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© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 6 Differential Equations
Section 6.1 Slope Fields and Euler’s Method.......................................................593
Section 6.2 Differential Equations: Growth and Decay.......................................605
Section 6.3 Separation of Variables and Logistic Equation.................................616
Section 6.4 First-Order Linear Differential Equations .........................................629
Review Exercises ........................................................................................................639
Problem Solving .........................................................................................................649
© 2010 Brooks/Cole, Cengage Learning 593
C H A P T E R 6 Differential Equations
Section 6.1 Slope Fields and Euler’s Method
1. Differential equation: 4y y′ =
Solution: 4xy Ce=
Check: 44 4xy Ce y′ = =
2. Differential Equation: 23 5 xy y e−′ + = −
Solution: 2
22
x
x
y e
y e
−
−
=
′ = −
Check: ( ) ( )2 2 23 2 5x x xe e e− − −− + = −
3. Differential equation: 2 22xyy
x y′ =
−
Solution: 2 2x y Cy+ =
Check:
( )
( )
2
2 2 2
2 2
2 2
2 22
2
22
22
2
2
x yy Cyxy
y C
xyyy Cy
xyy x y
xyy x
xyx y
′ ′+ =
−′ =−
−′ =−
−=
− +
−=
−
=−
4. Differential Equation: 2 1dy xydx y
=−
Solution: 2 22 lny y x− =
Check:
2
22 2
1
1
1
yy y xy
y y xy
xyy
yxyy
y
′ ′− =
⎛ ⎞′− =⎜ ⎟
⎝ ⎠
′ =−
′ =−
5. Differential Equation: 0y y′′ + =
Solution: 1 2
1 2
1 2
sin coscos sin
sin cos
y C x C xy C x C xy C x C x
= −
′ = +′′ = − +
Check: ( ) ( )1 2 1 2sin cos sin cos 0y y C x C x C x C x′′ + = − + + − =
6. Differential equation: 2 2 0y y y′′ ′+ + =
Solution: 1 2cos sinx xy C e x C e x− −= +
Check: ( ) ( )
( ) ( )( ) ( )( ) ( )
1 2 1 2
1 2
1 2
1 2 1 2 1 2
1 1 2 2 2 1 2 1
sin cos
2 sin 2 cos
2 2 2 sin 2 cos
2 sin cos 2 cos sin
2 2 2 2 sin 2 2 2 2 cos 0
x x
x x
x x
x x x x
x x
y C C e x C C e x
y C e x C e x
y y y C e x C e x
C C e x C C e x C e x C e x
C C C C e x C C C C e x
− −
− −
− −
− − − −
− −
′ = − + + − +
′′ = −
′′ ′+ + = − +
− + + − + + +
= − − + + − − + + =
594 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
7. Differential Equation: tany y x′′ + =
Solution:
( ) ( )( ) ( )( )
( ) ( )( )( )
2
2
cos ln sec tan
1cos sec tan sec sin ln sec tansec tan
cossec tan sec sin ln sec tan
sec tan
1 sin ln sec tan
1sin sec tan sec cos ln sec tansec tan
sin sec cos ln sec
y x x x
y x x x x x x xx x
xx x x x x x
x x
x x x
y x x x x x x xx x
x x x x
= − +
′ = − ⋅ + + ++
−= + + +
+
= − + +
′′ = ⋅ + + ++
= + + tan x
Check: ( )( )sin sec cos ln sec tan cos ln sec tan tan .y y x x x x x x x x x′′ + = + + − + =
8. Differential Equation: 4 2 xy y e′′ ′+ =
Solution: ( )
( )
4
4 4
4
252 8 245 5 532 25 5
x x
x x x x
x x
y e e
y e e e e
y e e
−
− −
−
= +
′ = − + = − +
′′ = +
Check: 4 432 2 8 2 2 84 4 25 5 5 5 5 5
x x x x x xy y e e e e e e− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞′′ ′+ = + + − + = + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
9. 2
2 2
2
sin cos cos
sin cos 2 cos sin
1 2 cos sin 2
y x x x
y x x x x
x x
= −
′ = − + +
= − + +
Differential Equation:
( ) ( )2 22 2 sin cos cos 1 2 cos sin 2
2 sin cos 1 sin 22 sin 2 1
y y x x x x x
x x xx
′+ = − + − + +
= − +
= −
Initial condition , 04π⎛ ⎞⎜ ⎟⎝ ⎠
:
2
2 2 2 2sin cos cos 04 4 4 2 2 2π π π ⎛ ⎞
− = ⋅ − =⎜ ⎟⎜ ⎟⎝ ⎠
10. 212 2 cos 3
2 sin
y x x
y x x
= − −
′ = +
Differential equation: 2 siny x x′ = +
Initial condition ( )0, 5 :− 0 2 cos 0 3 5− − = −
11.
( )
26
2 26 6
4
4 12 48
x
x x
y e
y e x xe
−
− −
=
′ = − = −
Differential equation:
( )2 26 612 12 4 48x xy xy x e xe− −′ = − = − = −
Initial condition ( )0, 4 : 04 4e =
12.
( )
cos
cos cossin sin
x
x x
y e
y e x x e
−
− −
=
′ = = ⋅
Differential Equation:
( )cossin sin sinxy x e x y y x−′ = ⋅ = =
Initial condition , 1 :2π⎛ ⎞⎜ ⎟⎝ ⎠
( )cos 2 0 1e eπ− = =
Section 6.1 Slope Fields and Euler's Method 595
© 2010 Brooks/Cole, Cengage Learning
In Exercises 13–20, the differential equation is ( )4 16 = 0.y y−
13. ( )
( )
4
4
3 cos
3 cos
16 45 cos 0,
y x
y x
y y x
=
=
− = − ≠
No
14. ( )
( ) ( )
4
4
2 sin
2 sin
16 2 sin 16 2 sin 0
y x
y x
y y x x
=
=
− = − ≠
No
15. ( )
( )
4
4
3 cos 2
48 cos 2
16 48 cos 2 48 cos 2 0,
y x
y x
y y x x
=
=
− = − =
Yes
16. ( )
( ) ( )
4
4
3 sin 2
48 sin 2
16 48 sin 2 16 3 sin 2 0
y x
y x
y y x x
=
=
− = − =
Yes
17. ( )
( )
2
4 2
4 2 2
16
16 16 16 0,
x
x
x x
y e
y e
y y e e
−
−
− −
=
=
− = − =
Yes
18.
( )
( )
44
44
5 ln30
3016 80 ln 0,
y x
yx
y y xx
=
= −
− = − − ≠
No
19. ( )
( )
2 21 2 3 4
4 2 21 2 3 4
4
sin 2 cos 2
16 16 16 sin 2 16 cos 2
16 0,
x x
x x
y C e C e C x C x
y C e C e C x C x
y y
−
−
= + + +
= + + +
− =
Yes
20. ( )
( ) ( ) ( )
2
4 2
4 2 2
3 4 sin 2
48 64 sin 2
16 48 64 sin 2 16 3 4 sin 2 0,
x
x
x x
y e x
y e x
y y e x e x
= −
= −
− = − − − =
Yes
In Exercises 21–28, the differential equation is 32 = .xxy y x e′ −
21.
( ) ( )2
2 3
, 2
2 2 2 0 ,x
y x y x
xy y x x x x e
′= =
′ − = − = ≠
No
22.
( )3 2
2 3 3 3
, 3
2 3 2 x
y x y x
xy y x x x x x e
′= =
′ − = − = ≠
No
23. ( )( )( ) ( )
2 2 2
2 2 3
, 2 2
2 2 2 ,
x x x x
x x x
y x e y x e xe e x x
xy y x e x x x e x e
′= = + = +
′ − = + − =
Yes
24. ( ) ( ) ( )2 22 , 2 2x x xy x e y x e x e′= + = + +
2 2 2
3
2 2 4 2 2
,
x x x
x
xy y x x e xe x x e x
x e
′ ⎡ ⎤ ⎡ ⎤− = + + − +⎣ ⎦ ⎣ ⎦
=
Yes
25.
( ) ( ) 3
sin , cos
2 cos 2 sin ,x
y x y x
xy y x x x x e
′= =
′ − = − ≠
No
26.
( ) 3
cos , sin
2 sin 2 cos x
y x y x
xy y x x x x e
′= = −
′ − = − − ≠
No
27.
3
1ln ,
12 2 ln ,x
y x yx
xy y x x x ex
′= =
⎛ ⎞′ − = − ≠⎜ ⎟⎝ ⎠
No
28. 2 2 25 , 2 10x x xy x e x y x e xe x′= − = + −
2 2 2
3
2 2 10 2 5
,
x x x
x
xy y x x e xe x x e x
x e
′ ⎡ ⎤ ⎡ ⎤− = + − − −⎣ ⎦ ⎣ ⎦
=
Yes
596 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
29. 2xy Ce−= passes through ( )0, 3 .
03 3Ce C C= = ⇒ =
Particular solution: 23 xy e−=
30. ( )2y x y C+ = passes through ( )0, 2 .
( )2 0 2 4C C+ = ⇒ =
Particular solution: ( )2 4y x y+ =
31. 2 3y Cx= passes through ( )4, 4 .
( ) 1416 64C C= ⇒ =
Particular solution: 2 314y x= or 2 34y x=
32. 2 22x y C− = passes through ( )3, 4 .
( )2 9 16 2C C− = ⇒ =
Particular solution: 2 22 2x y− =
33. Differential equation: 4 0yy x′ − =
General solution: 2 24y x C− =
Particular solutions: 0, Two intersecting lines1, 4, Hyperbolas
CC C
=
= ± = ±
34. Differential equation: 0yy x′ + =
General solution: 2 2x y C+ =
Particular solutions: 0,C = Point 1, 4,C C= = Circles
35. Differential equation: 2 0y y′ + =
General solution: 2xy Ce−=
( ) ( )2 22 2 2 0x xy y C e Ce− −′ + = − + =
Initial condition ( ) 00, 3 : 3 Ce C= =
Particular solution: 23 xy e−=
36. Differential equation: 3 2 0x yy′+ =
General solution: 2 23 2x y C+ =
( )6 4 0
2 3 2 0
3 2 0
x yy
x yy
x yy
′+ =
′+ =
′+ =
Initial condition ( )1, 3 :
( ) ( )2 23 1 2 3 3 18 21 C+ = + = =
Particular solution: 2 23 2 21x y+ =
−3 3
−2
C = 0
2
−3 3
−2
C = 1
2
−3 3
−2
C = −1
2
−3 3
−2
C = 4
2
−3 3
−2
C = −4
2
x1 2
1
2
y
Section 6.1 Slope Fields and Euler's Method 597
© 2010 Brooks/Cole, Cengage Learning
37. Differential equation: 9 0y y′′ + =
General solution: 1 2sin 3 cos 3y C x C x= +
( ) ( )
1 2
1 2
1 2 1 2
3 cos 3 3 sin 3 ,9 sin 3 9 cos 3
9 9 sin 3 9 cos 3 9 sin 3 cos 3 0
y C x C xy C x C x
y y C x C x C x C x
′ = −
′′ = − −
′′ + = − − + + =
Initial conditions , 26π⎛ ⎞⎜ ⎟⎝ ⎠
and 1y′ = when :6
x π=
1 2 1
1 2
1 2 2 2
2 sin cos 22 2
3 cos 3 3 sin 3
11 3 cos 3 sin 32 2 3
C C C
y C x C x
C C C C
π π
π π
⎛ ⎞ ⎛ ⎞= + ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
′ = −
⎛ ⎞ ⎛ ⎞= − = − ⇒ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Particular solution: 12 sin 3 cos 33
y x x= −
38. Differential equation: 0xy y′′ ′+ =
General solution: 1 2 lny C C x= +
2 2 2
2 22
1 1,
1 1 0
y C y Cx x
xy y x C Cx x
⎛ ⎞ ⎛ ⎞′ ′′= = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞′′ ′+ = − + =⎜ ⎟⎝ ⎠
Initial conditions ( )2, 0 and 12
y′ = when 2:x =
1 2
2
22 1
0 ln 2
1 1, ln 22 2
C CCyx
C C C
= +
′ =
= ⇒ = = −
Particular solution: ln 2 ln ln2xy x= − + =
39. Differential equation: 2 3 3 0x y xy y′′ ′− + =
General solution: 31 2y C x C x= +
( ) ( ) ( )2
1 2 2
2 2 2 32 1 2 1 2
3 , 6
3 3 6 3 3 3 0
y C C x y C x
x y xy y x C x x C C x C x C x
′ ′′= + =
′′ ′− + = − + + + =
Initial conditions ( )2, 0 and 4y′ = when 2:x =
1 2
21 2
1 2
0 2 8
34 12
C C
y C C xC C
= +
′ = +
= +
1 22 1
1 2
12
4 0, 2
12 4C C
C CC C
+ = ⎫= = −⎬
+ = ⎭
Particular solution: 3122y x x= − +
598 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
40. Differential equation: 9 12 4 0y y y′′ ′− + =
General solution: ( )2 31 2
xy e C C x= +
( ) ( )( ) ( )
( )( ) ( )( ) ( )( )
2 3 2 3 2 31 2 2 1 2 2
2 3 2 3 2 31 2 2 2 1 2 2
2 3 2 3 2 31 2 2 1 2 2 1 2
2 2 23 3 3
2 2 2 2 2 2 23 3 3 3 3 3 3
2 2 2 2 23 3 3 3 3
2
9 12 4 9 2 12 4 0
x x x
x x x
x x x
y e C C x C e e C C C x
y e C C C x e C e C C C x
y y y e C C C x e C C C x e C C x
′ = + + = + +
′′ = + + + = + +
′′ ′− + = + + − + + + + =
Initial conditions ( )0, 4 and ( )3, 0 :
( )( )( )( )
21 2
1 1
22 2
43
0 3
4 1 0 4
0 4 3
e C C
C C
e C C
= +
= + ⇒ =
= + ⇒ = −
Particular solution: ( )2 3 434xy e x= −
41. 2
2 3
6
6 2
dy xdx
y x dx x C
=
= = +∫
42.
( )
3
43 2
2 3
32 32 2
dy x xdx
xy x x dx x C
= −
= − = − +∫
43.
( )( )
2
22
2
11 ln 1
1 21 , 2
dy xdx x
xy dx x Cx
u x du x dx
=+
= = + ++
= + =
∫
44.
( )4
ln 44
x
x
xx
x
dy edx e
ey dx e Ce
=+
= = + ++∫
45.
2
2 21
21
2 ln ln
dy xdx x x
y dxx
x x C x x C
−= = −
⎛ ⎞= −⎜ ⎟⎝ ⎠
= − + = − +
∫
46.
( ) ( )( )
2
2 2
2
cos
1cos sin2
, 2
dy x xdx
y x x dx x C
u x du x dx
=
= = +
= =
∫
47.
( )
sin 2
1sin 2 cos 22
2 , 2
dy xdx
y x dx x C
u x du dx
=
= = − +
= =
∫
48.
( )
2 2
2
tan sec 1
sec 1 tan
dy x xdxy x dx x x C
= = −
= − = − +∫
49. 6dy x xdx
= −
Let 6,u x= − then 2 6x u= + and 2 .dx u du=
y = ( )( )( )
( )
( ) ( )
( ) ( )
( ) ( )
2
4 2
53
5 2 3 2
3 2
3 2
6 6 2
2 6
2 25
2 6 4 652 6 6 1052 6 45
x x dx u u u du
u u du
u u C
x x C
x x C
x x C
− = +
= +
⎛ ⎞= + +⎜ ⎟
⎝ ⎠
= − + − +
= − − + +
= − + +
∫ ∫∫
Section 6.1 Slope Fields and Euler's Method 599
© 2010 Brooks/Cole, Cengage Learning
50. 2 3dy x xdx
= −
Let 3 ,u x= − then 23x u= − and 2dx u du= −
( ) ( )
( )
( ) ( )
( ) ( )
( ) ( )
2
4 2
53
5 2 3 2
3 2
3 2
2 3 2 3 2
4 12
4 45
4 3 4 354 3 3 55
4 3 25
y x x dx u u u du
u u du
u u C
x x C
x x C
x x C
= − = − −
= −
= − +
= − − − +
= − − − +
= − − + +
∫ ∫∫
51. 2
2 212
x
x x
dy xedx
y xe dx e C
=
= = +∫
( )2 , 2u x du x dx= =
52. 25 xdy edx
−=
( )2 2 215 5 2 102
x x xy e dx e dx e C− − −⎛ ⎞= = − − = − +⎜ ⎟⎝ ⎠∫ ∫
53.
54.
55.
56.
57. sin 2dy xdx
=
For 0, 0.dyxdx
= = Matches (b).
58. 1 cos2
dy xdx
=
For 10, .2
dyxdx
= = Matches (c).
59. 2xdy edx
−=
As , 0.dyxdx
→ ∞ → Matches (d).
60. 1dydx x
=
For 0, dyxdx
= is undefined (vertical tangent). Matches (a).
