Chapter 09

January 27, 2005 11:46 L24-ch09 Sheet number 1 Page number 395 black

CHAPTER 9
Mathematical Modeling with Differential
Equations

EXERCISE SET 9.1
3
1. y = 9x2 ex = 3x2 y and y(0) = 3 by inspection.

2. y = x3 − 2 sin x, y(0) = 3 by inspection.

dy dy
3. (a) ﬁrst order; = c; (1 + x) = (1 + x)c = y
dx dx
(b) second order; y = c1 cos t − c2 sin t, y + y = −c1 sin t − c2 cos t + (c1 sin t + c2 cos t) = 0

dy c
4. (a) ﬁrst order; 2 + y = 2 − e−x/2 + 1 + ce−x/2 + x − 3 = x − 1
dx 2
(b) second order; y = c1 et − c2 e−t , y − y = c1 et + c2 e−t − c1 et + c2 e−t = 0

1 dy dy dy dy y3
5. = y 2 + 2xy , (1 − 2xy 2 ) = y 3 , =
y dx dx dx dx 1 − 2xy 2

dy
6. 2x + y 2 + 2xy = 0, by inspection.
dx
d
7. (a) IF: µ = e3 dx
ye3x = 0, ye3x = C, y = Ce−3x
= e3x ,
dx
dy
separation of variables: = −3dx, ln |y| = −3x + C1 , y = ±e−3x eC1 = Ce−3x
y
including C = 0 by inspection
d
(b) IF: µ = e−2 dt
[ye−2t ] = 0, ye−2t = C, y = Ce2t
= e−2t ,
dt
dy
separation of variables: = 2dt, ln |y| = 2t + C1 , y = ±eC1 e2t = Ce2t
y

d
ye−2x = 0, y = Ce2x
2 2
8. (a) IF: µ = e−4 = e−2x ,
2
x dx
dx
dy 2 2
separation of variables: = 4x dx, ln |y| = 2x2 + C1 , y = ±eC1 e2x = Ce2x
y
d
(b) IF: µ = e dt
yet = 0, y = Ce−t
= et ,
dt
dy
separation of variables: = −dt, ln |y| = −t + C1 , y = ±eC1 e−t = Ce−t
y

9. µ = e 4dx
= e4x , e4x y = ex dx = ex + C, y = e−3x + Ce−4x

d 1 2 1
yex = xex , yex = ex + C, y = + Ce−x
2 2 2 2 2
10. µ = e2 x dx
= ex ,
dx 2 2

395


396 Chapter 9

11. µ = e dx
= ex , ex y = ex cos(ex )dx = sin(ex ) + C, y = e−x sin(ex ) + Ce−x

dy 1 1 2x 1 1
12. + 2y = , µ = e 2dx
= e2x , e2x y = e dx = e2x + C, y = + Ce−2x
dx 2 2 4 4

dy x 2 1 2
13. + y = 0, µ = e (x/(x +1))dx = e 2 ln(x +1) = x2 + 1,
dx x2 + 1
d C
y x2 + 1 = 0, y x2 + 1 = C, y = √
dx x2 + 1

dy 1 ex
14. +y = − ,µ=e dx
= ex , ex y = − dx = ln |1−ex |+C, y = e−x ln |1−ex |+Ce−x
dx 1 − ex 1 − ex

1 1 y y
15. dy = dx, ln |y| = ln |x| + C1 , ln = C1 , = ±eC1 = C, y = Cx
y x x x

dy
16. = 2x dx, tan−1 y = x2 + C, y = tan x2 + C
1 + y2

dy x √ √
= −√ dx, ln |1 + y| = − 1 + x2 + C1 , 1 + y = ±e− 1+x eC1 = Ce− 1+x ,
2 2
17.
1+y 1 + x2
√
y = Ce− 1+x2
− 1, C = 0

x3 dx y 2 1
18. y dy = , = ln(1 + x4 ) + C1 , 2y 2 = ln(1 + x4 ) + C, y = ± [ln(1 + x4 ) + C]/2
1 + x4 2 4

2(1 + y 2 )
19. dy = ex dx, 2 ln |y| + y 2 = ex + C; by inspection, y = 0 is also a solution
y

dy
= −x dx, ln |y| = −x2 /2 + C1 , y = ±eC1 e−x /2 = Ce−x /2 , including C = 0 by inspection
2 2
20.
y

sin x
21. ey dy = dx = sec x tan x dx, ey = sec x + C, y = ln(sec x + C)
cos2 x

dy x2
22. = (1 + x) dx, tan−1 y = x + + C, y = tan(x + x2 /2 + C)
1 + y2 2

dy dx 1 1 y−1
23. = , − + dy = csc x dx, ln = ln | csc x − cot x| + C1 ,
y2 −y sin x y y−1 y
y−1 1
= ±eC1 (csc x − cot x) = C(csc x − cot x), y = , C = 0;
y 1 − C(csc x − cot x)
by inspection, y = 0 is also a solution, as is y = 1.

1
24. dy = cos x dx, ln |y| = sin x + C, y = C1 esin x
y


Exercise Set 9.1 397

dy 1 (1/x)dx d 1
25. + y = 1, µ = e = eln x = x, [xy] = x, xy = x2 + C, y = x/2 + C/x
dx x dx 2
1 3
(a) 2 = y(1) = + C, C = , y = x/2 + 3/(2x)
2 2
(b) 2 = y(−1) = −1/2 − C, C = −5/2, y = x/2 − 5/(2x)

dy x2 2 2
26. = x dx, ln |y| = + C1 , y = ±eC1 ex /2 = Cex /2
y 2
2
(a) 1 = y(0) = C so C = 1, y = ex /2

1 1 2
(b) = y(0) = C, so y = ex /2
2 2

= e−x , e−x y = 2xe−x dx = −e−x + C,
2 2 2
27. µ = e−2
2
x dx

2 2
y = −1 + Cex , 3 = −1 + C, C = 4, y = −1 + 4ex

28. µ = e dt
= et , et y = 2et dt = 2et + C, y = 2 + Ce−t , 1 = 2 + C, C = −1, y = 2 − e−t

29. (2y + cos y) dy = 3x2 dx, y 2 + sin y = x3 + C, π 2 + sin π = C, C = π 2 ,
y 2 + sin y = x3 + π 2

dy 1
30. = (x + 2)ey , e−y dy = (x + 2)dx, −e−y = x2 + 2x + C, −1 = C,
dx 2
1 2 1 2 1
−e−y = x + 2x − 1, e−y = − x − 2x + 1, y = − ln 1 − 2x − x2
2 2 2

