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C H A P T E R 6Differential Equations
Section 6.1 Slope Fields and Euler’s Method . . . . . . . . . . . . . . 518
Section 6.2 Differential Equations: Growth and Decay . . . . . . . . 529
Section 6.3 Differential Equations: Separation of Variables . . . . . . 539
Section 6.4 The Logistic Equation . . . . . . . . . . . . . . . . . . . 553
Section 6.5 First-Order Linear Differential Equations . . . . . . . . . 559
Section 6.6 Predator-Prey Differential Equations . . . . . . . . . . . . 570
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586
C H A P T E R 6Differential Equations
Section 6.1 Slope Fields and Euler’s Method
518
1. Differential equation:
Solution:
Check: y� � 4Ce4x � 4y
y � Ce4x
y� � 4y
3. Differential equation:
Solution:
Check:
�2xy
x2 � y2
��2xy
y2 � x2
��2xy
2y2 � �x2 � y2�
y� ��2xy
2y2 � Cy
y� ��2x
�2y � C�
2x � 2yy� � Cy�
x 2 � y 2 � Cy
y� �2xy
x 2 � y 2
2. Differential equation:
Solution:
Check: 3��e�x� � 4�e�x� � �3e�x � 4e�x � e�x
y� � �e�x
y � e�x
3y� � 4y � e�x
4. Differential equation:
Solution:
Check:
y� �xy
y2 � 1
y� �x
y �1y
�y �1y�y� � x
2yy� �2y
y� � 2x
y2 � 2 ln y � x2
dydx
�xy
y2 � 1
5. Differential equation:
Solution:
Check:
y� � y � �C1 cos x � C2 sin x � C1 cos x � C2 sin x � 0
y� � �C1 cos x � C2 sin x
y� � �C1 sin x � C2 cos x
y � C1 cos x � C2 sin x
y� � y � 0
6. Differential equation:
Solution:
Check:
� �2C1 � 2C1 � 2C2 � 2C2�e�x sin x � ��2C2 � 2C1 � 2C2 � 2C1�e�x cos x � 0
2���C1 � C2�e�x sin x � ��C1 � C2�e�x cos x� � 2�C1e�x cos x � C2e
�x sin x�
y� � 2y� � 2y � 2C1e�x sin x � 2C2e
�x cos x �
y� � 2 C1e�x sin x � 2C2e
�x cos x
y� � ��C1 � C2�e�x sin x � ��C1 � C2�e�x cos x
y � C1e�x cos x � C2e
�x sin x
y� � 2y� � 2y � 0
Section 6.1 Slope Fields and Euler’s Method 519
7. Differential equation:
Solution:
Check:
� tan x.
y� � y � �sin x��sec x� � cos x ln�sec x � tan x� � cos x ln�sec x � tan x� � �sin x��sec x� � cos x ln�sec x � tan x�
y� � �sin x� 1sec x � tan x
�sec x � tan x � sec2 x� � cos x ln�sec x � tan x�
� �1 � sin x ln�sec x � tan x�
���cos x�
sec x � tan x �sec x��tan x � sec x� � sin x ln�sec x � tan x�
y� � ��cos x� 1sec x � tan x
�sec x � tan x � sec2 x� � sin x ln�sec x � tan x�
y � �cos x ln�sec x � tan x�y� � y � tan x
8. Solution:
Check: y� � 2y� �23�4e�2x � ex� � 2�2
3���2e�2x � ex � � 2ex.
y� �23 �4e�2x � ex�
y� �23 ��2e�2x � ex �
y �23 �e�2x � ex �
9.
Differential equation:
Initial condition:
y��
4� � sin �
4 cos
�
4� cos2
�
4�
�22
��22
� ��22 �2
� 0
� 2 sin 2x � 1
� 2 sin x cos x � 1 � sin 2x
2y � y� � 2�sin x cos x � cos2 x� � ��1 � 2 cos2 x � sin 2x�
� �1 � 2 cos2 x � sin 2x
y� � �sin2 x � cos2 x � 2 cos x sin x
y � sin x cos x � cos2 x
10.
Differential equation:
Initial condition:
y�0� �12�0�2 � 4 cos�0� � 2 � �4 � 2 � �2
y� � x � 4 sin x
y� � x � 4 sin x
y �12x2 � 4 cos x � 2 11.
Differential equation:
Initial condition:
y�0� � 6e�2�0�2� 6e0 � 6�1� � 6
y� � �24xe�2x2� �4x�6e�2x2� � �4xy
y� � 6e�2x2��4x� � �24xe�2x2
y � 6e�2x2
520 Chapter 6 Differential Equations
12.
Differential equation:
Initial condition:
y��
2� � e�cos���2� � e0 � 1
y� � sin x � e�cos x � sin x�y� � y sin x
y� � e�cos x�sin x� � sin x � e�cos x
y � e�cos x
In Exercises 13–18, the differential equation is y�4 � 16y � 0.
13.
No
y�4� � 16y � 3 cos x � 16�3 cos x� � �45 cos x � 0,
y�4� � 3 cos x
y � 3 cos x
15.
Yes
y�4� � 16y � 16e�2x � 16e�2x � 0,
y�4� � 16e�2x
y � e�2x
14.
Yes
y�4� � 16y � 48 cos 2x � 48 cos 2x � 0,
y�4� � 48 cos 2x
y � 3 cos 2x
19.
No
xy� � 2y � x�2x� � 2�x2� � 0 � x3ex,
y � x 2, y� � 2x
21.
Yes
xy� � 2y � xx2e x � 2xe x � 4x� � 2x2e x � 2x2� � x3e x,
y � x 2�2 � e x �, y� � x 2�e x� � 2x�2 � e x�
16.
No
y�4� � 16y � �30x4 � 80 ln x � 0,
y�4� � �30x4
y � 5 ln x
18.
Yes
y�4� � 16y � �48e2x � 64 sin 2x� � 16�3e2x � 4 sin 2x� � 0,
y�4� � 48e2x � 64 sin 2x
y � 3e2x � 4 sin 2x
17.
Yes
y�4� � 16y � 0,
y�4� � 16C1e2x � 16C2e
�2x � 16C3 sin 2x � 16C4 cos 2x
y � C1e2x � C2e
�2x � C3 sin 2x � C4 cos 2x
In 19–24, the differential equation is xy� � 2y � x3ex.
20.
Yes
xy� � 2y � x�e x�x2 � 2x�� � 2�x2e x� � x3ex,
y � x 2ex, y� � x2e x � 2xex � e x�x2 � 2x�
22.
No
xy� � 2y � x�cos x� � 2�sin x� � x3ex,
y � sin x, y� � cos x
Section 6.1 Slope Fields and Euler’s Method 521
23.
No
xy� � 2y � x�1x� � 2 ln x � x3e x,
y � ln x, y� �1x 24.
Yes
xy� � 2y � xx2e x � 2xe x � 10x� � 2x2ex � 5x2� � x3e x,
y � x2e x � 5x2, y� � x2ex � 2xe x � 10x
25. passes through
Particular solution: y � 3e�x�2
3 � Ce0 � C ⇒ C � 3
�0, 3�.y � Ce�x�2 26. passes through
Particular solution: y�x2 � y� � 4
2�0 � 2� � C ⇒ C � 4
�0, 2�.y�x2 � y� � C
28. passes through
Particular solution: 2x 2 � y 2 � 2
2�9� � 16 � C ⇒ C � 2
�3, 4�.2x 2 � y 2 � C27. passes through
Particular solution: y2 �14x3 or 4y 2 � x3
16 � C�64� ⇒ C �14
�4, 4�.y 2 � Cx3
29. Differential equation:
General solution:
Particular solutions: Two intersecting linesHyperbolasC � ±1, C � ±4,
C � 0,
4y 2 � x2 � C
4yy� � x � 0 30. Differential equation:
General solution:
Particular solutions: Point Circles
x1 2
1
2
y
C � 1, C � 4,C � 0,
x 2 � y 2 � C
yy� � x � 0
31. Differential equation:
General solution:
Initial condition:
Particular solution: y � 3e�2x
y�0� � 3, 3 � Ce0 � C
y� � 2y � C��2�e�2x � 2�Ce�2x � � 0
y � Ce�2x
y� � 2y � 0 32. Differential equation:
General solution:
Initial condition:
Particular solution: 3x2 � 2y2 � 21
� 3 � 18 � 21 � C
y�1� � 3: 3�1�2 � 2�3�2
3x � 2yy� � 0
2�3x � 2yy� � � 0
6x � 4yy� � 0
3x2 � 2y2 � C
3x � 2yy� � 0
33. Differential equation:
General solution:
� 9�C1 sin 3x � C2 cos 3x� � 0
y� � 9y � ��9C1 sin 3x � 9C2 cos 3x�
y� � �9C1 sin 3x � 9C2 cos 3x
y� � 3C1 cos 3x � 3C2 sin 3x,
y � C1 sin 3x � C2 cos 3x
y� � 9y � 0 Initial conditions:
Particular solution: y � 2 sin 3x �13
cos 3x
� �3C2 ⇒ C2 � �13
1 � 3C1 cos��
2� � 3C2 sin��
2�y� � 3C1 cos 3x � 3C2 sin 3x
2 � C1 sin��
2� � C2 cos��
2� ⇒ C1 � 2
y��
6� � 2, y���
6� � 1
522 Chapter 6 Differential Equations
34. Differential equation:
General solution:
xy� � y� � x��C21x2� � C2
1x
� 0
y� � �C2� 1x2�
y� � C2�1x�
y � C1 � C2 ln x
xy� � y� � 0 Initial conditions:
Particular solution: y � �ln 2 � ln x � ln x2
12
�C2
2 ⇒ C2 � 1, C1 � �ln 2
y� �C2
x
0 � C1 � C2 ln 2
y�2� � 0, y��2� �12
35. Differential equation:
General solution:
Initial conditions:
Particular solution: y � �2x �12x3
C1 � 4C2 � 0 C1 � 12C2 � 4� C2 �
12, C1 � �2
4 � C1 � 12C2
y� � C1 � 3C2x2
0 � 2C1 � 8C2
y�2� � 0, y��2� � 4
x2y� � 3x y� � 3y � x2�6C2 x� � 3x�C1 � 3C2x2� � 3�C1x � C2x3� � 0
y� � C1 � 3C2 x2, y� � 6C2 x
y � C1x � C2 x3
x2y� � 3xy� � 3y � 0
36. Differential equation:
General solution:
Initial conditions:
Particular solution: y � e2x�3�4 �43x�
0 � e2�4 � 3C2� ⇒ C2 � �43
4 � �1��C1 � 0� ⇒ C1 � 4
0 � e2�C1 � 3C2�
y�0� � 4, y�3� � 0
9y� � 12y� � 4y � 9�23e2x�3��2
3C1 � 2C2 �23C2 x� � 12�e2x�3��2
3C1 � C2 �23C2 x� � 4�e2x�3��C1 � C2 x� � 0
y� �23 e2x�3�2
3C1 � C2 �23C2 x� � e2x�3 2
3C2 �23e2x�3�2
3C1 � 2C2 �23C2x�
y� �23e2x�3�C1 � C2x� � C2e
2x�3 � e2x�3�23C1 � C2 �
23C2x�
y � e2x�3�C1 � C2 x�
9y� � 12y� � 4y � 0
37.
