∣ =2A0

t ln32

+2A0

ln 2 =2A0

ln

[2

(32

)t]

so that

2A0 − 12C

A0 − 12C

= 2(

32

)t

and therefore C(t) = 4A03t − 2t

2(3t) − 2t.

(c) By the hint ∫dC(

A0 −m

m + nC

) (A0 −

n

m + nC

) =∫

k dt

∫1

A0(m− n)

⎡⎢⎣ m

A0 −m

m + nC

− n

A0 −n

m + nC

⎤⎥⎦ dC =

∫k dt

1A0(m− n)

[−(m + n) ln

∣∣∣∣A0 −m

m + nC

∣∣∣∣ + (m + n) ln∣∣∣∣A0 −

n

m + nC

∣∣∣∣]

= kt + K

m + n

A0(m− n)ln

∣∣∣∣∣∣∣A0 −

n

m + nC

A0 −m

m + nC

∣∣∣∣∣∣∣ = kt + K.

First, C(0) = 0 =⇒ K =m + n

A0(m− n)ln

∣∣∣∣A0

A0

∣∣∣∣ = 0. Then,

C(1) = A0 =⇒ k =m + n

A0(m− n)ln

∣∣∣∣∣∣∣A0 −

n

m + nA0

A0 −m

m + nA0

∣∣∣∣∣∣∣ =m + n

A0(m− n)ln

m

n.



502 SECTION 9.2

Thus,

m + n

A0(m− n)ln

∣∣∣∣∣∣∣A0 −

n

m + nC

A0 −m

m + nC

∣∣∣∣∣∣∣ =m + n

A0(m− n)ln

(m

n

)(t) + 0

so that

A0 −n

m + nC

A0 −m

m + nC

=(m

n

)t

and therefore C(t) = A0(m + n)[

mt − nt

mt+1 − nt+1

].

PROJECT 9.2

1. (a) 2x + 3y = C =⇒ 2 + 3y′ = 0 =⇒ y′ = − 23

The orthogonal trajectories are the solutions of:

y′ =32.

y′ = 32 =⇒ y = 3

2 x + C

(b) Curves: y = Cx, y′ = C =y

x

orthogonal trajectories: y′ = −x

y∫y dy +

∫x dx = K1; x2 + y2 = K (= 2K1)



SECTION 9.2 503

(c) xy = C =⇒ y + xy′ = 0 =⇒ y′ = − y

x


y′ =x

y.

y′ =x

y∫x dx =

∫y dy

12 x

2 = 12 y

2 + C

or x2 − y2 = C

(d) y = Cx3, y′ = 3Cx2 =3yx

orthogonal trajectories: y′ = − x

3y∫3y dy +

∫x dx = K1; 3y2 + x2 = K (= 2K1)

(e) y = C ex =⇒ y′ = C ex = y


y′ = − 1y.

y′ = − 1y∫

y dy = −∫

dx

12 y

2 = −x + K

or y2 = −2x + C

(f) x = Cy4, 1 = 4Cy3y′; y′ =y

4x

orthogonal trajectories:dy

dx= −4x

y∫y dy +

∫4x dx = K1; y2 + 4x2 = K (= 2K1)



504 SECTION 9.3

2. (a) Curves: y2 − x2 = C; 2yy′ − 2x = 0 =⇒ y′ =x

y

orthogonal trajectories: y′ = −y

x;

∫1ydy = −

∫1xdx; y =

C

x

(b) Curves: y2 = Cx3, 2yy′ = 3Cx2; y2 =2xyy′

3=⇒ y′ =

3y2x

orthogonal trajectories: y′ = −2x3y

;∫

3y dy +∫

2x dx = C; x2 +32y2 = C or 2x2 + 3y2 = C

(c) Curves: y =Cex

x, xy = Cex; xy′ + y = Cex =⇒ y′ =

y(x− 1)x

orthogonal trajectories: y′ =x

y(1 − x);

∫y dy =

∫x

1 − xdx =

∫ (1

1 − x− 1

)dx;

12y

2 = − ln |1 − x| − x + C or y2 + 2x + ln (1 − x)2 = C

(d) Curves: ex sin y = C; ex cos y y′ + ex sin y = 0 =⇒ y′ = − sin y

cos y

orthogonal trajectories: y′ =cos y

sin y;

∫sin y

cos ydy =

∫dx; ln | sec y| = x + C or sec y = Cex

3. (a) A differential equation for the given family is:

y2 = 2xyy′ + y2(y′)2

A differential equation for the family of orthogonal trajectories is found by replacing y′ by −1/y′.