61. (a), (b)
(c) As ,x y→ ∞ → −∞
As ,x y→ −∞ → −∞
62. (a), (b)
(c) As ,x y→ ∞ → ∞
As ,x y→ −∞ → −∞
x –4 –2 0 2 4 8
y 2 0 4 4 6 8
dy dx –4 Undef. 0 1 43 2
x –4 –2 0 2 4 8
y 2 0 4 4 6 8
dy dx 6 2 4 2 2 0
x –4 –2 0 2 4 8
y 2 0 4 4 6 8
dy dx 2 2− –2 0 0 2 2− –8
x –4 –2 0 2 4 8
y 2 0 4 4 6 8
dy dx 3 0 3− 3− 0 3
8
5
y
x
(4, 2)
−2
4
−4
4
x
y
600 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
63. (a), (b)
(c) As ,x y→ ∞ → −∞
As ,x y→ −∞ → −∞
64. (a), (b)
(c) As ,x y→ ∞ → −∞
As ,x y→ −∞ → −∞
65. (a) ( )1, 1, 0yx
′ =
As ,x y→ ∞ → ∞
[Note: The solution is ln .y x= ]
(b) ( )1, 2, 1yx
′ = −
As ,x y→ ∞ → ∞
66. (a) ( )1 , 0, 1yy
′ =
As ,x y→ ∞ → ∞
(b) ( )1 , 1, 1yy
′ =
As ,x y→ ∞ → ∞
67. ( )0.25 , 0 4dy y ydx
= =
(a), (b)
68. ( )4 , 0 6dy y ydx
= − =
(a), (b)
69. ( ) ( )0.02 10 , 0 2dy y y ydx
= − =
(a), (b)
4
−3
5
x
y(2, 2)
−4
y
x
2
−2
−4
−6
2−2−4
(0, −4)
6−1
−2
−3
1
2
3
x
y
(1, 0)
6−1
−2
−3
1
2
3
x
y
(2, −1)
y
x
(0, 1)
3
3
−3
y
x
(1, 1)
3
3
−3
−6 6
−4
12
−50
10
5
−12 48
−2
12
Section 6.1 Slope Fields and Euler's Method 601
© 2010 Brooks/Cole, Cengage Learning
70. ( ) ( )0.2 2 , 0 9dy x y ydx
= − =
(a), (b)
71. ( ) ( )0.4 3 , 0 1dy y x ydx
= − =
(a), (b)
72. ( )81 sin , 0 22 4
xdy ye ydx
π−= =
(a), (b)
73. ( )( ) ( )( )( ) ( )( )
1 0 0 0
2 1 1 1
, 0 2, 10, 0.1
, 2 0.1 0 2 2.2
, 2.2 0.1 0.1 2.2 2.43, etc.
y x y y n h
y y hF x y
y y hF x y
′ = + = = =
= + = + + =
= + = + + =
74. ( )( ) ( )( )( ) ( )( )
1 0 0 0
2 1 1 1
, 0 2, 20, 0.05
, 2 0.05 0 2 2.1
, 2.1 0.05 0.05 2.1 2.2075, etc.
y x y y n h
y y hF x y
y y hF x y
′ = + = = =
= + = + + =
= + = + + =
The table shows the values for 0, 2, 4, , 20.n = …
75. ( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )
1 0 0 0
2 1 1 1
3 2 , 0 3, 10, 0.05
, 3 0.05 3 0 2 3 2.7
, 2.7 0.05 3 0.05 2 2.7 2.4375, etc.
y x y y n h
y y hF x y
y y hF x y
′ = − = = =
= + = + − =
= + = + + =
n 0 1 2 3 4 5 6 7 8 9 10
nx 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
ny 2 2.2 2.43 2.693 2.992 3.332 3.715 4.146 4.631 5.174 5.781
n 0 2 4 6 8 10 12 14 16 18 20
nx 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
ny 2 2.208 2.447 2.720 3.032 3.387 3.788 4.240 4.749 5.320 5.960
n 0 1 2 3 4 5 6 7 8 9 10
nx 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
ny 3 2.7 2.438 2.209 2.010 1.839 1.693 1.569 1.464 1.378 1.308
−5 50
10
8
−2
−2
8
3
−3
−3
5
602 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
76. ( ) ( )( ) ( ) ( )( )( )( ) ( ) ( )( )( )
1 0 0 0
2 1 1 1
0.5 3 , 0 1, 5, 0.4
, 1 0.4 0.5 0 3 1 1
, 1 0.4 0.5 0.4 3 1 1.16, etc.
y x y y n h
y y hF x y
y y hF x y
′ = − = = =
= + = + − =
= + = + − =
77. ( )( ) ( ) ( )
( ) ( ) ( )( )
0 11 0 0 0
0.1 1.12 1 1 1
, 0 1, 10, 0.1
, 1 0.1 1.1
, 1.1 0.1 1.2116, etc.
xyy e y n h
y y hF x y e
y y hF x y e
′ = = = =
= + = + =
= + = + ≈
78. ( )( ) ( )( )( ) ( ) ( ) ( )( )
1 0 0 0
2 1 1 1
cos sin , 0 5, 10, 0.1
, 5 0.1 cos 0 sin 5 5.0041
, 5.0041 0.1 cos 0.1 sin 5.0041 5.0078, etc.
y x y y n h
y y hF x y
y y hF x y
′ = + = = =
= + = + + ≈
= + = + + ≈
79. ( ), 3 , 0, 3xdy y y edx
= =
80. ( )22 , 2 4, 0, 2dy x y xdx y
= = +
n 0 1 2 3 4 5
nx 0 0.4 0.8 1.2 1.6 2.0
ny 1 1 1.16 1.454 1.825 2.201
n 0 1 2 3 4 5 6 7 8 9 10
nx 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
ny 1 1.1 1.212 1.339 1.488 1.670 1.900 2.213 2.684 3.540 5.958
n 0 1 2 3 4 5 6 7 8 9 10
nx 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
ny 5 5.004 5.008 5.010 5.010 5.007 4.999 4.985 4.965 4.938 4.903
x 0 0.2 0.4 0.6 0.8 1
( )y x (exact) 3 3.6642 4.4755 5.4664 6.6766 8.1548
( ) ( )0.2y x h = 3 3.6000 4.3200 5.1840 6.2208 7.4650
( ) ( )0.1y x h = 3 3.6300 4.3923 5.3147 6.4308 7.7812
x 0 0.2 0.4 0.6 0.8 1
( )y x (exact) 2 2.0199 2.0785 2.1726 2.2978 2.4495
( ) ( )0.2y x h = 2 2.000 2.0400 2.1184 2.2317 2.3751
( ) ( )0.1y x h = 2 2.0100 2.0595 2.1460 2.2655 2.4131
Section 6.1 Slope Fields and Euler's Method 603
© 2010 Brooks/Cole, Cengage Learning
81. ( ) ( )1cos , sin cos , 0, 02
xdy y x y x x edx
= + = − +
82. As h increases (from 0.1 to 0.2), the error increases.
83. ( ) ( )1 72 , 0, 140 , 0.12
dy y hdt
= − − =
(a)
(b) 272 68 exactty e−= +
(c) ( ) ( )1 72 , 0, 140 , 0.052
dy y hdt
= − − =
The approximations are better using 0.05.h =
84. When 0, 0,x y′= = therefore (d) is not possible.
When , 0, 0x y y′> < (decreasing function) therefore (c) is the equation.
85. The general solution is a family of curves that satisfies the differential equation. A particular solution is one member of the family that satisfies given conditions.
86. A slope field for the differential equation ( ),y F x y′ = consists of small line segments at various
points ( ),x y in the plane. The line segment equals the
slope ( ),y F x y′ = of the solution y at the point ( ), .x y
87. Consider ( ) ( )0 0, , .y F x y y x y′ = = Begin with a point
( )0 0,x y that satisfies the initial condition, ( )0 0.y x y=
Then, using a step size of h, find the point ( ) ( )( )1 1 0 0 0 0, , , .x y x h y hF x y= + + Continue
generating the sequence of points ( ) ( )( )1 1, , , .n n n n n nx y x h y hF x y+ + = + +
88. kx
kx
y Cedy Ckedx
=
=
Because 0.07 ,dy dx y= you have 0.07 .kx kxCke Ce=
So, 0.07.k =
C cannot be determined.
89. False. Consider Example 2. 3y x= is a solution to
3 0,xy y′ − = but 3 1y x= + is not a solution.
90. True
91. True
92. False. The slope field could represent many different differential equations, such as 2 4 .y x y′ = +
t 0 1 2 3
Euler 140 112.7 96.4 86.6
x 0 0.2 0.4 0.6 0.8 1
( )y x (exact) 0 0.2200 0.4801 0.7807 1.1231 1.5097
( ) ( )0.2y x h = 0 0.2000 0.4360 0.7074 1.0140 1.3561
( ) ( )0.1y x h = 0 0.2095 0.4568 0.7418 1.0649 1.4273
t 0 1 2 3
Exact 140 113.24 97.016 87.173
t 0 1 2 3
Euler 140 112.98 96.7 86.9
604 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
93. ( ) 22 , 0 4, 4 xdy y y y edx
−= − = =
(a)
(b) If h is halved, then the error is approximately halved ( )0.5 .r ≈
(c) When 0.05,h = the errors will again be approximately halved.
94. ( ), 0 1, 1 2 xdy x y y y x edx
−= − = = − +
(a)
(b) If h is halved, then the error is halved ( )0.5 .r ≈
(c) When 0.05,h = the error will again be approximately halved.
95. (a) ( )
( )
4 12 24
1 24 12 6 34
dIL RI E tdt
dI Idt
dI I Idt
+ =
+ =
= − = −
(b) As , 2.t I→ ∞ → That is, ( )lim 2.t
I t→∞
= In fact,
2I = is a solution to the differential equation.
96.
( )
2
2
2
16 0
16 0
16 0 because 0
4
kt
kt
kt
kt kt
kt
y e
y ke
y k ey y
k e e
k e
k
=
′ =
′′ =
′′ − =
− =
− = ≠
= ±
97.
2
sincos
sin16 0
y A ty A t
y A ty y
ωω ω
ω ω
=
′ =
′′ = −
′′ + =
2
2
sin 16 sin 0
sin 16 0
A t A t
A t
ω ω ω
ω ω
− + =
⎡ ⎤− =⎣ ⎦
If 0,A ≠ then 4ω = ±
x 0 0.2 0.4 0.6 0.8 1
y 4 2.6813 1.7973 1.2048 0.8076 0.5413
1y 4 2.5600 1.6384 1.0486 0.6711 0.4295
2y 4 2.4000 1.4400 0.8640 0.5184 0.3110
1e 0 0.1213 0.1589 0.1562 0.1365 0.1118
2e 0 0.2813 0.3573 0.3408 0.2892 0.2303
r 0.4312 0.4447 0.4583 0.4720 0.4855
x 0 0.2 0.4 0.6 0.8 1
y 1 0.8375 0.7406 0.6976 0.6987 0.7358
1y 1 0.8200 0.7122 0.6629 0.6609 0.6974
2y 1 0.8000 0.6800 0.6240 0.6192 0.6554
1e 0 0.0175 0.0284 0.0347 0.0378 0.0384
2e 0 0.0375 0.0606 0.0736 0.0795 0.0804
r 0.47 0.47 0.47 0.48 0.48
−3 3
−3
3
t
I
Section 6.2 Differential Equations: Growth and Decay 605
© 2010 Brooks/Cole, Cengage Learning
98. ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2
22 2
, 0
2 2 2
2
f x f x xg x f x g x
f x f x f x f x xg x f x
d f x f x x g x f xdx
′′ ′+ = − ≥
′ ′ ′′ ′+ = − ⎡ ⎤⎣ ⎦
⎡ ⎤′ ′+ = − ⎡ ⎤⎣ ⎦⎣ ⎦
For ( ) ( ) 20, 2 0x x g x f x′< − ⎡ ⎤ ≥⎣ ⎦
For ( ) ( ) 20, 2 0x x g x f x′> − ⎡ ⎤ ≤⎣ ⎦
So, ( ) ( )2 2f x f x′+ is increasing for 0x < and decreasing for 0.x >
( ) ( )2 2f x f x′+ has a maximum at 0.x = So, it is bounded by its value at ( ) ( )2 20, 0 0 .x f f ′= + So, f (and f ′ ) is bounded.
99. Let the vertical line x k= cut the graph of the solution ( )y f x= at ( ), .k t The tangent line at ( ),k t is
( )( )y t f k x k′− = −
Because ( ) ( ),y p x y q x′ + = you have
( ) ( ) ( )y t q k p k t x k− = ⎡ − ⎤ −⎣ ⎦
For any value of t, this line passes through the point ( )
( )( )
1 , .q k
kp k p k
⎛ ⎞+⎜ ⎟⎜ ⎟
⎝ ⎠
To see this, note that
( )( ) ( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( ) ( )
( )
?
?
1
.
q kt q k p k t k k
p k p k
q k q kq k k p k tk t kq k p k kt t
p k p k
⎛ ⎞− = ⎡ − ⎤ + −⎜ ⎟⎣ ⎦⎜ ⎟
⎝ ⎠
= − + − − + = −
Section 6.2 Differential Equations: Growth and Decay
1.
( )2
3
3 32
dy xdx
xy x dx x C
= +
= + = + +∫
2.
( )2
6
6 62
dy xdx
xy x dx x C
= −
= − = − +∫
3.
1
1
3
31
3
ln 3
3
3
x C x
x
dy ydx
dy dxy
dy dxy
y x C
y e Ce
y Ce
+
= +
=+
=+
+ = +
+ = =
= −
∫ ∫
4.
1
1
6
61
6
ln 6
6
6
x C x
x
dy ydx
dy dxy
dy dxy
y dy x C
y e Ce
y Ce
− + −
−
= −
=−
−= −
−
− = − +
− = =
= −
∫ ∫
5.
2 21
2 2
5
5
5
5
1 52 25
xyy
yy x
yy dx x dx
y dy x dx
y x C
y x C
′ =
′ =
′ =
=
= +
− =
∫ ∫∫ ∫
606 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
6.
23 2
1
2 3 2
7
7
7
7 22 3
21 4
xyy
yy x
yy dx x dx
y x C
y x C
′ =
′ =
′ =
= +
− =
∫ ∫
7.
( )
( )
( )
3 21
3 22 3 1
3 22 31
3 22 3
2ln3
x C
xC
x
y x yy xy
y dx x dxy
dy x dxy
y x C
y e
e e
Ce
+
′ =
′=
′=
=
= +
=
=
=
∫ ∫
∫ ∫
8. ( )
( )
( )
2
1
2 2 1
2 21
2 2
1
1
1
1
ln 12
1
1
1
x C
C x
x
y x y
y xy
y dx x dxydy x dx
y
xy C
y e
y e e
Ce
+
′ = +
′=
+′
=+
=+
+ = +
+ =
= −
= −
∫ ∫
∫ ∫
9. ( )
( )( )
( )( )
2
2
2
2
2
21
2
2
2
1 2 0
21
21
21
21
ln ln 1
ln ln 1 ln
ln ln 1
1
x y xy
xyyx
y xy x
y xdx dxy x
dy x dxy x
y x C
y x C
y C x
y C x
′+ − =
′ =+
′=
+′
=+
=+
= + +
= + +
⎡ ⎤= +⎣ ⎦
= +
∫ ∫
∫ ∫
10.
( )
( )
( )
( )
2
1
2
1
2 2 1
2 21
2 2
100
100 100
100
1001
100
ln 1002
ln 1002
100
100
100
x C
xC
x
xy y x
y x xy x y
y xy
y dx x dxy
dy x dxy
xy C
xy C
y e
y e e
y Ce
− −
−−
−
′+ =
′ = + = −
′=
−′
=−
=−
− − = +
− = − −
− =
− = −
= −
∫ ∫
∫ ∫
11. 2
2
dQ kdt t
dQ kdt dtdt t
kdQ CtkQ Ct
=
=
= − +
= − +
∫ ∫
∫
12. ( )
( )
( )
( )
2
2
25
25
252
252
dP k tdt
dP dt k t dtdt
kdP t C
kP t C
= −
= −
= − − +
= − − +
∫ ∫
∫
Section 6.2 Differential Equations: Growth and Decay 607
© 2010 Brooks/Cole, Cengage Learning
13. ( )
( )
( )
( )
2
2
500
500
5002
5002
dN k sds
dN ds k s dsds
kdN s C
kN s C
= −
= −
= − − +
= − − +
∫ ∫
∫
14. ( )
( )
( )
2
1
2
1
2 2 1
2 21
2 2
1
1
12
ln2
kx C
kxC
kx
dy kx L ydxdy kx
L y dxdy dx kx dx
L y dx
kxdy CL y
kxL y C
L y e
y L e e
y L Ce
− −
−−
−
= −
=−
=−
= +−
− − = +
− =
− = − +
= −
∫ ∫
∫
15. (a)
(b) ( ) ( )
2
2 22 21
2 21
6 , 0, 0
6
ln 62
6
6
x C x
x
dy x ydx
dy x dxy
xy C
y e C e
y C e
− + −
−
= −
= −−
−− = +
− = =
= +
(0, 0): 1 1
2 2
0 6 6
6 6 x
C C
y e−= + ⇒ = −
= −
16. (a)
(b)
2
2 22 21
01 1
2 2
1, 0,2
ln2
1 1 10, :2 2 2
12
x C x
x
dy xydxdy x dxy
xy C
y e C e
C e C
y e
+
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
= +
= =
⎛ ⎞ = ⇒ =⎜ ⎟⎝ ⎠
=
17. ( )
( )
2
2
2
1 , 0, 102
12
14110 0 1041 104
dy tdt
dy t dt
y t C
C C
y t
=
=
= +
= + ⇒ =
= +
∫ ∫
18. ( )
( )
3 2
3 2
3 2
3 , 0, 104
34
12110 0 1021 102
dy tdt
dy t dt
y t C
C C
y t
= −
= −
= − +
= − + ⇒ =
= − +
∫ ∫
x−5 −1
9
5
y
(0, 0)
−6 6
−1
7
x
)(0, 12
y
−4
−4
4
4
4
−1
−4
(0, 10)
16
0 10
−5
(0, 10)
15
608 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
19. ( )
( )
1
2 2 21 1
0
2
1 , 0, 102
12
1ln2
10 10
10
t C C t t
t
dy ydtdy dty
y t C
y e e e Ce
Ce C
y e
− + − −
−
= −
= −
= − +
= = =
= ⇒ =
=
∫ ∫
20. ( )
( )
( )
1
3 4 1
3 4 3 41
0
3 4
3 , 0, 104
34
3ln4
10 10
10
t C
tC t
t
dy ydtdy dty
y t C
y e
e e Ce
Ce C
y e
+
=
=
= +
=
= =
= ⇒ =
=
∫ ∫
21.
( )Theorem 6.1kx
dy kydx
y Ce
=
=
( ) ( )
( ) ( )
0
4
0, 6 : 6
1 54, 15 : 15 6 ln4 2
k
k
Ce C
e k
= =
⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠
( )1 4 ln 5 2 0.22916 6x xy e e⎡ ⎤⎣ ⎦= ≈
When
( )( ) ( )21 4 ln 5 2 8 ln 5 2
8,
25 756 6 64 2
x
y e e
=
⎛ ⎞= = = =⎜ ⎟⎝ ⎠
22.
( )Theorem 6.1kt
dN kNdtN Ce
=
=
( )
( )
0, 250 : 250
400 81, 400 : 400 250 ln ln250 5
k
C
e k
=
= ⇒ = =
( )ln 8 5 0.4700250 250t tN e e= ≈
When ( ) ( )44 ln 8 5 ln 8 5
4
4, 250 250
8 8192250 .5 5
t N e e= = =
⎛ ⎞= =⎜ ⎟⎝ ⎠
23.
( )Theorem 6.1kt
dV kVdtV Ce
=
=
( )
( ) 4
0, 20,000 : 20,000
1 54, 12,500 : 12,500 20,000 ln4 8
k
C
e k
=
⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠
( )1 4 ln 5 8 0.117520,000 20,000t tV e e⎡ ⎤ −⎣ ⎦= ≈
When ( ) ( )( ) ( )3 21 4 ln 5 8 6 ln 5 8
3 2
6, 20,000 20,000
520,000 9882.118.8
t V e e= = =
⎛ ⎞= ≈⎜ ⎟⎝ ⎠
24.