y � 3x 2 dx � x3 � C
dydx
� 3x2 38.
y � �x3 � 4x� dx �x4
4� 2x2 � C
dydx
� x3 � 4x
39.
�u � 1 � x 2, du � 2x dx�
y � x1 � x2 dx �
12
ln�1 � x2� � C
dydx
�x
1 � x2 40.
y � ex
1 � ex dx � ln�1 � ex� � C
dydx
�ex
1 � ex
50.
Section 6.1 Slope Fields and Euler’s Method 523
42.
�u � x 2, du � 2x dx�
y � x cos�x2� dx �12
sin�x2� � C
dydx
� x cos x 2
44.
y � �sec2 x � 1� dx � tan x � x � C
dydx
� tan2 x � sec2 x � 1
41.
� x � 2 ln�x� � C � x � ln x 2 � C
y � �1 �2x� dx
dydx
�x � 2
x� 1 �
2x
43.
�u � 2x, du � 2 dx�
y � sin 2x dx � �12
cos 2x � C
dydx
� sin 2x
45.
Let
�25
�x � 3�5�2 � 2�x � 3�3�2 � C
� 2 �u4 � 3u 2� du � 2�u5
5� u3� � C
y � x�x � 3 dx � �u2 � 3��u��2u� du
u � �x � 3, then x � u2 � 3 and dx � 2u du.
dydx
� x�x � 3 46.
Let
� �103
�5 � x�3�2 �25
�5 � x�5�2 � C
��10u3
3�
2u5
5� C
� ��10u2 � 2u4� du
y � x�5 � x dx � �5 � u2�u��2u� du
u � �5 � x, u2 � 5 � x, dx � �2u du.
dydx
� x�5 � x
47.
�u � x 2, du � 2x dx�
y � xe x2 dx �12
ex2� C
dydx
� xex2 48.
� �10e�x�2 � C
y � 5e�x�2 dx � 5��2� e�x�2��12� dx
dydx
� 5e�x�2
51.
49.x 0 2 4 8
y 2 0 4 4 6 8
0�2�2�4�2�6dy�dx
�2�4x 0 2 4 8
y 2 0 4 4 6 8
Undef. 0 123
12�2dy�dx
�2�4
x 0 2 4 8
y 2 0 4 4 6 8
0 0 �8�2�2�2�2�2dy�dx
�2�452.
x 0 2 4 8
y 2 0 4 4 6 8
0 0 �3��3��3�3dy�dx
�2�4
53.
For Matches (b).dydx
� 1.x � �,
dydx
� cos�2x� 54.
For Matches (c).dydx
� 0.x � 0,
dydx
�12
sin x
524 Chapter 6 Differential Equations
55.
As Matches (d).dydx
→ 0.x →,
dydx
� e�2x 56.
For is undefined (vertical tangent). Matches (a).dydx
x � 0,
dydx
�1x
57. (a), (b) (c) As
As x → �, y → �.
6
−5
5
y
x
(2, 4)
−4
x →, y → �. 58. (a), (b) (c) As
As x → �, y → �.
4
−4
4
x
y
x →, y →.
59. (a), (b) (c) As
As x → �, y → �.
4
−4
4
x
y
(1, 1)
−4
x →, y → �. 60. (a), (b) (c) As
As x → �, y →.
−4 4
8
x
y
x →, y →.
61. (a)
As
Note: The solution is
(b)
As x →, y →.
6−1
−2
−3
1
2
3
x
y
(2, −1)
y� �1x, y�2� � �1
y � ln x.�x →, y →.
6−1
−2
−3
1
2
3
x
y
(1, 0)
y� �1x, y�1� � 0 62. (a)
As
(b)
As x →, y →.
(1, 1)
1 2 3 4 5 6
1
2
3
4
5
6
y
x
y� �1y, y�1� � 1
x →, y →.
(0, 1)
1 2 3 4 5 6
1
2
3
4
5
6
y
x
y� �1y, y�0� � 1
Section 6.1 Slope Fields and Euler’s Method 525
65.
−12 48
−2
12
dydx
� 0.02y�10 � y�, y�0� � 2 66.
−5 50
10
dydx
� 0.2x�2 � y�, y�0� � 9
63.
−6 6
−4
12
dydx
� 0.5y, y�0� � 6 64.
−5
−1
5
9
dydx
� 2 � y, y�0� � 4
67. Slope field for with solution passingthrough
8
−2
−2
8
�0, 1�y� � 0.4y�3 � x� 68. Slope field for with solution passing
through
3
−3
−3
5
�0, 2�
y� �12
e�x�8 sin �y4
70.
The table shows the values for n � 0, 2, 4, . . . , 20.
y2 � y1 � hF�x1, y1� � 2.1 � �0.05��0.05 � 2.1� � 2.2075, etc.
y1 � y0 � hF�x0, y0� � 2 � �0.05��0 � 2� � 2.1
y� � x � y, y�0� � 2, n � 20, h � 0.05
0 2 4 6 8 10 12 14 16 18 20
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
2 2.208 2.447 2.720 3.032 3.387 3.788 4.240 4.749 5.320 5.960yn
xn
n
69.
y2 � y1 � hF�x1, y1� � 2.2 � �0.1��0.1 � 2.2� � 2.43, etc.
y1 � y0 � hF�x0, y0� � 2 � �0.1��0 � 2� � 2.2
y� � x � y, y�0� � 2, n � 10, h � 0.1
0 1 2 3 4 5 6 7 8 9 10
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
2 2.2 2.43 2.693 2.992 3.332 3.715 4.146 4.631 5.174 5.781yn
xn
n
526 Chapter 6 Differential Equations
72.
y2 � y1 � hF�x1, y1� � 1 � �0.4��0.5�0.4��3 � 1�� � 1.16, etc.
y1 � y0 � hF�x0, y0� � 1 � �0.4��0.5�0��3 � 1�� � 1
y� � 0.5x�3 � y�, y�0� � 1, n � 5, h � 0.4
0 1 2 3 4 5
0 0.4 0.8 1.2 1.6 2.0
1 1 1.16 1.454 1.825 2.201yn
xn
n
71.
y2 � y1 � hF�x1, y1� � 2.7 � �0.05��3�0.05� � 2�2.7�� � 2.4375, etc.
y1 � y0 � hF�x0, y0� � 3 � �0.05��3�0� � 2�3�� � 2.7
y� � 3x � 2y, y�0� � 3, n � 10, h � 0.05
0 1 2 3 4 5 6 7 8 9 10
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
3 2.7 2.438 2.209 2.010 1.839 1.693 1.569 1.464 1.378 1.308yn
xn
n
73.
y2 � y1 � hF�x1, y1� � 1.1 � �0.1�e�0.1��1.1� � 1.2116, etc.
y1 � y0 � hF�x0, y0� � 1 � �0.1�e0�1� � 1.1
y� � exy, y�0� � 1, n � 10, h � 0.1
0 1 2 3 4 5 6 7 8 9 10
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
1 1.1 1.212 1.339 1.488 1.670 1.900 2.213 2.684 3.540 5.958yn
xn
n
74.
y2 � y1 � hF�x1, y1� � 5.0041 � �0.1��cos�0.1� � sin�5.0041�� � 5.0078, etc.
y1 � y0 � hF�x0, y0� � 5 � �0.1��cos 0 � sin 5� � 5.0041
y� � cos x � sin y, y�0� � 5, n � 10, h � 0.1
0 1 2 3 4 5 6 7 8 9 10
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
5 5.004 5.008 5.010 5.010 5.007 4.999 4.985 4.965 4.938 4.903yn
xn
n
75.dydx
� y, y � 3ex, y�0� � 3 x 0 0.2 0.4 0.6 0.8 1.0
3 3.6642 4.4755 5.4664 6.6766 8.1548(exact)
3 3.6000 4.3200 5.1840 6.2208 7.4650
3 3.6300 4.3923 5.3147 6.4308 7.7812�h � 0.1�y�x�
�h � 0.2�y�x�
y�x�
Section 6.1 Slope Fields and Euler’s Method 527
76.dydx
�2xy
, y � �2x2 � 4, y�0� � 2 x 0 0.2 0.4 0.6 0.8 1.0
2 2.0199 2.0785 2.1726 2.2978 2.4495(exact)
2 2.000 2.0400 2.1184 2.2317 2.3751
2 2.0100 2.0595 2.1460 2.2655 2.4131�h � 0.1�y�x�
�h � 0.2�y�x�
y�x�
77.dydx
� y � cos x, y �12
�sin x � cos x � ex�, y�0� � 0 78. As h increases (from 0.1 to 0.2), the error increases.
