1. SECTION 1.4
SEPARABLE EQUATIONS AND APPLICATIONS
Of course it should be emphasized to students that the possibility of separating the variables is
the first one you look for. The general concept of natural growth and decay is important for all
differential equations students, but the particular applications in this section are optional.
Torricelli's law in the form of Equation (24) in the text leads to some nice concrete examples and
problems.
⌠ dy
= − ∫ 2 x dx; ln y = − x 2 + c; y ( x) = e − x +c
= C e− x
2 2
1. 
⌡ y
⌠ dy ⌠ 1 1
2.  2 = −  2 x dx; − = − x 2 − C; y ( x) =
⌡y ⌡ y x +C
2
⌠ dy
3. 
⌡ y
= ∫ sin x dx; ln y = − cos x + c; y( x) = e− cos x + c = C e− cos x
⌠ dy
= ⌠
4 dx
4.   ; ln y = 4 ln(1 + x ) + ln C; y ( x ) = C (1 + x) 4
⌡ y ⌡ 1+ x
⌠ dy
5. 
⌡ 1− y 2
= ⌠

dx
⌡2 x
; sin −1 y = x + C; y ( x) = sin ( x +C )
⌠ dy
∫3 y ( x) = ( x3/ 2 + C )
2
6.  = x dx; 2 y = 2 x3/ 2 + 2C ;
⌡ y
3/ 2
⌠ dy
7.  1/ 3 = ∫4x
1/ 3
dx; 3
y 2/3
= 3x 4/3
+ C;
3
y ( x) = ( 2 x 4/3
+C)
⌡y
2 2
8. ∫ cos y dy = ∫ 2 x dx; sin y = x 2 + C ; y ( x) = sin −1 ( x 2 + C )
⌠ dy ⌠ 2 dx ⌠ 1 1 
9.  =  =  +  dx (partial fractions)
⌡ y ⌡ 1− x ⌡  1+ x 1− x 
2
1+ x
ln y = ln(1 + x) − ln(1 − x) + ln C; y ( x) = C
1− x
⌠ dy ⌠ dx 1 1 1 + C (1 + x)
10.  =  ; − = − −C = −
⌡ (1 + y ) ⌡ (1 + x) 1+ y 1+ x 1+ x
2 2
Section 1.4 1

2. 1+ x 1+ x x − C (1 + x)
1+ y = ; y ( x) = −1 =
1 + C (1 + x) 1 + C (1 + x) 1 + C (1 + x)
⌠ dy ⌠
y ( x ) = (C − x 2 )
1 x2 C −1/ 2
11.  3 =  x dx; − = − ;
⌡y ⌡ 2 y2 2 2
⌠ y dy
= ∫ x dx; ln ( y 2 + 1) = x 2 + 1 ln C ; y2 +1 = C ex
2
 2
1 1
12.
⌡ y +1
2 2 2
⌠ y 3 dy
13.  4 = ∫ cos x dx; 1
ln ( y 4 + 1) = sin x + C
⌡ y +1
4
14. ∫ (1 + )
y dy = ∫ (1 + )
x dx; y + 2 y 3/ 2 = x + 2 x3/ 2 + C
3 3
⌠ 2 1  ⌠1 1  2 1 1
15.   y 2 − y 4  dy =   x − x 2  dx; − + 3 = ln x + + C
⌡  ⌡  y 3y x
⌠ sin y dy
= ⌠ ln (1 + x 2 ) + ln C
x dx
16.   ; − ln(cos x) = 1
⌡ cos y ⌡ 1+ x2 2
sec y = C 1 + x 2 ; (
y ( x) = sec −1 C 1 + x 2 )
17. y′ = 1 + x + y + xy = (1 + x)(1 + y )
⌠ dy