( )Theorem 6.1kt
dP kPdtP Ce
=
=
( )
( )
0, 5000 : 5000
191, 4750 : 4750 5000 ln20
k
C
e k
=
⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠
( )ln 19 20 0.05135000 5000t tP e e−= ≈
When ( )( )ln 19 20 5
5
5, 5000
195000 3868.905.20
t P e= =
⎛ ⎞= ≈⎜ ⎟⎝ ⎠
25. ( )
( ) ( )
5
ln 10 5 5 0.4605
1, 0, , 5, 52
1212152ln 10
51 1 110 or2 2 2
kt
kt
k
t t t
y Ce
C
y e
e
k
y e y e⎡ ⎤⎣ ⎦
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
=
=
=
= = ≈
10
−1
−1
(0, 10)
16
−5 5
−5
(0, 10)
40
Section 6.2 Differential Equations: Growth and Decay 609
© 2010 Brooks/Cole, Cengage Learning
26. ( )
( )
5
0.4159
1, 0, 4 , 5,2
4
41 42
ln 1 80.4159
54
kt
kt
k
t
y Ce
C
y e
e
k
y e−
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
=
=
= ≈ −
=
27. ( ) ( )
5
, 1, 5 , 5, 2
5 10 2
2 10 5
kt
k k
k k
y Ce
Ce Ce
Ce Ce
=
= ⇒ =
= ⇒ =
5
5
4
1 4
2 5
2 525
1 2 2ln ln4 5 5
k k
k k
k
Ce Ce
e e
e
k
=
=
=
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( )
( )
1 4 1 41 4 ln 2 5
1 41 4 ln 2 5 0.2291
2 55 5 5 55 2
55 6.28722
k
t t
C e e
y e e
−−−
⎡ ⎤ −⎣ ⎦
⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞= ≈⎜ ⎟⎝ ⎠
28. ( )
3 3
4 4
1, 3, , 4, 52
1 1 22
15 15
kt
k k
k k
y Ce
Ce Ce
Ce Ce
⎛ ⎞= ⎜ ⎟⎝ ⎠
= ⇒ =
= ⇒ =
3 4
3 4
125
10
10ln 10 2.3026
k k
k k
k
Ce Ce
e e
ek
=
=
=
= ≈
( )
.3026
2.3026 4
2.3026
50.0005
0.0005
t
t
y Ce
CeC
y e
2=
=
≈
=
29. In the model ,kty Ce C= represents the initial value of y (when 0t = ). k is the proportionality constant.
30. dyy kydt
′ = =
31. 12
dy xydx
=
0dydx
> when 0.xy > Quadrants I and III.
32. 212
dy x ydx
=
0dydx
> when 0.y > Quadrants I and II.
33. Because the initial quantity is 20 grams,
20 .kty e=
Because the half-life is 1599 years,
( )
( )1599
1 11599 2
10 20
ln .
ke
k
=
=
So, ( )ln 1 2 159920 .ty e⎡ ⎤⎣ ⎦=
When ( ) ( )ln 1 2 1599 10001000, 20 12.96 .t y e g⎡ ⎤⎣ ⎦= = ≈
When 10,000, 0.26g.t y= ≈
34. Because the half-life is 1599 years,
( )
( )
159912
1 11599 2
1
ln .
ke
k
=
=
Because there are 1.5 g after 1000 years,
( ) ( )ln 1 2 1599 10001.52.314.Ce
C
⎡ ⎤⎣ ⎦=
≈
So, the initial quantity is approximately 2.314 g.
When ( ) ( )ln 1 2 1599 10,00010,000, 2.3140.03 g.
t y e⎡ ⎤⎣ ⎦= =
≈
35. Because the half-life is 1599 years,
( )
( )
159912
1 11599 2
1
ln .
ke
k
=
=
Because there are 0.1 gram after 10,000 years,
( ) ( )ln 1 2 1599 10,0000.17.63.Ce
C
⎡ ⎤⎣ ⎦=
≈
So, the initial quantity is approximately 7.63 g.
When ( ) ( )ln 1 2 1599 10001000, 7.634.95 g.
t y e⎡ ⎤⎣ ⎦= =
≈
610 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
36. Because the half-life is 5715 years,
( )
( )
571512
1 15715 2
1
ln .
ke
k
=
=
Because there are 3 grams after 10,000 years,
( ) ( )ln 1 2 5715 10,000310.089.Ce
C
⎡ ⎤⎣ ⎦=
≈
So, the initial quantity is approximately 10.09 g.
When ( ) ( )ln 1 2 5715 10001000, 10.098.94 g.
t y e⎡ ⎤⎣ ⎦= =
≈
37. Because the initial quantity is 5 grams, 5.C =
Because the half-life is 5715 years,
( )
( )5715
1 15715 2
2.5 5
ln .
ke
k
=
=
When ( ) ( )ln 1 2 5715 10001000 years, 5 4.43 g.t y e⎡ ⎤⎣ ⎦= = ≈
When ( ) ( )ln 1 2 5715 10,00010,000 years, 51.49 g.
t y e⎡ ⎤⎣ ⎦= =
≈
38. Because the half-life is 5715 years,
( )
( )
571512
1 15715 2
1
ln .
ke
k
=
=
Because there are 1.6 grams when 1000t = years,
( ) ( )ln 1 2 5715 10001.61.806.Ce
C
⎡ ⎤⎣ ⎦=
≈
So, the initial quantity is approximately 1.806 g.
When ( ) ( )ln 1 2 5715 10,00010,000, 1.8060.54 g.
t y e⎡ ⎤⎣ ⎦= =
≈
39. Because the half-life is 24,100 years,
( )
( )
24,10012
1 124,100 2
1
ln .
ke
k
=
=
Because there are 2.1 grams after 1000 years,
( ) ( )ln 1 2 24,100 10002.12.161.Ce
C
⎡ ⎤⎣ ⎦=
≈
So, the initial quantity is approximately 2.161 g.
When ( ) ( )ln 1 2 24,100 10,00010,000, 2.1611.62 g.
t y e⎡ ⎤⎣ ⎦= =
≈
40. Because the half-life is 24,100 years,
( )
( )
24,10012
1 124,100 2
1
ln .
ke
k
=
=
Because there are 0.4 grams after 10,000 years,
( ) ( )ln 1 2 24,100 10,0000.40.533.Ce
C
⎡ ⎤⎣ ⎦=
≈
So, the initial quantity is approximately 0.533 g.
When ( ) ( )ln 1 2 24,100 10001000, 0.5330.52 g.
t y e⎡ ⎤⎣ ⎦= =
≈
41. ( )
( )
159912
1 11599 2ln
kt
k
y Ce
C Ce
k
=
=
=
When ( ) ( )ln 1 2 1599 100100,0.9576 C
t y Ce⎡ ⎤⎣ ⎦= =
≈
Therefore, 95.76% remains after 100 years.
42.
( )
( )
( )
5715
[ln 1 2 5715]
12
1 1ln5715 2
0.15
1ln2ln 0.15
571515,641.8 years
kt
k
t
y Ce
C Ce
k
C Ce
t
t
=
=
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
⎛ ⎞⎜ ⎟⎝ ⎠=
≈
43. Because 0.064000 ,tA e= the time to double is given by
0.06
0.06
8000 4000
2ln 2 0.06
ln 2 11.55 years.0.06
t
t
e
et
t
=
=
=
= ≈
Amount after 10 years: ( )( )0.06 104000 $7288.48A e= ≈
44. Because 0.05518,000 ,tA e= the time to double is given by
0.055
0.055
36,000 18,000
2ln 2 0.055
ln 2 12.6 years.0.055
t
t
e
et
t
=
=
=
= ≈
Amount after 10 years: ( )( )0.055 1018,000 $31,198.55A e= ≈
Section 6.2 Differential Equations: Growth and Decay 611
© 2010 Brooks/Cole, Cengage Learning
45. Because 750 rtA e= and 1500A = when 7.75,t = you have the following.
7.751500 750ln 2 0.0894 8.94%7.75
re
r
=
= ≈ =
Amount after 10 years: ( )0.0894 10750 $1833.67A e= ≈
46. Because 12,500 rtA e= and 25,000A = when 5,t = you have the following.
525,000 12,500ln 2 0.1386 13.86%
5
re
r
=
= ≈ =
Amount after 10 years: ( ) ( )ln 2 5 1012,500 $50,000A e⎡ ⎤⎣ ⎦= =
47. Because 500 rtA e= and 1292.85A = when 10,t = you have the following.
( )
101292.85 500
ln 1292.85 5000.0950 9.50%
10
re
r
=
= ≈ =
The time to double is given by
0.09501000 500ln 2 7.30 years.
0.095
te
t
=
= ≈
48. Because 2000 rtA e= and 5436.56A = when 10,t = you have the following.
( )
105436.56 2000
ln 5436.56 20000.10 10%
10
re
r
=
= ≈ =
The time to double is given by
0.104000 2000ln 2 6.93 years.0.10
te
t
=
= ≈
49. ( )( )12 20
240
0.0751,000,000 112
0.0751,000,000 112
$224,174.18
P
P−
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
≈
50. ( )( )
( )
12 40
480
0.061,000,000 112
1,000,000 1.005 $91,262.08
P
P −
⎛ ⎞= +⎜ ⎟⎝ ⎠
= ≈
51. ( )( )12 35
420
0.081,000,000 112
0.081,000,000 112
$61,377.75
P
P−
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
=
52. ( )( )12 25
300
0.091,000,000 112
0.091,000,000 112
$106,287.83
P
P−
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
≈
53. (a) ( )2000 1000 1 0.07
2 1.07ln 2 ln 1.07
ln 2 10.24 yearsln 1.07
t
t
t
t
= +
=
=
= ≈
(b)
( )( )
12
12
0.072000 1000 112
0.0072 112
0.07ln 2 12 ln 112
ln 2 9.93 years12 ln 1 0.07 12
t
t
t
t
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
= ≈+
(c)
( )( )
365
365
0.072000 1000 1365
0.072 1365
0.07ln 2 365 ln 1365
ln 2 9.90 years365 ln 1 0.07 365
t
t
t
t
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
= ≈+
(d) ( )0.07
0.07
2000 1000
2ln 2 0.07
ln 2 9.90 years0.07
t
t
e
et
t
=
=
=
= ≈
612 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
54. (a) ( )2000 1000 1 0.6
2 1.06ln 2 ln 1.06
ln 2 11.90 yearsln 1.06
t
t
t
t
= +
=
=
= ≈
(b) 12
12
0.062000 1000 112
0.062 112
0.06ln 2 12 ln 112
1 ln 2 11.58 years0.0612 ln 112
t
t
t
t
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
= ≈⎛ ⎞+⎜ ⎟⎝ ⎠
(c) 365
365
0.062000 1000 1365
0.062 1365
0.06ln 2 365 ln 1365
1 ln 2 11.55 years0.06365 ln 1365
t
t
t
t
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
= ≈⎛ ⎞+⎜ ⎟⎝ ⎠
(d) 0.06
0.06
2000 1000
2ln 2 0.06
ln 2 11.55 years0.06
t
t
e
et
t
=
=
=
= ≈
55. (a) ( )2000 1000 1 0.085
2 1.085ln 2 ln 1.085
ln 2 8.50 yearsln 1.085
t
t
t
t
= +
=
=
= ≈
(b) 12
12
0.0852000 1000 112
0.0852 112
0.085ln 2 12 ln 112
1 ln 2 8.18 years0.08512 ln 1
12
t
t
t
t
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
= ≈⎛ ⎞+⎜ ⎟⎝ ⎠
(c) 365
365
0.0852000 1000 1365
0.0852 1365
0.085ln 2 365 ln 1365
1 ln 2 8.16 years0.085365 ln 1365
t
t
t
t
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
= ≈⎛ ⎞+⎜ ⎟⎝ ⎠
(d) 0.085
0.085
2000 1000
2ln 2 0.085
ln 2 8.15 years0.085
t
t
e
et
t
=
=
=
= ≈
56. (a) ( )2000 1000 1 0.055
2 1.055ln 2 ln 1.055
ln 2 12.95 yearsln 1.055
t
t
t
t
= +
=
=
= ≈
(b) 12
12
0.0552000 1000 112
0.0552 112
0.055ln 2 12 ln 112
1 ln 2 12.63 years0.05512 ln 1
12
t
t
t
t
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
= ≈⎛ ⎞+⎜ ⎟⎝ ⎠
(c) 365
365
0.0552000 1000 1365
0.0552 1365
0.055ln 2 365 ln 1365
1 ln 2 12.60 years0.055365 ln 1365
t
t
t
t
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
= ≈⎛ ⎞+⎜ ⎟⎝ ⎠
(d) 0.055
0.055
2000 1000
2ln 2 0.055
ln 2 12.60 years0.055
t
t
e
et
t
=
=
=
= ≈
Section 6.2 Differential Equations: Growth and Decay 613
© 2010 Brooks/Cole, Cengage Learning
57. (a)
( ) ( )
0.006
0.006 7
0.006
7 2.3 2.40
2.40
kt t
t
P Ce Ce
P Ce C
P e
−
−
−
= =
= = ⇒ ≈
=
(b) For ( )0.006 1515, 2.40 2.19 milliont P e−= = ≈
(c) Because 0,k < the population is decreasing.
58. (a)
( ) ( )
0.017
0.017 7
0.017
7 80.3 71.29
71.29
kt t
t
P Ce Ce
P Ce C
P e
= =
= = ⇒ ≈
=
(b) For ( )0.017 1515, 71.29 92.0 milliont P e= = ≈
(c) Because 0,k > the population is increasing.
59. (a)
( ) ( )
0.024
0.024 7
0.024
7 6.7 5.66
5.66
kt t
t
P Ce Ce
P Ce C
P e
= =
= = ⇒ ≈
=
(b) For ( )0.024 1515, 5.66 8.11 milliont P e= = ≈
(c) Because 0,k > the population is increasing.
60. (a)
( ) ( )
0.003
0.003 7
0.003
7 10.0 10.21
10.21
kt t
t
P Ce Ce
P Ce C
P e
−
−
−
= =
= = ⇒ ≈
=
(b) For ( )0.003 1515, 10.21 9.76 milliont P e−= = ≈
(c) Because 0,k < the population is decreasing.
61. (a)
( ) ( )
0.036
0.036 7
0.036
7 30.3 23.55
23.55
kt t
t
P Ce Ce
P Ce C
P e
= =
= = ⇒ ≈
=
(b) For ( )0.036 1515, 23.55 40.41 milliont P e= = ≈
(c) Because 0,k > the population is increasing.
62. (a) Because the population increases by a constant each month, the rate of change from month to month will always be the same. So, the slope is constant, and the model is linear. (b) Although the percentage increase is constant each
month, the rate of growth is not constant. The rate of change of y is given by
dy rydt
=
which is an exponential model.
63. (a) ( )100.1596 1.2455 tN =
(b) 400N = when 6.3t = hours (graphing utility)
Analytically,
( )400 100.1596 1.2455
4001.2455 3.9936100.1596
ln 1.2455 ln 3.9936ln 3.9936 6.3 hours.ln 1.2455
t
t
t
t
=
= =
=
= ≈
64. (a) Let .kty Ce=
At time 2: ( )2 2125 125k kCe C e−= ⇒ =
At time 4: ( ) ( )( )4 2 4
2145
145
1 142 5
350 350 125
2 ln
ln 0.5148
k k k
k
Ce e e
e
k
k
−= ⇒ =
=
=
= ≈
( ) ( )
( )
2
2 1 2 ln 14 5
5 62514 14
125
125
125 44.64
kC e
e
−
−
=
=
= = ≈
Approximately 45 bacteria at time 0.
(b) ( ) ( )1 2 ln 14 5 0.514862514 44.64t ty e e= ≈
(c) When 8,t = ( ) ( ) ( )41 2 ln 14 5 8625 625 14
14 14 5 2744.y e= = =
(d) ( ) ( )1 2 ln 14 56251425,000 12.29 hourste t= ⇒ ≈
65. (a) ( )
( )
20
20
19 30 1
30 11
ln 11 300.0502
20
k
k
e
e
k
= −
=
= ≈ −
( )0.050230 1 tN e−≈ −
(b) ( )0.0502
0.0502
25 30 1
16
ln 6 36 days0.0502
t
t
e
e
t
−
−
= −
=
−= ≈
−
614 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
66. (a) ( )
( )
30
30
20 30 1
30 10
ln 1 3 ln 3 0.036630 30
k
k
e
e
k
= −
=
−= = ≈ −
( )0.036630 1 tN e−≈ −
(b) ( )0.0366
0.0366
25 30 1
16
ln 6 49 days0.0366
t
t
e
e
t
−
−
= −
=
−= ≈
−
67. (a)
( )1
10 205181
1811205 181 ln 0.01245
10
kt kt
k
P Ce e
e k
= =
= ⇒ = ≈
( )0.012451 181 181 1.01253 ttP e≈ ≈
(b) Using a graphing utility, ( )2 182.3248 1.01091 tP ≈
(c)
The model 2P fits the data better.
(d) Using the model 2,P
( )
( )
( )( )
320 182.3248 1.01091
320 1.01091182.3248
ln 320 182.3248ln 1.01091
51.8 years, or 2011.
t
t
t
=
=
=
≈
68. (a) ( ) 0.0573
4 3 2
1654.2353 1.0590 1654.2353
0.01942 1.3690 21.970 93.66 435.6
t tR e
I t t t t
= =
= − + − +
(b)
According to the model, ( ) 0.057394.79 tR t e′ ≈
(c)
(d) ( ) IP tR
=
69. ( ) 1610 0
010 log , 10II I
Iβ −= =
(a) ( )14
1410 16
1010 10 log 20 decibels10
β−
−−= =
(c) ( )6.5
6.510 16
1010 10 log 95 decibels10
β−
−−= =
(b) ( )9
910 16
1010 10 log 70 decibels10
β−
−−= =
(d) ( )4
410 16
1010 10 log 120 decibels10
β−
−−= =
00
4000
11
00
800
11
00
0.3
11
150500
P1P2
300
Section 6.2 Differential Equations: Growth and Decay 615
© 2010 Brooks/Cole, Cengage Learning
70. ( )
( )
10 1016
6.710
10 1016
810
93 10 log 10 log 1610
6.7 log 10
80 10 log 10 log 1610
8 log 10
I I
I II I
I I
−
−
−
−
= = +
− = ⇒ =
= = +
− = ⇒ =
Percentage decrease: ( )6.7 8
6.710 10 100 95%
10
− −
−
⎛ ⎞−≈⎜ ⎟
⎝ ⎠
71. ( ) ( ) 0.10
0.8 0.10
0.8 0.10
0.8 0.10
100,000
100,000
0.4100,000 0.10 0 when 16.
t
t t
t t
t t
A t V t e
e e
e
dA edt t
−
−
−
−
=
=
=
⎛ ⎞= − =⎜ ⎟⎝ ⎠
The timber should be harvested in the year 2024, ( )2008 16 .+ Note: You could also use a graphing utility
to graph ( )A t and find the maximum of ( ).A t Use the
viewing rectangle 0 30x≤ ≤ and 0 600,000.y≤ ≤
72. 0 ln 10ln ln ln 0, 10ln 10 ln 10
R RI I IR I e− −= = = =
(a) 0
8.3
ln ln8.3ln 10
10 199,526,231.5
I I
I
−=
= ≈
(b)
( ) ( )
0
2 22 ln 10 2 ln 10 ln 10
ln ln2ln 10
10R R R R
I IR
I e e e
−=
= = = =
Increases by a factor of 2 ln 10Re or 10 .R
(c)
( )
0ln lnln 101
ln 10
I IR
dRdI I
−=
=
73. Because ( )80dy k ydt
= −
( )
180
ln 80 .
dy k dty
y kt C
=−
− = +
∫ ∫
When 0, 1500.t y= = So, ln 1420.C =
When 1, 1120.t y= = So,
( ) ( )1 ln 1420 ln 1120 80
104ln 1040 ln 1420 ln .142
k
k
+ = −
= − =
So, ( )ln 104 1421420 80.ty e⎡ ⎤⎣ ⎦= +
When 5, 379.2 F.t y= ≈ °
74. ( )
( )( )
( )
( ) ( )
( )
0
5
5
1 5 ln 2 7
ln 2 7
20
20 See Example 6
160 20 140
60 20 14027
1 2ln 0.250555 7
30 20 140
1 214 71 2ln ln
14 5 715 ln 5 ln 1414 10.53 minutes2 7ln ln
7 2
kt
k
k
k
t
t st s
dy k ydt
y Ce
Ce C
e
e
k
e
e
t
t
= −
= +
= + ⇒ =
= +
=
⎛ ⎞= ≈ −⎜ ⎟⎝ ⎠
= +
⎛ ⎞= = ⎜ ⎟⎝ ⎠
=
= = ≈
It will take 10.53 5 5.53 minutes− = longer.