79.
(a)
(b) , exacty � 72 � 68e�t�2
h � 0.1y�0� � 140,dydt
� �12
�y � 72�,
t 0 1 2 3
Euler 140 112.7 96.4 86.6
t 0 1 2 3
Exact 140 113.24 97.016 87.173
80.
(a)
(b) , exact
The approximations are better using h � 0.05.
y � 72 � 68e�t�2
h � 0.05y�0� � 140,dydt
� �12
�y � 72�,
t 0 1 2 3
Euler 140 112.98 96.7 86.9
t 0 1 2 3
Exact 140 113.24 97.016 87.173
x 0 0.2 0.4 0.6 0.8 1.0
0 0.2200 0.4801 0.7807 0.1231 0.5097(exact)
0 0.2000 0.4360 0.7074 0.0140 0.3561
0 0.2095 0.4568 0.7418 0.0649 0.4273�h � 0.1�y�x�
�h � 0.2�y�x�
y�x�
83. Consider Begin with a point that satisfies the initial condition, Then using a stepsize of h, find the point Continue generating the sequence of points�xn�1, yn�1� � �xn � h, yn � hF�xn, yn��.
�x0 � h, y0 � hF�x0, y0��.�x1, y1� �y�x0� � y0.
�x0, y0�y� � F�x, y�, y�x0� � y0.
82. A slope field for the differential equation consists of small line segments at various points in the plane. The line segment equals the slope
of the solution y at the point �x, y�.y� � F�x, y�
�x, y�y� � F�x, y�81. A general solution of order n has n arbitrary constants while in
a particular solution initial conditions are given in order to solvefor all these constants.
84.
Since we have Thus, cannot be determined.Ck � 0.07.
Ckekx � 0.07Cekx.dy�dx � 0.07y,
dydx
� Ckekx
y � Cekx
85. False. Consider Example 2. is a solution tobut is not a solution.y � x3 � 1xy� � 3y � 0,
y � x3 86. True
87. True 88. False. The slope field could represent many differ-ent differential equations, such as y� � 2x � 4y.
528 Chapter 6 Differential Equations
89. solution
(a)
dydx
� �2y, y�0� � 4, y � 4e�2x,
x 0 0.2 0.4 0.6 0.8 1.0
y 4 2.6813 1.7973 1.2048 0.8076 0.5413
4 2.5600 1.6384 1.0486 0.6711 0.4295
4 2.4000 1.4400 0.8640 0.5184 0.3110
0 0.1213 0.1589 0.1562 0.1365 0.1118
0 0.2813 0.3573 0.3408 0.2892 0.2303
r 0.4312 0.4447 0.4583 0.4720 0.4855
e2
e1
y2
y1
90. solution
(a)
dydx
� x � y, y�0� � 1, y � x � 1 � 2e�x,
x 0 0.2 0.4 0.6 0.8 1.0
y 1 0.8375 0.7406 0.6976 0.6987 0.7358
1 0.8200 0.7122 0.6629 0.6609 0.6974
1 0.8000 0.6800 0.6240 0.6192 0.6554
0 0.0175 0.0284 0.0347 0.0378 0.0384
0 0.0375 0.0606 0.0736 0.0795 0.0804
r 0.47 0.47 0.47 0.48 0.48
e2
e1
y2
y1
(b) If h is halved, then the error is approximatelyhalved
(c) When the errors will again be approximately halved.
h � 0.05,
�r � 0.5�.
(b) If h is halved, then the error is halved
(c) When the error will again be approximately halved.
h � 0.05,
�r � 0.5�.
91. (a)
(b) As That is,
In fact, is a solution to the differential equation.I � 2
limt →
I�t� � 2.I → 2.t →,
� 6 � 3I
dIdt
�14
�24 � 12I�
4 dIdt
� 12 I � 24
−3 3
−3
3
t
I L dIdt
� RI � E�t� 92.
k � ±4
�since ekt � 0� k2 � 16 � 0
k2ekt � 16ekt � 0
y� � 16y � 0
y� � k2ekt
y� � kekt
y � ekt
93.
If then radians/sec. � ±4A � 0,
A sin t16 � 2� � 0
�A2 sin t � 16A sin t � 0
y� � 16y � 0
y� � �A2 sin t
y� � A cos t
y � A sin t 94.
For
For
Thus, is increasing for and decreasingfor
has a maximum at Thus, it is bounded by its value at Thus,f is bounded.�and f � �
f �0�2 � f��0�2.x � 0,x � 0. f �x�2 � f��x�2
x > 0.x < 0 f �x�2 � f��x�2
�2xg�x� f � �x��2 ≤ 0.x > 0,
�2xg�x� f ��x��2 ≥ 0.x < 0,
ddx
f �x�2 � f��x�2� � �2xg�x� f ��x��2
2f �x�f��x� � 2f��x�f � �x� � �2xg�x� f��x��2
f �x� � f � �x� � �xg�x�f � �x�, g�x� ≥ 0
Section 6.2 Differential Equations: Growth and Decay 529
95. Let the vertical line cut the graph of the solution at The tangent line at is
Since we have
For any value of this line passes through the point
To see this, note that
�q�k�p�k� � t.
�?
q�k�k � p�k�tk �q�k�p�k� � t � kq�k� � p�k�kt
q�k�p�k� � t �
? �q�k� � p�k�t��k �1
p�k� � k�
�k �1
p�k�, q�k�p�k��.
t,
y � t � �q�k� � p�k�t��x � k�.y� � p�x�y � q�x�,
y � t � f��k��x � k�.
�k, t��k, t�.y � f �x�x � k
Section 6.2 Differential Equations: Growth and Decay
1.
y � ��x � 2� dx �x2
2� 2x � C
dy
dx� x � 2 2.
y � ��4 � x� dx � 4x �x2
2� C
dy
dx� 4 � x
6.
9y2 � 4x3�2 � C
3y2
2�
23
x3�2 � C1
�3yy� dx � �x dx
3yy� � x
y� �x3y 7.
� Ce�2�3�x3�2
� eC1e�2�3�x3�2
y � e�2�3�x3�2�C1
lny �23 x3�2 � C1
�dyy
� �x dx
�y�
y dx � �x dx
y�
y� x
y� � x y 8.
� Cex2�2 � 1
y � eC1ex2�2 � 1
1 � y � e�x2�2��C1
ln�1 � y� �x 2
2� C1
� dy1 � y
� �x dx
� y�
1 � y dx � �x dx
y�
1 � y� x
y� � x�1 � y�
3.
y � Cex � 2
y � 2 � ex�C1 � Cex
lny � 2 � x � C1
� 1y � 2
dy � �dx
dy
y � 2� dx
dydx
� y � 2 4.
y � 4 � Ce�x
4 � y � e�x�C1 � Ce�x
ln4 � y dy � �x � C1
� �14 � y
dy � ��dx
dy
4 � y� dx
dydx
� 4 � y 5.
y 2 � 5x 2 � C
12
y 2 �52
x 2 � C1
�y dy � �5x dx
�yy� dx � �5x dx
yy� � 5x
y� �5xy
530 Chapter 6 Differential Equations
9.
y � C�1 � x2�
lny � ln �C�1 � x2��
lny � ln�1 � x 2� � ln C
lny � ln�1 � x2� � C1
�dyy
� � 2x1 � x2 dx
�y�
y dx � � 2x
1 � x 2 dx
y�
y�
2x1 � x2
y� �2xy
1 � x2
�1 � x2�y� � 2xy � 0 10.
y � 100 � Ce�x2�2
�y � e�C1e�x2�2 � 100
100 � y � e��x2�2��C1
ln�100 � y� � �x2
2� C1
�ln�100 � y� �x 2
2� C1
� 1100 � y
dy � �x dx
� y�
100 � y dx � �x dx
y�
100 � y� x
y� � 100x � xy � x�100 � y�
xy � y� � 100x
11.
Q � �kt
� C
�dQ � �kt
� C
�dQdt
dt � � kt 2 dt
dQdt
�kt2
12.
P � �k2
�10 � t�2 � C
�dP � �k2
�10 � t�2 � C
�dPdt
dt � �k�10 � t� dt
dPdt
� k�10 � t� 13.