The result is:

y2 = − 2xyy′

+y2

(y′)2which simplifies to y2 = 2xyy′ + y2(y′)2

Thus, the given family is self-orthogonal.

(b)x2

C2+

y2

C2 − 4= 1 =⇒ 2x

C2+

2yy′

C2 − 4= 0 =⇒ C2 =

4xx + yy′

A differential equation for the given family is:

x2 + xyy′ − xy

y′− y2 = 4

A differential equation for the family of orthogonal trajectories is found by replacing y′ by −1/y′.

The result is:

x2 − xy1y′

+ xyy′ − y2 = 4

Thus, the given family is self-orthogonal.



SECTION 9.3 505

SECTION 9.3

1. The characteristic equation is:

r2 + 2r − 8 = 0 or (r + 4)(r − 2) = 0.

The roots are: r = −4, 2. The general solution is:

y = C1e−4x + C2e

2x.

2. r2 − 13r + 42 = 0 =⇒ r = 6, 7; y = C1e6x + C2e

7x


r2 + 8r + 16 = 0 or (r + 4)2 = 0.

There is only one root: r = −4. By Theorem 9.3.6 II, the general solution is:

y = C1e−4x + C2xe

−4x.

4. r2 + 7r + 3 = 0 =⇒ r = −72±

√372

; y = C1e−7+

√37

2 x + C2e−7−

√37

2 x.

5. The characteristic equation is: r2 + 2r + 5 = 0.

The roots are complex: r = −1 ± 2i. By Theorem 9.3.6 III, the general solution is:

y = e−x (C1 cos 2x + C2 sin 2x) .

6. r2 − 3r + 8 = 0 =⇒ r =32±

√232

i; y = e3x/2

(C1 cos

√232

x + C2 sin√

232

x

)


2r2 + 5r − 3 = 0 or (2r − 1)(r + 3) = 0.

The roots are: r = 12 , −3. The general solution is:

y = C1ex/2 + C2e

−3x.

8. r2 − 12 = 0 =⇒ r = ±2√

3; y = C1e2√

3x + C2e−2

√3x.


r2 + 12 = 0.

The roots are complex: r = ±2√

3 i. The general solution is:

y = C1 cos 2√

3x + C2 sin 2√

3x.



506 SECTION 9.3

10. r2 − 3r +94

= 0 =⇒ r =32; y = C1e

32x + C2xe

32x .


5r2 + 114 r − 3

4 = 0 or 20r2 + 11r − 3 = (5r − 1)(4r + 3) = 0.

The roots are: r = 15 , − 3

4 . The general solution is:

y = C1ex/5 + C2e

−3x/4.

12. 2r2 + 3r = 0 =⇒ r = 0,−32; y = C1 + C2e

− 32x.


r2 + 9 = 0.

The roots are complex: r = ±3i. The general solution is:

y = C1 cos 3x + C2 sin 3x.

14. r2 − r − 30 = 0 =⇒ r = 6,−5; y = C1e6x + C2e

−5x.


2r2 + 2r + 1 = 0.

The roots are complex: r = − 12 ± 1

2 i. The general solution is:

y = e−x/2 [C1 cos(x/2) + C2 sin(x/2)] .

16. r2 − 4r + 4 = 0 =⇒ r = 2; y = C1e2x + C2xe

2x.


8r2 + 2r − 1 = 0 or (4r − 1)(2r + 1) = 0.

The roots are: r = 14 , − 1

2 . The general solution is:

y = C1ex/4 + C2e

−x/2.

18. 5r2 − 2r + 1 = 0 =⇒ r =15± 2

5i; y = ex/5

(C1 cos

2x5

+ C2 sin2x5

).


r2 − 5r + 6 = 0 or (r − 3)(r − 2) = 0.

The roots are: r = 3, 2. The general solution and its derivative are:

y = C1e3x + C2e

2x, y′ = 3C1e3x + 2C2e

2x.

The conditions: y(0) = 1, y′(0) = 1 require that

C1 + C2 = 1 and 3C1 + 2C2 = 1.

Solving these equations simultaneously gives C1 = −1, C2 = 2.