75. False. If , constant.kt kty Ce y Cke′= = ≠
76. True
77. True
78. True
616 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
Section 6.3 Separation of Variables and the Logistic Equation
1.
2 2
1
2 2
2 2
dy xdx y
y dy x dx
y x C
y x C
=
=
= +
− =
∫ ∫
2. 2
2
2 2
33
1
3 3
3
3
33
dy xdx y
y dy x dx
y x C
y x C
=
=
= +
− =
∫ ∫
3. 2
2
2
2 3
1
2 3
5 0
5
5
52 3
15 2
dyx ydxdyy xdx
y dy x dx
y x C
y x C
+ =
= −
= −
−= +
+ =
∫ ∫
4.
( )
2
2
2 2
33
1
3 3
36
6 3
2 33
6 9
dy xdx y
y dy x dx
xy x C
y x x C
−=
= −
= − +
− + =
∫ ∫
5.
1
0.75 1
0.75
0.75
0.75
ln 0.75s C
s
dr rdsdr dsrr s C
r e
r Ce
+
=
=
= +
=
=
∫ ∫
6.
2
2
0.75
0.75
0.752
0.375
dr sdsdr s ds
sr C
r s C
=
=
= +
= +
∫ ∫
7. ( )
( )
( )
3
3
2 3
32
ln 3 ln 2 ln ln 2
2
x y y
dy dxy x
y x C C x
y C x
′+ =
=+
= + + = +
= +
∫ ∫
8.
ln ln ln ln
xy ydy dxy xy x C Cxy Cx
′ =
=
= + =
=
∫ ∫
9.
2
1
2
4 sin
4 sin
4 sin
4 cos2
8 cos
yy xdyy xdx
y dy x dx
y x C
y C x
′ =
=
=
= − +
= −
∫ ∫
10. ( )
( )
( )( )
( )
2
2
8 cos
8 cos
8 cos
8 sin2
16 sin
yy x
dyy xdx
y dy x dx
xy C
y x C
π
π
π
ππ
ππ
′ = −
= −
= −
−= +
−= +
∫ ∫
11.
( ) ( )
2
2
2
1 22
2
1 4
1 4
1 41 1 4 881 1 44
x y xxdy dx
xxdy dx
x
x x dx
y x C
−
′− =
=−
=−
= − − −
= − − +
∫ ∫
∫
12. 2
2
2
2
16 1111
1611
16
11 16
x y xdy xdx x
xdy dxx
y x C
′− =
=−
=−
= − +
∫ ∫
Section 6.3 Separation of Variables and the Logistic Equation 617
© 2010 Brooks/Cole, Cengage Learning
13.
( )
( )( ) ( )
21
2 21 2 ln ln 21
ln 0
ln ln ,
1ln ln2
x C x
y x xy
dy x dxdx u x duy x x
y x C
y e Ce+
′− =
⎛ ⎞= = =⎜ ⎟⎝ ⎠
= +
= =
∫ ∫
14.
2
12 7 0
12 7
12 7
6 7
x
x
x
x
yy edyy edx
y dy e dx
y e C
′ − =
=
=
= +
∫ ∫
15.
2
2 0
2
2
22
x
x
x
x
yy edyy edx
y dy e dx
y e C
′ − =
=
=
= +
∫ ∫
Initial condition ( )0, 3 : 9 522 2
C C= + ⇒ =
Particular solution: 2
2
522 2
4 5
x
x
y e
y e
= +
= +
16. 1 2 1 2
3 2 3 21
3 2 3 2
0
2 23 3
x y y
y dy x dx
y x C
y x C
′+ =
= −
= − +
+ =
∫ ∫
Initial condition ( )1, 9 :
( ) ( )3 2 3 29 1 27 1 28 C+ = + = =
Particular solution: 3 2 3 2 28y x+ =
17. ( )
( )
( )
( )
2
1
21 2
1 0
1
1ln
2x
y x y
dy x dxy
xy C
y Ce− +
′+ + =
= − +
+= − +
=
∫ ∫
Initial condition ( ) 1 2 1 22, 1 : 1 ,Ce C e−− = =
Particular solution: ( ) ( )2 21 1 2 2 2x x x
y e e⎡ ⎤− + − +⎢ ⎥⎣ ⎦= =
18.
( )
2
2
2 ln 0
2 2 ln
ln
ln2
xy xdyx xdx
xdy dxx
xy C
′ − =
=
=
= +
∫ ∫
Initial condition ( )1, 2 : 2 C=
Particular solution: ( )21 ln 22
y x= +
19. ( ) ( )
( ) ( )
( ) ( ) ( )( )
2 2
2 2
2 21
2 2 2
2 2
1 1
1 11 1ln 1 ln 12 2
ln 1 ln 1 ln ln 1
1 1
y x y x y
y xdy dxy x
y x C
y x C C x
y C x
′+ = +
=+ +
+ = + +
⎡ ⎤+ = + + = +⎣ ⎦
+ = +
Initial condition ( )0, 3 : 1 3 4C C+ = ⇒ =
Particular solution: ( )2 2
2 2
1 4 1
3 4
y x
y x
+ = +
= +
20.
( ) ( )( ) ( )
2 2
1 2 1 22 2
1 2 1 22 2
1 1
1 1
1 1
dyy x x ydx
y y dy x x dx
y x C
− −
− = −
− = −
− − = − − +
∫ ∫
Initial condition ( )0, 1 : 0 1 1C C= − + ⇒ =
Particular solution: 2 21 1 1y x− = − −
21.
( )
2
2
21
2cos 2
sin
sin
1ln cos2
v
du uv vdvdu v v dvu
u v C
u Ce−
=
=
= − +
=
∫ ∫
Initial condition: ( ) 1 21 210 1:u C e
e−= = =
Particular solution: ( )21 cos 2vu e
−=
618 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
22. 2
2
212
r s
r s
r s
dr eds
e dr e ds
e e C
−
− −
− −
=
=
− = − +
∫ ∫
Initial condition:
( ) 1 10 0: 12 2
r C C= − = − + ⇒ = −
Particular solution: 2
2
22
2
1 12 2
1 12 2
1 1 1ln ln2 2 2
2ln1
r s
r s
ss
s
e e
e e
er e
re
− −
− −
−−
−
− = − −
= +
⎛ ⎞+⎛ ⎞− = + = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞= ⎜ ⎟+⎝ ⎠
23.
1
0
lnkt
dP kP dtdP k dtPP kt C
P Ce
− =
=
= +
=
∫ ∫
Initial condition: ( ) 00 00 ,P P P Ce C= = =
Particular solution: 0ktP P e=
24. ( )
( ) 1
70 0
70ln 70
70 kt
dT k T dt
dT k dtTT kt C
T Ce−
+ − =
= −−− = − +
− =
∫ ∫
Initial condition: ( ) 00 140: 140 70 70T Ce C= − = = =
Particular solution: ( )70 70 , 70 1kt ktT e T e− −− = = +
25.
22
4
4
22
dy xydx y
y dy x dx
xy C
′ = =
=
= +
∫ ∫
Initial condition ( ) ( )20, 2 : 2 2 0 8C C= + ⇒ =
Particular solution: 2
2
2 2
2 82
4 16
xy
y x
= +
− =
26.
2 2
916
16 9
982
dy xdx y
y dy x dx
y x C
−=
= −
−= +
∫ ∫
Initial condition ( ) 9 251, 1 : 8 ,2 2
C C= − + =
Particular solution: 2 2
2 2
9 2582 2
16 9 25
y x
y x
−= +
+ =
27.
1
2
22 1
2 ln ln ln ln
dy yydx x
dy dxy x
y x C x C
y Cx
′ = =
=
= + = +
=
∫ ∫
Initial condition ( ) 19, 1 : 1 99
C C= ⇒ =
Particular solution: 2
2
19
9 013
y x
y x
y x
=
− =
=
28.
3 2
3 2
23
3 2
ln ln ln
dy ydx x
dy dxy x
y x C
y Cx
=
=
= +
=
∫ ∫
Initial condition ( ) ( )3 2 18, 2 : 2 8 ,8
C C= =
Particular solution: 3 2 2 318 ,2
y x y x= =
29. ( )
1
2
02 2
12
1ln2
x
dy y ymdx x x
dy dxy
y x C
y Ce−
−= = = −
+ −
= −
= − +
=
∫ ∫
Section 6.3 Separation of Variables and the Logistic Equation 619
© 2010 Brooks/Cole, Cengage Learning
30. 00
dy y ymdx x x
dy dxy x
−= = =
−
=∫ ∫
1ln ln ln ln lny x C x C Cxy Cx= + = + =
=
31. ( )( )
( )
3 2 3
3 3 2 2 3 3
3 3 2 3
, 4
, 4
4
f x y x xy y
f tx ty t x txt y t y
t x xy y
= − +
= − +
= − +
Homogeneous of degree 3
32. ( )( )
3 2 2 2
3 3 4 2 2 2 2
, 3 2
, 3 2
f x y x x y y
f tx ty t x t x y t y
= + −
= + −
Not homogeneous
33. ( )
( )
2 2
2 2
4 2 2 2 23
2 2 2 2 2 2
,
,
x yf x yx y
t x y x yf tx ty tt x t y x y
=+
= =+ +
Homogeneous of degree 3
34. ( )
( )
2 2
2 2 2 2
2
2 2 2 2
,
,
xyf x yx y
tx tyf tx tyt x t y
t xy xytt x y x y
=+
=+
= =+ +
Homogeneous of degree 1
35. ( )( ) [ ]
( )2 2
, 2 ln
, 2 ln
2 ln 2 ln ln
f x y xy
f tx ty txty
t xy t xy
=
=
⎡ ⎤= = +⎣ ⎦
Not homogeneous
36. ( ) ( )( ) ( ) ( )
, tan
, tan tan
f x y x y
f tx ty tx ty t x y
= +
= + = ⎡ + ⎤⎣ ⎦
Not homogeneous
37. ( )
( )
, 2 ln
, 2 ln 2 ln
xf x yytx xf tx tyty y
=
= =
Homogeneous of degree 0
38. ( )
( )
, tan
, tan tan
yf x yxty yf tx tytx x
=
= =
Homogeneous of degree 0
39.
( )
( )
( )
( )
( )
2
2
2
2
2
2
,2
21 1
2 2
21
ln 1 ln ln ln
11
1
1 /
x yy y vxx
dv x vxv xdx xdv v vx vdx
dv dxv x
v x C Cx
Cxv
Cxy x
x Cxx y
x C x y
+′ = =
++ =
+ −= − =
=−
− − = + =
=−
=⎡ − ⎤⎣ ⎦
=−
= −
∫ ∫
40. ( )
( )
( ) ( ) ( )( )
3 3
2
2 3 3
2 33
4 2 3 3 3 3 3
2
,
x yy
xy
xy dy x y dx
y vx dy x dv v dx
x vx x dv v dx x vx dx
x v dv x v dx x dx v x dx
xv dv dx
+′ =
= +
= = +
+ = +
+ = +
=
2
3
3
3 3 3
1
ln3
3 ln
3 ln
v dv dxx
v x C
y x Cx
y x x Cx
=
= +
⎛ ⎞ = +⎜ ⎟⎝ ⎠
= +
∫ ∫
620 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
41.
2
2
121
22
2
2 2
2 2
,
111 1 21 1
12 1
1 ln 2 1 ln ln ln2
2 1
2 1
2
x yy y vxx y
dv x xvv xdx x xv
vv dx x dv dxvv v vx dv v dx dxv v
v dxdvv v x
Cv v x Cx
Cv vx
y y Cx x x
y xy x C
−′ = =+
−+ =
+−
+ =+− − −⎛ ⎞= − =⎜ ⎟+ +⎝ ⎠
+= −
+ −
+ − = − + =
+ − =
+ − =
+ − =
∫ ∫
42.
( )
2 2
2 2 2
2
2
2
2
2
2
2
2 2
,2
212 2
21
ln 1 ln ln ln
1
1
x yy y vxxy
dv x v xv xdx x v
vv dx x dv dxv
v dxdvv x
Cv x Cx
Cvx
y Cx x
y x Cx
+′ = =
++ =
++ =
= −−
− = − + =
− =
− =
− =
∫ ∫
43. 2 2
2
2 2 2
2
3
2 2
2
3
1 12
12
2
12
2 22
,
1
1 1
1
1 ln ln ln ln2
1 ln2
ln2
x y
xyy y vxx y
dv x vv xdx x x v
vv dx x dv dxvv vx dv v dx dx
v v
v dxdvv x
v x C C xv
C xvvx C yy
y Ce−
′ = =−
+ =−
+ =−
⎛ ⎞⎛ ⎞= − = ⎜ ⎟⎜ ⎟− −⎝ ⎠ ⎝ ⎠
−=
− − = + =
−=
−=
=
∫ ∫
44.
2 2
2
2
2
3
2 3 ,
2 3 2 3
2 2 21
ln 1 ln ln ln
1
1
1
x yy y vxx
dv x vxv x vdx xdv dv dxx vdx v xv x C x C
v x Cy x Cxy x Cxy Cx x
+′ = =
++ = = +
= + ⇒ =+
+ = + =
+ =
+ =
= −
= −
∫ ∫
Section 6.3 Separation of Variables and the Logistic Equation 621
© 2010 Brooks/Cole, Cengage Learning
45. ( )( ) ( )
/2 0,
2 0
y x
v
x dy xe y dx y vx
x v dx x dv xe vx dx
−
−
− + = =
+ − + =
21
/ 21
/ 2
2
ln
ln ln
ln
v
v
y x
y x
e dv dxx
e C x
e C x
e C x
=
=
= +
= +
∫ ∫
Initial condition ( )1, 0 : 1 C=
Particular solution: / 21 lny xe x= +
46. ( )( )( )
2
2 2 2 2
0,
0
y dx x x y dy y vx
x v dx x x v v dx x dv
− + + = =
− + + + =
11
1
1
1
1
ln ln ln ln
ln
v
y x
y x
v dxdvv x
Cv v x CxCvxv
C evxC ey
y Ce−
+= −
+ = − + =
=
=
=
=
∫ ∫
Initial condition ( ) 11, 1 : 1 Ce C e−= ⇒ =
Particular solution: 1 y xy e −=
47.
( ) ( )( )
sec 0,
sec 0
sec
yx y dx x dy y vxx
x v xv dx x v dx x dv
v v dx v dx x dv
⎛ ⎞+ − = =⎜ ⎟⎝ ⎠
+ − + =
+ = +
( )
1
sin
sin
cos
sin ln lnv
y x
dxv dvx
v x C
x Ce
Ce
=
= +
=
=
∫ ∫
Initial condition ( ) 01, 0 : 1 Ce C= =
Particular solution: ( )sin y xx e=
48. ( )2 22 0x y dx xy dy+ + =
Let , .y vx dy x dv v dx= = +
( ) ( )( )
( )( )
( )
( )( )
( )
( )
2 2 2
2 2 2 3
2
2
21
1 22 2
1 22 2
1 22 1 22 22 2
1 22 2
2 0
2 2 0
2 2
2112 ln ln 12
ln ln 1 ln
1
1 1
1
x v x dx x vx x dv v dx
x x v dx x v dv
v dx xv dv
vdx dvx v
x v C
x v C
x C v
y CC x yx x x
C x yx
−
−
+ + + =
+ + =
+ = −
−=
+
− = + +
= + +
= +
⎛ ⎞= + = +⎜ ⎟
⎝ ⎠
= +
Initial condition ( ) ( )1, 0 : 1 1 0 1C C= + ⇒ =
Particular solution: 2 2
2 2
1
1
x yx
x x y
= +
= +
49. dy xdx
=
212
y x dx x C= = +∫
50. dy xdx y
= −
2 2
1
2 2
2 2
y dy x dx
y x C
y x C
= −
−= +
+ =
4
2
−4
−2x
y
x−4 −2 2 4
−4
−2
2
4
y
622 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
51. 4dy ydx
= −
1
1
4
ln 4
4
4
x C
x
dy dxy
y x C
y e
y Ce
− +
−
=−
− = − +
− =
= +
∫ ∫
52. ( )
( ) ( )
( )
21
2 21 8 1 81
21 8
0.25 4
0.254
10.254 4
1ln 48
4
4
C x x
x
dy x ydx
dy x dxy
dy x dx x dxy
y x C
y e Ce
y Ce
− −
−
= −
=−
= − = −−
− = − +
− = =
= +
∫ ∫ ∫
53. (a) Euler's Method gives 0.1602y ≈ when 1.x =
(b)
( )
21
23
23
6
6
ln 3
0 5 5
5
x
x
dy xydxdy xy
y x C
y Ce
y C
y e
−
−
= −
= −
= − +
=
= ⇒ =
=
∫ ∫
(c) At ( )3 11, 5 0.2489.x y e−= = ≈
Error: 0.2489 0.1602 0.0887− ≈
54. (a) Euler's Method gives 0.2622y ≈ when 1.x =
(b)
2
2
21
2
22
6
6
1 3
131 13
31 3
1 9 133
dy xydxdy x dxy
x Cy
yx C
CC
yxx
= −
= −
−= − +
=+
= ⇒ =
= =++
∫ ∫
(c) At ( )
3 31, 0.3.9 1 1 10
x y= = = =+
Error: 0.3 0.2622 0.0378− =
55. (a) Euler's Method gives 3.0318y ≈ when 2.x =
(b) 22 123 4
dy xdx y
+=
−
( ) ( )2
3 2
3 4 2 12
4 12
y dy x dx
y y x x C
− = +
− = + +
∫ ∫
( ) ( )3
3 2
1 2: 2 4 2 1 12 13
4 12 13
y C C
y y x x
= − = + + ⇒ = −
− = + −
(c) At 2,x =
( )
( )( )
3 2
3
2
4 2 12 2 13 15
4 15 0
3 3 5 0 3.
y y
y y
y y y y
− = + − =
− − =
− + + = ⇒ =
Error: 3.0318 3 0.0318− =
56. (a) Euler's Method gives 1.7270y ≈ when 1.5.x =
(b) ( )
( )( )
( )
2
2
2
2
2
2
2 1
21
arctan
arctan 0 1 1
arctan 1
tan 1
dy x ydx
dy x dxy
y x C
C C
y x
y x
= +
=+
= +
= + ⇒ = −
= −
= −
∫ ∫
(c) At ( )21.5, tan 1.5 1 3.0096.x y= = − ≈
Error: 1.7270 3.0096 1.2826− = −
8
4321−3−4x
y
x−4 −2
−2
2
8
4
4
y
Section 6.3 Separation of Variables and the Logistic Equation 623
© 2010 Brooks/Cole, Cengage Learning
57. , ktdy ky y Cedt
= =
Initial amount: ( ) 00y y C= =
Half-life: ( )15990021 1ln
1599 2
ky y e
k
=
⎛ ⎞= ⎜ ⎟⎝ ⎠
( )ln 1 2 1599 ty Ce⎡ ⎤⎣ ⎦=
When 50, 0.9786t y C= = or 97.86%.