N � �k2
�250 � s�2 � C
�dN � �k2
�250 � s�2 � C
�dNds
ds � �k�250 � s� ds
dNds
� k�250 � s�
14.
y � L � Ce�kx2�2
�y � �L � e�C1e�kx2�2
L � y � e��kx2�2��C1
�ln�L � y� �kx2
2� C1
� 1L � y
dy �kx2
2� C1
� 1L � y
dydx
dx � �kx dx
1
L � y dydx
� kx
dydx
� kx�L � y� 15. (a)
(b)
−6 6
−1
7
�0, 0�: 0 � 6 � C1 ⇒ C1 � �6 ⇒ y � 6 � 6e�x2�2
y � 6 � C1e�x2�2
y � 6 � e�x2�2�C � C1e�x2�2
lny � 6 ��x2
2� C
dy
y � 6� �x dx
dydx
� x�6 � y�, �0, 0�
x−5 −1
9
5
y
(0, 0)
Section 6.2 Differential Equations: Growth and Decay 531
16. (a)
(b)
�0, 12�:
12
� C1e0 ⇒ C1 �
12
⇒ y �12
ex2�2
y � ex2�2�C � C1ex2�2
lny �x2
2� C
dyy
� x dx
dydx
� xy, �0, 12�
x
)(0, 12
y
−4
−4
4
4
17.
4
−1
−4
(0, 10)
16
y �14
t2 � 10
10 �14
�0�2 � C ⇒ C � 10
y �14
t2 � C
�dy � �12
t dt
dydt
�12
t, �0, 10�
19.
10
−1
−1
(0, 10)
16
y � 10e�t�2
10 � Ce0 ⇒ C � 10
y � e��t�2��C1 � eC1e�t�2 � Ce�t�2
lny � �12
t � C1
�dyy
� ��12
dt
dydt
� �12
y, �0, 10�18.
0 10
−5
(0, 10)
15
y � �12
t3�2 � 10
10 � �12
�0�3�2 � C ⇒ C � 10
y � �12
t3�2 � C
�dy � ��34
t dt
dydt
� �34
t, �0, 10�
20.
y � 10e3t�4
10 � Ce0 ⇒ C � 10
� eC1e�3�4�t � Ce3t�4
y � e�3�4�t�C1
ln y �34
t � C1
�dyy
� �34
dt−5 5
−5
(0, 10)
40 dydt
�34
y, �0, 10� 21.
When
� 4�52�
2
� 25.
y � 4e1�3 ln�5�2��6� � 4eln�5�2�2x � 6,
�3, 10�: 10 � 4e3k ⇒ k �13
ln�52�
�0, 4�: 4 � Ce0 � C
y � Cekx
dydx
� ky
532 Chapter 6 Differential Equations
22.
When
� 250�85�
4
�8192
5.
N � 250e4 ln�8�5� � 250eln�8�5�4t � 4,
�1, 400�: 400 � 250ek ⇒ k � ln 400250
� ln 85
�0, 250�: C � 250
N � Cekt
dNdt
� kN
24.
When
� 5000�1920�
5
� 3868.905.
P � 5000eln�19�20��5�t � 5,
�1, 4750�: 4750 � 5000ek ⇒ k � ln�1920�
�0, 5000�: C � 5000
P � Cekt
dPdt
� kP
23.
When
� 20,000�58�
3�2
� 9882.118.
V � 20,000e1�4 ln�5�8��6� � 20,000eln�5�8�3�2t � 6,
�4, 12,500�: 12,500 � 20,000e4k ⇒ k �14
ln�58�
�0, 20,000�: C � 20,000
V � Cekt
dVdt
� kV
25.
or y �12
e0.4605t y �12
e�ln 10�5�t �12
�10t�5�
k �ln 10
5
5 �12
e5k
y �12
ekt
C �12
y � Cekt, �0, 12�, �5, 5�
27.
y � 0.6687e0.4024t
�C � 5�1�4� C � 0.6687
1 � Ce0.4024
y � Ce0.4024t
k �ln 5
4� 0.4024
5 � e4k
5ek � e5k
5Cek � Ce5k
5 � Ce5k
1 � Cek
y � Cekt, �1, 1�, �5, 5�26.
y � 4e�0.4159t
k �ln�1�8�
5� �0.4159
12
� 4e5k
y � 4ekt
C � 4
y � Cekt, �0, 4�, �5, 12� 28.
y � 0.0005e2.3026t
C � 0.0005
5 � Ce2.3026�4�
y � Ce2.3026t
k � ln 10 � 2.3026
10 � ek
10e3k � e4k
2Ce3k �15
Ce4k
5 � Ce4k
12
� Ce3k
y � Cekt, �3, 12�, �4, 5�
29. In the model C represents the initial value of yk is the proportionality constant.�when t � 0�.
y � Cekt,
31.
when Quadrants I and III.xy > 0;dydx
> 0
dydx
�12
xy
30. y� �dydt
� ky
32.
when Quadrants I and II.y > 0;dydx
> 0
dydx
�12
x2y
Section 6.2 Differential Equations: Growth and Decay 533
33. Since the initial quantity is 10 grams,
Since the half-life is 1599 years,
Thus,
When
When y � 0.13 g.t � 10,000,
y � 10e�ln�1�2��1599��1000� � 6.48 g.t � 1000,
y � 10e�ln�1�2��1599�t.
k �1
1599 ln�12�.
5 � 10ek�1599�
y � 10ekt.
34. Since the half-life is 1599 years,
Since there are 1.5 g after 1000 years,
Hence, the initial quantity is approximately 2.314 g.
When y � 2.314e�ln�1�2��1599��10,000� � 0.03 g.t � 10,000,
C � 2.314.
1.5 � Ce�ln�1�2��1599��1000�
k �1
1599 ln�12�.
12 � 1ek�1599�
35. Since the half-life is 1599 years,
Since there are 0.5 gram after 10,000 years,
Hence, the initial quantity is approximately 38.158 g.
When y � 38.158e�ln�1�2��1599��1000� � 24.74 g.t � 1000,
C � 38.158.
0.5 � Ce�ln�1�2��1599��10,000�
k �1
1599 ln�12�.
12 � 1ek�1599�
38. Since the half-life is 5715 years,
Since there are 3.2 grams when years,
Thus, the initial quantity is approximately 3.613 g.
When y � 3.613e�ln�1�2��5715��10,000� � 1.07 g.t � 10,000,
C � 3.613.
3.2 � Ce�ln�1�2��5715��1000�
t � 1000
k �1
5715 ln�12�.
12 � 1ek�5715�
36. Since the half-life is 5715 years,
Since there are 2 grams after 10,000 years,
Hence, the initial quantity is approximately 6.726 g.
When y � 6.726e�ln�1�2��5715��1000� � 5.96 g.t � 1000,
C � 6.726.
2 � Ce�ln�1�2��5715��10,000�
k �1
5715 ln�12�.
12 � 1ek�5715�
37. Since the initial quantity is 5 grams,
Since the half-life is 5715 years,
When years,
When years, y � 5e�ln�1�2��5715��10,000� � 1.49 g.t � 10,000
y � 5e�ln�1�2��5715��1000� � 4.43 g.t � 1000
k �1
5715 ln�12�.
2.5 � 5ek�5715�
C � 5.
39. Since the half-life is 24,100 years,
Since there are 2.1 grams after 1000 years,
Thus, the initial quantity is approximately 2.161 g.
When y � 2.161e�ln�1�2��24,100��10,000� � 1.62 g.
t � 10,000,
C � 2.161.
2.1 � Ce�ln�1�2��24,100��1000�
k �1
24,100 ln�12�.
12 � 1ek�24,100�
40. Since the half-life is 24,100 years,
Since there are 0.4 grams after 10,000 years,
Thus, the initial quantity is approximately 0.533 g.
When y � 0.533e�ln�1�2��24,100��1000� � 0.52 g.t � 1000,
C � 0.533.
0.4 � Ce�ln�1�2��24,100��10,000�
k �1
24,100 ln�12�.
12 � 1ek�24,100�
534 Chapter 6 Differential Equations
41.
When
Therefore, 95.76% remains after 100 years.
� 0.9576 C
y � Ce�ln�1�2��1599��100�t � 100,
k �1
1599 ln�1
2�
12
C � Cek�1599�
y � Cekt 42.
t � 15,641.8 years
ln�0.15� �ln�1�2�t
5715
0.15C � Ce�ln�1�2��5715�t
k �1
5715 ln�1
2�
12
C � Cek�5715�
y � Cekt
43. Since the time to double is given byand we have
Amount after 10 years: A � 1000e�0.06��10� � $1822.12
t �ln 20.06
� 11.55 years.
ln 2 � 0.06t
2 � e0.06t
2000 � 1000e0.06t
A � 1000e0.06t, 44. Since the time to double is given byand we have
Amount after 10 years: A � 20,000e�0.055��10� � $34,665.06
t �ln 2
0.055� 12.6 years.
ln 2 � 0.055t
2 � e0.055t
40,000 � 20,000e0.055t
A � 20,000e0.055t,
45. Since and when we havethe following:
Amount after 10 years: A � 750e0.0894�10� � $1833.67
r �ln 27.75
� 0.0894 � 8.94%
1500 � 750e7.75r
t � 7.75,A � 1500A � 750ert 46. Since and when wehave the following:
Amount after 10 years: A � 10,000e��ln 2��5��10� � $40,000
r �ln 2
5� 0.1386 � 13.86%
20,000 � 10,000e5r
t � 5,A � 20,000A � 10,000ert
48. Since and when wehave the following:
The time to double is given by
t �ln 20.10
� 6.93 years.