The solution of the initial value problem is: y = 2e2x − e3x.



SECTION 9.3 507

20. r2 + 2r + 1 = 0 =⇒ r = −1; y = C1e−x + C2xe

−x

1 = y(2) = C1e−2 + 2C2e

−2, 2 = y′(2) = −C1e−2 − C2e

−2

=⇒ C1 = −5e2, C2 = 3e2 =⇒ y = −5e2−x + 3xe2−x.


r2 + 14 = 0.

The roots are: r = ± 12 i. The general solution and its derivative are:

y = C1 cos(x/2) + C2 sin(x/2) y′ = − 12 C1 sin(x/2) + 1

2 C2 cos(x/2).

The conditions: y(π) = 1, y′(π) = −1 require that

C2 = 1 and C1 = 2.

The solution of the initial value problem is: y = 2 cos(x/2) + sin(x/2).

22. r2 − 2r + 2 = 0 =⇒ r = 1 ± i; y = ex(C1 cosx + C2 sinx).

−1 = y(0) = C1, −1 = y′(0) = C1 + C2 =⇒ C2 = 0, y = −ex cosx


r2 + 4r + 4 = 0 or (r + 2)2 = 0.

There is only one root: r = −2. The general solution and its derivative are:

y = C1e−2x + C2xe

−2x y′ = −2C1e−2x + C2e

−2x − 2C2xe−2x.

The conditions: y(−1) = 2, y′(−1) = 1 require that

C1e2 − C2e

2 = 2 and − 2C1e2 + 3C2e

2 = 1.

Solving these equations simultaneously gives C1 = 7e−2, C2 = 5e−2.

The solution of the initial value problem is: y = 7e−2e−2x + 5e−2xe−2x = 7e−2(x+1) + 5xe−2(x+1).

24. r2 − 2r + 5 = 0 =⇒ r = 1 ± 2i; y = ex(C1 cos 2x + C2 sin 2x).

0 = y(π/2) = eπ/2(−C1) =⇒ C1 = 0; 2 = y′(π/2) = eπ/2(−2C2) =⇒ C2 = −e−π/2

=⇒ y = −ex−π/2 sin 2x.


r2 − r − 2 = 0 or (r − 2)(r + 1) = 0.

The roots are: r = 2, −1. The general solution and its derivative are:

y = C1e2x + C2e

−x y′ = 2C1e2x − C2e

−x.

(a) y(0) = 1 =⇒ C1 + C2 = 1 =⇒ C2 = 1 − C1.

Thus, the solutions that satisfy y(0) = 1 are: y = Ce2x + (1 − C)e−x.



508 SECTION 9.3

(b) y′(0) = 1 =⇒ 2C1 − C2 = 1 =⇒ C2 = 2C1 − 1.

Thus, the solutions that satisfy y′(0) = 1 are: y = Ce2x + (2C − 1)e−x.

(c) To satisfy both conditions, we must have 2C − 1 = 1 − C =⇒ C = 23 .

The solution that satisfies y(0) = 1, y′(0) = 1 is:

y = 23 e

2x + 13 e

−x.

26. r2 − ω2 = 0 =⇒ r = ±ω; y = A1eωx + A2e

−ωx

Since eωx = coshωx + sinhωx and e−ωx = coshωx− sinhωx, we can write

y = C1 coshωx + C2 sinhωx (with C1 = A1 + A2, C2 = A1 −A2).

27. α =r1 + r2

2, β =

r1 − r22

;

y = k1er1x + k2e

r2x = eαx (C1 coshβx + C2 sinhβx) , where k1 =C1 + C2

2, k2 =

C1 − C2

2.

28. r2 + ω2 = 0 =⇒ r = ±ωi; y = C1 cosωx + C2 sinωx.

Assuming that C21 + C2

2 > 0, we have

C1 cosωx + C2 sinωx =√C1

2 + C22

(C1√

C12 + C2

2cosωx +

C2√C1

2 + C22

sinωx

)

= A (sinφ0 cosωx + cosφ0 sinωx) = A sin(ωx + φ0),

where A =√C1

2 + C22 and φ0, φ0 ∈ [0, 2π), is the angle such that

sinφ0 =C1√

C12 + C2

2and cosφ0 =

C2√C1

2 + C22

29. (a) Let y1 = eαx, y2 = xeαx. Then

W (x) = y1y′2 − y2y

′1 = eαx [eαx + αxeαx] − xeαx [αeαx] = e2αx �= 0

(b) Let y1 = eαx cosβx, y2 = eαx sinβx, β �= 0. Then

W (x) = y1y′2 − y2y

′1

= eαx cosβx [αeαx sinβx + βeαx cosβx] − eαx sinβx [αeαx cosβx− βeαx sinβx]

= βe2αx �= 0

30. Characteristic equation: r2 + 103r +1C

= 0; roots: r =−103 ±

√106 − 4/C2

.