58. , ktdy ky y Cedt
= =
Initial conditions: ( ) ( )0
0 40, 1 35
40
35 407ln8
k
y y
Ce C
e
k
= =
= =
=
=
Particular solution: ( )ln 7 840 ty e=
When 75% has been changed:
( )
( )
( )( )
ln 7 8
ln 7 8
10 4014
ln 1 410.38 hours
ln 7 8
t
t
e
e
t
=
=
= ≈
59. (a) ( )4dy k ydx
= −
(b) The direction field satisfies ( ) 0dy dx = along
4;y = but not along 0.y = Matches (a).
60. (a) ( )4dy k xdx
= −
(b) The direction field satisfies ( ) 0dy dx = along
4.x = Matches (b).
61. (a) ( )4dy ky ydx
= −
(b) The direction field satisfies ( ) 0dy dx = along
0y = and 4.y = Matches (c).
62. (a) 2dy kydx
=
(b) The direction field satisfies ( ) 0dy dx = along
0,y = and grows more positive as y increases. Matches (d).
63. ( )
1
1
1200
1200ln 1200
1200
1200
kt C kt
kt
dw k wdt
dw k dtww kt C
w e Ce
w Ce
− + −
−
= −
=−− = − +
− = =
= −
∫ ∫
( )0 60 1200 1200 60 1140
1200 1140 kt
w C C
w e−= = − ⇒ = − =
= −
(a)
0.8k = 0.9k = 1k =
(b) 0.8: 1.31 years0.9: 1.16 years1.0: 1.05 years
k tk tk t
= =
= =
= =
(c) Maximum weight: 1200 pounds lim 1200
xw
→∞=
0
1400
1000
1400
1000
1400
100
624 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
64. From Exercise 63:
( )
( )0 0
0
1200 , 1
1200
0 1200 1200
1200 1200
kt
t
t
w Ce k
w Ce
w w C C w
w w e
−
−
−
= − =
= −
= = − ⇒ = −
= − −
65. Given family (circles): 2 2
2 2 0x y C
x yyxyy
+ =
′+ =
′ = −
Orthogonal trajectory (lines):
ln ln ln
yyx
dy dxy x
y x K
y Kx
′ =
=
= +
=
∫ ∫
66. Given family (hyperbolas): 2 222 4 0
2
x y Cx yy
xyy
− =
′− =
′ =
Orthogonal trajectory:
22
2
2
ln 2 ln ln
yyx
dy dxy xy x k
ky kxx
−
−′ =
= −
= − +
= =
∫ ∫
67. Given family (parabolas): 2
2
22 2 2
x Cyx Cy
x x yyC x y x
=
′=
′ = = =
Orthogonal trajectory (ellipses):
22
1
2 2
2
2
22
xyy
y dy x dx
xy K
x y K
′ = −
= −
= − +
+ =
∫ ∫
68. Given family (parabolas): 2
2
22 2
12 2
y Cxyy C
C y yyy x y x
=′ =
⎛ ⎞′ = = =⎜ ⎟⎝ ⎠
Orthogonal trajectory (ellipse):
22
1
2 2
2
2
22
xyy
y dy x dx
y x K
x y K
′ = −
= −
= − +
+ =
∫ ∫
69. Given family: 2 3
2
2 2 2
3
2 3
3 3 32 2 2
y Cxyy Cx
Cx x y yyy y x x
=
′ =
⎛ ⎞′ = = =⎜ ⎟
⎝ ⎠
Orthogonal trajectory (ellipses):
22
1
2 2
23
3 2
32
3 2
xyy
y dy x dx
y x K
y x K
′ = −
= −
= − +
+ =
∫ ∫
−6 6
−4
4
−3
−2
3
2
−6 6
−4
4
− 6 6
−4
4
−6 6
−4
4
Section 6.3 Separation of Variables and the Logistic Equation 625
© 2010 Brooks/Cole, Cengage Learning
70. Given family (exponential functions): x
x
y Ce
y Ce y
=
′ = =
Orthogonal trajectory (parabolas):
2
1
2
1
22
yy
y dy dx
y x K
y x K
′ = −
= −
= − +
= − +
∫ ∫
71. 121 xy
e−=
+
Because ( )0 6,y = it matches (c) or (d).
Because (d) approaches its horizontal asymptote slower than (c), it matches (d).
72. 121 3 xy
e−=
+
Because ( ) 120 3,4
y = = it matches (a).
73. 12112
xy
e−=
+
Because ( ) 120 8,32
y = =⎛ ⎞⎜ ⎟⎝ ⎠
it matches (b).
74. 212
1 xye−
=+
Because ( )0 6,y = it matches (c) or (d).
Because y approaches 12L = faster for (c), it matches (c).
75. ( ) 0.752100
1 29 tP te−
=+
(a) 0.75k =
(b) 2100L =
(c) ( ) 21000 701 29
P = =+
(d) 0.75
0.75
0.75
210010501 29
1 29 2129
10.75 ln ln 2929
ln 29 4.4897 yr0.75
t
t
t
ee
e
t
t
−
−
−
=+
+ =
=
⎛ ⎞− = = −⎜ ⎟⎝ ⎠
= ≈
(e) ( )0.75 1 , 0 702100
dP PP Pdt
⎛ ⎞= − =⎜ ⎟⎝ ⎠
76. ( ) 0.25000
1 39 tP te−
=+
(a) 0.2k =
(b) 5000L =
(c) ( ) 50000 1251 39
P = =+
(d) 0.2
0.2
0.2
500025001 39
1 39 2139
10.2 ln ln 3939
ln 39 18.31780.2
t
t
t
ee
e
t
t
−
−
−
=+
+ =
=
⎛ ⎞− = = −⎜ ⎟⎝ ⎠
= ≈
(e) ( )0.2 1 , 0 1255000
dP PP Pdt
⎛ ⎞= − =⎜ ⎟⎝ ⎠
77. 3 1100
dP PPdt
⎛ ⎞= −⎜ ⎟⎝ ⎠
(a) 3k =
(b) 100L =
(c)
−6 6
−4
4
00
120
5
626 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
(d) 2
2 3 1 3100 100
3 23 3 1 1 3 1 9 1 1 9 1 1100 100 100 100 100 100 100 100 100
d P P PP Pdt
P P P P P P P P PP P P P
′−⎛ ⎞ ⎛ ⎞′= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − − − − = − − − = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦ ⎣ ⎦
2
2 0d Pdt
= for 50,P = and by the first Derivative Test, this is a maximum. 100Note: 502 2LP⎛ ⎞= = =⎜ ⎟
⎝ ⎠
78.
( )
20.1 0.0004
0.1 1 0.004
0.1 1250
dP P Pdt
P P
PP
= −
= −
⎛ ⎞= −⎜ ⎟⎝ ⎠
(a) 10.110
k = =
(b) 250L =
(c)
(d) 250 125.2
P = = (Same argument as in Exercise 77)
79. ( )1 , 0 436
1, 3636
1 1kt t
dy yy ydtk L
Lybe be− −
⎛ ⎞= − =⎜ ⎟⎝ ⎠
= =
= =+ +
( ) 360, 4 : 4 81
bb
= ⇒ =+
Solution: 361 8 ty
e−=
+
80. ( )
( )
2.8
2.8 1 , 0 710
2.8, 1010
1 110 10 30, 7 : 7 1
1 7 7
kt t
dy yy ydtk L
Lybe be
b bb
− −
⎛ ⎞= − =⎜ ⎟⎝ ⎠
= =
= =+ +
= ⇒ + = ⇒ =+
Solution: 2.8
10317
ty
e−=
⎛ ⎞+ ⎜ ⎟⎝ ⎠
81. ( )
( )
2
0.8
4 4 1 , 0 85 150 5 120
4 0.8, 1205
1201 1
1200, 8 : 8 141
kt t
dy y y yy ydt
k L
Lybe be
bb
− −
⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
= = =
= =+ +
= ⇒ =+
Solution: 0.8120
1 14 tye−
=+
82. ( )
( )
( )
2
3 20
3 3 1 , 0 1520 1600 20 240
3 , 24020
2401 1
2400, 15 : 15 151
kt t
dy y y yy ydt
k L
Lybe be
bb
− −
⎛ ⎞= − = − =⎜ ⎟⎝ ⎠
= =
= =+ +
= ⇒ =+
Solution: ( )3 20240
1 15ty
e−
=+
83. (a) ( ), 200, 0 251 kt
LP L Pbe−
= = =+
( )2
2
2
20025 71
200391 72001 7392339
1 23 1 39ln ln 0.26402 39 2 23
k
k
k
bb
e
e
e
k
−
−
−
= ⇒ =+
=+
+ =
=
⎛ ⎞ ⎛ ⎞= − = ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
0.2640200
1 7 tPe−
=+
(b) For 5, 70 panthers.t P= ≈
00
300
100
Section 6.3 Separation of Variables and the Logistic Equation 627
© 2010 Brooks/Cole, Cengage Learning
(c) 0.264
0.264
2001001 7
1 7 2
10.264 ln7
7.37 years
t
t
ee
t
t
−
−
=+
+ =
⎛ ⎞− = ⎜ ⎟⎝ ⎠
≈
(d)
( )
1
0.264 1 , 0 25200
dP PkPdt L
PP P
⎛ ⎞= −⎜ ⎟⎝ ⎠
⎛ ⎞= − =⎜ ⎟⎝ ⎠
Using Euler's Method, 65.6P ≈ when 5.t =
(e) P is increasing most rapidly where 200 2 100,P = = corresponds to 7.37 years.t ≈
84. (a)
( ) ( )
2
2
2
0.7791
, 20, 0 1, 2 41
201 191
2041 19
1 19 5
19 4
1 4 1 19ln ln 0.77912 19 2 4
201 19
kt
k
k
k
t
Ly L y ybe
bb
ee
e
k
ye
−
−
−
−
−
= = = =+
= ⇒ =+
=+
+ =
=
⎛ ⎞ ⎛ ⎞= − = ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=+
(b) For 5, 14.43 gramst y= ≈
(c) 0.7791
0.7791
0.7791
0.7791
20181 1920 101 1918 911991
1711 1ln 6.60 hours
0.7791 171
t
t
t
t
e
e
e
e
t
−
−
−
−
=+
+ = =
=
=
− ⎛ ⎞= ≈⎜ ⎟⎝ ⎠
(d) 1 191 ln 12 4 20
dy y yky ydt L
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(e) The weight is increasing most rapidly when 2 20 2 10,y L= = = corresponding to 3.78 hours.t ≈
t 0 1 2 3 4 5
Exact 1 2.06 4.00 7.05 10.86 14.43
Euler 1 1.74 2.98 4.95 7.86 11.57
628 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
85. A differential equation can be solved by separation of variables if it can be written in the form
( ) ( ) 0.dyM x N ydx
+ =
To solve a separable equation, rewrite as,
( ) ( )M x dx N y dy= −
and integrate both sides.
86. ( ) ( ), , 0,M x y dx N x y dy+ = where M and N are
homogeneous functions of the same degree. See Example 7a.
87. Two families of curves are mutually orthogonal if each curve in the first family intersects each curve in the second family at right angles.
88. Answers will vary. Sample answer: There might be limits on available food or space.
89.
( )( )
( ) ( )
( ) ( )
( ) ( )
2
11
1
1
1 1
1 11 1
11 111
kt
kt
kt
kt
kt kt
kt
kt kt
ktkt
ybe
y bkebe
k bebe be
k bebe be
k ky ybebe
−
−
−
−
− −
−
− −
−−
=+
−′ = −+
= ⋅+ +
+ −= ⋅
+ +
⎛ ⎞= ⋅ − = −⎜ ⎟++ ⎝ ⎠
90. (a) ( )
1lnkt
dv k W vdt
dv k dtW vW v kt C
v W Ce−
= −
=−
− − = +
= −
∫ ∫
Initial conditions: 20, 0 when 0W v t= = = and 10v =
when 0.5t = so, 20, ln 4.C k= =
Particular solution:
( )( )ln 4 120 1 20 14
ttv e−
⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
or
( )1.38620 1 tv e−= −
(b) ( ) ( )1.386 1.38620 1 20 0.7215t ts e dt t e C− −= − ≈ + +∫
Because ( )0 0, 14.43s C= ≈ − and you have
( )1.38620 14.43 1 .ts t e−≈ + −
91. False. dy xdx y
= is separable, but 0y = is not a solution.
92. True
( )( )2 1dy x ydx
= − +
93. False
( )
( )
2 2 2 2 2
2
, 4 6 1
,
f tx ty t x xyt t y
t f x y
= − + +
≠
94. True
2 2 2 22 2x y Cy x y Kxdy x dy K xdx C y dx y
+ = + =
−= =
−
2
2
2
2
2 2 2
2 2 2
2 2
2 2
2 22 2
22
1
x K x Kx xC y y Cy y
Kx xCy y
x y xx y y
y xx y
− −⋅ =
− −
−=
−
+ −=
+ −
−=
−
= −
95.
( )Product Rule
0
0
fg gf f g
f f g gf
fg gf f
′ ′ ′ ′+ =
′ ′ ′− + =
′′ + =
′−
Need ( )2 2 22 1 2 0,x x xf f e xe x e′− = − = − ≠ so
avoid 1.2
x =
( )
( )
( )
2
2
1
1 2
2 112 12 1
1ln ln 2 12
2 1
x
x
x
g f xeg f f xx e
g x x x C
g x Ce x
′ ′= = = +
′ − −−
= + − +
= −
So there exists g and interval ( ), ,a b as long as
( )1 , .2
a b∉
Section 6.4 First-Order Linear Differential Equations 629
© 2010 Brooks/Cole, Cengage Learning
Section 6.4 First-Order Linear Differential Equations
1.
( )
3
2 3
11 1 1
x
x
x y xy e
y y ex x
′ + = +
′ + = +
Linear
2.
( ) ( )( )
2 ln
ln 1 2 0
1 20
ln
xy y x y
x y x y
xy y
x
′− =
′ + − =
−′ + =
Linear
3. 2siny y x xy′ − =
Not linear, because of the 2-term.xy
4. 2 5
2 55 2
y xy
y xyy xy
′−=
′− =
′ + =
Linear
5. 1 6 2dy y xdx x
⎛ ⎞+ = +⎜ ⎟⎝ ⎠
Integrating factor: ( )1 lnx dx xe e x∫ = =
( ) 3 2
2
6 2 2
2
xy x x dx x x C
Cy x xx
= + = + +
= + +
∫
6. 2 3 5dy y xdx x
+ = −
Integrating factor: 22 ln 2x dx xe e x∫ = =
( )
32 2 4
22
3 53 54 3
3 54 3
xx y x x dx x C
Cy x xx
= − = + +
= + +
∫
7. 16y y′ − =
Integrating factor: 1dx xe e− −∫ =
16
16 16
16
x x x
x x x
x
e y e y e
ye e dx e C
y Ce
− − −
− − −
′ − =
= = − +
= − +
∫
8. 2 10y xy x′ + =
Integrating factor: 22x dx xe e∫ =
2 2 2
2
10 5
5
x x x
x
ye xe dx e C
y Ce−
= = +
= +
∫
9. ( )( )( )
1 cos
1 cos cos cos
cos cos
y x dx dy
y y x y x x
y x y x
+ =
′ = + = +
′ − =
Integrating factor: cos sinx dx xe e− −∫ =
( ) ( )( )
sin sin sin
sin sin
sin
sin
cos cos
cos
1
x x x
x x
x
x
y e x e y x e
ye x e dx
e C
y Ce
− − −
− −
−
′ − =
=
= − +
= − +
∫
10. ( )( )1 sin 0
sin sin
y x dx dy
y x y x
⎡ − ⎤ − =⎣ ⎦′ − = −
Integrating factor: sin cosx dx xe e−∫ =
cos cos cos
cos
sin
1
x x x
x
ye xe dx e C
y Ce−= − = +
= +
∫
11. ( ) 21 1
1 11
x y y x
y y xx
′− + = −
⎛ ⎞′ + = +⎜ ⎟−⎝ ⎠
Integrating factor: ( )1 1 ln 1 1x dx xe e x−⎡ ⎤ −⎣ ⎦∫ = = −
( ) ( )
( )
2 31
3
11 13
33 1
y x x dx x x C
x x Cyx
− = − = − +
− +=
−
∫
12. 33 xy y e′ + =
Integrating factor: 3 3dx xe e∫ =
3 3 3 6 6
3 3
16
16
x x x x x
x x
ye e e dx e dx e C
y e Ce−
= = = +
= +
∫ ∫
630 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
13. 323 xy x y e′ − =
Integrating factor: 2 33x dx xe e− −∫ =
( )
3 3 3
3
x x x
x
ye e e dx dx x C
y x C e
− −= = = +
= +
∫ ∫
14. tan secy y x x′ + =
Integrating factor: tan ln cos secx dx xe e x−∫ = =
2sec sec tan
sin cos
y x x dx x C
y x C x
= = +
= + ⋅∫
15. (a) Answers will vary.
(b)
Integrating factor:
x
dxx x
dy e ydxdy y e e edx
= −
∫+ = =
( )2
2
2
2
x x x
x x
x x
e y e y e
ye e dx
ye e C
′ + =
=
1= +
∫
( ) 1 10 1 12 2
y C C= ⇒ = + ⇒ =
( )
21 12 21 1 12 2 2
x x
x x x x
ye e
y e e e e− −
= +
= + = +
(c)
16. (a)
(b) ( ) ( )
( ) ( )
2 2
1 ln
1 1sin , , sin
x dx x
y y x P x Q x xx x
u x e e x
′ + = = =
∫= = =
2
2 2
2
2
sin1sin cos2
1 1 cos2
1 1 10 cos2 2
1 1 1cos2 2
y x y x x
yx x x dx x C
y x Cx
C C
y xx
ππ
′ + =
= = − +
⎡ ⎤= − +⎢ ⎥⎣ ⎦
⎡ ⎤= − + ⇒ = −⎢ ⎥⎣ ⎦
⎡ ⎤= − −⎢ ⎥⎣ ⎦
∫
(c)
17.