⌡ 1+ y
= ∫ (1 + x) dx; ln 1 + y = x + 1 x 2 + C
2
18. x 2 y′ = 1 − x 2 + y 2 − x 2 y 2 = (1 − x 2 )(1 + y 2 )
⌠ dy ⌠ 1  1  1 
 =   2 − 1 dx; tan −1 y = − − x + C; y ( x) = tan  C − − x 
⌡ 1+ y ⌡ x   
2
x x
⌠ dy
= ∫e ln y = e x + ln C ; y ( x) = C exp(e x )
x
19.  dx;
⌡ y
y (0) = 2e implies C = 2 so y ( x) = 2 exp(e x )
⌠ dy
20.  = ∫ 3x
2
dx; tan −1 y = x3 + C ; y ( x) = tan ( x 3 + C )
⌡ 1+ y
2
y (0) = 1 implies C = tan −1 1 = π / 4 so y ( x) = tan ( x3 + π / 4 )
Section 1.4 2

3. ⌠ ⌠ x dx
21.  2 y dy =  ; y2 = x 2 − 16 + C
⌡ ⌡ x − 16
2
y (5) = 2 implies C = 1 so y 2 = 1 + x 2 − 16
⌠ dy
= ∫ (4x − 1) dx; ln y = x 4 − x + ln C ; y ( x) = C exp( x 4 − x)
3
22. 
⌡ y
y (1) = − 3 implies C = − 3 so y ( x ) = − 3exp( x 4 − x)
⌠ dy
23. 
⌡ 2 y −1
= ∫ dx; 1
2 ln (2 y − 1) = x + 1 ln C ;
2 2 y − 1 = C e2 x
y (1) = 1 implies C = e −2 so y ( x) = 1
2 (1 + e )2 x−2
⌠ dy ⌠ cos x dx
24.  =  ; ln y = ln(sin x) + ln C ; y ( x) = C sin x
⌡ y ⌡ sin x
y ( π ) = π implies C = π so y ( x) =
2 2 2
π
2 sin x
⌠ dy ⌠1 
25.  =   + 2x  ; ln y = ln x + x 2 + ln C ; y ( x) = C x exp( x 2 )
⌡ y ⌡ x 
y (1) = 1 implies C = e −1 so y ( x) = x exp( x 2 − 1)
⌠ dy ⌠1 
25.  =   + 2x  ; ln y = ln x + x 2 + ln C ; y ( x) = C x exp( x 2 )
⌡ y ⌡ x 
y (1) = 1 implies C = e −1 so y ( x) = x exp( x 2 − 1)
⌠ dy ⌠ −1
 2 =  ( 2 x + 3x ) ;
1
26. 2
− = x 2 + x3 + C; y ( x) =
⌡y ⌡ y x + x3 + C
2
1
y (1) = − 1 implies C = − 1 so y ( x) =
1 − x 2 − x3
27. ∫e
y
dy = ∫6e
2x
dx; e y = 3 e2 x + C ; y ( x) = ln (3 e2 x + C )
y (0) = 0 implies C = − 2 so y ( x) = ln (3 e 2 x − 2 )
28.
⌠ 2
⌡
⌠ dx
 sec y dy = 
⌡2 x
; tan y = x + C; y ( x) = tan −1 ( x +C )
Section 1.4 3