(a) r = 100(−5 ±√

5); y = C1e100(−5+

√5)t + C2e

100(−5−√

5)t

(b) r = −500; y = C1e−500t + C2te

−500t

(c) r = 500(−1 ± i); y = e−500t (C1 cos 500t + C2 sin 500t)



SECTION 9.3 509

31. (a) The solutions y1 = e2x, y2 = e−4x imply that the roots of the characteristic equation

are r1 = 2, r2 = −4. Therefore, the characteristic equation is:

(r − 2)(r + 4) = r2 + 2r − 8 = 0

and the differential equation is: y′′ + 2y′ − 8y = 0.

(b) The solutions y1 = 3e−x, y2 = 4e5x imply that the roots of the characteristic equation

are r1 = −1, r2 = 5. Therefore, the characteristic equation is

(r + 1)(r − 5) = r2 − 4r − 5 = 0

and the differential equation is: y′′ − 4y′ − 5y = 0.

(c) The solutions y1 = 2e3x, y2 = xe3x imply that 3 is the only root of the characteristic

equation. Therefore, the characteristic equation is

(r − 3)2 = r2 − 6r + 9 = 0

and the differential equation is: y′′ − 6y′ + 9y = 0.

32. (a) We want r = ±2i, so r2 = −4. Differential equation: y′′ + 4y = 0

(b) We want r = −2 ± 3i, so (r + 2)2 = −9. Differential equation: y′′ + 4y′ + 13y = 0

33. (a) Let y = eαxu. Then

y′ = αeαxu + eαxu′ and y′′ = α2eαxu + 2αeαxu′ + eαxu′′

Now,

y′′ − 2αy + α2y =(α2eαxu + 2αeαxu′ + eαxu′′) − 2α (αeαxu + eαxu′) + α2eαxu

= eαxu′′

Therefore, y′′ − 2αy + α2y = 0 =⇒ eαxu′′ =⇒ u′′ = 0.

(b) y′′ − 2αy′ +(α2 + β2

)y = y′′ − 2αy′ + α2y + β2y.

From part (a) y = eαxu =⇒ y′′ − 2αy′ + α2y = eαxu′′. Therefore,

y′′ − 2αy′ +(α2 + β2

)y = 0 =⇒ eαxu′′ + β2eαxu = 0 =⇒ u′′ + β2u = 0.

34. r2 + ar + b = 0 =⇒ r1, r2 =−a±

√a2 − 4b

2.

If a2 − 4b > 0, then√a2 − 4b < a, so

−a±√a2 − 4b

2is negative, and the solutions:

y = C1er1x + C2e

r2x → 0 as x → ∞.

If a2 − 4b = 0, then r = r1 = r2 = −a/2 < 0, and the solutions:

y = C1erx + C2xe

rx → 0 as x → ∞.



510 SECTION 9.3

If a2 − 4b < 0, then y = e−ax/2(C1 cos 1

2

√b2 − 4a x + C2 sin 1

2

√b2 − 4a x

)satisfies

|y| < e−ax/2 =⇒ y → 0 as x → ∞.

35. (a) If a = 0, b > 0, then the general solution of the differential equation is:

y = C1 cos√b x + C2 sin

√b x = A cos

(√b x + φ

)where A and φ are constants. Clearly | y(x) | ≤ |A | for all x.

(b) If a > 0, b = 0, then the general solution of the differential equation is:

y = C1 + C2e−ax and lim

x→∞y(x) = C1.

The solution which satisfies the conditions: y(0) = y0, y′(0) = y1 is:

y = y0 +y1

a− y1

ae−ax and lim

x→∞y(x) = y0 +

y1

a; k = y0 +

y1

a.

