Section 1.4: Separable Equations and Applications 35
(c) If 1b (and similarly if 1b ), then we can pick any ca and define the solution
1if
sec if 2
xc
yx xc c xc
.
34. The population growth rate is
ln 6 10 0.17918k
, so the population after t hours is
given by
0.17918
0
t
Pt Pe. To find how long it takes for the population to double, we
therefore need only solve the equation 0.17918
00
2t
PPe for t, finding
ln 2 0.17198 3.87t
hours.
36. As in Problem 35, the number of 14 C atoms after t years is given by
10 0.0001216
5.0 10 t
Nt e
. Hence we need only solve the equation
10 10 0.0001216
4.6 10 5.0 10 t
e
for the age t of the relic, finding
ln 5.0 4.6 0.0001216 686t
years. Thus it appears not to be a genuine relic of the
time of Christ 2000 years ago.
38. When the book has been overdue for t years, the fine owed is given in dollars by
0.05
0.30 t
At e. Hence the amount owed after 100 years is given by
0.05 100
100 0.30 44.52Ae
dollars.
36 Chapter 1: First-Order Differential Equations
40. To find the decay rate of radioactive cobalt, we solve the equation 5.27
1
2
k
e
(half-life
5.27 years) for
ln 2 5.27 0.13153k
. Thus the amount of radioactive cobalt left af-
ter t years is given by
0.13153
0
t
At Ae
. We therefore solve the equation
0.13153
00
0.01
t
At Ae A
for t, finding
ln100 0.13153 35.01t
years. Thus it will
be about 35 years until the region is again inhabitable.
41. Taking 0t when the body was formed and tT now, we see that the amount
Qt of
42. Taking 0t when the rock contained only potassium and tT now, we see that the
amount
Qt of potassium in the rock at time t (in years) is given by
0
kt
Qt Qe
,
43. Because 0A in Newton’s law of cooling, the differential equation reduces to TkT
44. The amount of sugar remaining undissolved after t minutes is given by
0
kt
At Ae
; we
Section 1.4: Separable Equations and Applications 37
45. (a) The light intensity at a depth of x meters is given by
1.4
0
x
Ix Ie
. We solve the
equation
1.4 1
00
2
x
Ix Ie I
for x, finding
ln 2 1.4 0.495x
meters.
46. Solving the initial value problem shows that the pressure at an altitude of x miles is given
by
0.2
29.92
x
px e
inches of mercury.
47. If
Nt denotes the number of people (in thousands) who have heard the rumor after t
days, then the initial value problem is
100Nk N
,
00N. Separating variables
48. Let
8
Nt
and
5
Nt
be the numbers of 238 U and 235 U atoms, respectively, at time t (in
billions of years after the creation of the universe). Then
80
kt
Nt Ne
and
49. Newton’s law of cooling gives
70
dT kT
dt
, and separating variables and integrating
lead to
ln 70TktC
. The initial condition
0 210T gives ln140C, and
38 Chapter 1: First-Order Differential Equations
50. (a) The initial condition implies that
10 kt
At e. The fact that
At triples every 7.5
years implies that 15 2
15
2
30 ( ) 10 k
Ae , which gives 15 2
3
k
e
, or 215
2ln3 ln 3
k .
51. (a) The initial condition gives
15 kt
At e
, and then
510A implies that 15 10
kt
e,
or 3
2
kt
e, or 13
ln
52
k. Thus
55
33 2
15exp ln 15 15
52 2 3
tt
t
At
.
52. If
Lt denotes the number of human language families at time t (in years), then
ln(3 2)
that the original human language was spoken about 120 thousand years ago.
53. As in Problem 52, if
Lt denotes the number of Native American language families at
time t (in years), then
kt
Lt e for some constant k, and the condition that
Section 1.4: Separable Equations and Applications 39
54. With
Ay constant, Equation (30) in the text takes the form dy ky
dt , which we read-
55. With 2
3A
and 2
()
112
a
, and taking 2
32 ft/secg, Equation (30) reduces to
56. The radius of the cross-section of the cone at height y is proportional to y, so
Ay is
proportional to 2
y
. Therefore Equation (30) takes the form 2
yy k y
, and a general
y
57. The solution of yky
is given by 2yktC . The initial condition
0yh
(the height of the cylinder) yields 2Ch. Then substituting tT and 0y gives
58. Since 34
x
y, the cross-sectional area is
232
Ay x y
. Hence the general equa-
tion
2
Ay
a
gy
y reduces to the differential equation
yy
k
with general solu-
40 Chapter 1: First-Order Differential Equations
59. (a) Since 2
x
by, the cross-sectional area is
2
Ay x by
. Hence equation (30)
becomes
1/2 2yy k ab g
, with general solution 32
2
60. With 2
32 ft secg and 2
1
12
()a
, Equation (30) simplifies to
18
dy
A
yy
dt
. If z
denotes the distance from the center of the cylinder down to the fluid surface, then
61.