4. y (4) = π implies C = − 1 so y ( x ) = tan −1
4 ( x −1)
29. The population growth rate is k = ln(30000 / 25000) /10 ≈ 0.01823, so the population
of the city t years after 1960 is given by P(t ) = 25000 e0.01823t . The expected year 2000
population is then P(40) = 25000 e0.01823×40 ≈ 51840.
30. The population growth rate is k = ln(6) /10 ≈ 0.17918, so the population after t
hours is given by P(t ) = P0 e0.17918 t . To find how long it takes for the population to
double, we therefore need only solve the equation 2 P = P0 e0.17918 t for
t = (ln 2) / 0.17918 ≈ 3.87 hours.
31. As in the textbook discussion of radioactive decay, the number of 14C atoms after t
years is given by N (t ) = N 0 e−0.0001216 t . Hence we need only solve the equation
1
6
N 0 = N 0 e −0.0001216 t for t = (ln 6) / 0.0001216 ≈ 14735 years to find the age of the
skull.
32. As in Problem 31, the number of 14C atoms after t years is given by
N (t ) = 5.0 × 1010 e −0.0001216 t . Hence we need only solve the equation
4.6 × 1010 = 5.0 × 1010 e −0.0001216 t for the age t = ( ln (5.0 / 4.6) ) / 0.0001216 ≈ 686 years
of the relic. Thus it appears not to be a genuine relic of the time of Christ 2000 years
ago.
33. The amount in the account after t years is given by A(t ) = 5000 e0.08 t . Hence the
amount in the account after 18 years is given by A(20) = 5000 e0.08×20 ≈ 21,103.48
dollars.
34. When the book has been overdue for t years, the fine owed is given in dollars by
A(t ) = 0.30 e0.05 t . Hence the amount owed after 100 years is given by
A(100) = 0.30 e0.05×100 ≈ 44.52 dollars.
35. To find the decay rate of this drug in the dog's blood stream, we solve the equation
2 =
1
e −5 k (half-life 5 hours) for k = (ln 2) / 5 ≈ 0.13863. Thus the amount in the dog's
bloodstream after t hours is given by A(t ) = A0 e −0.13863t . We therefore solve the
equation A(1) = A0 e −0.13863 = 50 × 45 = 2250 for A0 ≈ 2585 mg, the amount to
anesthetize the dog properly.
36. To find the decay rate of radioactive cobalt, we solve the equation 1 = e −5.27 k (half-life
2
5.27 years) for k = (ln 2) / 5.27 ≈ 0.13153. Thus the amount of radioactive cobalt left
after t years is given by A(t ) = A0 e −0.13153t . We therefore solve the equation
Section 1.4 4

5. A(t ) = A0 e −0.13153t = 0.01 A0 for t = (ln100) / 0.13153 ≈ 35.01 and find that it will be
about 35 years until the region is again inhabitable.
37. Taking t = 0 when the body was formed and t = T now, the amount Q(t) of lead in
the body at time t (in years) is given by Q(t) = Q0e–kt, where k = (ln 2)/(4.51×109).
The given information tells us that
Q(T )
= 0.9 .
Q0 − Q(T )
After substituting Q(T) = Q0e–kT, we solve readily for ekT = 19/9, so
t = (1/k)ln(19/9) ≈ 4.86×109. Thus the body was formed approximately 4.86 billion
years ago.
38. Taking t = 0 when the rock contained only potassium and t = T now, the amount
Q(t) of potassium in the rock at time t (in years) is given by Q(t) = Q0e–kt, where
k = (ln 2)/(1.28×109). The given information tells us that the amount A(t) of argon at
time t is
A(t ) = 1
9 [Q0 − Q(t )]
and also that A(T) = Q(T). Thus
Q0 − Q(T ) = 9 Q(T ).
After substituting Q(T ) = Q0 e − kT we readily solve for
T = (ln 10 / ln 2)(1.28 × 10 9 ) ≈ 4.25 × 10 9 .
Thus the age of the rock is about 1.25 billion years.
39. Because A = 0 the differential equation reduces to T' = kT, so T(t) = 25e–kt. The
fact that T(20) = 15 yields k = (1/20)ln(5/3), and finally we solve
5 = 25e–kt for t = (ln 5)/k ≈ 63 min.
40. The amount of sugar remaining undissolved after t minutes is given by A(t ) = A0 e− kt ;
we find the value of k by solving the equation A(1) = A0 e− k = 0.75 A0 for
k = − ln 0.75 ≈ 0.28768. To find how long it takes for half the sugar to dissolve, we solve
the equation A(t ) = A0 e− kt = 1 A0 for t = (ln 2) / 0.28768 ≈ 2.41 minutes.
2
41. (a) The light intensity at a depth of x meters is given by I ( x ) = I 0e −1.4 x . We solve
the equation I ( x) = I 0 e −1.4 x = 1 I 0 for x = (ln 2) /1.4 ≈ 0.495 meters.
2
Section 1.4 5