36. Let y1 and y2 be solutions of the homogeneous equation.

Suppose that y2 = k y1 for some scalar k. Then

W (y1, y2) =

∣∣∣∣∣y1 k y1

y′1 k y′1

∣∣∣∣∣ = 0

Now suppose that W (y1, y2) = 0, and suppose that y1 is not identically 0. Let I be an interval on

which y1(x) �= 0. Then,

(y2

y1

)′=

y1y′2 − y2y

′1

(y1)2=

W (y1, y2)(y1)2

= 0

Therefore,y2

y1= k constant on I. Finally, y2 = ky1 on I implies y2 = ky1 for all x by the uniqueness

theorem.

37. Let W be the Wronskian of y1 and y2. Then

W (a) =

∣∣∣∣∣ 0 0

y′1(a) y′2(a)

∣∣∣∣∣ = 0

Therefore one of the solutions is a multiple of the other (see the Supplement to this Section).

38. From the hint,dy

dx=

dy

dz

1x. Differentiating with respect to x again, we have

d2y

dx2=

d2y

dz2

dz

dx

1x

+dy

dz

(− 1x2

)=

1x2

(d2y

dz2− dy

dz

).



SECTION 9.3 511

Substituting into the differential equation x2y′′ + αxy′ + βy = 0, we get

(d2y

dz2− dy

dz

)+ α

dy

dz+ βy = 0, or

d2y

dz2+ a

dy

dz+ by = 0,

where a = α− 1, b = β.

39. From Exercise 38, the change of variable z = lnx transforms the equation

x2y′′ − xy′ − 8y = 0

into the differential equation with constant coefficients

d2y

dz2− 2

dy

dz− 8y = 0.

The characteristic equation is:

r2 − 2r − 8 = 0 or (r − 4)(r + 2) = 0

The roots are: r = 4, r = −2, and the general solution (in terms of z) is:

y = C1e4z + C2e

−2z.

Replacing z by lnx we get

y = C1e4 lnx + C2e

−2 lnx = C1x4 + C2x

−2.

40. Using the result of Exercise 38, we get

d2y

dz2− 3

dy

dz+ 2y = 0, so r2 − 3r + 2 = 0 =⇒ r = 1, 2.

=⇒ y = C1ez + C2e

2z. Substituting z = lnx, we get y = C1x + C2x2.

41. From Exercise 38, the change of variable z = lnx transforms the equation

x2y′′ − 3xy′ + 4y = 0

into the differential equation with constant coefficients

d2y

dz2− 4

dy

dz+ 4y = 0.

The characteristic equation is:

r2 − 4r + 4 = 0 or (r − 2)2 = 0.

The only root is: r = 2, and the general solution (in terms of z) is:

y = C1e2z + C2ze

2z.

Replacing z by lnx we get

y = C1e2 lnx + C2 lnx e2 lnx = C1x

2 + C2x2 lnx.

42. From Exercise 38, we getd2y

dz2− 2

dy

dz+ 5y = 0

r2 − 2r + 5 = 0 =⇒ r = 1 ± 2i; and y = ez (C1 cos 2z + C2 sin 2z) .

Substituting z = lnx we get: y = x [C1 cos(2 lnx) + C2 sin(2 lnx)] .



512 REVIEW EXERCISES

REVIEW EXERCISES

1. First calculate the integrating factor eH(x):

H(x) =∫

1dx = x and eH(x) = ex

Multiplication by ex gives

exy′ + exy = 2e−x which isd

dx(exy) = 2e−x

Integrating this equation, we get

exy = −2e−x + C

and

y = −2e−2x + Ce−x

2. The equation can be written

−2 cos(2x) +2y2 + 1

yy′ = 0.

The equation is separable:∫−2 cos(2x)dx +

∫2y2 + 1

ydy = C and − sin(2x) + y2 + ln |y| = C


cos2 xdx− y

y2 + 1dy = 0.

The equation is separable:∫cos2 xdx−

∫y

y2 + 1dy = C and

14

sin(2x) +12x− 1

2ln(y2 + 1) = C

or sin(2x) + 2x− 2 ln(y2 + 1) = C.


xex − (y ln y)y′ = 0.

The equation is separable:∫xexdx−

∫y ln ydy = C and ex(x− 1) − 1

2y2 ln y +

14y2 + C


y′ +3xy =

sin 2xx2

.



REVIEW EXERCISES 513

Calculate the integrating factor eH(x):

H(x) =∫

3xdx = lnx3 and eH(x) = x3.