31. 2(y − 1) dy = (2t + 1) dt, y 2 − 2y = t2 + t + C, 1 + 2 = C, C = 3, y 2 − 2y = t2 + t + 3

sinh x (sinh x/ cosh x)dx
32. y + y = cosh x, µ = e = eln cosh x = cosh x,
cosh x
1 1 1 1 1
(cosh x)y = cosh2 x dx =
(cosh 2x + 1)dx = sinh 2x + x + C = sinh x cosh x + x + C,
2 4 2 2 2
1 1 1 1 1 1
y = sinh x + x sech x + C sech x, = C, y = sinh x + x sech x + sech x
2 2 4 2 2 4

dy dx 1
33. (a) = , ln |y| = ln |x| + C1 , x = –0.5y2
2
y
y 2x 2 x = –1.5y2
x = y2 x = 2y2
|y| = C|x|1/2 , y 2 = Cx;
x = 2.5y2
by inspection y = 0 is also a solution. x = – 3y2 y=0 x
–2 2
2 2
(b) 1 = C(2) , C = 1/4, y = x/4

–2


398 Chapter 9

y
y2 x2 3 y = √9 – x 2
34. (a) y dy = −x dx, = − + C1 , y = ± C 2 − x2
2 2 y = √2.25 – x 2
√
(b) y = 25 − x2 y = √0.25 – x 2
x
–3 3
y = – √1 – x 2
y = – √4 – x 2
–3
y = – √6.25 – x 2

dy x dx
35. =− 2 , 36. y + 2y = 3et , µ = e2 dt = e2t ,
y x +4
d
1 ye2t = 3e3t , ye2t = e3t + C,
ln |y| = − ln(x2 + 4) + C1 , dt
2
C y = et + Ce−2t
y=√
x2 + 4 100
C=2
1.5
C=1
C=0
–2 2
C=2
C=0 C=1
C = –1
–2 2 C = –2
C = –1
C = –2 –100

–1

1 y2
37. (1 − y 2 ) dy = x2 dx, 38. +y dy = dx, ln |y| + = x + C1 ,
y 2
y3 x3 2
y− = + C1 , x3 + y 3 − 3y = C yey /2
= ±eC1 ex = Cex including C = 0
3 3
y y
3
2

x
–5 5
x
–2 2 –3

–2

1 1
39. Of the solutions y = , all pass through the point 0, − and thus never through (0, 0).
2x2 − C C
A solution of the initial value problem with y(0) = 0 is (by inspection) y = 0. The methods of
Example 4 fail because the integrals there become divergent when the point x = 0 is included in
the integral.

1 1
40. If y0 = 0 then, proceeding as before, we get C = 2x2 − , C = 2x2 − , and
0
y y0
1
y= 2 − 2x2 + 1/y
, which is deﬁned for all x provided 2x2 is never equal to 2x2 − 1/y0 ; this
0
2x 0 0
last condition will be satisﬁed if and only if 2x0 − 1/y0 < 0, or 0 < 2x0 y0 < 1. If y0 = 0 then y = 0
2 2

is, by inspection, also a solution for all real x.



dy x2
41. = xe−y , ey dy = x dx, ey = + C, x = 2 when y = 0 so 1 = 2 + C, C = −1, ey = x2 /2 − 1
dx 2

dy 3x2
42. = , 2y dy = 3x2 dx, y 2 = x3 + C, 1 = 1 + C, C = 0, 2
dx 2y
y 2 = x3 , y = x3/2 passes through (1, 1).

0 1.6
0

dy
43. = rate in − rate out, where y is the amount of salt at time t,
dt
dy y 1 dy 1
= (4)(2) − (2) = 8 − y, so + y = 8 and y(0) = 25.
dt 50 25 dt 25
(1/25)dt
µ=e = et/25 , et/25 y = 8et/25 dt = 200et/25 + C,

y = 200 + Ce−t/25 , 25 = 200 + C, C = −175,

(a) y = 200 − 175e−t/25 oz (b) when t = 25, y = 200 − 175e−1 ≈ 136 oz

dy y 1 dy 1
44. = (5)(20) − (20) = 100 − y, so + y = 100 and y(0) = 0.
dt 200 10 dt 10
1
µ=e 10 dt = et/10 , et/10 y = 100et/10 dt = 1000et/10 + C,

y = 1000 + Ce−t/10 , 0 = 1000 + C, C = −1000;

(a) y = 1000 − 1000e−t/10 lb (b) when t = 30, y = 1000 − 1000e−3 ≈ 950 lb

45. The volume V of the (polluted) water is V (t) = 500 + (20 − 10)t = 500 + 10t;
if y(t) is the number of pounds of particulate matter in the water,
dy y 1 dy 1 dt
then y(0) = 50, and = 0 − 10 = − y, + y = 0; µ = e 50+t = 50 + t;
dt V 50 + t dt 50 + t
d
[(50 + t)y] = 0, (50 + t)y = C, 2500 = 50y(0) = C, y(t) = 2500/(50 + t).
dt
The tank reaches the point of overﬂowing when V = 500 + 10t = 1000, t = 50 min, so
y = 2500/(50 + 50) = 25 lb.

46. The volume of the lake (in gallons) is V = 264πr2 h = 264π(15)2 3 = 178,200π gals. Let y(t) denote
dy y y
the number of pounds of mercury salts at time t, then = 0 − 103 = − lb/h and
dt V 178.2π
dy dt t
y0 = 10−5 V = 1.782π lb; =− , ln y = − + C1 , y = Ce−t/(178.2π) , and
y 178.2π 178.2π
C = y(0) = y0 10−5 V = 1.782π, y = 1.782πe−t/(178.2π) lb of mercury salts.

t 1 2 3 4 5 6 7 8 9 10 11 12
y(t) 5.588 5.578 5.568 5.558 5.549 5.539 5.529 5.519 5.509 5.499 5.489 5.480


400 Chapter 9

dv c d gm ct/m
47. (a) + v = −g, µ = e(c/m) dt = ect/m , vect/m = −gect/m , vect/m = − e + C,
dt m dt c
gm gm gm gm gm −ct/m
v=− +Ce−ct/m , but v0 = v(0) = − +C, C = v0 + ,v = − + v0 + e
c c c c c
mg
(b) Replace with vτ and −ct/m with −gt/vτ in (23).
c
vτ
(c) From Part (b), s(t) = C − vτ t − (v0 + vτ ) e−gt/vτ ;
g
vτ vτ vτ
s0 = s(0) = C −(v0 +vτ ) , C = s0 +(v0 +vτ ) , s(t) = s0 −vτ t+ (v0 +vτ ) 1 − e−gt/vτ
g g g

48. Given m = 240, g = 32, vτ = mg/c: with a closed parachute vτ = 120 so c = 64, and with an open
parachute vτ = 24, c = 320.
(a) Let t denote time elapsed in seconds after the moment of the drop. From Exercise 47(b),
while the parachute is closed
v(t) = e−gt/vτ (v0 + vτ ) − vτ = e−32t/120 (0 + 120) − 120 = 120 e−4t/15 − 1 and thus
v(25) = 120 e−20/3 − 1 ≈ −119.85, so the parachutist is falling at a speed of 119.85 ft/s
120
when the parachute opens. From Exercise 47(c), s(t) = s0 − 120t + 120 1 − e−4t/15 ,
32
s(25) = 10000 − 120 · 25 + 450 1 − e−20/3 ≈ 7449.43 ft.