4000 � 2000e0.10t
r �ln�5436.56�2000�
10� 0.10 � 10%
5436.56 � 2000e10r
t � 10,A � 5436.56A � 2000ert47. Since and when we havethe following:
The time to double is given by
t �ln 2
0.095� 7.30 years.
1000 � 500e0.0950t
r �ln�1292.85�500�
10� 0.0950 � 9.50%
1292.85 � 500e10r
t � 10,A � 1292.85A � 500ert
49.
P � 500,000�1 �0.075
12 ��240� $112,087.09
500,000 � P�1 �0.075
12 ��12��20�
51.
� $30,688.87
P � 500,000�1 �0.0812 �
�420
500,000 � P�1 �0.0812 �
�12��35�
50.
P � 500,000�1.005��480 � $45,631.04
500,000 � P�1 �0.0612 ��12��40�
52.
� $53,143.92
P � 500,000�1 �0.0912 �
�300
500,000 � P�1 �0.0912 �
�12��25�
Section 6.2 Differential Equations: Growth and Decay 535
53. (a)
(b)
t �ln 2
12 ln�1 � �0.07�12�� � 9.93 years
ln 2 � 12t ln�1 �0.0712 �
2 � �1 �0.007
12 �12t
2000 � 1000�1 �0.0712 �12t
t �ln 2
ln 1.07� 10.24 years
ln 2 � t ln 1.07
2 � 1.07t
2000 � 1000�1 � 0.07�t (c)
(d)
t �ln 20.07
� 9.90 years
ln 2 � 0.07t
2 � e0.07t
2000 � 1000e�0.07�t
t �ln 2
365 ln�1 � �0.07�365�� � 9.90 years
ln 2 � 365t ln�1 �0.07365 �
2 � �1 �0.07365 �
365t
2000 � 1000�1 �0.07365 �
365t
54. (a)
(b)
t �1
12
ln 2
ln�1 �0.0612 �
� 11.58 years
ln 2 � 12t ln�1 �0.0612 �
2 � �1 �0.0612 �
12t
2000 � 1000�1 �0.0612 �
12t
t �ln 2
ln 1.06� 11.90 years
ln 2 � t ln1.06
2 � 1.06t
2000 � 1000�1 � 0.6�t (c)
(d)
t �ln 20.06
� 11.55 years
ln 2 � 0.06t
2 � e0.06t
2000 � 1000e0.06t
t �1
365
ln 2
ln�1 �0.06365 �
� 11.55 years
ln 2 � 365t ln�1 �0.06365 �
2 � �1 �0.06365 �
365t
2000 � 1000�1 �0.06365 �
365t
55. (a)
(b)
t �1
12
ln 2
ln�1 �0.085
12 �� 8.18 years
ln 2 � 12t ln�1 �0.085
12 �
2 � �1 �0.085
12 �12t
2000 � 1000�1 �0.085
12 �12t
t �ln 2
ln 1.085� 8.50 years
ln 2 � t ln1.085
2 � 1.085t
2000 � 1000�1 � 0.085�t (c)
(d)
t �ln 2
0.085� 8.15 years
ln 2 � 0.085t
2 � e0.085t
2000 � 1000e0.085t
t �1
365
ln 2
ln�1 �0.085365 �
� 8.16 years
ln 2 � 365t ln�1 �0.085365 �
2 � �1 �0.085365 �
365t
2000 � 1000�1 �0.085365 �
365t
536 Chapter 6 Differential Equations
56. (a)
(b)
t �1
12
ln 2
ln�1 �0.055
12 �� 12.63 years
ln 2 � 12t ln�1 �0.055
12 �
2 � �1 �0.055
12 �12t
2000 � 1000�1 �0.055
12 �12t
t �ln 2
ln 1.055� 12.95 years
ln 2 � t ln1.055
2 � 1.055t
2000 � 1000�1 � 0.055�t (c)
(d)
t �ln 2
0.055� 12.60 years
ln 2 � 0.055t
2 � e0.055t
2000 � 1000e0.055t
t �1
365
ln 2
ln�1 �0.055365 �
� 12.60 years
ln 2 � 365t ln�1 �0.055365 �
2 � �1 �0.055365 �
365t
2000 � 1000�1 �0.055365 �
365t
57. (a)
(b) For
(c) If the population is decreasing.k < 0,
P � 7.7696e�0.009�15� � 6.79 million.t � 15,
P � 7.7696e�0.009t
C � 7.7696
P�1� � 7.7 � Ce�0.009�1�
P � Cekt � Ce�0.009t
59. (a)
(b) For
(c) For the population is increasing.k > 0,
P � 5.0665e0.026�15� � 7.48 million.t � 15,
P � 5.0665e0.026t
C � 5.0665
5.2 � P�1� � Ce0.026�1�
P � Cekt � Ce0.026t 60. (a)
(b) For
(c) For the population is decreasing.k < 0,
P � 3.6072e�0.002�15� � 3.5 million.t � 15,
P � 3.6072e�0.002t
C � 3.6072
P�1� � 3.6 � Ce�0.002�1�
P � Cekt � Ce�0.002t
58. (a)
(b) For
(c) For the population is increasing.k > 0,
P � 12.4734e0.018�15� � 16.34 million.t � 15,
P � 12.4734e0.018t
C � 12.4734
P�1� � 12.7 � Ce0.018�1�
P � Cekt � Ce0.018t
61. (a)
(b) when hours (graphing utility)
Analytically,
t �ln 3.9936ln 1.2455
� 6.3 hours.
t ln 1.2455 � ln 3.9936
1.2455t �400
100.1596� 3.9936
400 � 100.1596�1.2455�t
t � 6.3N � 400
N � 100.1596�1.2455�t
Section 6.2 Differential Equations: Growth and Decay 537
62. (a) Let
At time 2:
At time 4:
Approximately 45 bacteria at time 0.
(b)
(c) When
(d) 25,000 �62514 e�1�2�ln�14�5�t ⇒ t � 12.29 hours
y �62514 e�1�2�ln�14�5�8 �
62514 �14
5 �4� 2744.t � 8,
y �62514 e�1�2�ln�14�5�t � 44.64e0.5148t
C � 125e�2k � 125e�2�1�2�ln�14�5� � 125� 514� �
62514 � 44.64
k �12 ln 14
5 � 0.5148
2k � ln 145
350 � Cek�4� ⇒ 350 � �125e�2k��e4k� ⇒ 145 � e2k
125 � Cek�2� ⇒ C � 125e�2k
y � Cekt.
65. (a)
(c)
The model fits the data better.P2
300
150500
P1P2
P1 � 181e0.01245t � 181�1.01253�t
205 � 181e10k ⇒ k �1
10 ln�205
181� � 0.01245
P1 � Cekt � 181ekt (b) Using a graphing utility,
(d) Using the model
or 2011. t �ln�320�182.3248�
ln�1.01091� � 51.8 years,
320
182.3248� �1.01091�t
320 � 182.3248�1.01091�t
P2,
P2 � 182.3248�1.01091�t
66. (a)
(c)
00 12
500
I � 0.14164t4 � 3.9288t3 � 36.599t2 � 120.82t � 417.0
� 941.6088e0.0729t
R � 941.6088�1.0756�t
64. (a)
(b)
t ��ln 6
�0.0366� 49 days
e�0.0366t �16
25 � 30�1 � e�0.0366t �
N � 30�1 � e�0.0366t �
k �ln�1�3�
30�
�ln 330
� �0.0366
30e30k � 10
20 � 30�1 � e30k �63. (a)
(b)
t ��ln 6
�0.0502� 36 days
e�0.0502t �16
25 � 30�1 � e�0.0502t �
N � 30�1 � e�0.0502t �
k �ln�11�30�
20� �0.0502
30e20k � 11
19 � 30�1 � e20k �
(b) According to the model,
(d)
00
1
12
P�t� �IR
R��t� � 68.6e0.0729t.2500
00 12
538 Chapter 6 Differential Equations
67.
(a)
(b)
(c)
(d) ��10�4� � 10 log10 10�4
10�16 � 120 decibels
��10�6.5� � 10 log10 10�6.5
10�16 � 95 decibels
��10�9� � 10 log10 10�9
10�16 � 70 decibels
��10�14� � 10 log10 10�14
10�16 � 20 decibels
��I � � 10 log10 II0
, I0 � 10�16 68.
Percentage decrease: �10�6.7 � 10�8
10�6.7 ��100� � 95%
�8 � log10 I ⇒ I � 10�8
80 � 10 log10 I
10�16 � 10�log10 I � 16�
�6.7 � log10 I ⇒ I � 10�6.7
93 � 10 log10 I
10�16 � 10�log10 I � 16�
69.
The timber should be harvested in the year 2014, Note: You could also use a graphing utility to graph and find the maximum of Use the viewing rectangle and 0 ≤ y ≤ 600,000.0 ≤ x ≤ 30A�t�.A�t�
�1998 � 16�.
dAdt
� 100,000�0.4t
� 0.10�e0.8t�0.10t � 0 when t � 16.