Multiplying by x3 gives

x3y′ + 3x2y = x sin 2x which isd

dx(x3y) = x sin 2x.


x3y =∫

x sin 2xdx + C = −12x cos 2x +

14

sin 2x + C.

and

y = − 12x2

cos 2x +1

4x3sin 2x +

C

x3.


y′ +2xy =

2xex

2.


H(x) =∫


Multiplication by x2 gives

x2y′ + 2xy = 2xex2

which isd

dx(x2y) = 2xex

2.

Integrating the equation, we get

x2y = ex2+ C.

and

y =1x2

(ex2+ C).


1 + x2 − 11 + y2

y′ = 0.

The equation is separable:∫(1 + x2)dx−

∫1

1 + y2dy = C and x +

x3

3− arctan y = C

or arctan y = x + x3

3 + C.


x2 − 1 − y + 1y

y′ = 0




The equation is separable:∫(x2 − 1)dx−

∫y + 1y

dy = C and13x3 − x− y − ln |y| = C.


y′ +2xy = x2.


H(x) =∫


Multiplication by x2 gives

x2y′ + 2xy = x4 which isd

dx(x2y) = x4.


x2y =15x5 + C and y =

15x3 +

C

x2.


x√

1 + x2 − 1y2

y′ = 0

The equation is separable:∫x√

1 + x2dx−∫

1y2

dy = C and13(1 + x2)3/2 +

1y

= C.

Solving for y, we have

y =−3

(1 + x2)3/2 + C.


y′ +1xy =

2x2

+ 1

The integrating factor is

eH(x) = elnx = x.

Multiplication by x gives

xy′ + y =2x

+ x which isd

dx(xy) =

2x

+ x.


xy = lnx2 +12x2 + C and y =

1x

(lnx2 +12x2 + C).

Applying the initial condition y(1) = 2, we have

ln 1 +12

+ C = 2 and C =32.




Therefore

y =1x

(lnx2 +

12x2 +

32

).


4x− y√y2 + 1

y′ = 0.

The equation is separable:∫4xdx−

∫y√

y2 + 1dy = C and 2x2 − (y2 + 1)1/2 = C.

To find the solution that satisfies y(0) = 1, we set x = 0, y = 1 and solve for C:

C = −(1 + 1)12 = −

√2.

Therefore 2x2 − (y2 + 1)1/2 +√

2 = 0 is the solution.


e2x +1

2y − 1y′ = 0.

The equation is separable:∫e2xdx +

∫1

2y − 1dy = C and

12e2x +

12

ln |2y − 1| = C.

Solving for y, we get

y =12

+ Ce−e2x .

To find the solution that satisfies y(0) = 12 + 1

e , we set x = 0, y = 12 + 1

e and solve for C.

We have C = 1. Therefore y = 12 + e−e2x .


tanx− cos yy′ = 0.

The equation is separable:∫tanxdx−

∫cos ydy = C and ln | secx| − sin y = C.

Applying the initial condition: y(0) = π2 , we have C = −1. Therefore sin y = ln | secx| + 1.

15. The characteristic equation is

r2 − 2r + 2 = 0.




The roots are: r = 1 ± i.

The general solution is

y = C1ex cosx + C2e

x sinx


r2 + r +14

= 0.

The roots are: r = − 12 with multiplicity 2.


y = C1e−x/2 + C2xe

−x/2


r2 − r − 2 = 0.

The roots are: r = 2,−1.


y = C1e2x + C2e

−x


r2 − 4r = 0.

The roots are: r = 0, 4. The general solution is

y = C1 + C2e4x


r2 − 6r + 9 = 0.

The roots are: r = 3 with multiplicity 2.


y = C1e3x + C2xe

3x


r2 + 4 = 0

The roots are: r = ±2i.





y = C1 cos 2x + C2 sin 2x


r2 + 4r + 13 = 0

The roots are: r = −2 ± 3i.


y = e−2x(C1 cos 3x + C2 sin 3x)


3r2 − 5r − 2 = 0.

The roots are: r = 2,− 13 .


y = C1e2x + C2e

−x/3.


r2 − r = 0.

The roots are: r = 0, 1.


y = C1 + C2ex.

Applying the initial conditions y(0) = 1 and y′(0) = 0, we have

C1 + C2 = 1, C2 = 0 =⇒ C1 = 1.