(b) If t denotes time elapsed after the parachute opens, then, by Exercise 47(c),
24
s(t) = 7449.43 − 24t + (−119.85 + 24) 1 − e−32t/24 = 0, with the solution (Newton’s
32
Method) t = 307.4 s, so the sky diver is in the air for about 25 + 307.4 = 332.4 s.

dI R V (t) d Rt/L V (t) Rt/L
49. + I= , µ = e(R/L) dt
= eRt/L , (e I) = e ,
dt L L dt L
t t
1 1 −Rt/L
IeRt/L = I(0) + V (u)eRu/L du, I(t) = I(0)e−Rt/L + e V (u)eRu/L du.
L 0 L 0
t t
1 −2t
(a) I(t) = e 20e2u du = 2e−2t e2u = 2 1 − e−2t A.
5 0 0

(b) lim I(t) = 2 A
t→+∞

50. From Exercise 49 and Endpaper Table #42,
t t
1 e2u
I(t) = 15e−2t + e−2t 3e2u sin u du = 15e−2t + e−2t (2 sin u − cos u)
3 0 5 0

−2t 1 1
= 15e + (2 sin t − cos t) + e−2t .
5 5
dv ck
51. (a) = − g, v = −c ln(m0 − kt) − gt + C; v = 0 when t = 0 so 0 = −c ln m0 + C,
dt m0 − kt
m0
C = c ln m0 , v = c ln m0 − c ln(m0 − kt) − gt = c ln − gt.
m0 − kt
(b) m0 − kt = 0.2m0 when t = 100 so
m0
v = 2500 ln − 9.8(100) = 2500 ln 5 − 980 ≈ 3044 m/s.
0.2m0



dv dv dx dv dv dv
52. (a) By the chain rule, = = v so m = mv .
dt dx dt dx dt dx
mv m
(b) dv = −dx, ln(kv 2 + mg) = −x + C; v = v0 when x = 0 so
kv 2 + mg 2k
2
m m m m kv0 + mg
C= 2
ln(kv0 + mg), ln(kv 2 + mg) = −x + 2
ln(kv0 + mg), x = ln 2 .
2k 2k 2k 2k kv + mg
(c) x = xmax when v = 0 so
2
m kv0 + mg 3.56 × 10−3 (7.3 × 10−6 )(988)2 + (3.56 × 10−3 )(9.8)
xmax = ln = ln ≈ 1298 m
2k mg 2(7.3 × 10−6 ) (3.56 × 10−3 )(9.8)

dh √ π √
53. (a) A(h) = π(1)2 = π, π = −0.025 h, √ dh = −0.025dt, 2π h = −0.025t + C; h = 4 when
dt h
√ √ 0.025
t = 0, so 4π = C, 2π h = −0.025t + 4π, h = 2 − t, h ≈ (2 − 0.003979 t)2 .
2π
(b) h = 0 when t ≈ 2/0.003979 ≈ 502.6 s ≈ 8.4 min.

54. (a) A(h) = 6 2 4 − (h − 2)2 = 12 4h − h2 , 2√4 − (h − 2)2
dh √ √
12 4h − h2 = −0.025 h, 12 4 − h dh = −0.025dt,
dt
h−2
−8(4 − h)3/2
= −0.025t + C; h = 4 when t = 0 so C = 0, 2

(4 − h) 3/2
= (0.025/8)t, 4 − h = (0.025/8)
2/3 2/3
t , h

h ≈ 4 − 0.021375t2/3 ft
8
(b) h = 0 when t = (4 − 0)3/2 = 2560 s ≈ 42.7 min
0.025
dv 1 1 1 1 1 1
55. = − v 2 , 2 dv = − dt, − = − t + C; v = 128 when t = 0 so − = C,
dt 32 v 32 v 32 128
1 1 1 128 dx dx 128
− =− t− ,v = cm/s. But v = so = , x = 32 ln(4t + 1) + C1 ;
v 32 128 4t + 1 dt dt 4t + 1
x = 0 when t = 0 so C1 = 0, x = 32 ln(4t + 1) cm.

dv √ 1 √
56. = −0.02 v, √ dv = −0.02dt, 2 v = −0.02t + C; v = 9 when t = 0 so 6 = C,
dt v
√ dx dx
2 v = −0.02t + 6, v = (3 − 0.01t)2 cm/s. But v = so = (3 − 0.01t)2 ,
dt dt
100 100
x=− (3 − 0.01t)3 + C1 ; x = 0 when t = 0 so C1 = 900, x = 900 − (3 − 0.01t)3 cm.
3 3
dy
= − sin x + e−x , y(0) = 1.
2
57. Diﬀerentiate to get
dx
1 dP d
58. (a) Let y = [H(x) + C] where µ = eP (x) , = p(x), H(x) = µq, and C is an arbitrary
µ dx dx
constant. Then
dy 1 µ p
+ p(x)y = H (x) − 2 [H(x) + C] + p(x)y = q − [H(x) + C] + p(x)y = q
dx µ µ µ


402 Chapter 9

1
(b) Given the initial value problem, let C = µ(x0 )y0 −H(x0 ). Then y = [H(x)+C] is a solution
µ
of the initial value problem with y(x0 ) = y0 . This shows that the initial value problem has
a solution.
To show uniqueness, suppose u(x) also satisﬁes (5) together with u(x0 ) = y0 . Following the
1
arguments in the text we arrive at u(x) = [H(x) + C] for some constant C. The initial
µ
condition requires C = µ(x0 )y0 − H(x0 ), and thus u(x) is identical with y(x).

dH dy dH dG
59. Suppose that H(y) = G(x) + C. Then = G (x). But = h(y) and = g(x), hence
dy dx dy dx
y(x) is a solution of (10).

60. (a) y = x and y = −x are both solutions of the given initial value problem.

(b) y dy = − x dx, y 2 = −x2 + C; but y(0) = 0, so C = 0. Thus y 2 = −x2 , which is

impossible.

EXERCISE SET 9.2
y
1. y 1 = xy/4 2.
−2 ≤ x ≤ +2 4

−2 ≤ y ≤ +2 3
y
2
2
1
x
x
1 2 3 4
–2 2

–2

y
3. y(0) = 2
2

y(0) = 1

x
5

–1 y(0) = –1

dy y
4. + y = 1, µ = e dx = ex , 5.
dx 10
d
[yex ] = ex , y(1) = 1
dx y(–1) = 0
yex = ex + C, y = 1 + Ce−x x
-2 2
(a) −1 = 1 + C, C = −2, y = 1 − 2e−x y(0) = –1
(b) 1 = 1 + C, C = 0, y = 1
(c) 2 = 1 + C, C = 1, y = 1 + e−x –10



dy d
6. − 2y = −x, µ = e−2 dx = e−2x , ye−2x = −xe−2x ,
dx dx
1 1
ye−2x = (2x + 1)e−2x + C, y = (2x + 1) + Ce2x
4 4
1 1
(a) 1 = 3/4 + Ce2 , C = 1/(4e2 ), y = (2x + 1) + e2x−2
4 4
1 5
(b) −1 = 1/4 + C, C = −5/4, y = (2x + 1) − e2x
4 4
1 1
(c) 0 = −1/4 + Ce−2 , C = e2 /4, y = (2x + 1) + e2x+2
4 4

+∞ if y0 ≥ 1/4
7. lim y = 1 8. lim y =
x→+∞ x→+∞ −∞, if y0 < 1/4

9. (a) IV, since the slope is positive for x > 0 and negative for x < 0.
(b) VI, since the slope is positive for y > 0 and negative for y < 0.
(c) V, since the slope is always positive.
(d) II, since the slope changes sign when crossing the lines y = ±1.
(e) I, since the slope can be positive or negative in each quadrant but is not periodic.
(f ) III, since the slope is periodic in both x and y.