A�t� � V�t�e�0.10t � 100,000e0.8t e�0.10t � 100,000e0.8t�0.10t
70. R �ln I � ln I0
ln 10�
ln I � 0ln 10
, I � eR ln 10 � 10R
71. Since
When Thus,
When Thus,
Thus,
When t � 5, y � 379.2�.
y � 1420e�ln�104�142��t � 80.
k � ln 1040 � ln 1420 � ln 104142
.
k�1� � ln 1420 � ln�1120 � 80�
t � 1, y � 1120.
C � ln 1420.t � 0, y � 1500.
ln�y � 80� � kt � C.
� 1y � 80
dy � �k dt
dydt
� k�y � 80�,
(a)
I � 108.3 � 199,526,231.5
8.3 �ln I � ln I0
ln 10(b)
Increases by a factor of e2R ln 10 or 10R.
� e2R ln 10 � �eR ln 10�2 � �10R�2
I � e2R ln 10
2R �ln I � ln I0
ln 10(c)
dRdI
�l
I ln �10�
R �ln I � ln I0
ln 10
72.
(See Example 6)
It will take minutes longer.10.53 � 5 � 5.53
t �5 ln
114
ln 27
�5 ln 14
ln 72
� 10.53 minutes
ln 1
14�
t5
ln 27
114
� eln�2�7� t�5� �2
7�t�5
30 � 20 � 140e�1�5� ln�2�7�t
k �15
ln�27� � �0.25055
27
� e5k
60 � 20 � 140ek�5�
160 � 20 � Cek�0� ⇒ C � 140
y � 20 � Cekt
dydt
� k�y � 20�
73. False. If y � Cekt, y� � Ckekt � constant.
74. True 75. True 76. True
Section 6.3 Differential Equations: Separation of Variables
Section 6.3 Differential Equations: Separation of Variables 539
1.
y 2 � x 2 � C
y 2
2�
x 2
2� C1
�y dy � �x dx
dydx
�xy
2.
y3 �x3
3� 2x � C
�3y 2 dy � ��x2 � 2� dx
dydx
�x 2 � 2
3y23.
r � e0.05s�C1 � Ce0.05s
ln�r� � 0.05s � C1
�drr
� �0.05 ds
drds
� 0.05r
4.
r � 0.025s2 � C
�dr � �0.05s ds
drds
� 0.05s 5.
y � C�x � 2�3
� ln�C�2 � x�3� ln�y� � 3 ln�2 � x� � ln C
�dyy
� � 32 � x
dx
�2 � x�y� � 3y 6.
y � Cx
ln�y� � ln�x� � ln C � ln�Cx�
�dyy
� �dxx
xy� � y
8.
y2 �12�
sin �x � C
y2
2�
6�
sin �x � C1
�y dy � �6 cos �x dx
y dydx
� 6 cos �x7.
y 2 � �2 cos x � C
y 2
2� �cos x � C1
�y dy � �sin x dx
yy� � sin x
9.
y � �14
�1 � 4x2�1�2 � C
� �18
��1 � 4x2��1�2��8x dx�
�dy � � x�1 � 4x2 dx
dy �x
�1 � 4x2 dx
�1 � 4x2 y� � x 10.
y � 5�x2 � 9�1�2 � C
�dy � � 5x�x2 � 9
dx
�x2 � 9 dydx
� 5x
12.
2y2 � 3ex � C
�4y dy � �3ex dx
4y dydx
� 3ex11.
y � e�1�2��ln x�2�C1 � Ce�ln x�2�2
ln�y� �12
�ln x�2 � C1
�dyy
� �ln xx
dx �u � ln x, du �dxx �
y ln x � xy� � 0
540 Chapter 6 Differential Equations
13.
Initial condition:
Particular solution: y 2 � 2ex � 14
y�0� � 4, 16 � 2 � C, C � 14
y 2 � 2ex � C
y 2
2� ex � C1
�y dy � �ex dx
yy� � ex � 0 14.
Initial condition:
Particular solution: y3�2 � x3�2 � 9
�4�3�2 � �1�3�2 � 8 � 1 � 9 � Cy�1� � 4,
y3�2 � x3�2 � C
23
y3�2 � �23
x3�2 � C1
�y1�2 dy � ��x1�2 dx
�x � �y y� � 0
16.
y �12
�ln x�2 � 2
y�1� � 2: 2 � C
y ��ln x�2
2� C
�dy � �ln xx
dx
2x dydx
� 2 ln x
2xy� � ln x2 � 015.
Initial condition:
Particular solution: y � e1��x�1�2�2 � e��x2�2x��2
y��2� � 1, 1 � Ce�1�2, C � e1�2
y � Ce��x�1�2�2
ln�y� � ��x � 1�2
2� C1
�dyy
� ���x � 1� dx
y�x � 1� � y� � 0
17.
y 2 � 3 � 4x 2
1 � y2 � 4�1 � x2�
y�0� � �3: 1 � 3 � C ⇒ C � 4
1 � y 2 � C�1 � x2�
ln�1 � y 2� � ln�1 � x2� � ln C � lnC�1 � x2�
12
ln�1 � y 2� �12
ln�1 � x 2� � C1
y
1 � y 2 dy �x
1 � x 2 dx
y�1 � x2� y� � x�1 � y 2� 18.
�1 � y2 � �1 � x2 � 1
0 � �1 � C ⇒ C � 1y�0� � 1:
��1 � y2�1�2 � ��1 � x2�1�2 � C
��1 � y2��1�2y dy � ��1 � x2��1�2x dx
y�1 � x2 dydx
� x�1 � y 2
20.
r � ln� 21 � e�2s�
�r � ln�12
e�2s �12� � ln�1 � e�2s
2 �
e�r �12
e�2s �12
�e�r � �12
e�2s �12
r�0� � 0: �1 � �12
� C ⇒ C � �12
�e�r � �12
e�2s � C
�e�rdr � �e�2s ds
drds
� er�2s19.
Initial condition:
Particular solution: u � e�1�cos v2��2
u�0� � 1, C �1
e�1�2 � e1�2
u � Ce��cos v2��2
ln�u� � �12
cos v 2 � C1
�duu
� �v sin v 2 dv
dudv
� uv sin v 2
Section 6.3 Differential Equations: Separation of Variables 541
21.
Initial condition:
Particular solution: P � P0ekt
P�0� � P0, P0 � Ce0 � C
P � Cekt
ln�P� � kt � C1
�dPP
� k�dt
dP � kP dt � 0 22.
Initial condition:
Particular solution: T � 70 � 70e�kt, T � 70�1 � e�kt �
140 � 70 � 70 � Ce0 � CT�0� � 140;
T � 70 � Ce�kt
ln�T � 70� � �k t � C1
� dTT � 70
� �k�dt
dT � k�T � 70� dt � 0
23.
Initial condition:
Particular solution:
16y 2 � 9x 2 � 25
8y2 ��92
x2 �252
y�1� � 1, 8 � �92
� C, C �252
8y 2 ��92
x 2 � C
�16y dy � ��9x dx
dydx
��9x16y
25.
y � Ce�x�2
ln�y� � �12
x � C1
�dyy
� ��12
dx
m �dydx
�0 � y
�x � 2� � x� �
y2
24.
Initial condition:
Particular solution: 8y3 � x2, y �12
x2�3
y�8� � 2, 23 � C�82�, C �18
y3 � Cx2
ln y3 � ln x2 � ln C
�3y dy � �2
x dx
dydx
�2y3x
26.
y � Cx
ln y � ln x � C1 � ln x � ln C � ln Cx
�dyy
� �dxx
m �dydx
�y � 0x � 0
�yx
27.
Homogeneous of degree 3
� t3�x3 � 4xy 2 � y3�
f �t x, t y� � t 3x3 � 4txt 2y 2 � t3y3
f �x, y� � x3 � 4xy 2 � y 3 28.
Not homogeneous
f �tx, ty� � t 3x3 � 3t 4x2y2 � 2t2y2
f �x, y� � x3 � 3x2y2 � 2y2
30.
Homogeneous of degree 1
�t 2 xy
t�x 2 � y 2� t
xy�x 2 � y 2
f �t x, t y� �t x t y
�t 2 x2 � t 2 y 2
f �x, y� �xy
�x2 � y2
32.
Not homogeneous
f �t x, t y� � tan�tx � t y� � tant �x � y�
f �x, y� � tan�x � y�31.
Not homogeneous
� 2 lnt 2xy � 2�ln t 2 � ln xy� f �t x, t y� � 2 lntxty
f �x, y� � 2 ln xy
29.
Homogeneous of degree 3
f �tx, ty� �t 4x2y2
�t2x 2 � t2y2� t3 x2y2
�x2 � y2
f �x, y� �x2y2
�x2 � y2
542 Chapter 6 Differential Equations
33.
Homogeneous degree 0
f �t x, t y� � 2 ln t xty
� 2 ln xy
f �x, y� � 2 ln xy
34.
Homogeneous of degree 0
f �t x, t y� � tan t ytx
� tan yx
f �x, y� � tan yx
35.
�x� � C�x � y�2
x 2
�x � y�2 � �Cx�
11 � �y�x�2 � �Cx�
1
�1 � v�2 � �Cx�
�ln�1 � v�2 � ln�x� � ln C � ln�Cx�
2� dv1 � v
� �dxx
xdvdx
�1 � v
2� v �
1 � v2
v � x dvdx
�x � vx
2x
y� �x � y
2x, y � vx 36.
y3 � 3x3 ln�x� � Cx3
�yx�
3� 3 ln�x� � C
v3
3� ln�x� � C
�v 2 dv � �1x dx
xv 2 dv � dx
x4 v2 dv � x3 v3 dx � x3 dx � v3 x3 dx
x�vx�2�x dv � v dx� � �x3 � �vx�3� dx
y � vx, dy � x dv � v dx
xy 2 dy � �x3 � y3� dx
y� ��x3 � y3�
xy 2
37.