( )2
2 2
cos 1 0
sec sec
y x y
y x y x
′ + − =
′ + =
Integrating factor: 2sec tanx dx xe e∫ =
tan 2 tan tan
tan
sec
1
x x x
x
ye xe dx e C
y Ce−= = +
= +
∫
Initial condition: ( )0 5, 4y C= =
Particular solution: tan1 4 xy e−= +
18. 23 1
213 3
2
2 1
x
x
x y y e
y y ex x
′ + =
⎛ ⎞′ + =⎜ ⎟⎝ ⎠
Integrating factor: ( ) ( )3 22 1x dx xe e
−∫ =
2113 2
2212
1 121
2
x
x
ye dx Cx x
Cxy ex
− = = − +
⎛ ⎞−= ⎜ ⎟
⎝ ⎠
∫
Initial condition: ( )1 , 3y e C= =
Particular solution: 221
23 1
2x xy e
x⎛ ⎞−
= ⎜ ⎟⎝ ⎠
x−4
−3
4
5
y
−6
−2
6
6
4
−4
−4
4
x
y
4
4
−4
−4
Section 6.4 First-Order Linear Differential Equations 631
© 2010 Brooks/Cole, Cengage Learning
19. tan sec cosy y x x x′ + = +
Integrating factor: tan ln sec secx dx xe e x∫ = =
( )sec sec sec cos tan
sin cos cos
y x x x x dx x x C
y x x x C x
= + = + +
= + +∫
Initial condition: ( )0 1, 1y C= =
Particular solution: ( )sin 1 cosy x x x= + +
20. sec secy y x x′ + =
Integrating factor: sec ln sec tan sec tanx dx x xe e x x+∫ = = +
( ) ( )sec tan sec tan sec
sec tan
1sec tan
y x x x x x dx
x x CCy
x x
+ = +
= + +
= ++
∫
Initial condition: ( )0 4, 4 1 , 31 0
Cy C= = + =+
Particular solution:
3 3 cos1 1sec tan 1 sin
xyx x x
= + = ++ +
21. 1 0y yx
⎛ ⎞′ + =⎜ ⎟⎝ ⎠
Integrating factor: ( )1 lnx dx xe e x∫ = =
Separation of variables:
1 1
ln ln lnln ln
dy ydx x
dy dxy x
y x Cxy Cxy C
= −
= −
= − +
=
=
∫ ∫
Initial condition: ( )2 2, 4y C= =
Particular solution: 4xy =
22. ( )2 1 0y x y′ + − =
Integrating factor: ( ) 22 1x dx x xe e− −∫ =
2
2
x x
x x
ye C
y Ce
−
−
=
=
Separation of variables:
( )
21
21
2
1 1 2
ln lnx x
x x
dy x dxy
y C x x
yC e
y Ce
−
−
= −
+ = −
=
=
∫ ∫
Initial condition: ( )1 2, 2y C= =
Particular solution: 22 x xy e −=
23. ( )2
2 21
x dy x y dx
dy x y ydx x x x
= + +
+ += = + +
( ) ( )1
1 21 Linear
1x dx
dy ydx x x
u x ex
−
− = +
∫= =
22 1 1 21
2ln
2 ln
y x dx x dxx x x x
x x Cx
x x Cx
⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
−⎡ ⎤= + +⎢ ⎥⎣ ⎦= − + +
∫ ∫
( )1 10 2 12
2 ln 12
y C C
y x x x
= = − + ⇒ =
= − + +
24.
( ) ( )
3
2
1 21 2
2
1 1 Linear2 2 2
1x dx
xy y x x
dy xydx x
u x ex
−
′ − = −
− = −
∫= =
2 3 2 1 21 2 1 2
1 2
5 21 2 1 2
3
1 12 2 2 2
5
5
x x xy x dx x dxx
xx x C
x x C x
−⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎡ ⎤= − +⎢ ⎥
⎣ ⎦
= − +
∫ ∫
( )
3
64 174 2 4 25 5
175 5
y C C
xy x x
= = − + ⇒ = −
= − −
632 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
25.
( )
( )
1
2
2
3
, constant
1
1 ln
lnkt C
kt
kt
dP kP N Ndt
dP dtkP N
dP dtkP N
kP N t Ck
kP N kt C
kP N e
C e NPk
NP Cek
+
= +
=+
=+
+ = +
+ = +
+ =
−=
= −
∫ ∫
0
0 0
0
When 0:
kt
t P PN NP C C Pk k
N NP P ek k
= =
= − ⇒ = +
⎛ ⎞= + −⎜ ⎟⎝ ⎠
26.
( )
( )
1
2
2
3
1 ln
lnrt C
rt
rt
dA rA Pdt
dA dtrA P
dA dtrA P
rA P t Cr
rA P rt C
rA P e
C e PAr
PA Cer
+
= +
=+
=+
+ = +
+ = +
+ =
−=
= −
∫ ∫
( )
When 0: 0
0
1rt
t AP PC Cr r
PA er
= =
= − ⇒ =
= −
27. (a) ( )( )( )0.08 10
1
275,000 1 $4,212,796.940.06
rtPA er
A e
= −
= − ≈
(b) ( )( )0.059 25550,000 1 $31,424,909.750.05
A e= − ≈
28. ( )
( )
0.08
0.08
125,0001,000,000 10.08
1.64
ln 1.646.18 years
0.08
t
t
e
e
t
= −
=
= ≈
29. (a) , constantdQ q kQ qdt
= −
(b) Q kQ q′ + =
Let ( ) ( ), ,P t k Q t q= = then the integrating factor
is ( ) .ktu t e=
kt kt kt kt ktq qQ e qe dt e e C Cek k
− − −⎛ ⎞= = + = +⎜ ⎟⎝ ⎠∫
0
0 0
0
When 0:
kt
t Q Qq qQ C C Qk kq qQ Q ek k
−
= =
= + ⇒ = −
⎛ ⎞= + −⎜ ⎟⎝ ⎠
(c) limt
qQk→∞
=
30. (a) ( )75dN k Ndt
= −
(b) 75N kN k′ + =
Integrating factor: k dt kte e∫ =
( )75
75
75 75
75
kt kt kt
kt kt
kt kt kt
kt
N e kNe ke
Ne ke
Ne ke e C
N Ce−
′ + =
′ =
= = +
= +
∫
(c) For 1, 20:
20 75 55k k
t N
Ce Ce− −
= =
= + ⇒ − =
For 20 20
20, 35:
35 75 40k k
t N
Ce Ce− −
= =
= + ⇒ − =
19
2055 11 1 11ln40 8 19 8
0.0168
kk
kCe e k
Ce
−
−⎛ ⎞= ⇒ = ⇒ = ⎜ ⎟⎝ ⎠
≈
55
55 55.9296
k
k
Ce
C e
− = −
= − ≈ −
0.016875 55.9296 tN e−= −
Section 6.4 First-Order Linear Differential Equations 633
© 2010 Brooks/Cole, Cengage Learning
31. Let Q be the number of pounds of concentrate in the solution at any time t. Because the number of gallons of solution in the tank at any time t is ( )0 1 2v r r t+ − and
because the tank loses 2r gallons of solution per minute, it must lose concentrate at the rate
( ) 2
0 1 2.Q r
v r r t⎡ ⎤⎢ ⎥
+ −⎢ ⎥⎣ ⎦
The solution gains concentrate at the rate 1 1.r q Therefore, the net rate of change is
( )1 1 2
0 1 2
dQ Qq r rdt v r r t
⎡ ⎤= − ⎢ ⎥
+ −⎢ ⎥⎣ ⎦
or
( )
21 1
0 1 2.dQ r Q q r
dt v r r t+ =
+ −
32. From Exercise 31, and using 1 2 ,r r r= =
10
.dQ rQ q rdt v
+ =
33. (a) ( )
( )
2
1 10 1 2
0 0 1 0
1 2
0 , 25, 0, 200,
110, 10, 020
r QQ q rv r r t
Q q q q v
r r Q Q
′ + =+ −
= = = =
′= = + =
( )
1
1 20
1 120
1ln ln20
t
dQ dtQ
Q t C
Q Ce−
= −
= − +
=
∫ ∫
Initial condition: ( )0 25, 25Q C= =
Particular solution: ( )1 2025 tQ e−=
(b) ( )1 2015 25
3 1ln5 20
320 ln 10.2 min5
te
t
t
−=
⎛ ⎞ = −⎜ ⎟⎝ ⎠
⎛ ⎞= − ≈⎜ ⎟⎝ ⎠
(c) ( )1 20lim 25 0t
te−
→∞=
34. (a) ( )
( )
2
1 10 1 2
0 1 0
1 2
0 25, 0.04, 200,
110, 10, 0.420
r QQ q rv r r t
Q q q v
r r Q Q
′ + =+ −
= = = =
′= = + =
Integrating factor: 1 20te
( )
( )
( )( )
1 201 20 1 20
1 2
1 20
0.4 8
8
0 25 8 17
8 17
tt t
t
t
Qe e dt e C
Q Ce
Q C C
Q e
−
−
−
= = +
= +
= = = + ⇒ =
= +
∫
(b) ( )1 20
1 20
15 8 17
7 17
t
t
e
e
−
−
= +
=
7 1 7ln 20 ln 17.75 min17 20 17
t t⎛ ⎞ ⎛ ⎞= − ⇒ = − ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
(c) ( )lim 8 lbst
Q t→∞
=
35. (a) The volume of the solution in the tank is given by ( )0 1 2 .v r r t+ − Therefore, ( )100 5 3 200t+ − = or
50t = minutes.
(b) ( )
( )
21 1
0 1 2
0 0 1 0 1
2
0 , 0, 0.5, 100, 5,
33, 2.5100 2
r QQ q rv r r t
Q q q q v r
r Q Qt
′ + =+ −
= = = = =
′= + =+
Integrating factor: ( ) ( )3 23 100 2 50t dte t+⎡ ⎤⎣ ⎦∫ = +
( ) ( ) ( )
( ) ( )
3 2 3 2 5 2
3 2
50 2.5 50 50
50 50
Q t t dt t C
Q t C t −
+ = + = + +
= + + +
∫
Initial condition:
( ) ( )3 2 5 20 0, 0 50 50 , 50Q C C−= = + = −
Particular solution:
( ) ( )
( ) ( )
3 25 2
3 25 2
50 50 50
50 100 50 100
25100 82.32 lbs2
Q t t
Q
−−
−
= + − +
= −
= − ≈
634 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
(c) The volume of the solution is given by ( ) ( )0 1 2 100 5 3 200 50v r r t t t+ − = + − = ⇒ =
minutes.
( )
( )
21 1
0 1 2
0 1 0 1 20 0, 1, 100, 5, 3
3 5100 2
r QQ q rv r r t
Q q q v r r
QQt
′ + =+ −
= = = = = =
′ + =+
Integrating factor is ( )3 250 ,t+ as in #43.
( ) ( ) ( )
( ) ( )
3 2 3 2 5 2
3 2
50 5 50 2 50
2 50 50
Q t t dt t C
Q t C t −
+ = + = + +
= + + +
∫
( )
( ) ( ) ( )
( ) ( ) ( )
3 2 3 2 5 2
5 2 3 2
0 0:
0 100 50 100 50 2 50
2 50 2 50 50
Q
C C
Q t t
−
−
=
= + ⇒ = − = −
= + − +
When 50,t =
( ) ( )5 2 3 2 50200 2 50 100 2002
164.64 lbs (double the answer to part (b))
Q −= − = −
≈
36. ( ) ( )y P x y Q x′ + =
Integrating factor: ( )P x dxu e∫=
( ) ( )
( ) ( )
y u P x yu Q x u
uy Q x u
′ + =
′ =
so ( ) ( )u x P x u′ =
Answer (a)
37. From Example 6,
( )1 , Solutionkt m
dv kv gdt m
mgv ek
−
+ =
= −
( ) 8 132, 8, 5 101,4
g mg v mg−
= − = − = − = =
implies that ( )( )5 1 48101 1 .kek
−−− = −
Using a graphing utility, 0.050165,k ≈ and
( )0.2007159.47 1 .tv e−= − −
As , 159.47 ft/sec.t v→ ∞ → − The graph of v is shown below.
38. ( ) ( )
( )
( )( )
0.2007
0.2007
0.2007
159.47 1
159.47 794.57
0 5000 794.57 5794.57
159.47 794.57 5794.57
t
t
t
s t v t dt
e dt
t e C
s C C
s t t e
−
−
−
=
= − −
= − − +
= = − + ⇒ =
= − − +
∫∫
The graph of ( )s t is shown below.
( ) 0s t = when 36.33 sec.t ≈
39. 00,dI R EL RI E I I
dt L L′+ = + =
Integrating factor: ( )R L dt Rt Le e∫ =
0 0
0
Rt L Rt L Rt L
Rt L
E EI e e dt e CL R
EI CeR
−
= = +
= +
∫
40. ( ) 00 0, 120 volts, 600 ohms,I E R= = =
4 henrysL =
( )
( ) ( )
( )( )
0
150
150
150
150
120 10600 51 15 51lim amp5
1 10.90 0.18 15 5
0.9 1
0.1
150 ln 0.1
ln 0.10.0154 sec
150
Rt L
t
t
t
t
t
EI CeR
C C
I e
I
e
e
e
t
t
−
−
→∞
−
−
−
= +
= + ⇒ = −
= −
=
= = −
= −
=
− =
= ≈−
41. ( ) ( )
( ) ( )
Standard form
Integrating factorP x dx
dy P x y Q xdx
u x e
+ =
∫=
50
−200
0 40
0 40
−500
6000
Section 6.4 First-Order Linear Differential Equations 635
© 2010 Brooks/Cole, Cengage Learning
42. ( ) ( ) Standard formny P x y Q x y′ + =
Let ( )1 0, 1 .nz y n−= ≠ Multiplying by
( )1 nn y−− produces
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
11 1 1
1 1 . Linear
n nn y y n P x y n Q x
z n P x z n Q x
− −′− + − = −
′ + − = −
43.
2
2 0
2
y x
dy x dx
y x C
′ − =
=
= +
∫ ∫
Matches c.
44.
1
2
2 0
2
ln 2x
y ydy dxyy x C
y Ce
′ − =
=
= +
=
∫ ∫
Matches d.
45.
21
2
2 0
2
lnx
y xydy x dxy
y x C
y Ce
′ − =
=
= +
=
∫ ∫
Matches a.
46.
( ) 21
22
2
2
2 11 1ln 2 12 2
2 112
x
x
y xy xdy x dx
y
y x C
y C e
y Ce
′ − =
=+
+ = +
+ =
= − +
∫ ∫
Matches b.
47. 2 2 3
2 2
3
3, , 3
y x y x y
n Q x P x
′ + =
= = =
( ) ( ) ( )2 22 3 2 32 2
3 32 2 2 2
3 32 2 2
32 2
322
2
2
1313
1 13
x dx x dx
x x
x x
x
x
y e x e dx
y e x e dx
y e e C
y Ce
Cey
− −−
− − −
− − −
−
∫ ∫= −
= −
= +
= +
+ +
∫∫
48. 1
221, , , x dx x
y xy xy
n Q x P x e e
−′ + =
∫= − = = =
2 2 22
22
2
1
x x x
x
y e xe dx e C
y Ce−
= = +
= +
∫
49.
( )
2
1
1 ln 1
1
2, ,x dx x
y y xyx
n Q x P x
e e x
−
− − −
⎛ ⎞′ + =⎜ ⎟⎝ ⎠
= = =
∫ = =
( )1 1 1
2
2
1
1
y x x x dx x C
x Cxy
yCx x
− − −= − = − +
= − +
=−
∫
50.
( )( ) ( )
1
1 2 1 1 2 ln
1
1, ,2x dx x
y y x yx
n Q x P x
e e x
−
⎛ ⎞′ + =⎜ ⎟⎝ ⎠
= = =
= =
( )
( )
1 2 1 2 1 2
5 25 2
1
25 2
12
15 5
25
y x x x dx
x Cx C
x Cy
x
=
+= + =
+=
∫
51. 3
3
22 ln 2
1
13, 1, ,dx xx
xy y xy
y y yx
n Q P e e xx
−− −
′ + =
′ + =
∫= = = = =
2 2 2 1
2 2
2 22 2
2 2
21 1or 2
2
y x x dx C x C
y x Cx
y x Cxx Cx y
− − − −
−
= − + = +
= +
= = ++
∫
52. ( )
3
2 1 23, 1, 1, dx x
y y y
n P Q e e− −
′ − =
∫= = − = =
( )2 2 2 2
2 2
22
2
11
1
x x x
x
x
y e e dx e C
y Ce
yCe
−
− −
−
= − = − +
= − +
=− +
∫
636 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
53.
( ) ( )
3
2 3 2 3
13, , , 1x x
dx x
y y e y n Q e P
e e− −
′ − = = = = −
∫ =
( ) ( ) ( )
( ) ( )
2 3 2 3 1 32 3
2 3 1 32 3
2 3 2 3
2 23 3
2
2
x x xx
x x
x x
y e e e dx e dx
y e e C
y e Ce
− −
−
= =
= +
= +
∫ ∫
54. 2
1
22
x
x
yy y ey y e y−
′ − =
′ − =
1, , 2xn Q e P= − = = −
( )2 2 4
2 4 4 3
2 4
23
23
2
dx x
x x x x
x x
e e
y e e e dx e C
y e Ce
− −
− − −
∫ =
= = − +
= − +
∫
55. (a)
(b) 21dy y xdx x
− =
Integrating factor: 1 ln 1x dx xe ex
− −= =
2
2
3
1 1
12
2
y y xx x
xy x dx Cx
xy Cx
′ − =
⎛ ⎞ = = +⎜ ⎟⎝ ⎠
= +
∫
( ) ( )
( ) ( )
32
32
8 12, 4 : 4 2 4 4 82 2 2
8 12, 8 : 8 2 2 2 42 2 2
xC C y x x x
xC C y x x x
−− = − ⇒ = − ⇒ = − = −
= + ⇒ = ⇒ = + = +
(c)
56. (a)
(b) 3 34y x y x′ + =
Integrating factor: 3 44x dx xe e∫ =
4 4 43 3
4 4 43
4
14
14
4x x x
x x x
x
y e x ye x e
ye x e dx e C
y Ce−
′ + =
= = +
= +
∫
( )( )
4
4
7 7 13 131 12 2 4 4 4 4
3 31 1 1 12 2 4 4 4 4
0, :
0, :
x
x
C C y e
C C y e
−
−
= + ⇒ = ⇒ = +
− − = + ⇒ = − ⇒ = −
(c)
− 4
−6
10
4
−6
10
− 4 4
−1
−1
5
3
−1
−1
5
3
Section 6.4 First-Order Linear Differential Equations 637
© 2010 Brooks/Cole, Cengage Learning
57. (a)
(b) ( )cot 2y x y′ + =
Integrating factor: cot ln sin sinx dx xe e x∫ = =
( )sin cos 2 sin
sin 2 sin 2 cos
2 cot csc
y x x y x
y x x dx x C
y x C x
′ + =
= = − +
= − +∫
( )
( )
1 2 cot 11,1 : 1 2 cot 1 csc 1 sin 1 2 cos 1csc 1
2 cot sin 1 2 cos 1 csc
C C
y x x
+= − + ⇒ = = +
= − + +
( )
( )
2 cot 3 13, 1 : 1 2 cot 3 csc 3 2 cos 3 sin 3csc 3
2 cot 2 cos 3 sin 3 csc
C C
y x x
−− − = − + ⇒ = = −
= − + −
(c)
58. (a)
(b) 22y xy xy′ + =
Bernoulli equation, 2n = letting 1 2 1,z y y− −= = you obtain 22x dx xe e− −= and ( ) 2 211 .