The solution of the initial-value problem is: y = 1.


r2 + 7r + 12 = 0.

The roots are: r = −3,−4.


y = C1e−3x + C2e

−4x.




Applying the initial conditions y(0) = 2, y′(0) = 8, we have

C1 + C2 = 2, −3C1 − 4C2 = 8 =⇒ C1 = 16, C2 = −14.

The solution of the initial-value problem is: y = 16e−3x − 14e−4x.


r2 − 6r + 13 = 0.

The roots are: r = 3 ± 2i.


y = e3x(C1 cos 2x + C2 sin 2x).

Applying the initial conditions y(0) = 2, y′(0) = 2, we have

C1 = 2, 3C1 + 2C2 = 2 =⇒ C1 = 2, C2 = −2.

The solution of the initial-value problem is: y = e3x(2 cos 2x− 2 sin 2x).


r2 + 4r + 4 = 0.

The roots are: r = −2 with multiplicity 2.


y = C1e−2x + C2xe

−2x.

Applying the initial conditions y(−1) = 2, y′(−1) = 1, we have

C1e2 − C2e

2 = 2, −2C2e2 + 3C2e

2 = 1 =⇒ C1 = 7e−2, C2 = 5e−2.

The solution of the initial-value problem is: y = 7e−2x−2 + 5xe−2x−2.

27. Curves: y = Ce2x; y′ = 2Ce2x =⇒ y′ = 2y

Orthogonal trajectories: y′ = − 12y

;∫

2y dy = −∫

dx; y2 = C − x.

28. Curves: y =C

1 + x2; y′ = − 2xC

(1 + x2)2=⇒ y′ = − 2xy

1 + x2

Orthogonal trajectories: y′ =1 + x2

2xy;

∫2y dy =

∫1 + x2

xdx =

∫ (1x

+ x

)dx;

and

y2 = ln |x| + 12x2 + C




29. Substituting y = xr into the equation, we get

r(r − 1)xr + 4rxr + 2xr = 0 or xr(r2 + 3r + 2) = 0 =⇒ r2 + 3r + 2 = (r + 2)(r + 1) = 0.

The solutions are: r = −1,−2.

30. Substituting y = xr into the equation, we get

r(r − 1)xr − rxr − 8xr = 0 or xr(r2 − 2r − 8) = 0 =⇒ r2 − 2r − 8 = (r + 2)(r − 4) = 0.

The solutions are: r = 4,−2.

31. Let y(t) be the value of the business (measured in millions) at time t. Then y(t) satisfies

dy

dt= ky2

The general solution of this equation is

y(t) =1

−kt + C

Applying the given conditions, y(0) = 1 and y(1) = 1.5, to find C and k, we get C = 1 and

k = 1/3.

1 year from now the business will be worth:

y(2) =1

− 13 + 1

= 3 million.

1.5 years from now the business will be worth:

y(2.5) =1

− 13

(52

)+ 1

= 6 million.

2 years from now the business will be worth:

y(3) =1

− 13 (3) + 1

= ∞.

Obviously the business cannot continue to grow at a rate proportional to its value squared.

32. Let y(t) be the value of the business (measured in millions) at the time t. Then y(t) satisfies

dy

dt= k

√y


y(t) =(k

2t + C

)2

Applying the given conditions, y(0) = 1 and y(2) = 1.44, to find C and k, we get C = 1 and

k = 0.2 so

y(t) =(

110 t + 1

)2.




5 years from now the business will be worth:

y(7) = (0.7 + 1)2 = 2.89 million.

Solve

y(t) =(

t

10+ 1

)2

= 4

t = 10. The business will be worth 4 million 8 years from now.

33. (a) The general solution of the differential equation is

y =a

b+ Ce−bt.

Applying the initial condition y(0) = 0, we get C = −a

b, and

y =a

b(1 − e−bt)

(b) limt→∞

y(t) =a

b

(c) Setting y = 0.9a

b, we have

0.9a

b=

a

b(1 − e−bt).

The solution to this equation is t =ln 10b

hours.

34. Let T (t) be the temperature of the bar at time t. It follows from Newton’s law of cooling that

T (t) = τ + Ce−kt.

By the conditions given in the problem, we have

τ = 0, T (0) = 100, T (20) = 50.

Applying these conditions, we get C = 100 and k =ln 220

. Therefore

T (t) = 100e−(t/20) ln 2 = 100(2)−t/20.