6. (b) At depth 10 meters the intensity is I (10) = I 0e −1.4×10 ≈ (8.32 × 10−7 ) I 0 .
(c) We solve the equation I ( x ) = I 0e −1.4 x = 0.01I 0 for x = (ln100) /1.4 ≈ 3.29 meters.
42. (a) The pressure at an altitude of x miles is given by p( x) = 29.92 e−0.2 x . Hence the
pressure at altitude 10000 ft is p (10000 / 5280) ≈ 20.49 inches, and the pressure at
altitude 30000 ft is p (30000 / 5280) ≈ 9.60 inches.
(b) To find the altitude where p = 15 in., we solve the equation 29.92 e −0.2 x = 15 for
x = (ln 29.92 /15) / 0.2 ≈ 3.452 miles ≈ 18, 200 ft.
43. (a) A′ = rA + Q
(b) The solution of the differential equation with A(0) = 0 is given by
r A + Q = Q er t .
When we substitute A = 40 (thousand), r = 0.11, and t = 18, we find that
Q = 0.70482, that is, $704.82 per year.
44. Let N8 (t ) and N 5 (t ) be the numbers of 238U and 235U atoms, respectively, at time t (in
billions of years after the creation of the universe). Then N8 (t ) = N 0 e − k t and
N 5 (t ) = N 0 e− ct , where N 0 is the initial number of atoms of each isotope. Also,
k = (ln 2) / 4.51 and c = (ln 2) / 0.71 from the given half-lives. We divide the equations
for N8 and N 5 and find that when t has the value corresponding to "now",
N8
e( c − k ) t = = 137.7.
N5
Finally we solve this last equation for t = (ln137.7) /(c − k ) ≈ 5.99. Thus we get an
estimate of about 6 billion years for the age of the universe.
45. The cake's temperature will be 100° after 66 min 40 sec; this problem is just like
Example 6 in the text.
46. (b) By separating the variables we solve the differential equation for
c – r P(t) = (c – r P0) er t.
With P(t) = 0 this yields
Section 1.4 6

7. c = r P0 er t / (er t – 1).
With P0 = 10,800, t = 60, and r = 0.010 we get $239.37 for the monthly payment
at 12% annual interest. With r = 0.015 we get $272.99 for the monthly payment at
18% annual interest.
47. If N(t) denotes the number of people (in thousands) who have heard the rumor after t
days, then the initial value problem is
N′ = k(100 – N), N(0) = 0
and we are given that N(7) = 10. When we separate variables ( dN /(100 − N ) = k dt )
and integrate, we get ln(100 − N ) = − kt + C , and the initial condition N (0) = 0 gives
C = ln 100. Then 100 − N = 100 e − kt , so N (t ) = 100 (1 − e − kt ). We substitute t = 7,
N = 10 and solve for the value k = ln(100 / 90) / 7 ≈ 0.01505. Finally, 50 thousand
people have heard the rumor after t = (ln 2) / k ≈ 46.05 days.
48. With A(y) constant, Equation (19) in the text takes the form
dy
= k y
dt
We readily solve this equation for 2 y = kt + C . The condition y(0) = 9 yields
C = 6, and then y(1) = 4 yields k = 2. Thus the depth at time t (in hours) is
y(t) = (3 – t)2, and hence it takes 3 hours for the tank to empty.
49. With A = π (3)2 and a = π (1 / 12)2 , and taking g = 32 ft/sec2, Equation (20) reduces
to 162 y′ = – y . The solution such that y = 9 when t = 0 is given by
324 y = –t + 972. Hence y = 0 when t = 972 sec = 16 min 12 sec.
50. The radius of the cross-section of the cone at height y is proportional to y, so A(y) is
proportional to y2. Therefore Equation (20) takes the form
y2 y′ = − k y ,
and a general solution is given by
2y5/2 = –5kt + C.
The initial condition y(0) = 16 yields C = 2048, and then y(1) = 9 implies that
5k = 1562. Hence y = 0 when
t = C/5k = 2048/1562 ≈ 1.31 hr.
Section 1.4 7