11. (a) y0 = 1, n 0 1 2 3 4 5
yn+1 = yn + (xn + yn )(0.2) = (xn + 6yn )/5 xn 0 0.2 0.4 0.6 0.8 1.0
yn 1 1.20 1.48 1.86 2.35 2.98

d
(b) y − y = x, µ = e−x , ye−x = xe−x , xn 0 0.2 0.4 0.6 0.8 1.0
dx
ye−x = −(x + 1)e−x + C, 1 = −1 + C, y(xn) 1 1.24 1.58 2.04 2.65 3.44
C = 2, y = −(x + 1) + 2ex abs. error 0 0.04 0.10 0.19 0.30 0.46
perc. error 0 3 6 9 11 13
y
(c)
3

x
0.2 0.4 0.6 0.8 1

12. h = 0.1, yn+1 = (xn + 11yn )/10
n 0 1 2 3 4 5 6 7 8 9 10
xn 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
yn 1.00 1.10 1.22 1.36 1.53 1.72 1.94 2.20 2.49 2.82 3.19

In Exercise 11, y(1) ≈ 2.98; in Exercise 12, y(1) ≈ 3.19; the true solution is y(1) ≈ 3.44; so the
absolute errors are approximately 0.46 and 0.25 respectively.


404 Chapter 9

1/3
13. y0 = 1, yn+1 = yn + 1 yn
2
y
n 0 1 2 3 4 5 6 7 8 9
xn 0 0.5 1 1.5 2 2.5 3 3.5 4
yn 1 1.50 2.11 2.84 3.68 4.64 5.72 6.91 8.23

x
1 2 3 4

14. y0 = 1, yn+1 = yn + (xn − yn )/4
2

y
n 0 1 2 3 4 5 6 7 8
xn 2
0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
yn 1 0.75 0.67 0.68 0.75 0.86 0.99 1.12 1.24
1.5

1

0.5
x
0.5 1 1.5 2

1
15. y0 = 1, yn+1 = yn + cos yn
2
y
n 0 1 2 3 4
tn 0 0.5 1 1.5 2 3
yn 1 1.27 1.42 1.49 1.53

t
3

16. y0 = 0, yn+1 = yn + e−yn /10
y
n 0 1 2 3 4 5 6 7 8 9 10 1
tn 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
yn 0 0.10 0.19 0.27 0.35 0.42 0.49 0.55 0.60 0.66 0.71

t
1

1
17. h = 1/5, y0 = 1, yn+1 = yn + sin(πn/5)
5
n 0 1 2 3 4 5
tn 0 0.2 0.4 0.6 0.8 1.0
yn 1 1 1.12 1.31 1.50 1.62



dy
= e−x and y(0) = 0.
2
18. (a) By inspection,
dx
(b) yn+1 = yn + e−xn /20 = yn + e−(n/20) /20 and y20 = 0.7625. From a CAS, y(1) = 0.7468.
2 2

√
19. (b) y dy = −x dx, y 2 /2 = −x2 /2 + C1 , x2 + y 2 = C; if y(0) = 1 then C = 1 so y(1/2) = 3/2.
√
20. (a) y0 = 1, yn+1 = yn + ( yn /2)∆x
√
∆x = 0.2 : yn+1 = yn + yn /10; y5 ≈ 1.5489
√
∆x = 0.1 : yn+1 = yn + yn /20; y10 ≈ 1.5556
√
∆x = 0.05 : yn+1 = yn + yn /40; y20 ≈ 1.5590
dy 1 √
(c) √ = dx, 2 y = x/2 + C, 2 = C,
y 2
√
y = x/4 + 1, y = (x/4 + 1)2 ,
y(1) = 25/16 = 1.5625

21. (a) The slope field does not vary with t, hence along a given parallel line all values are equal
since they only depend on the height y.
(b) As in part (a), the slope field does not vary with t; it is independent of t.
d 1 dy d dy
(c) From G(y) − x = C we obtain (G(y) − x) = −1= C = 0, i.e. = f (y).
dx f (y) dx dx dx

dy √
22. (a) Separate variables: √ = dx, 2 y = x + C, y = (x/2 + C1 )2 is a parabola that opens up,
y
and is therefore concave up.
√
(b) A curve is concave up if its derivative is increasing, and y = y is increasing.

dy dy dy y 3 − 2xy
23. (a) By implicit differentiation, y 3 + 3xy 2 − 2xy − x2 = 0, = 2 .
dx dx dx x − 3xy 2
(b) If y(x) is an integral curve of the slope field in part (a), then
d
{x[y(x)]3 − x2 y(x)} = [y(x)]3 + 3xy(x)2 y (x) − 2xy(x) − x2 y (x) = 0, so the integral curve
dx
must be of the form x[y(x)]3 − x2 y(x) = C.
(c) x[y(x)]3 − x2 y(x) = 2

dy dy dy ey + yex
24. (a) By implicit differentiation, ey + xey + ex + yex = 0, =− y
dx dx dx xe + ex
(b) If y(x) is an integral curve of the slope field in part (a), then
d
{xey(x) + y(x)ex } = ey(x) + xy (x)ey(x) + y (x)ex + y(x)ex = 0 from part (a). Thus
dx
xey(x) + y(x)ex = C.
(c) Any integral curve y(x) of the slope field above satisfies xey(x) + y(x)ex = C; if it passes
through (1, 1) then e + e = C, so xey(x) + y(x)ex = 2e defines the curve implicitly.

25. Euler’s Method is repeated application of local linear approximation, each step dependent on the
previous step.

26. (a) For any n, yn is the value of the discrete approximation at the right endpoint, that, is an
approximation of y(1). By increasing the number of subdivisions of the interval [0, 1] one
might expect more accuracy, and hence in the limit y(1).