�y2 � 2xy � x2��C
�y 2
x 2 � 2 yx
� 1� �Cx2
�v 2 � 2v � 1� �Cx 2
12
ln�v 2 � 2v � 1� � �ln�x� � ln C1 � ln�C1
x � � v � 1
v2 � 2v � 1 dv � ��dx
x
x dv � �1 � v1 � v
� v� dx �1 � 2v � v2
1 � v dx
v dx � x dv �1 � v1 � v
dx
v � x dvdx
�x � xvx � xv
y� �x � yx � y
, y � vx 38.
y 2 � x2 � Cx
y 2
x2 � 1 �Cx
v2 � 1 �Cx
ln�v 2 � 1� � �ln x � ln C � ln Cx
� 2vv2 � 1
dv � ��dxx
2v dx � 2x dv �1 � v 2
vdx
v � x dvdx
�x2 � v 2 x2
2x2 v
y� �x2 � y2
2xy, y � vx
Section 6.3 Differential Equations: Separation of Variables 543
39.
y � Ce�x2�2y2
�x2
2y 2 � ln�C1y�
�12v 2 � ln�C1xv�
�1
2v 2 � ln�v� � ln�x� � ln C1 � ln�C1 x�
�1 � v 2
v3 dv � �dxx
x dv � � v1 � v2 � v� dx � � v3
1 � v2� dx
v dx � x dv �v
1 � v 2 dx
v � x dvdx
�x2 v
x2 � x2 v 2
y� �xy
x2 � y 2, y � vx 40.
y � Cx3 � x
yx
� x2C � 1
1 �yx
� x2C
1 � v � x2C
ln�1 � v� � ln x2 � ln C � ln x2C
x dvdx
� 2 � 2v ⇒ � dv1 � v
� 2�dxx
v � x dvdx
�2x � 3vx
x� 2 � 3v
y� �2x � 3y
x, y � vx
41.
Initial condition:
Particular solution: ey�x � 1 � ln x2
y�1� � 0, 1 � C
ey�x � C � ln x2
ey�x � ln C1 � ln x2
ev � ln C1 x 2
�ev dv � �2x dx
x�v dx � x dv� � �2xe�v � vx� dx � 0
x dy � �2xe�y�x � y� dx � 0, y � vx 42.
Initial condition:
Particular solution: y � e1�y�x
y�1� � 1, 1 � Ce�1 ⇒ C � e
y � Ce�y�x
C1
y� ey�x
C1
vx� ev
v � ln C1
xv
v � ln v � �ln x � ln C1 � ln C1
x
�1 � vv
dv � ��dxx
�x 2 v 2 dx � �x 2 � x 2 v��v dx � x dv� � 0
�y 2 dx � x�x � y� dy � 0, y � vx
43.
Initial condition:
Particular solution: x � esin� y�x�
y�1� � 0, 1 � Ce0 � C
x � Cesin v � Cesin� y�x�
sin v � ln x � ln C1
�cos v dv � �dxx
�sec v � v� dx � v dx � x dv
�x sec v � xv� dx � x�v dx � x dv� � 0
�x sec yx
� y� dx � x dy � 0, y � vx
544 Chapter 6 Differential Equations
45.
y � �x dx �12
x 2 � C
4
2
4
2x
y dydx
� x
44.
Let
1 � x�x2 � y2
1x
� �x2 � y2
y�1� � 0: 1 � C�1 � 0� ⇒ C � 1
1x
� C�x2 � y2�1�2
1x2 � C�1 �
y2
x2�1�2
�Cx
�x2 � y2�1�2
x�2 � C�1 � v2�1�2
ln x�2 � ln�1 � v 2�1�2 � ln C
�2 ln x �12
ln�1 � v 2� � C1
�2x
dx �v
1 � v 2 dv
�2 � 2v 2� dx � �xv dv
�2x2 � 2x2v 2� dx � x3v dv � 0
�2x 2 � v 2x 2� dx � x�vx��x dv � v dx� � 0
y � vx, dy � x dv � v dx.
�2x 2 � y2� dx � xy dy � 0
46.
y 2 � x2 � C
y 2
2�
�x 2
2� C1
y dy � �x dx
x−4 −2 2 4
−4
−2
2
4
y dydx
� �xy
47.
y � 4 � Ce�x
4 � y � e�x �C1
ln �4 � y� � �x � C1
� dy4 � y
� �dx
8
4321−3−4x
y dydx
� 4 � y 48.
y � 4 � Ce��1�8�x2
y � 4 � eC1��1�8�x2� Ce��1�8�x2
ln�y � 4� � �18
x 2 � C1
� �14
�x dx
� dyy � 4
� ��0.25x dx
dy
4 � y� 0.25x dx
x−4 −2
−2
2
8
4
4
y
dydx
� 0.25x�4 � y�
Section 6.3 Differential Equations: Separation of Variables 545
49. (a) Euler’s Method gives
(b)
(c) At
Error: 0.2489 � 0.1602 � 0.0887
y � 5e�3�1� � 0.2489.x � 1,
y � 5e�3x2
y�0� � 5 ⇒ C � 5
y � Ce�3x2
ln�y� � �3x2 � C1
�dyy
� ��6x dx
dydx
� �6xy
y�1� � 0.1602. 50. (a) Euler’s Method gives
(b)
(c) At
Error: 0.3 � 0.2622 � 0.0378
y �3
9�1� � 1�
310
� 0.3.x � 1,
y �1
3x2 � �1�3� �3
9x2 � 1
3 �1C
⇒ C �13
y �1
3x2 � C
�1y
� �3x2 � C1
�dyy2 � ��6x dx
dydx
� �6xy2
y�1� � 0.2622.
51. (a) Euler’s Method gives
(b)
(c) For
Error: 3.0318 � 3 � 0.0318
�y � 3��y2 � 3y � 5� � 0 ⇒ y � 3.
y3 � 4y � 15 � 0
y3 � 4y � 22 � 12�2� � 13 � 15
x � 2,
y3 � 4y � x2 � 12x � 13
23 � 4�2� � 1 � 12 � C ⇒ C � �13y�1� � 2:
y3 � 4y � x2 � 12x � C
��3y2 � 4� dy � ��2x � 12� dx
dydx
�2x � 123y2 � 4
y�2� � 3.0318. 52. (a) Euler’s Method gives
(b)
(c) At y � tan�1.52 � 1� � 3.0096.x � 1.5,
y � tan�x2 � 1�
arctan�y� � x2 � 1
arctan�0� � 12 � C ⇒ C � �1
arctan y � x2 � C
� dy1 � y2 � �2x dx
dydx
� 2x�1 � y2�
y�1.5� � 1.7270.
53.
When or 98.9%.y � 0.989Ct � 25,
y � Celn�1�2��1599t
k �1
1599 ln�1
2�
y0
2� y0ek�1599�
y�0� � y0 � C, initial amount
dydt
� ky, y � Cekt 54.
Initial conditions:
Particular solution:
When 75% has been changed:
t �ln�1�4�ln�4�5� � 6.2 hr
14
� et ln�4�5�
5 � 20et ln�4�5�
y � 20et ln�4�5�
k � ln 45
16 � 20ek
20 � Ce0 � C
y�0� � 20, y�1� � 16
dydt
� ky, y � Cekt
546 Chapter 6 Differential Equations
56.
The direction field satisfies along Matches (b).
x � 4:�dy�dx� � 0
dydx
� k�x � 4�55.
The direction field satisfies along butnot along . Matches (a).y � 0
y � 4;�dy�dx� � 0
dydx
� k�y � 4�
57.
The direction field satisfies along andMatches (c).y � 4.
y � 0�dy�dx� � 0
dydx
� ky�y � 4� 58.
The direction field satisfies along andgrows more positive as y increases. Matches (d).
y � 0,�dy�dx� � 0
dydx
� ky 2
59.