2
8
Ayy
y
as in Example 6 in the text, but now 144
a
in Equation (30), so
that the initial value problem is
2
18 8
yyy y
,
08y. Separating variables
62. Here
2
1
Ay
y
and the area of the bottom hole is 4
10a
, so Equation (30)
leads to the initial value problem
24
11029.8
dy
yy
,
01y, or
Section 1.4: Separable Equations and Applications 41
63. (a) As in Example 6, the initial value problem is
2
8dy
yy ky
dt
,
04y,
where 22
0.6 2 4.8krgr
. Separating variables and applying the initial condition just
64. The given rate of fall of the water level is 1
4 in hr ft sec
10800
dy
dt . With
2
Ay x
(where
yfx) and 2
ar
, Equation (30) becomes
65. The temperature
Tt of the body satisfies the differential equation
70
dT kT
dt
.
Separating variables gives 1
70 dT k dt
T
, or (since
70Tt for all t)
42 Chapter 1: First-Order Differential Equations
which simplify to
28.6 10
ka
e
ln 2
66. (a) Let 0t when it began to snow, and let 0
tt at 7:00 a.m. Also let 0x where the
snowplow begins at 7:00 a.m. If the constant rate of snowfall is given by c, then the
(b) Separating variables gives 1
kdx dt
t
, or lnkx t C
, and then solving for t
gives kx
tCe
. The initial condition
00xt gives 0
Ct
. We are further given that
67. We still have 0
kx
tte, but now the given information yields the conditions
4
00
7
00
1
2
k
k
tte
tte
at 8 a.m. and 9 a.m., respectively. Elimination of 0
t gives the equation 47
210
kk
ee
,
Section 1.4: Separable Equations and Applications 43
68. (a) Note first that if
denotes the angle between the tangent line and the horizontal, then
2
, so
cot cot tan
2yx
. It follows that
(b) The substitution 2
2sin
y
at, 4sincosd
y
at tdt now gives
2
2
22sin cos
4sincos 2sin sin
aa t t
at tdt dx dx
at t
, 2
4sindx a t dt
Integration now gives
69. Substitution of vdydx in the differential equation for
yyx gives 2
1
dv
av
dx ,
and separation of variables then yields 2
11
1
dv dx
a
v
, or 1
1
sinh x
vC
a
, or
44 Chapter 1: First-Order Differential Equations
.
SECTION 1.5
LINEAR FIRST-ORDER EQUATIONS
1. An integrating factor is given by
exp 1
x
dx e
, and multiplying the differential
equation by
gives 2
x
xx
ey ey e
, or
2
x
x
x
De
y
e . Integrating then leads to
x
y
2. An integrating factor is given by
2
exp 2
x
dx e
, and multiplying the differen-
tial equation by gives 22
23
xx
ey ey
, or
x
y
23
x
. Integrating then leads to
3. An integrating factor is given by
3
exp 3
x
dx e
, and multiplying the differential
equation by
gives
32
x
x
D
y
ex
. Integrating then leads to 32x
y
exC, and
thus to the general solution
23
x
y
xxCe
.
x
x
y
5. We first rewrite the differential equation for 0x as 23yy
x
. An integrating factor
Section 1.5: Linear First-Order Equations 45
6. We first rewrite the differential equation for 0x as 57yyx
x
. An integrating fac-
7. We first rewrite the differential equation for 0x as 15
2
yy
x
x
. An integrating
x
8. We first rewrite the differential equation for 0x as 14
3
yy
x
. An integrating fac-
9. We first rewrite the differential equation for 0x as 11yy
. An integrating factor
46 Chapter 1: First-Order Differential Equations
10. We first rewrite the differential equation for 0x as 2
39
22
yyx
x
. An integrating
11. We first collect terms and rewrite the differential equation for 0x as 130yy
x
. An integrating factor is given by
x
12. We first rewrite the differential equation for 0x as 4
32yyx
x
. An integrating fac-
tor is given by 3ln 3
3
exp x
dx e x
x
, and multiplying by
gives
13. An integrating factor is given by
exp 1
x
dx e
, and multiplying by
gives
x
Section 1.5: Linear First-Order Equations 47
14. We first rewrite the differential equation for 0x as 2
3
yyx
x
. An integrating factor
x
15. An integrating factor is given by
2
exp 2
x
x
dx e
, and multiplying by
gives
16. We first rewrite the differential equation as
cos cosyxyx
. An integrating factor
x
sin
x
17. We first rewrite the differential equation for 1x as 1cos
11
x
yy
x
x
. An integrat-