406 Chapter 9

1 n+1
(b) For a ﬁxed value of n we have, for k = 1, 2, . . . , n, yk = yk−1 + yk−1 = yk−1 In partic-
n n
2 n n
n+1 n+1 n+1 n+1
ular yn = yn−1 = yn−2 = . . . = y0 = . Consequently,
n n n n
n
n+1
lim yn = lim = e, which is the (correct) value y = ex .
n→+∞ n→+∞ n x=1

EXERCISE SET 9.3
dy dy
1. (a) = ky 2 , y(0) = y0 , k > 0 (b) = −ky 2 , y(0) = y0 , k > 0
dt dt

ds 1 d2 s ds
3. (a) = s (b) =2
dt 2 dt2 dt

2
dv d2 s ds
4. (a) = −2v 2 (b) = −2
dt dt2 dt

dy
5. (a) = 0.02y, y0 = 10,000 (b) y = 10,000e2t/100
dt
1
(c) T = ln 2 ≈ 34.657 h (d) 45,000 = 10,000e2t/100 ,
0.02
45,000
t = 50 ln ≈ 75.20 h
10,000

1 1
6. k = ln 2 = ln 2
T 20
dy
(a) = ((ln 2)/20)y, y(0) = 1 (b) y(t) = et(ln 2)/20 = 2t/20
dt
(c) y(120) = 26 = 64 (d) 1,000,000 = 2t/20 ,
ln 106
t = 20 ≈ 398.63 min
ln 2

dy 1 ln 2
7. (a) = −ky, y(0) = 5.0 × 107 ; 3.83 = T = ln 2, so k = ≈ 0.1810
dt k 3.83
(b) y = 5.0 × 107 e−0.181t
(c) y(30) = 5.0 × 107 e−0.1810(30) ≈ 219,000
ln 0.1
(d) y(t) = (0.1)y0 = y0 e−kt , −kt = ln 0.1, t = − = 12.72 days
0.1810
1 1 dy
8. (a) k = ln 2 = ln 2 ≈ 0.0050, so = −0.0050y, y0 = 10.
T 140 dt
(b) y = 10e−0.0050t
(c) 10 weeks = 70 days so y = 10e−0.35 ≈ 7 mg.
ln 0.3
(d) 0.3y0 = y0 e−kt , t = − ≈ 240.8 days
0.0050



1
9. 100e0.02t = 10,000, e0.02t = 100, t = ln 100 ≈ 230 days
0.02
1
10. y = 10,000ekt , but y = 12,000 when t = 5 so 10,000e5k = 12,000, k = ln 1.2. y = 20,000 when
5
ln 2 ln 2
2 = ekt , t = =5 ≈ 19, in the year 2017.
k ln 1.2
1 3.5 1
11. y(t) = y0 e−kt = 10.0e−kt , 3.5 = 10.0e−k(5) , k = − ln ≈ 0.2100, T = ln 2 ≈ 3.30 days
5 10.0 k
1
12. y = y0 e−kt , 0.7y0 = y0 e−5k , k = − ln 0.7 ≈ 0.07
5
ln 2
(a) T = ≈ 9.90 yr
k
y
(b) y(t) ≈ y0 e−0.07t , ≈ e−0.07t , so e−0.07t × 100 percent will remain.
y0

ln 2
13. (a) k = ≈ 0.1155; y ≈ 3e0.1155t (b) y(t) = 4e0.02t
6
1
(c) y = y0 ekt , 1 = y0 ek , 200 = y0 e10k . Divide: 200 = e9k , k = ln 200 ≈ 0.5887,
9
y ≈ y0 e0.5887t ; also y(1) = 1, so y0 = e−0.5887 ≈ 0.5550, y ≈ 0.5550e0.5887t .
ln 2
(d) k = ≈ 0.1155, 2 = y(1) ≈ y0 e0.1155 , y0 ≈ 2e−0.1155 ≈ 1.7818, y ≈ 1.7818e0.1155t
T
ln 2
14. (a) k = ≈ 0.1386, y ≈ 10e−0.1386t (b) y = 10e−0.015t
T
1
(c) 100 = y0 e−k , 1 = y0 e−10k . Divide: e9k = 100, k = ln 100 ≈ 0.5117;
9
y0 = e10k ≈ e5.117 ≈ 166.83, y = 166.83e−0.5117t .
ln 2
(d) k = ≈ 0.1386, 10 = y(1) ≈ y0 e−0.1386 , y0 ≈ 10e0.1386 ≈ 11.4866, y ≈ 11.4866e−0.1386t
T

16. (a) None; the half-life is independent of the initial amount.
(b) kT = ln 2, so T is inversely proportional to k.

ln 2
17. (a) T = ; and ln 2 ≈ 0.6931. If k is measured in percent, k = 100k,
k
ln 2 69.31 70
then T = ≈ ≈ .
k k k
(b) 70 yr (c) 20 yr (d) 7%

18. Let y = y0 ekt with y = y1 when t = t1 and y = 3y1 when t = t1 + T ; then y0 ekt1 = y1 (i) and
1
y0 ek(t1 +T ) = 3y1 (ii). Divide (ii) by (i) to get ekT = 3, T = ln 3.
k

y(t) ln 0.27
19. From (11), y(t) = y0 e−0.000121t . If 0.27 = = e−0.000121t then t = − ≈ 10,820 yr, and
y0 0.000121
y(t) ln 0.30
if 0.30 = then t = − ≈ 9950, or roughly between 9000 B.C. and 8000 B.C.
y0 0.000121


408 Chapter 9

20. (a) 1 (b) t = 1988 yields
y/y0 = e−0.000121(1988) ≈ 79%.

0 50000
0

ln 2
21. (a) Let T1 = 5730 − 40 = 5690, k1 = ≈ 0.00012182; T2 = 5730 + 40 = 5770, k2 ≈ 0.00012013.
T1
1 y 1
With y/y0 = 0.92, 0.93, t1 = − ln = 684.5, 595.7; t2 = − ln(y/y0 ) = 694.1, 604.1; in
k1 y0 k2
1988 the shroud was at most 695 years old, which places its creation in or after the year 1293.

(b) Suppose T is the true half-life of carbon-14 and T1 = T (1 + r/100) is the false half-life. Then
ln 2 ln 2
with k = , k1 = we have the formulae y(t) = y0 e−kt , y1 (t) = y0 e−k1 t . At a certain
T T1
point in time a reading of the carbon-14 is taken resulting in a certain value y, which in the
case of the true formula is given by y = y(t) for some t, and in the case of the false formula
is given by y = y1 (t1 ) for some t1 .
1 y
If the true formula is used then the time t since the beginning is given by t = − ln . If
k y0
1 y
the false formula is used we get a false value t1 = − ln ; note that in both cases the
k1 y0
value y/y0 is the same. Thus t1 /t = k/k1 = T1 /T = 1 + r/100, so the percentage error in
the time to be measured is the same as the percentage error in the half-life.
dp
22. (a) = −kp, p(0) = p0
dh
(b) p0 = 1, so p = e−kh , but p = 0.83 when h = 5000 thus e−5000k = 0.83,
ln 0.83
k=− ≈ 0.0000373, p ≈ e−0.0000373h atm.
5000
23. (a) y = y0 bt = y0 et ln b = y0 ekt with k = ln b > 0 since b > 1.
(b) y = y0 bt = y0 et ln b = y0 e−kt with k = − ln b > 0 since 0 < b < 1.