8. 51. The solution of y′ = –k y is given by
2 y = –kt + C.
The initial condition y(0) = h (the height of the cylinder) yields C = 2 h . Then
substitution of t = T, y = 0 gives k = (2 h )/T. It follows that
y = h(1 – t/T)2.
If r denotes the radius of the cylinder, then
V ( y) = π r2 y = π r 2 h(1 − t / T ) 2 = V0 (1 − t / T )2.
52. Since x = y3/4, the cross-sectional area is A( y ) = π x 2 = π y 3/ 2 . Hence the
general equation A( y) y ′ = − a 2 gy reduces to the differential equation yy′ = − k
with general solution
(1/2)y2 = –kt + C.
The initial condition y(0) = 12 gives C = 72, and then y(1) = 6 yields k = 54.
Upon separating variables and integrating, we find that the the depth at time t is
y (t ) = 144 − 108 t y(t).
Hence the tank is empty after t = 144/108 hr, that is, at 1:20 p.m.
53. (a) Since x2 = by, the cross-sectional area is A( y ) = π x 2 = π by. Hence the
equation A( y ) y′ = − a 2 gy reduces to the differential equation
y1/ 2 y′ = − k = − (a / π b) 2 g
with the general solution
(2/3)y3/2 = –kt + C.
The initial condition y(0) = 4 gives C = 16/3, and then y(1) = 1 yields k = 14/3.
It follows that the depth at time t is
y(t) = (8 – 7t)2/3.
(b) The tank is empty after t = 8/7 hr, that is, at 1:08:34 p.m.
Section 1.4 8

9. We see above that k = (a/πb)√(2g) = 14/3. Substitution of a = π r , b = 1,
2
(c)
g = (32)(3600)2 ft/hr2 yields r = (1/60) √(7/12) ft ≈ 0.15 in for the radius of the
bottom-hole.
54. With g = 32 ft/sec2 and a = π (1 / 12)2 , Equation (24) simplifies to
dy π
A( y) = − y.
dt 18
If z denotes the distance from the center of the cylinder down to the fluid surface, then
y = 3 – z and A(y) = 10(9 – z2)1/2. Hence the equation above becomes
dz π
0 (9 − z 2 )1/ 2 = (3 − z )1/ 2 ,
dt 18
80 (3 + z )1/ 2 dz = π dt,
and integration yields
20(3 + z )1/ 2 = π t + C.
Now z = 0 when t = 0, so C = 120(3)3/2. The tank is empty when z = 3 (that is,
when y = 0) and thus after
t = (120/π)(63/2 – 33/2) ≈ 362.90 sec.
It therefore takes about 6 min 3 sec for the fluid to drain completely.
55. A( y) = π (8 y − y 2 ) as in Example 7 in the text, but now a = π / 144 in Equation (24),
so the initial value problem is
18(8y – y2)y′ = – y , y(0) = 8.
We seek the value of t when y = 0. The answer is t ≈ 869 sec = 14 min 29 sec.
56. The cross-sectional area function for the tank is A = π (1 − y 2 ) and the area of the
bottom-hole is a = 10 −4 π , so Eq. (24) in the text gives the initial value problem
dy
π (1 − y 2 ) = − 10 −4 π 2 × 9.8 y , y(0) = 1.
dt
Simplification gives
y −1 / 2 − y 3 / 2 % dy
dt
= − 1.4 × 10 −4 10
Section 1.4 9