(a) T (30) = 100(2)−3/2 ∼= 35.4◦.

(b) Solve 25 = 100(2)−t/20 for t:

25 = 100(2)−t/20, − t

20ln 2 = ln(1/4), t =

20 ln 4ln 2

= 40.

It will take 40 minutes for the bar to reach 25◦.

35. Let T (t) be the temperature of the object at time t. It follows from Newton’s law of cooling that

T (t) = τ + Ce−kt.




By the conditions given in the problem, we have

τ = 70, T (10) = 20, T (20) = 35.

Applying these conditions, we get

C = −5007

and k =110

ln(10/7).

(a) The temperature of the object at time t is

T = 70 − 5007

e−(t/10) ln(10/7) = 70 − 5007

(710

)t/10

(b) T (0) = 70 − 5007

= −107

36. (a) Let T be the length of time needed to fill the tank. Then

600 + (6 − 4)T = 1200

and T = 300 minutes.

(b) Let S(t) be the amount of salt dissolved in the tank at time t. Then

dS

dt= rate in − rate out =

12× 6 − S

600 + 2t× 4 = 3 − 2S

300 + t

The equation can be rewritten

S′ +2S

300 + t= 3.

The general solution to this equation is

S = 300 + t +C

(300 + t)2.

Since S(0) = 40, we get C = −3002 · 260. Therefore, the amount of salt in the tank at time t is:

S(t) = 300 + t− 3002 · 260(300 + t)2

(c) S(T ) = 300 + 300 − 3002 · 260(300 + t)2

= 535 pounds.

37. (a) Let T be the length of time needed to empty the tank. Then

80 − (8 − 4)T = 0 and T = 20 minutes.

(b) Let S(t) be the amount of salt in the tank at the time t. Then

dS

dt= 1 × 4 − S

80 − 4t× 8 = 4 − 2S

20 − t, S(0) =

18× 80 = 10

The solution to this initial-value problem is

S(t) = 4(20 − t) − 740

(20 − t)2 (1)




(c) Let t0 be the time that the tank contains exactly 40 gallons. Then

80 − 4t0 = 40 and t0 = 10.

Substituting t = 10 into (1), we get S(10) = 22.5 pounds.

38. (a) The differential equation is separable and can be written as

dP

P (10−1 − 10−5P )= dt.

With the initial condition P (0) = 2000, the solution is

P (t) =2500e0.1t

1 + 0.25e0.1t.

Then

limt→∞

P (t) = 104.

(b) Setting P = 0.9 × 104, we get

0.9 × 104 =2500e0.1t

1 + 0.25e0.1t.

The solution to this equation is t = 35.835 ∼= 36 months.

39. Let P (t) be the number of people that have heard the rumor at time t. Then P (t) satisfies:

dP

dt= kP (20, 000 − P )


P (t) =20, 000

1 + Ce−20,000kt.

Now, P (0) =20, 0001 + C

= 500 =⇒ C = 39;

P (10) =20, 000

1 + 39e−20,000(10k)= 1200 =⇒ 20, 000k ∼= −.0912.

Therefore, P (t) =20, 000

1 + 39e−0.09120t.

(a) P (20) =20, 000

1 + 39e−0.0912(20)∼= 2742

(b) The rumor will be spreading fastest when the number of people who have heard it is equal to the

number of people who have not heard it:

20, 0001 + 39e−0.0912t

= 10, 000.

The solution of this equation is: t ∼= 40 days.




40. (a) From (2),

du√1 + u2

=1adx

Integrating, we get

ln∣∣∣u +

√1 + u2

∣∣∣ =x

a+ C.

Applying the initial condition y′(0) = u(0) = 0 =⇒ C = 0. Thus, ln∣∣u +

√1 + u2

∣∣ =x

a.

(b) Set u = y′:

ln∣∣∣y′ + √

1 + (y′)2∣∣∣ =

x

a, y′ +

√1 + (y′)2 = ex/a and

√1 + (y′)2 = ex/a − y′.

Squaring both sides and simplyfying, we get

y′ =ex/a − e−x/a

2= sinh (x/a) =⇒ y = a cosh (x/a) + C.

Applying the initial condition y(0) = a, we get C = 0.

Therefore y(x) = a cosh(x/a), a catenary.

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Calculus one and several variables 10E Salas solutions manual ch09

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