(c) y = 4(2t ) = 4et ln 2 (d) y = 4(0.5t ) = 4et ln 0.5 = 4e−t ln 2

ln(y1 /y0 )
24. If y = y0 ekt and y = y1 = y0 ekt1 then y1 /y0 = ekt1 , k = ; if y = y0 e−kt and
t1
ln(y1 /y0 )
y = y1 = y0 e−kt1 then y1 /y0 = e−kt1 , k = − .
t1
1
25. (a) If y = y0 ekt , then y1 = y0 ekt1 , y2 = y0 ekt2 , divide: y2 /y1 = ek(t2 −t1 ) , k = ln(y2 /y1 ),
t2 − t1
ln 2 (t2 − t1 ) ln 2
T = = . If y = y0 e−kt , then y1 = y0 e−kt1 , y2 = y0 e−kt2 ,
k ln(y2 /y1 )
1 ln 2 (t2 − t1 ) ln 2
y2 /y1 = e−k(t2 −t1 ) , k = − ln(y2 /y1 ), T = =− .
t2 − t1 k ln(y2 /y1 )
(t2 − t1 ) ln 2
In either case, T is positive, so T = .
ln(y2 /y1 )
ln 2
(b) In Part (a) assume t2 = t1 + 1 and y2 = 1.25y1 . Then T = ≈ 3.1 h.
ln 1.25



26. (a) In t years the interest will be compounded nt times at an interest rate of r/n each time. The
value at the end of 1 interval is P + (r/n)P = P (1 + r/n), at the end of 2 intervals it is
P (1 + r/n) + (r/n)P (1 + r/n) = P (1 + r/n)2 , and continuing in this fashion the value at the
end of nt intervals is P (1 + r/n)nt .
(b) Let x = r/n, then n = r/x and
lim P (1 + r/n)nt = lim P (1 + x)rt/x = lim P [(1 + x)1/x ]rt = P ert .
n→+∞ +
x→0 + x→0
rt
(c) The rate of increase is dA/dt = rP e = rA.

27. (a) A = 1000e(0.08)(5) = 1000e0.4 ≈ $1, 491.82
(b) P e(0.08)(10) = 10, 000, P e0.8 = 10, 000, P = 10, 000e−0.8 ≈ $4, 493.29
(c) From (11), with k = r = 0.08, T = (ln 2)/0.08 ≈ 8.7 years.

28. Let r be the annual interest rate when compounded continuously and r1 the eﬀective annual
interest rate. Then an amount P invested at the beginning of the year is worth P er = P (1 + r1 )
at the end of the year, and r1 = er − 1.

dT dT
29. (a) = −k(T − 21), T (0) = 95, = −k dt, ln(T − 21) = −kt + C1 ,
dt T − 21
T = 21 + eC1 e−kt = 21 + Ce−kt , 95 = T (0) = 21 + C, C = 74, T = 21 + 74e−kt
t
64 32 32
(b) 85 = T (1) = 21 + 74e−k , k = − ln = − ln , T = 21 + 74et ln(32/37) = 21 + 74 ,
74 37 37
t
30 32 ln(30/74)
T = 51 when = , t= ≈ 6.22 min
74 37 ln(32/37)

dT
30. = k(70 − T ), T (0) = 40; − ln(70 − T ) = kt + C, 70 − T = e−kt e−C , T = 40 when t = 0, so
dt
70 − 52 5
30 = e−C , T = 70 − 30e−kt ; 52 = T (1) = 70 − 30e−k , k = − ln = ln ≈ 0.5,
30 3
T ≈ 70 − 30e−0.5t

31. Let T denote the body temperature of McHam’s body at time t, the number of hours elapsed after
dT dT
10:06 P.M.; then = −k(T − 72), = −kdt, ln(T − 72) = −kt + C, T = 72 + eC e−kt ,
dt T − 72
3.6
77.9 = 72 + eC , eC = 5.9, T = 72 + 5.9e−kt , 75.6 = 72 + 5.9e−k , k = − ln ≈ 0.4940,
5.9
ln(26.6/5.9)
T = 72+5.9e−0.4940t . McHam’s body temperature was last 98.6◦ when t = − ≈ −3.05,
0.4940
so around 3 hours and 3 minutes before 10:06; the death took place at approximately 7:03 P.M.,
while Moore was on stage.

dT dT
32. If T0 < Ta then = k(Ta − T ) where k > 0. If T0 > Ta then = −k(T − Ta ) where k > 0;
dt dt
both cases yield T (t) = Ta + (T0 − Ta )e−kt with k > 0.

dy y
33. (a) Both y(t) = 0 and y(t) = L are solutions of the logistic equation = k 1− y as both
dt L
sides of the equation are then zero.
dy
(b) If y is very small relative to L then y/L ≈ 0, and the logistic equation becomes ≈ ky,
dt
which is a form of the equation for exponential growth.


410 Chapter 9

(c) All the terms on the right-hand-side of the logistic equation are positive, except perhaps
y
1 − , which is positive if y < L and negative if y > L.
L
(d) The rate of change of y is a function of only one variable, y itself. The right-hand-side of the
diﬀerential equation is a quadratic equation in y, which can be thought of as a parabola in
y which opens down and crosses the y-axis at y = 0 and y = L. The parabola thus takes its
maximum midway between the two y-intercepts, namely at y = L/2.

dy y 1 1 1 k
34. (a) Given = k 1− y, separation of variables yields + dy = dt so that
dt L L y L−y L
ln y − ln(L − y) = kt + C. The initial condition yields
y0 y y0
ln y0 − ln(L − y0 ) = C, C = ln and thus ln = kt + ln ,
L − y0 L−y L − y0
y y0 y0 L
= ekt + , with solution y(t) = .
L−y L − y0 y0 + (L − y0 )e−kt
y0 L
(b) From part (a), lim y(t) = =L
t→+∞ y0 + (L − y0 ) lim e−kt
t→+∞

35. (a) k = L = 1, y0 = 2 (b) k = L = y0 = 1
2 2

0 2 0 2
0 0

(c) k = y0 = 1, L = 2 (d) k = y0 = 1, L = 4
2 4

0 5 0 12
0 0

36. y0 = 400, L ≈ 1000, and y(300) ≈ 700; solve to obtain k ≈ 0.042

2·8 16
37. y0 ≈ 2, L ≈ 8; since the curve y = −kt
passes through the point (2, 4), 4 = ,
2 + 6e 2 + 6e−2k
1
6e−2k = 2, k = ln 3 ≈ 0.5493.
2



400,000
38. y0 ≈ 400, L ≈ 1000; since the curve y = passes through the point (200, 600),
400 + 600e−kt
400,000 800 1
600 = , 600e−200k = , k= ln 2.25 ≈ 0.00405.
400 + 600e−200k 3 200

39. (a) y0 = 5 (b) L = 12 (c) k = 1
60
(d) L/2 = 6 = , 5 + 7e−t = 10, t = − ln(5/7) ≈ 0.3365
5 + 7e−t
dy 1
(e) = y(12 − y), y(0) = 5
dt 12

40. (a) y0 = 1 (b) L = 1000 (c) k = 0.9
1000 1
(d) 750 = −0.9t
, 3(1 + 999e−0.9t ) = 4, t = ln(3 · 999) ≈ 8.8949
1 + 999e 0.9
dy 0.9
(e) = y(1000 − y), y(0) = 1
dt 1000