48 Chapter 1: First-Order Differential Equations
18. We first rewrite the differential equation for 0x as 2
2cosyyxx
x
. An integrating
19. For 0x an integrating factor is given by
ln(sin )
exp cot sin
x
x
dx e x
, and mul-
tiplying by
gives
sin cos sin cos
x
yxyxx
, or
sin sin cos
x
D
yx xx . In-
20. We first rewrite the differential equation as
11yxyx
. An integrating factor is
21. We first rewrite the differential equation for 0x as 3
3cosyyxx
x
. An integrating
Section 1.5: Linear First-Order Equations 49
22. We first rewrite the differential equation as 2
2
23
x
yxyxe
. An integrating factor is
given by
2
exp 2
x
x
dx e
, and multiplying by
gives 22
2
23
xx
eyxeyx
,
23. We first rewrite the differential equation for 0x as 3
3
24yyx
x
. An integrat-
x
x
24. We first rewrite the differential equation as 22
3
44
x
x
yy
xx
. An integrating factor
is given by
32
22
2
33
exp exp ln 4 4
42
xdx x x
x
, and multiplying by
gives
32 12 12
222
434 4xyxxyxx
, or
25. We first rewrite the differential equation as
2
3
3
2
22
36
11
x
xx
yye
xx
. An integrating
50 Chapter 1: First-Order Differential Equations
or (as can be verified using the product rule twice, together with some algebra)
2
3
3/2 5/2
22
2
161
x
x
Dyx e xx
. Integrating then leads to
2
3
3/2 5/2 3/2
222
2
16121
x
y
xexxdxx C
,
x
26. At points
,
x
y with 2
14 0xy and
0y
, rewriting the differential equation as
3
2
y
dy y
shows that
2
3
14dx xy
, or (putting
x
for dx
) 3
41
xx
, a linear
27. At points
,
x
y with 0
y
xye, rewriting the differential equation as 1
y
dy
dx x
y
e
y
x
Section 1.5: Linear First-Order Equations 51
28. At points
,
x
y with 12 0xy
, rewriting the differential equation as
2
1
12
dy y
dx xy
shows that 2
12
1
dx x
y
d
yy
, or (putting
x
for dx
d
y
) 22
21
11
y
xx
yy
, a linear equation
y
29. We first rewrite the differential equation as 21yxy
. An integrating factor is given
by
2
exp 2
x
x
dx e
, and multiplying by
gives
222
2
x
xx
eyxeye
, or
22
x
x
x
D
ye e
. Integrating then leads to 22
xx
y
eedx
. Any antiderivative of
x
x
30. We first rewrite the differential equation for 0x as 1cos
yyx
. An integrating
52 Chapter 1: First-Order Differential Equations
12 32 12
1cos
2
x
yxyx x
, or
12 12cos
x
Dx
y
xx
. Integrating then leads to
12 12cos
x
yx xdx
y
x
x
differs by a constant (call it C)
31. (a) The fundamental theorem of calculus implies, for any value of C, that
Pxdx
cc
yx Ce Px Pxyx
,
and thus that
0
cc
yx Pxyx
. Therefore c
y
is a general solution of
dy Pxy Qx
dx
.
(c) The stated assumptions imply that
cp cp
yxPxyyxyxPxyxyx
32. (a) Substituting
p
yx
into the given differential equation gives
cos sin sin cos 2sinAxBx AxBx x
,
Section 1.5: Linear First-Order Equations 53
that is
sin cos 2sinAB x AB x x
,
for all x. It follows that 2
A
B
and 0AB
, and solving this system gives 1
A
and 1B . Thus
sin cos
p
yx x x.
33. Let
x
t denote the amount of salt (in kg) in the tank after t seconds. We want to know
when
10xt . In the notation of Equation (18) of the text, the differential equation for
x
t is
34. Let
x
t denote the amount of pollutants in the reservoir after t days, measured in mil-
lions of cubic feet (mft3). The volume of the reservoir is 8000 mft3, and the initial
amount
0x of pollutants is
3
0.25% 8000 20 mft. We want to know when
x
54 Chapter 1: First-Order Differential Equations
x
35. The only difference from the Example 4 solution in the textbook is that 3
1640 kmV and
3
410 km yrr
for Lake Ontario, so the time required is
36. (a) Let
x
t denote the amount of salt (in kg) in the tank after t minutes. Because the
volume of liquid in the tank is decreasing by 1 gallon each minute, the volume after t min
is 60 t gallons. Thus in the notation of Equation (18) of the text, the differential equa-
tion for
x
t is
x
(b) By part (a),
2
3
160
3600
x
tt
, which is zero when 60 20 3t . We ig-
nore 60 20 3t because the tank is empty after 60 min. The facts that
660 0
3600
xt t
for 060t and that 60 20 3t is the lone critical point