10. so integration yields
2 5/ 2
2 y1 / 2 − y = − 1.4 × 10 −4 10 t + C.
5
The initial condition y(0) = 1 implies that C = 2 - 2/5 = 8/5, so y = 0 after
t = (8 / 5) / (1.4 × 10 −4 10 ) ≈ 3614 seconds. Thus the tank is empty at about 14
seconds after 2 pm.
57. (a) As in Example 8, the initial value problem is
dy
π (8 y − y 2 ) = − π k y, y( 0 ) = 4
dt
where k = 0.6 r 2 2 g = 4.8 r 2 . Integrating and applying the initial condition just in the
Example 8 solution in the text, we find that
16 3 / 2 2 5 / 2 448
y − y = − kt + .
3 5 15
When we substitute y = 2 (ft) and t = 1800 (sec, that is, 30 min), we find that
k ≈ 0.009469. Finally, y = 0 when
448
t = ≈ 3154 sec = 53 min 34 sec.
15k
Thus the tank is empty at 1:53:34 pm.
(b) The radius of the bottom-hole is
r = k / 4.8 ≈ 0.04442 ft ≈ 0.53 in, thus about a half inch.
58. The given rate of fall of the water level is dy/dt = –4 in/hr = –(1/10800) ft/sec. With
A = πx 2 and a = πr 2 , Equation (24) is
(πx 2 )(1 / 10800 ) = − (πr 2 ) 2 gy = − 8πr 2 y .
Hence the curve is of the form y = kx4, and in order that it pass through (1, 4) we
must have k = 4. Comparing y = 2x2 with the equation above, we see that
(8r2)(10800) = 1/2,
so the radius of the bottom hole is r = 1 / (240 3 ) ft ≈ 1 / 35 in.
Section 1.4 10

11. 59. Let t = 0 at the time of death. Then the solution of the initial value problem
T' = k(70 – T), T(0) = 98.6
is
T (t ) = 70 + 28.6 e − kt .
If t = a at 12 noon, then we know that
T (t ) = 70 + 28.6 e − ka = 80,
T ( a + 1) = 70 + 28.6 e − k ( a +1) = 75.
Hence
28.6 e − ka = 10 and 28.6 e − ka e − k = 5.
It follows that e–k = 1/2, so k = ln 2. Finally the first of the previous two equations
yields
a = (ln 2.86)/(ln 2) ≈ 1.516 hr ≈ 1 hr 31 min,
so the death occurred at 10:29 a.m.
60. Let t = 0 when it began to snow, and t = t0 at 7:00 a.m. Let x denote distance along
the road, with x = 0 where the snowplow begins at 7:00 a.m. If y = ct is the snow
depth at time t, w is the width of the road, and v = dx/dt is the plow′s velocity, then
"plowing at a constant rate" means that the product wyv is constant. Hence our
differential equation is of the form
dx 1
k = .
dt t
The solution with x = 0 when t = t0 is
t = t0 ekx.
We are given that x = 2 when t = t0 + 1 and x = 4 when t = t0 + 3, so it follows
that
t0 + 1 = t0 e2k and t0 + 3 = t0 e4k.
Elimination of t0 yields the equation
e4k – 3e2k + 2 = (e2k – 1)(e2k – 2) = 0,
Section 1.4 11

12. so it follows (since k > 0) that e2k = 2. Hence t0 + 1 = 2t0, so t0 = 1. Thus it began
to snow at 6 a.m.
61. We still have t = t0 ekx, but now the given information yields the conditions
t0 + 1 = t0 e4k and t0 + 2 = t0 e7k
at 8 a.m. and 9 a.m., respectively. Elimination of t0 gives the equation
2e4k – e7k – 1 = 0,
which we solve numerically for k = 0.08276. Using this value, we finally solve one of
the preceding pair of equations for t0 = 2.5483 hr ≈ 2 hr 33 min. Thus it began to
snow at 4:27 a.m.
Section 1.4 12