41. Assume y(t) students have had the ﬂu t days after semester break. Then y(0) = 20, y(5) = 35.
dy
(a) = ky(L − y) = ky(1000 − y), y0 = 20
dt
20000 1000
(b) Part (a) has solution y = −kt
= ;
20 + 980e 1 + 49e−kt
1000 1000
35 = −5k
, k = 0.115, y ≈ .
1 + 49e 1 + 49e−0.115t
(c) t 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
y(t) 20 22 25 28 31 35 39 44 49 54 61 67 75 83 93

y
(d)
100

75

50

25
t
3 6 9 12

EXERCISE SET 9.4
1. (a) y = e−2x , y = −2e−2x , y = 4e−2x ; y + y − 2y = 0
y = ex , y = ex , y = ex ; y + y − 2y = 0.
(b) y = c1 e−2x + c2 ex , y = −2c1 e−2x + c2 ex , y = 4c1 e−2x + c2 ex ; y + y − 2y = 0

2. (a) y = e−2x , y = −2e−2x , y = 4e−2x ; y + 4y + 4y = 0
y = xe−2x , y = (1 − 2x)e−2x , y = (4x − 4)e−2x ; y + 4y + 4y = 0.
(b) y = c1 e−2x + c2 xe−2x , y = −2c1 e−2x + c2 (1 − 2x)e−2x ,
y = 4c1 e−2x + c2 (4x − 4)e−2x ; y + 4y + 4y = 0.


412 Chapter 9

3. m2 + 3m − 4 = 0, (m − 1)(m + 4) = 0; m = 1, −4 so y = c1 ex + c2 e−4x .

4. m2 + 5m + 6 = 0, (m + 2)(m + 3) = 0; m = −2, −3 so y = c1 e−2x + c2 e−3x .

5. m2 − 2m + 1 = 0, (m − 1)2 = 0; m = 1, so y = c1 ex + c2 xex .

6. m2 − 6m + 9 = 0, (m − 3)2 = 0; m = 3 so y = c1 e3x + c2 xe3x .

7. m2 + 1 = 0, m = ±i so y = c1 cos x + c2 sin x.
√ √ √
8. m2 + 5 = 0, m = ± 5 i so y = c1 cos 5 x + c2 sin 5 x.

9. m2 − m = 0, m(m − 1) = 0; m = 0, 1 so y = c1 + c2 ex .

10. m2 + 3m = 0, m(m + 3) = 0; m = 0, −3 so y = c1 + c2 e−3x .

11. m2 − 4m + 4 = 0, (m − 2)2 = 0; m = 2 so y = c1 e2t + c2 te2t .

12. m2 − 10m + 25 = 0, (m − 5)2 = 0; m = 5 so y = c1 e5t + c2 te5t .

13. m2 + 4m + 13 = 0, m = −2 ± 3i so y = e−2x (c1 cos 3x + c2 sin 3x).

14. m2 − 6m + 25 = 0, m = 3 ± 4i so y = e3x (c1 cos 4x + c2 sin 4x).

15. 8m2 − 2m − 1 = 0, (4m + 1)(2m − 1) = 0; m = −1/4, 1/2 so y = c1 e−x/4 + c2 ex/2 .

16. 9m2 − 6m + 1 = 0, (3m − 1)2 = 0; m = 1/3 so y = c1 ex/3 + c2 xex/3 .

17. m2 + 2m − 3 = 0, (m + 3)(m − 1) = 0; m = −3, 1 so y = c1 e−3x + c2 ex and y = −3c1 e−3x + c2 ex .
Solve the system c1 + c2 = 1, −3c1 + c2 = 9 to get c1 = −2, c2 = 3 so y = −2e−3x + 3ex .

18. m2 − 6m − 7 = 0, (m + 1)(m − 7) = 0; m = −1, 7 so y = c1 e−x + c2 e7x , y = −c1 e−x + 7c2 e7x .
Solve the system c1 + c2 = 5, −c1 + 7c2 = 3 to get c1 = 4, c2 = 1 so y = 4e−x + e7x .

19. m2 + 6m + 9 = 0, (m + 3)2 = 0; m = −3 so y = (c1 + c2 x)e−3x and y = (−3c1 + c2 − 3c2 x)e−3x .
Solve the system c1 = 2, −3c1 + c2 = −5 to get c1 = 2, c2 = 1 so y = (2 + x)e−3x .
√ √ √
20. m2 + 4m + 1 = 0, m = −2 ± 3 so y = c1 e(−2+ 3)x + c2 e(−2− 3)x ,
√ √ √ √
y = (−2 + 3)c1 e(−2+ 3)x + (−2 − 3)c2 e(−2− 3)x . Solve the system c1 + c2 = 5,
√ √ √ √
(−2 + 3)c1 + (−2 − 3)c2 = 4 to get c1 = 5 + 7 3, c2 = 5 − 7 3 so
2 3 2 3
√ √ √ √
y = 5 + 7 3 e(−2+ 3)x + 5 − 7 3 e(−2− 3)x .
2 3 2 3

21. m2 + 4m + 5 = 0, m = −2 ± i so y = e−2x (c1 cos x + c2 sin x),
y = e−2x [(c2 − 2c1 ) cos x − (c1 + 2c2 ) sin x]. Solve the system c1 = −3, c2 − 2c1 = 0
to get c1 = −3, c2 = −6 so y = −e−2x (3 cos x + 6 sin x).

22. m2 − 6m + 13 = 0, m = 3 ± 2i so y = e3x (c1 cos 2x + c2 sin 2x),
y = e3x [(3c1 + 2c2 ) cos 2x + (−2c1 + 3c2 ) sin 2x]. Solve the system c1 = −2, 3c1 + 2c2 = 0
to get c1 = −2, c2 = 3 so y = e3x (−2 cos 2x + 3 sin 2x).

23. (a) m = 5, −2 so (m − 5)(m + 2) = 0, m2 − 3m − 10 = 0; y − 3y − 10y = 0.
(b) m = 4, 4 so (m − 4)2 = 0, m2 − 8m + 16 = 0; y − 8y + 16y = 0.
(c) m = −1 ± 4i so (m + 1 − 4i)(m + 1 + 4i) = 0, m2 + 2m + 17 = 0; y + 2y + 17y = 0.