2x xxe dx e− −− =∫ The solution is:
2 21
22
2
12
1 1 1 22 2
2
1 2
x x
xx
x
y e e C
CeCey
yCe
− − −= +
+= + =
=+
( )
( )( )
22
2
2 2 10, 3 : 3 1 21 2 3 6
2 631 3
2 10,1 : 1 1 2 21 2 2
21
xx
x
C CC
yee
C CC
ye
= ⇒ + = ⇒ = −+
= =−−
= ⇒ + = ⇒ =+
=+
(c)
−2
−3
6
3
−2
−3
6
3
−5
−3
7
5
−5
−3
7
5
638 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
59. 2 0x y x ye dx e dy+ −− =
Separation of variables:
2
2
21
2
12
2
x y x y
x y
x y
x y
e e dx e e dy
e dx e dy
e e C
e e C
−
−
−
−
=
=
= − +
+ =
∫ ∫
60. ( )
34
dy xdx y y
−=
+
Separation of variables:
( ) ( )2
3 22
1
3 2 2
4 3
2 33 2
2 12 3 18
y y dy x dx
y xy x C
y y x x C
+ = −
+ = − +
+ = − +
∫ ∫
61. ( )cos cos 0y x x dx dy− + =
Separation of variables:
( )( )
sin
1cos1
sin ln 1 ln
ln 1 sin ln
1x
x dx dyy
x y C
y x C
y Ce−
−=
−
= − − +
− = − +
= +
∫ ∫
62. 22 1y x y′ = −
Separation of variables:
( )
2
2
2
1 21
arcsin
sin
dy x dxy
y x C
y x C
=−
= +
= +
∫ ∫
63. ( ) ( )2 23 4 2 0y xy dx xy x dy+ + + =
Homogeneous: ,y vx dy v dx x dv= = +
( ) ( )( )
( )
2 2 2 2 2
2
5 2
5 2
3 2 4
3 4 2 0
5 2 1 0
ln ln ln
v x vx dx vx x v dx x dv
vdx dvx v v
x v v C
x v v C
x y x y C
+ + + + =
+⎛ ⎞+ =⎜ ⎟+⎝ ⎠
+ + =
+ =
+ =
∫ ∫
64. ( ) 0x y dx x dy+ − =
Linear: 1 1y yx
′ − =
Integrating factor: ( ) 1ln1 1xx dxe ex
−−∫ = =
( )
1 1 ln
ln
y dx x Cx xy x x C
= = +
= +
∫
65. ( )2 0xy e dx x dy− + =
Linear: 2 1 xy y ex x
⎛ ⎞′ + =⎜ ⎟⎝ ⎠
Integrating factor: ( ) 22 ln 2x dx xe e x∫ = =
( )
( )
2 2
2 2
1 1
1
x x
x
yx x e dx e x Cx
e Cy xx x
= = − +
= − +
∫
66. ( )2 2 0y xy dx x dy+ − =
Homogeneous: ,y vx dy v dx x dv= = +
( ) ( )2 2 2 2
2
2
0
01 1
1ln
ln
v x vx dx x v dx x dv
v dx x dv
dx dvx v
x Cv
xyC x
+ − + =
− =
=
= − +
=−
∫ ∫
67. ( )2 4 3 3
3 3
1 0
1
x y dx x y dy
y y x yx
− −
− + =
⎛ ⎞′ + =⎜ ⎟⎝ ⎠
Bernoulli: 3 13, , ,n Q x P x− −= − = =
( ) 44 ln 4x dx xe e x∫ = =
( )( )4 4 3 4 2
4 4 2
4 2
2
y x x x dx x C
x y x C
−= = +
− =
∫
Review Exercises for Chapter 6 639
© 2010 Brooks/Cole, Cengage Learning
68. ( )3 4 0y dx x y dy+ + =
Homogeneous: ,x vy dx v dy y dv= = +
( ) ( )
( )( )
4
4
3
3 4 0
1 41
ln 1 ln ln
1
y v dy y dv vy y dy
dv dyv y
v y C
y v C
y x y C
+ + + =
= −+
+ = − +
+ =
+ =
∫ ∫
69. ( )2
2
3 4
3 12
3 12
y x dx x dy
dyx y xdx
y y xx
− = −
= − +
′ + =
Integrating factor: ( )3 3 ln 3x dx xe e x∫ = =
( )3 3 3 4
3 4 5
23
3 12 12
12125
125
y x x y x x xx
yx x dx x C
Cy xx
′ + = =
= = +
= +
∫
70. ( )( )2 1 0yx dx y e x dy+ + + =
Separation of variables:
( )
( )( )
2
2 21
2 2
11 1ln 12 2
ln 1 2
y
y
y
x dx y e dyx
x y e C
x y e C
= − ++
+ = − − +
+ + + =
∫ ∫
71. False. The equation contains .y
72. True. ( ) 0xy x e y′ + − = is linear.
Review Exercises for Chapter 6
1.
( ) ( )3 2
2 3 3
, 3
2 4 2 3 4 10 .
y x y x
xy y x x x x
′= =
′ + = + =
Yes, it is a solution.
2.
( )
2 sin 24 cos 2
8 sin 216 cos 2
8 16 cos 2 8 2 sin 2 0
y xy xy xy x
y y x x
=
′ =
′′ = −
′′′ = −
′′′ − = − − ≠
Not a solution
3.
( )
2
32
4 7
44 7 73
dy xdx
xy x dx x C
= +
= + = + +∫
4.
( )
3
3 4 2
3 8
33 8 44
dy x xdx
y x x dx x x C
= −
= − = − +∫
5. cos 2
1cos 2 sin 22
dy xdx
y x dx x C
=
= = +∫
6. 2 sin
2 sin 2 cos
dy xdx
y x dx x C
=
= = − +∫
7. 5
5
dy x xdx
y x x dx
= −
= −∫
Let 5, , 5u x du dx x u= − = = +
( )
( )
( ) ( )
( ) ( )
( ) ( )
3 2 1 2
5 2 3 2
5 2 3 2
3 2
3 2
5
5
2 105 32 105 55 31 5 6 5 50
151 5 6 20
15
y u u du
u u du
u u C
x x C
x x C
x x C
= +
= +
= + +
= − + − +
= − ⎡ − + ⎤ +⎣ ⎦
= − + +
∫∫
640 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
8. 2 7
2 7
dy x xdx
y x x dx
= −
= −∫
Let 7, , 7:u x du dx x u= − = = +
( )
( ) ( )
( ) ( )
1 2
5 2 3 2
5 2 3 2
3 2
2 7
4 285 34 287 75 34 7 3 14
15
y u u du
u u C
x x C
x x C
= +
= + +
= − + − +
= − + +
∫
9. 2
2 2
x
x x
dy edx
y e dx e C
−
− −
=
= = − +∫
10. 3
3 3
3
3 9
x
x x
dy edx
y e dx e C
−
− −
=
= = − +∫
11. 2dy x ydx
= −
12. sin4
dy yxdx
π⎛ ⎞= ⎜ ⎟⎝ ⎠
13. ( )3 , 2, 1y x′ = −
(a) and (b)
14. ( )22 , 0, 2y x x′ = −
(a) and (b)
15. ( )21 14 3 , 0, 3y x x′ = −
(a) and (b)
16. ( )4 , 1, 1y y x′ = + −
(a) and (b)
17. ( )2 , 0, 14
xyyx
′ =+
(a) and (b)
18. ( )2 , 0, 21
yyx
′ = −+
(a) and (b)
x –4 −2 0 2 4 8
y 2 0 4 4 6 8
dy dx –10 –4 –4 0 2 8
x –4 −2 0 2 4 8
y 2 0 4 4 6 8
dy dx –4 0 0 0 –4 0
x
y(2, 1)
4
8−2
y
x
(0, 2)
3−3
5
−1
−4 4
−4
4
x
y
(0, 3)
y
x3−3
2
−4
(−1, 1)
4
−4
4
x
y
(0, 1)
−4
y
x
(0, −2)
6−6
−10
2
Review Exercises for Chapter 6 641
© 2010 Brooks/Cole, Cengage Learning
19.
( )2
8
8 82
dy xdx
xy x dx x C
= −
= − = − +∫
20.
1
1
8
8
ln 8
8
8
x C x
x
dy ydx
dy dxy
y x C
y e Ce
y Ce
+
= +
=+
+ = +
+ = =
= − +
∫ ∫
21. ( )
( )
( )
2
2
1
3
3
3
13
13
dy ydx
y dy dx
y x C
yx C
yx C
−
−
= +
+ =
− + = +
−+ =
+
= − −+
∫ ∫
22.
( )
1 2
1 21
11 2
2
10
10
2 10
52
5
dy ydx
y dy dx
y x C
Cy x C C
y x C
−
=
=
= +
⎛ ⎞= + =⎜ ⎟⎝ ⎠
= +
∫ ∫
23.
( )
( )
( )( )
1
22
2 0
2
12
1 212
ln 2 ln 2
22
xx
x y xy
dyx xydx
xdy dxy x
dy dxy x
y x x C
Cey Ce xx
−
′+ − =
+ =
=+
⎛ ⎞= −⎜ ⎟+⎝ ⎠
= − + +
= + =+
24. ( )
( )
1
1 0
1
1
ln lnx
xy x y
dyx x ydxdy x dxy x
y x x C
y Cxe
′ − + =
= +
+=
= + +
=
∫ ∫
25. kty Ce=
( )( ) ( )
( )
5
5
3 34 4
34
203
2015 3
0, :
5, 5 : 5
ln
k
k
C
e
e
k
=
=
=
=
( )ln 20 3 5 0.3793 34 4
t ty e e⎡ ⎤⎣ ⎦= ≈
26. kty Ce=
( )( ) ( )
( )
2 2
4 2 4 2
2
3 3 32 2 2
3 32 2
10 1013 2 3
2, :
4, 5 : 5
ln
k k
k k k k
k
Ce C e
Ce e e e
e k
−
−
= ⇒ =
= = =
= ⇒ =
So, ( ) ( ) ( )2 1 2 ln 10 33 3 3 92 2 10 20.C e−= = =
Finally, ( )1 2 ln 10 3920 .ty e=
27. kty Ce=
( )0, 5 : 5C =
51 15, : 56 6
1 1 ln 30ln5 30 5
ke
k
⎛ ⎞ =⎜ ⎟⎝ ⎠
−⎛ ⎞= =⎜ ⎟⎝ ⎠
[ ]ln30 5 0.6805 5t ty e e− −= ≈
28. kty Ce=
( )( ) ( )
( )
6 6 5
1 25 9
1, 9 : 9 9
6, 2 : 2 2 9 9
ln 0.3008
k k
k k k k
Ce C e
Ce e e e
k
−
−
= ⇒ =
= ⇒ = =
= ≈ −
So, ( ) ( ) 1 51 5 ln 2 9 299 9 12.15864.C e
−−= = ≈
Finally, 0.300812.1586 .ty e−≈
29. ( ), 0 30dP kp Pdh
= =
( )( )
( )
( ) ( )
( ) ( )
18,000
ln 2 18,000
35,000 ln 2 18,000
30
18,000 30 15
ln 1 2 ln 218,000 18,000
30
35,000 30 7.79 inches
kh
k
h
P h e
P e
k
P h e
P e
−
−
=
= =
−= =
=
= ≈
642 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
30. ( )1599
15
7.5 15
kt kt
k
y Ce e
e
= =
=
( )1 11599 2ln 0.000433k = ≈ −
When ( )0.000433 750750, 15 10.84 .t y e g−= = ≈
31. k tS Ce=
(a)
( ) 1.7918
16
16
5 when 1
5
lim 30
5 30
ln 1.7918
30 30
k
k tt
k
t
S t
Ce
Ce C
e
k
S e
→∞
−
= =
=
= =
=
= ≈ −
= ≈
(b) When 5, 20.9646t S= ≈ which is 20,965 units.
(c)
32. ( )25 1 ktS e= −
(a) ( )( ) ( )( )
1
0.1744
4 21 2125 25 254 25 1 1 ln 0.1744
25 1
k k k
t
e e e k
S e−
= − ⇒ − = ⇒ = ⇒ = ≈ −
= −
(b) 25,000 units ( )lim 25t
S→∞
=
(c) When 5, 14.545t S= ≈ which is 14,545 units.
(d)
33. 0.0185
0.0185
0.0185
2
2ln 2 0.0185
ln 2 37.5 years0.0185
t
t
t
P Ce
C Ce
et
t
=
=
=
=
= ≈
34. (a)
1
0.012
0.012 , 50
10.012
1 ln0.012
s
dy y sdsdy dsy
y s C
y Ce−
= − >
−=
−= +
=
∫ ∫
When ( )0.012 50 0.650, 28 28s y Ce C e−= = = ⇒ =
0.6 0.01228 , 50.sy e s−= >
(b)
Speed (s) 50 55 60 65 70
Miles per Gallon (y) 28 26.4 24.8 23.4 22.0
00
40
30
0 80
25
Review Exercises for Chapter 6 643
© 2010 Brooks/Cole, Cengage Learning
35. 5
5
7 7
7 7 ln5
dy x xdx x x
xy x dx x Cx
4
4
+= = +
⎛ ⎞= + = + +⎜ ⎟⎝ ⎠∫
36.
( )
2
2
2 2
2 2
2
11 2
1 2 11 ln 12
x
x
x x
x x
x
dy edx e
e edy dx dxe e
y e C
−
−
− −
− −
−
=+
−= = −
+ +
= − + +
∫ ∫ ∫
37.
21
28 1
28
16 0
16
1 16
ln 8
x C
x
y xydy xydx
dy x dxy
y x C
e y
y Ce
+
′ − =
=
=
= +
=
=
∫ ∫
38.
( )
1
1
sin 0
sin
sin
cos1
cos
1ln ln coscos
y
y
y
y
y
y e xdy e xdx
e dy x dx
e x C
e C Cx C
y x Cx C
−
−
′ − =
=
=
− = − +
= = −+
= = − ++
∫ ∫
39. 2 2
2dy x ydx xy
+= (homogeneous differential equation)
( )2 2 2 0x y dx xy dy+ − =
Let , .y vx dy x dv v dx= = +
( ) ( )( )
( )( )
( )
( )
2 2 2
2 2 2 2 2 3
2 2 2 3
2
2
2 21
2
22 2 2
12 2 2 2
2 0
2 2 0
2
1 2
21
ln ln 1 ln 1 ln
1 1
1 or
x v x dx x vx x dv v dx
x v x x v dx x v dv
x x v dx x v dv
v dx xv dv
dx v dvx vx v C v C
C C Cxxv x yy x
Cx xCx y x y
+ − + =
+ − − =
− =
− =
=−
= − − + = − − +
= = =− −−
= =− −
∫ ∫
644 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
40. ( )3 x ydydx x
+= (homogeneous differential equation)
( )3 0x y dx x dy+ − =
Let , .y vx dy x dv v dx= = +
( ) ( )( )
( )
( )
( )
( )
( )
2
1 21 2
1 22
2
3
3
3 0
3 2 0
3 2
1 13 2
1ln ln 3 2 ln 3 2 ln2
3 2
3 2 3 2
3 2 3 2
32
x vx dx x x dv v dx
x vx dx x dv
v dx x dv
dx dvx v
x v C v C
x C v
yx C v Cx
x C x y Cx Cy
x CxyC
+ − + =
+ − =
+ =
=+
= + + = + +
= +
⎛ ⎞⎛ ⎞= + = +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + = +
−=
∫ ∫
41. 31 2
21 2
2
36
y C x C x
y C C xy C x
= +
′ = +
′′ =
( ) ( ) ( )2 2 2 3
2 1 2 1 2
3 3 32 1 2 1 2
3 3 6 3 3 3
6 3 9 3 3 0
x y xy y x C x x C C x C x C x
C x C x C x C x C x
′′ ′− + = − + + +
= − − + + =
1 2 1 22, 0: 0 2 8 4x y C C C C= = = + ⇒ = −
( )1 2
2 2 2 2 1
3
12
12
2, 4: 4 12
4 4 12 8 , 2
2
x y C C
C C C C C
y x x
′= = = +
= − + = ⇒ = = −
= − +
42. 9.8dv kvdt
= −
(a)
1
2
23
3
9.81 ln 9.8
ln 9.8
9.81 9.8
kt C kt
kt
dv dtkv
kv t Ck
kv kt C
kv e C e
v C ek
+
=−
− = +
− = +
− = =
⎡ ⎤= +⎣ ⎦
∫ ∫
At ( )
( )
0 3 3 0
0
10, 9.8 9.8
1 9.8 9.8 .kt
t v C C kvk
v kv ek
= = + ⇒ = −
⎡ ⎤= + −⎣ ⎦
Note that 0k < because the object is moving downward.
(b) ( ) 9.8limt
v tk→∞
=
Review Exercises for Chapter 6 645
© 2010 Brooks/Cole, Cengage Learning
(c) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )( )
0 0 02
0 0 02 2
0 0 0 0 02 2 2
1 1 1 9.8 19.8 9.8 9.8 9.8 9.8
1 10 9.8 9.8
9.8 1 1 9.8 19.8 9.8 9.8 1
kt kt kt
kt kt
ts t kv e dt t kv e C kv e Ck k k k k
s kv C C s kvk k
t ts t kv e s kv kv e sk k k k k
⎡ ⎤⎡ ⎤= + − = + − + = + − +⎢ ⎥⎣ ⎦ ⎣ ⎦
= − + ⇒ = − −
= + − + − − = + − − +
∫
43. 4dy xdx y
−=
2
21
2 2
4
22
4 ellipses
y dy x dx
y x C
x y C
= −
= − +
+ =
∫ ∫
44. 3 2dy ydx
= −
1
1
22
22
2
2 31 ln 2 32
ln 2 3 2 2
2 3
2 332
x
x
x
dy dxy
y x C
y x C
y C e
y C e
y Ce
−
−
−
= −−
− = − +
− = − +
− =
= +
= +
∫ ∫
45. ( ) 0.555250
1 34 tP te−
=+
(a) 0.55k =
(b) 5250L =
(c) ( ) 52500 1501 44
P = =+
(d) 0.55
0.55
0.55
525026251 34
1 34 2134
1 1ln 6.41 yr0.55 34
t
t
t
ee
e
t
−
−
−
=+
+ =
=
− ⎛ ⎞= ≈⎜ ⎟⎝ ⎠
(e) 0.55 15250
dP PPdt
⎛ ⎞= −⎜ ⎟⎝ ⎠
46. ( ) 0.154800
1 14 tP te−
=+
(a) 0.15k =
(b) 4800L =
(c) ( ) 48000 3201 14
P = =+
(d) 0.15
0.15
480024001 14
14 1
1 1ln 17.59 yr0.15 14
t
t
ee
t
−
−
=+
=
⎛ ⎞= − ≈⎜ ⎟⎝ ⎠
(e) 0.15 14800
dP PPdt
⎛ ⎞= −⎜ ⎟⎝ ⎠
47. ( )1 , 0, 880
dy yydt
⎛ ⎞= −⎜ ⎟⎝ ⎠
( )
1, 8080
1 1800 8: 8 9
1
kt t
k LLybe be
y bb
− −
= =
= =+ +
= = ⇒ =+
Solution: 801 9 ty
e−=
+
x
y
4
−4
−4 4
4x
y
646 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
48. ( )1.76 1 , 0, 38
dy yydt
⎛ ⎞= −⎜ ⎟⎝ ⎠
( )
1.76
1.76, 88
1 18 50 3: 3
1 3
kt t
k LLybe be
y bb
− −
= =
= =+ +
= = ⇒ =+
Solution: 1.76
8513
ty
e−=
⎛ ⎞+ ⎜ ⎟⎝ ⎠
49. (a)
( ) ( )
( )
( )
0.553
20,400, 0 1200, 1 2000
20,4001
20,4000 1200 16120,4001 2000
1 1646165
23 40ln ln 0.55340 23
20,4001 16
kt
k
k
t
L y y
ybe
y bb
ye
e
k
ye
−
−
−
−
= = =
=+
= = ⇒ =+
= =+
=
= − = ≈
=+
(b) ( )8 17,118 trouty ≈
(c) 0.55320,40010,000 4.94 yr
1 16 t te−
= ⇒ ≈+
50. ( )0.553 1 , 0 120020,400
dy yy ydt
⎛ ⎞= − =⎜ ⎟
⎝ ⎠
Use Euler's method with 1.h =
Euler's method gives ( )8 16,170 trout.y ≈
51.
( )
( )
10
1, ( ) 10dx x
y y
P x Q x
u x e e− −
′ − =
= − =
∫= =
( )
1 10
10
10
xx
x x
x
y e dxee e C
Ce
−−
−
=
= − +
= − +
∫
52. 4 1
4
x x
x
e y e y
y y e−′ + =
′ + =
( )
( ) 4 4
4, ( ) x
dx x
P x Q x e
u x e e
−= =
∫= =
4 4 3 441 1 1
3 3x x x x x x
xy e e dx e e C e Cee
− − − −⎛ ⎞= = + = +⎜ ⎟⎝ ⎠∫
53.
4
41 14 4
x y
x
y e y
y y e
′ = +
′ − =
( ) ( )
( ) ( ) ( )
4
1 4 1 4
1 1,4 4
x
dx x
P x Q x e
u x e e− −
= − =
∫= =
( )( )
( )
1 441 4
1 4
4 4
1 14
14
14
xxx
x
x x
y e e dxe
e x C
xe Ce
−−
=
⎛ ⎞= +⎜ ⎟⎝ ⎠
= +
∫
54. 2 25 1dy y
dx x x− =
( ) ( )
( ) ( )2 2
25 5
5 ,
x dx x
P x Q xx x
u x e e−
1= − =
∫= =
5 5 55 2 51 1 1 1 1
5 5x x x
x xy e dx e C Cee x e
−⎛ ⎞= = − + = − +⎜ ⎟⎝ ⎠∫
55. ( )2 1x y y′− + =
1 12 2
dy ydx x x
+ =− −
( ) ( )
( ) ( )1 2 ln 2
1 ,2 2
2x dx x
P x Q xx x
u x e e x− −
1= =
− −∫= = = −
( ) ( )1 1 122 2 2
y x dx x Cx x x
⎛ ⎞= − = +⎜ ⎟− − −⎝ ⎠∫
t 0 2 4 6 8
Exact 1200 3241 7414 12,915 17,117
Euler 1200 2743 5853 10,869 16,170
Review Exercises for Chapter 6 647
© 2010 Brooks/Cole, Cengage Learning
56. ( ) ( )23 2 2 3x y y x′+ + = +
( )2 2 33
dy y xdx x
+ = ++
( ) ( ) ( )
( ) ( ) ( ) ( )22 3 2 ln 3
2 , 2 33
3x dx x
P x Q x xx
u x e e x+ +
= = ++∫= = = +
( )( )( )
( )( )
( )( )
22
4
2
2
2
1 2 3 33
3123
32 3
y x x dxx
xC
x
x Cx
= + ++
⎡ ⎤+⎢ ⎥= +⎢ ⎥+ ⎣ ⎦
+= +
+
∫
57. ( )3 sin 2 0
3 sin 2
y x dx dy
y y x
+ − =
′ − =
Integrating factor: 3 3dx xe e− −∫ =
( )( )
3 3
3
3
113
113
sin2
3 sin 2 2 cos 2
3 sin 2 2 cos 2
x x
x
x
ye e x dx
e x x C
y x x Ce
− −
−
=
= − − +
= − + +
∫
58. ( )( )
tan 2
tan 2
x
x
dy y x e dx
dy x y edx
= +
− =
Integrating factor: tan ln cos cosx dx xe e x−∫ = =
( )
( )
cos 2 cos cos sin
1 tan sec
x x
x
y x e x dx e x x C
y e x C x
= = + +
= + +
∫
59. 55 xy y e′ + =
Integrating factor: 5 5dx xe e∫ =
5 10 10
5 5
110
110
x x x
x x
ye e dx e C
y e Ce−
= = +
= +
∫
60. 3ay y bxx
⎛ ⎞′ − =⎜ ⎟⎝ ⎠
Integrating factor: ( ) lna x dx a x ae e x− − −∫ = =
( )3 4
4
4
4
a a a
a
byx bx x dx x Ca
bxy Cxa
− − −= = +−
= +−
∫
61. 2 Bernoulli equationy y xy′ + =
1 2 1 22, let , .n z y y z y y− − −′ ′= = = = −
( ) ( )2 2
Linear equation
y y y y x
z z x
− −′− + − = −
′ − = −
( )
1
1
11
1
dx x
x x x xx
x
x
u x e e
z xe dx e xe e Ce
y x Ce
yx Ce
− −
− − −−
−
∫= =
⎡ ⎤= − = + +⎣ ⎦
= + +
=+ +
∫
62. 22 Bernoulli equationy xy xy′ + =
1 2 1 22, let , .n z y y z y y− − −′ ′= = = = −
( ) ( )2 22
2 Linear equation
y y xy y x
z xz x
− −′− + − = −
′ − = −
( ) 22x dx xu x e e− −∫= =
( ) 2 2 22
2
221
1 12
1 12
1 21 12
x x xx
x
xx
z x e dx e e Ce
Cey
yC eCe
− −
−
⎛ ⎞= − = +⎜ ⎟⎝ ⎠
= +
= =++
∫
63. 3
21 Bernoulli equationyy yx x
′ + =
1 3 2 33, let , 2 .n z y y z y y− − −′ ′= = = = −
( ) ( )3 3
2
2
1 22 2
2 2 Linear equation
y y y yx x
z zx x
− − −′− + − =
−′ − =
( ) ( )2 2 ln 2x dx xu x e e x− − −∫= = =
( )
32 2
2 2
2
1 2 23
1 23
xz x dx x Cx x
Cxy x
−−
−
2
⎛ ⎞−= = +⎜ ⎟
⎝ ⎠
= +
∫
648 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
64. 2
2
1 2 1 2
1 Bernoulli Equation
2, let , .
xy y xy
y y yx
n z y y z y y− − −
′ + =
′ + =
′ ′= = = = −
( ) ( )2 2 2 21
1 1 Linear equation
y y y y y yx
z zx
− − −′− + − = −
′ − = −
( ) ( )1 1x dxu x ex
−∫= =
1 ln
1 ln
1ln
z x dx x x Cx
x x Cxy
yCx x x
⎡ ⎤= − = − +⎣ ⎦
= − +
=−
∫
65. Answers will vary. Sample answer: ( )2 23 2 0x y dx xy dy+ − =
Solution: Let , .y vx dy x dv v dx= = +
( ) ( )( )
( )( )
( )
( )
2 2 2
2 2 2 3
2
2
21
22
2
3 2 2
3 2 0
2 0
1 2
21
ln ln 1
1 1
x v x dx x vx x dv v dx
x v x dx x v dv
v dx xv dv
dx v dvx vx v C
yx C v Cx
x C x y
+ − + =
+ − =
+ =
=+
= + +
⎛ ⎞= + = +⎜ ⎟
⎝ ⎠
= +
∫ ∫
66. Answers will vary. Sample answer: 140yy y⎛ ⎞′ = −⎜ ⎟
⎝ ⎠
Solution: 1, 40k L= =
401 1kt t
Lybe be− −= =
+ +
67. Answers will vary.
Sample answer: 3 2
3
2 12 1
x y x y
y yx x
′ + =
′ + =
( ) ( )
( )
2 2
22 3 2
1 1 1 ln
x dxu x e x
y x dx x Cx x x
∫= =
⎡ ⎤= = +⎣ ⎦∫
68. Answers will vary. Sample answer: 1y xy xy−′ + =
Solution: 221, , , x dx xn Q x P x e e∫= − = = =
2 2 22
22
2
1
x x x
x
y e xe dx e C
y Ce−
= = +
= +
∫
Problem Solving for Chapter 6 649
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 6
1. (a)
( )
1.01
1.01
0.01
1
0.01
0.01
100
0.011 0.01
10.01
10.01
dy ydt
y dy dt
y t C
t Cy
yC t
yC t
−
−
=
=
= +−
= − +
=−
=−
∫ ∫
( ) 10010 1: 1 1y C
C= = ⇒ =
So, ( )100
1 .1 0.01
yt
=−
For 100, lim .t T
T y−→
= = ∞
(b) ( )
( )
1
1
11
y dy k dt
y kt C
y kt C
yC kt
ε
ε
ε
ε
εε
ε
− +
−
−
=
= +−
= − +
=−
∫ ∫
( ) 10 1
0 0
1 1 10y y C CC y y
εε
ε
⎛ ⎞= = ⇒ = ⇒ = ⎜ ⎟
⎝ ⎠
So, 1
0
1 .1
y
kty
ε
ε ε
=⎛ ⎞
−⎜ ⎟⎝ ⎠
For 0
1 , .t yy kεε
→ → ∞
2. Because ( )20 ,dy k ydt
= −
120
ln 20
20.kt
dy k dty
y kt C
y Ce
=−
− = +
= +
∫ ∫
When 0, 72.t y= = Therefore, 52.C =
When 1, 48.t y= = Therefore, 28 748 52 20, ,52 13
k ke e= + = = and 7ln .13
k = So,
( )ln 7 1352 20.ty e⎡ ⎤⎣ ⎦= +
When ( )5 ln 7 135, 52 20 22.35 F.t y e= = + ≈ °
3. (a) ( )1dS k S L Sdt
= −
1 kt
LSCe−
=+
is a solution because
( ) ( )
( )
( )
2
2
1 1
1
1
1 1
1 1
, where .
kt kt
kt
kt
kt
kt kt
kt kt
dS L Ce Ckedt
LC ke
Ce
k L C LeL Ce Cek L LLL Ce Ce
kk S L S kL
−− −
−
−
−
− −
− −
= − + −
=+
⎛ ⎞= ⋅⎜ ⎟ + +⎝ ⎠⎛ ⎞ ⎛ ⎞= ⋅ −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
= − =
100.L = Also, 10S = when 0 9.t C= ⇒ =
And, 20S = when 41 ln .9
t k= ⇒ = −
Particular Solution: ( ) 0.8109ln 4 9100 100
1 91 9 ttSee −= =
++
(b) ( )
( )
( )
1
2
12
1
100
100
100 2
0 when 50 or 0.
dS k S Sdt
d S dS dSk S Sdt dt dt
dSk Sdt
dSSdt
= −
⎡ ⎤⎛ ⎞= − + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
= −
= = =
Choosing 50,S = you have:
( )
( )
( )( )
ln 4 9
ln 4 9
100501 9
2 1 9ln 1 9ln 4 9
2.7 months
t
tee
t
t
=+
= +
=
≈
(This is the point of inflection.) (c)
(d)
(e) Sales will decrease toward the line .S L=
00
10
125
t1 2 3 4
140
120
100
80
60
40
20
S
650 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
4. (a)
[ ][ ] 1
ln
ln ln
ln ln ln
ln kt
ktCe
ktCe
dy Lk ydt y
dy k dty L y
L y kt C
L CeyL ey
y Le
−
−
−−
⎛ ⎞= ⎜ ⎟
⎝ ⎠
=−
− = − +
=
=
= (b)
(c) As , ,t y L→ ∞ → the carrying capacity.
(d) 0 500 5000 10 ln 10C Cy e e C−= = ⇒ = ⇒ =
( )
2
2 2
2
ln
1ln
ln 1 ln ln 1
dy Lk ydt y
d y L dy L dyk kydt y dt L y y dt
dy L L Lk k ydt y y y
⎛ ⎞= ⎜ ⎟
⎝ ⎠
⎛ ⎞ ⎛ ⎞−= +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦
So, 2
2 0d ydt
= when
ln 1 .L L Le yy y e
⎛ ⎞= ⇒ = ⇒ =⎜ ⎟
⎝ ⎠
5000 1839.4Lye e
= = ≈ and 41.7.t ≈
The graph is concave upward on ( )0, 41.7 and
downward on ( )41.7, .∞
5. Let 1 ln .2
bu k tk
⎛ ⎞= −⎜ ⎟⎝ ⎠
( )( )
4
2
ln2 ln
21 tanh 11
u
u u u
k t b ku b kt kt
e eue e e
e e e e be
−
− −
− −− − −
−+ = + =
+ +
= = =
Finally,
[ ]1 1 ln1 tanh 1 tanh2 2 2
22 1
.1
kt
kt
b LL k t uk
Lbe
Lbe
−
−
⎡ ⎤⎛ ⎞⎛ ⎞+ − = +⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦
=+
=+
Notice the graph of the logistics function is just a shift of the graph of the hyperbolic tangent. (See section 5.8.)
6. ( ) ( ) ( ) ( )?
f x g x f x g x′ ′ ′⎡ ⎤ =⎣ ⎦
(a) Let ( ) ( ), 1,g x x g x′= = then
( ) ( )( ) ( ) ( )
( ) ( )
( )
( )
1
1
ln ln 1
11
f x x f x
f x x f x f x
df x f xdx
df dxf x
f x x
f xx
′ ′⎡ ⎤ =⎣ ⎦′ ′+ =
− = −
=−
= − −
=−
∫ ∫
(b) ( )
( )
ln
g dxg g
fg f g
f g fg f g
f g g fg
f gf g g
gf dxg g
f e′
′ −
′ ′ ′=
′ ′ ′ ′+ =
′ ′ ′− = −
′ ′=
′ −′
=′ −
∫=
∫
(c) If ( ) ,xg x e= then ( ) ( ) 0x xg x g x e e′ − = − =
Therefore, no f can exist.
2000
00 500
7000
00 500
Problem Solving for Chapter 6 651
© 2010 Brooks/Cole, Cengage Learning
7. 21
1232
k
g
π⎛ ⎞= ⎜ ⎟⎝ ⎠
=
( )
( )
22
22 2
6 36 Equation of tank
36 6 12
x y
x y y y
+ − =
= − − = −
Area of cross section: ( ) ( )212A h h h π= −
( )
( )
( )( )
( )
2
2 1 2
3 2 1 2
5 2 3 2
3 2
2
112 64144112
1818 216
36 1445
36 7205
dhA h k ghdtdhh h hdtdhh h hdt
h h dh dt
h h t C
h h t C
π π
= −
− = −
− = −
− =
− = +
− = +
∫ ∫
When 6, 0h t= = and ( )3 26 504 1481.45.5
C = − ≈ −
The tank is completely drained when 0 1481.45 sec 24 min, 41 sech t= ⇒ = ≈
8. (a) ( )
( )
2
1 22 2
1
1
2
64
8 8,
2
2 18 at 0, 18
dhA h k ghdt
dhr k hdt
k kh dh dt C dt Cr r
h Ct C
C t h
π
π π−
= −
= −
−= = − =
= − +
= = =
So, 2 6 2.h Ct= − +
At ( )30 60 1800, 12:t h= = =
2 12 1800 6 2
6 2 4 3 0.0008651800
C
C
= − +
−= ≈
So, 2 0.000865 6 2.h t= − +
( )
6 200.0008659809.1seconds 2 h, 43 min, 29 sec
h t= ⇒ =
≈
(b) ( )3600 sec 2 0.000865 3600 6 2
7.21 ft
t h
h
= ⇒ = − +
⇒ ≈
9.
( )
1 2
2
64 836
1288
2288
dhA h k ghdtdh hdt
h dh dt
th C
ππ
−
= −
−=
−=
−= +
∫ ∫
20: 2 20 4 5
2 4 5288
h Cth
= = =
−= +
( )0 4 5 288
2575.95 sec 42 min, 56 sec
h t= ⇒ =
≈ ≈
h
6 ft
x2 + (y − 6)2 = 36
x
y
h
r
18 ft
652 Chapter 6 Differential Equations
© 2010 Brooks/Cole, Cengage Learning
10. Let the radio receiver be located at ( )0 , 0 .x The tangent line to 2y x x= − joins ( )1, 1− and ( )0 , 0 .x
(a) If ( ),x y is the point of tangency on 2,y x x= − then
2
2 2
2
2
1 11 21 1
2 1 2 1
2 2 0
2 4 8 1 32
3 3 5.
y x xxx x
x x x x x
x x
x
y x x
− − −− = =
+ +− + − = − −
+ − =
⎛ ⎞− ± += = − +⎜ ⎟⎜ ⎟⎝ ⎠
= − = −
Then
( )( ) ( )0
0 0
0
1 0 1 3 3 5 6 3 31 1 1 3 3
3 1 6 3 3 6 3 3 6 3 3
4 3 6 1.155.6 3 3
x
x x
x
− − + −= =
− − − + − −
= + − = − + −
−= ≈
−
(b) Now let the transmitter be located at ( )1, .h−
( )( )
2
2 2
2
2
1 21 1
2 1 2
2 1 0
2 4 4 11 2
23 2 4
y h x x hxx x
x x x x x h
x x h
hx h
y x x h h
− − −− = =
+ +− + − = − −
+ − − =
− ± + += = − + +
= − = + − −
Then, ( )
( )0
0
0
3 2 40 2 4 3 21 21 1 2
1 22 4 3 2
2 1.2 4 3 2
h h hh h hx hh
x hh h h
h hxh h
− + − −− + − += =
− − − +− − − + +
+ +=
+ − +
+= −
+ − +
(c)
There is a vertical asymptote at 1,4
h = which is the height of the mountain.
xx
x0
−1 1
2
1
y x x= − 2
Transmitter
Radio
( 1, 1)−
y
0.25 3
−2
10
Problem Solving for Chapter 6 653
© 2010 Brooks/Cole, Cengage Learning
11. 3.5 0.019ds sdt
= −
(a) 3.5 0.019
ds dts
−= −
−∫ ∫
1
2
0.0193
0.0193
0.019
1 ln 3.5 0.0190.019
ln 3.5 0.019 0.019
3.5 0.0190.019 3.5
184.21
t
t
t
s t C
s t C
s C es C es Ce
−
−
−
− = − +
− = − +
− =
− = −
= −
(b)
(c) As 0.019, 0,tt Ce−→ ∞ → and 184.21.s →
12. (a)
1ln
Rt V
dC R dtC V
RC t KV
C Ke−
= −
= − +
=
∫ ∫
Since 0C C= when 0,t = it follows that 0K C= and the function is 0 .Rt VC C e−= (b) Finally, as ,t → ∞ we have
0lim lim 0.Rt Vt t
C C e−→∞ →∞
= =
13. From Exercises 12, you have 0 .Rt VC C e−=
(a) For 2, 0.5,V R= = and 0 0.6,C = you have 0.250.6 tC e−=
(b) For 2, 1.5,V R= = and 0 0.6,C = you have 0.750.6 .tC e−=
14. (a)
( )
( )( ) ( )
1
1
1
1 1
1 ln
1 1
R t V K
R t V K Rt V
dC dtQ RC V
tQ RC KR V
Q RC e
C Q e Q KeR R
− +⎡ ⎤⎣ ⎦
− +⎡ ⎤ −⎣ ⎦
=−
− − = +
− =
= − = −
∫ ∫
Because 0C = when 0,t = it follows that K Q= and you have ( )1 .Rt VQC eR
−= −
(b) As ,t → ∞ the limit of C is .Q R
00 200
400
00
4
0.8
00
4
0.8
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