24. c1 ex + c2 e−x is the general solution, but cosh x = 1 ex + 1 e−x and sinh x = 1 ex − 1 e−x
2 2 2 2
so cosh x and sinh x are also solutions.
√
25. m2 + km + k = 0, m = −k ± k 2 − 4k /2
(a) k 2 − 4k > 0, k(k − 4) > 0; k < 0 or k > 4
(b) k 2 − 4k = 0; k = 0, 4 (c) k 2 − 4k < 0, k(k − 4) < 0; 0 < k < 4

dy dy dz 1 dy
26. z = ln x; = = and
dx dz dx x dz
d2 y d dy d 1 dy 1 d2 y dz 1 dy 1 d2 y 1 dy
2
= = = 2 dx
− 2 = 2 2
− 2 ,
dx dx dx dx x dz x dz x dz x dz x dz
d2 y dy
substitute into the original equation to get 2
+ (p − 1) + qy = 0.
dz dz

d2 y dy
27. (a) +2 + 2y = 0, m2 + 2m + 2 = 0; m = −1 ± i so
dz 2 dz
1
y = e−z (c1 cos z + c2 sin z) = [c1 cos(ln x) + c2 sin(ln x)].
x
d2 y dy √
(b) 2
−2 − 2y = 0, m2 − 2m − 2 = 0; m = 1 ± 3 so
dz dz
√ √ √ √
y = c1 e(1+ 3)z
+ c2 e(1−3)z
= c1 x1+ 3
+ c2 x1−3

28. m2 + pm + q = 0, m = 1 (−p ± p2 − 4q ). If 0 < q < p2 /4 then y = c1 em1 x + c2 em2 x where
2
m1 < 0 and m2 < 0, if q = p2 /4 then y = c1 e−px/2 + c2 xe−px/2 , if q > p2 /4 then
y = e−px/2 (c1 cos kx + c2 sin kx) where k = 1 4q − p2 . In all cases lim y(x) = 0.
2 x→+∞

29. (a) Neither is a constant multiple of the other, since, e.g. if y1 = ky2 then em1 x = kem2 x ,
e(m1 −m2 )x = k. But the right hand side is constant, and the left hand side is constant only
if m1 = m2 , which is false.
(b) If y1 = ky2 then emx = kxemx , kx = 1 which is impossible. If y2 = y1 then xemx = kemx ,
x = k which is impossible.

30. y1 = eax cos bx, y1 = eax (a cos bx − b sin bx), y1 = eax [(a2 − b2 ) cos bx − 2ab sin bx] so
y1 + py1 + qy1 = eax [(a2 − b2 + ap + q) cos bx − (2ab + bp) sin bx]. But a = − 1 p and b = 1 4q − p2
2 2
so a2 − b2 + ap + q = 0 and 2ab + bp = 0 thus y1 + py1 + qy1 = 0. Similarly, y2 = eax sin bx is also
a solution.
Since y1 /y2 = cot bx and y2 /y1 = tan bx it is clear that the two solutions are linearly independent.

31. (a) The general solution is c1 eµx + c2 emx ; let c1 = 1/(µ − m), c2 = −1/(µ − m).
eµx − emx
(b) lim = lim xeµx = xemx .
µ→m µ−m µ→m

32. (a) If λ = 0, then y = 0, y = c1 + c2 x. Use y(0) = 0 and y(π) = 0 to get c1 = c2 = 0. If
λ < 0, then let λ = −a2 where a > 0 so y − a2 y = 0, y = c1 eax + c2 e−ax . Use y(0) = 0 and
y(π) = 0 to get c1 = c2 = 0.
√ √ √
(b) If λ > 0, then m2 + λ = 0, m2 = −λ = λi2 , m = ± λi, y √ c1 cos λ x + c2 sin λ x. If
√ = √
y(0) = √ and y(π) = 0, then c1 = 0 and c1 cos π λ + c2 sin π λ = 0 so c2 sin π λ = 0. But
0 √ √
c2 sin π λ = 0 for arbitrary values of c2 if sin π λ = 0, π λ = nπ, λ = n2 for n = 1, 2, 3, . . .,
otherwise c2 = 0.


414 Chapter 9

33. k/M = 0.5/2 = 0.25
(a) From (20), y = 0.4 cos(t/2) (b) T = 2π · 2 = 4π s, f = 1/T = 1/(4π) Hz
(c) y (d) y = 0 at the equilibrium position,
0.3
so t/2 = π/2, t = π s.
(e) t/2 = 2π at the maximum position below
x the equlibrium position, so t = 4π s.
o 4c

–0.3
√
34. 64 = w = −M g, M = 2, k/M = 0.25/2 = 1/8, k/M = 1/(2 2)
√
(a) From (20), y = cos(t/(2 2))
M √ √
(b) T = 2π = 2π(2 2) = 4π 2 s,
k
√
f = 1/T = 1/(4π 2) Hz
(c) y
1

t
2p 6p 10p

–1

√ √
(d) y = 0 at the equilibrium position, so t/(2 2) = π/2, t = π 2 s
√ √
(e) t/(2 2) = π, t = 2π 2 s

35. l = 0.05, k/M = g/l = 9.8/0.05 = 196 s−2
(a) From (20), y = −0.12 cos 14t. (b) T = 2π M/k = 2π/14 = π/7 s,
f = 7/π Hz
y
(c) (d) 14t = π/2, t = π/28 s
0.15
(e) 14t = π, t = π/14 s

t
π 2π
7 7

–0.15

36. l = 2, k/M = g/l = 32/2 = 16, k/M = 4

(a) From (20), y = −2 cos 4t. (b) T = 2π M/k = 2π/4 = π/2 s;
f = 1/T = 2/π Hz



(c) y (d) 4t = π/2, t = π/8 s
2
(e) 4t = π, t = π/4 s

t
3 6

–2

37. (a) From the graph it appears that the maximum velocity occurs at the equilibrium position,
t = π/8.
dv dy dv
To show this mathematically, vmax occurs when = 0. But v = = 8 sin 4t, =
dt dt dt
dv
32 cos 4t, = 0 when 4t = π/2, t = π/8.
dt
(b) From the graph it appears that the minimum velocity occurs at the equilibrium position as the
dv
block is falling, i.e. t = 3π/8. Mathematically, = 32 cos 4t = 0 when 4t = π/2, 3π/2, . . ..
dt
When 4t = 3π/2 the block is falling, so t = 3π/8.

M 4π 2 4π 2 w 4π 2 w 4π 2 w + 4
38. (a) T = 2π , k = 2 M = 2 , so k = = , 25w = 9(w + 4),
k T T g g 9 g 25
9 4π 2 w 4π 2 1 π2
25w = 9w + 36, w = ,k = = =
4 g 9 32 4 32

9
(b) From Part (a), w =
4

39. By Hooke’s Law, F (t) = −kx(t), since the only force is the restoring force of the spring. Newton’s
Second Law gives F (t) = M x (t), so M x (t) + kx(t) = 0, x(0) = x0 , x (0) = 0.

k k
40. 0 = v(0) = y (0) = c2 , so c2 = 0; y0 = y(0) = c1 , so y = y0 cos t.
M M

k M 2πt
41. y = y0 cos t, T = 2π , y = y0 cos
M k T
2π 2πt
(a) v = y (t) = − y0 sin has maximum magnitude 2π|y0 |/T and occurs when
T T
2πt/T = nπ + π/2, y = y0 cos(nπ + π/2) = 0.
4π 2 2πt
(b) a = y (t) = − y0 cos has maximum magnitude 4π 2 |y0 |/T 2 and occurs when
T2 T
2πt/T = jπ, y = y0 cos jπ = ±y0 .

d 1 1 k
42. k[y(t)]2 + M (y (t))2 = ky(t)y (t)+M y (t)y (t) = M y (t)[ y(t)+y (t)] = 0, as required.
dt 2 2 M

Chapter 09

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Chapter 09