(a)
00
10
1400
00
10
1400
00
10
1400
w � 1200 � 1140e�kt
w�0� � 60 � 1200 � C ⇒ C � 1200 � 60 � 1140
w � 1200 � Ce�kt
1200 � w � e�kt�C1 � Ce�kt
ln�1200 � w� � �kt � C1
� dw1200 � w
� �k dt
dwdt
� k�1200 � w�
(b)
(c) Maximum weight: 1200 pounds
limt→�
w � 1200
k � 1.0: t � 1.05 years
k � 0.9: t � 1.16 years
k � 0.8: t � 1.31 years
60. From Exercise 59:
w � 1200 � �1200 � w0�e�t
w�0� � w0 � 1200 � C ⇒ C � 1200 � w0
w � 1200 � Ce�t
w � 1200 � Ce�kt, k � 1
61. Given family (circles):
Orthogonal trajectory (lines):
y � Kx
ln�y� � ln�x� � ln K
�dyy
� �dxx
−6 6
−4
4 y� �yx
y� � �xy
2x � 2yy� � 0
x 2 � y 2 � C
62. Given family (hyperbolas):
Orthogonal trajectory:
y � kx�2 �kx2
ln y � �2 ln x � ln k
�dyy
� ��2x dx −3
−2
3
2 y� ��2y
x
y� �x
2y
2x � 4yy� � 0
x2 � 2y2 � C
Section 6.3 Differential Equations: Separation of Variables 547
63. Given family (parabolas):
Orthogonal trajectory (ellipses):
x2 � 2y2 � K
y2 � �x 2
2� K1
2�y dy � ��x dx −6 6
−4
4 y� � �x
2y
y� �2xC
�2x
x2�y�
2yx
2x � Cy�
x2 � Cy
64. Given family (parabolas):
Orthogonal trajectory (ellipse):
2x 2 � y2 � K
y 2
2� �x 2 � K1
�y dy � ��2x dx −6 6
−4
4 y� � �2xy
y� �Cy
�y 2
2x �1
y� �y
2x
2yy� � 2C
y2 � 2Cx
65. Given family:
Orthogonal trajectory (ellipses):
3y2 � 2x2 � K
3y 2
2� �x2 � K1
3�y dy � �2�x dx −6 6
−4
4 y� � �2x3y
y� �3Cx2
2y�
3x2
2y �y2
x3� �3y2x
2yy� � 3Cx2
y2 � Cx3
66. Given family (exponential functions):
Orthogonal trajectory (parabolas):
y2 � �2x � K
y2
2� �x � K1
�y dy � ��dx −6 6
−4
4 y� � �1y
y� � Cex � y
y � Cex
67.
When Thus, Thus,
When Thus,
Therefore, N �125e0.2452t
1 � 0.25e0.2452t �500
1 � 4e�0.2452t.
200 �125e2000k
1 � 0.25e2000k ⇒ k �ln�8�3�2000
� 0.00049.N � 200.t � 4,
N �125e500kt
1 � 0.25e500kt.100 �500C1 � C
⇒ C � 0.25.N � 100.t � 0,
N �500Ce500kt
1 � Ce500kt
N
500 � N� e500kt�C2 � Ce500kt
ln�N� � ln�500 � N� � 500�kt � C1�
1
500 � � 1
N�
1500 � N dN � � k dt
� dN
N�500 � N� � � k dt
dNdt
� kN�500 � N�
548 Chapter 6 Differential Equations
68. The differential equation is given by the following.
When Thus,
Therefore,
�10L
10 � �L � 10�e�Lkt
S �CL
C � e�Lkt �10��L � 10�L
10��L � 10� � e�Lkt
C �10
L � 10.S � 10.t � 0,
S �CLeLkt
1 � CeLkt �CL
C � e�Lkt
S
L � S� CeLkt
1L
ln�S� � ln�L � S� � kt � C1
� dS
S�L � S� � � k dt
dSdt
� kS�L � S�
69. The general solution is
Since when it follows that
and
Therefore,
Since when we have
Thus, y �45
1 � �41�8�t �360
8 � 41t.
4 �45
1 � 45k�2� ⇒ k � �41
360.
t � 2,y � 4
y � �1
kt � �1�45� �45
1 � 45kt.
C ��145
.45 ��1C
t � 0,y � 45
y ��1
kt � C.
030
45
70. The general solution is Since when we have Therefore,
Since when we have
Thus, we have
y �75
1 � 5.25t�
3004 � 21t
.
00
4
80
12 �75
1 � 75k ⇒ k � �
7100
.
t � 1,y � 12
y � �1
kt � �1�75� �75
1 � 75kt.
C � �1�75.t � 0,y � 75y � �1��kt � C�. 71. Since when it follows that
which implies that Therefore, we haveSince when it
follows that
Therefore, y is given by y � 500e�1.6904e�0.1451t.
� 0.1452.
k � �12
ln ln 0.3ln 0.2
e�2k �ln 0.3ln 0.2
300
15t
y 150 � 500e��ln 5�e�2k
t � 2,y � 150y � 500e��ln 5�e�kt.C � ln 5.
100 � 500e�C,t � 0,y � 100
72. The general solution is Since when it follows that which implies that Therefore, we have
Since when it follows that
Thus, we have
y � 5000e��2.3026�e��0.1019�t.
� 0.1019.
k � �ln� ln�1�8�ln�1�10��
e�k �ln�1�8�
ln�1�10�
252015105
4000
3000
2000
1000
5000
t
y 625 � 5000e��ln 10�e�k
t � 1,y � 625y � 5000e��ln 10�e�kt.
C � �ln 110 � ln 10.
500 � 5000e�Ct � 0,y � 500y � 5000e�Ce�kt
. 73. From Example 11, the general solution is
Since when
Since when
Thus,
When beavers.t � 10, y � 34
y � 60e�2.0149e�0.1246t.
k � �13
ln� ln 1�4�2.0149� � 0.1246.
ln 14
� �2.0149e�3k
14
� e�2.0149e�3k
15 � 60e�2.0149e�3k
t � 3,y � 15
8 � 60e�C ⇒ C � ln 152
� 2.0149.
t � 0,y � 8
y � 60e�Ce�kt.
Section 6.3 Differential Equations: Separation of Variables 549
74. From Example 11, the general solution is
Since when
Since when
Thus,
Finally, when y � 244 rabbits.t � 3,
y � 400e�2.5903e�0.5519t.
k � �ln� ln�9�40��2.5903� � 0.5519.
ln� 940� � �2.5903e�k
940
� e�2.5903e�k
90 � 400e�2.5903e�k
t � 1,y � 90
30 � 400e�C ⇒ C � ln�403 � � 2.5903.
t � 0,y � 30
y � 400e�Ce�kt.
75. Following Example 12, the differential equation is
and its general solution is
when
when
Hence, the particular solution is
Using a symbolic algebra utility or graphing utility, you find that when
and or 92%.y � 0.92,
y�2 � y��1 � y�2 � 3e0.4024�10�
t � 10,
y�2 � y��1 � y�2 � 3e0.4024t.
� 3e2k�4� ⇒ k �18
ln 5 � 0.2012.
t � 4 ⇒ �3�4��5�4�
�1�4�2 � 15 y � 0.75 �34
t � 0 ⇒ �1�2��3�2�
�1�2�2 � C ⇒ C � 3y �12
y�2 � y��1 � y�2 � Ce2kt
dydt
� ky�1 � y��2 � y�
76. Following Example 12, the differential equation is
and its general solution is
when
when
Hence, the particular solution is
Using a symbolic algebra utility or graphing utility, wefind that when y � 0.91.t � 8,
y�2 � y��1 � y�2 �
169
e0.5205t.
⇒ k �1
10 ln�27
2 � � 0.2603
⇒ 272
� e10k
t � 5 ⇒ �0.8��1.2��0.2�2 � 24 �
169
e2k�5�y � 0.8
t � 0 ⇒ �0.4��1.6��0.6�2 �
169
� Cy � 0.4
y�2 � y��1 � y�2 � Ce2kt.
dydt
� ky�1 � y��2 � y�
77. (a)
Since when we have thus, theparticular solution is
(b) When we have
minutes t � 10.217
�20 ln�35� � t
ln�35� � �
120
t
35
� e��1�20�t
15 � 25e��1�20�t.Q � 15,
Q � 25e��1�20�t.25 � C;t � 0,Q � 25
Q � e��1�20�t�C1 � Ce��1�20�t
ln�Q� � �1
20t � C1
� dQQ
� � �1
20 dt
dQdt
� �Q20
550 Chapter 6 Differential Equations
78. Since is a first-order linear differential equation with and we have the integrating factor and the general solution is
Since when we have and Finally, when we have Q � 38.843 lbs�gal.
t � 30,Q � 50�1 � e�0.05t�.C � �50t � 0,Q � 0
Q � e�0.05t � 52
e0.05t dt � e�0.05t�50e0.05t � C� � 50 � Ce�0.05t.
u�t� � e� �1�20� dt � e�1�20�t,R�x� �
52,P�x� �
120Q� �
120Q �
52
80.
Since when and when it follows that and whichimplies
and
Therefore, s is given by the following.
2 ≤ h ≤ 15 � 25 �13 ln�h�2�
ln 3,
� �1
ln 3�13 ln h2
� 25 ln 3
� �13
ln 3�ln h2
�2513
ln 3
s � �13
ln 3 ln�h
2e��25�13� ln 3
k �25
ln�2C� ��13ln 3
� �11.8331.
C �12
e��25�13�ln 3 � 0.0605
12 � k ln �6C�,25 � k ln �2C�h � 6,s � 12h � 2s � 25
s � k ln h � C1 � k ln Ch
� ds � � kh
dh
dsdh
�kh
81. The general solution is Since whenwe have
Thus, When we have
t � 3.15 hours.
ln 0.20 � �0.5108t
0.20C � Ce�0.5108t
y � 0.20C,y � Ce�0.5108t.0.60C � Cek ⇒ k � ln 0.60 � �0.5108.t � 1,
y � 0.60Cy � Cekt. 82.
y � Cx
ln�y� � ln�x� � C1 � ln�Cx�
� 1y dy � �
1x dx
� �1y dydt� dt � � �1
x dxdt� dt
83.
P � Cekt �Nk
kP � N � C2ekt
1k ln�kP � N� � t � C1
� 1kP � N
dP � � dt 84.
(a) 8364 feet 1.5841 miles
inches
(b) 20,320 feet 3.8485 miles
inchesy�3.8485� � 13.86
�
y�1.5841� � 21.80
�
y�0� � 29.92 ⇒ C � 29.92 ⇒ y � 29.92e�0.2x
y � Ce�0.2x
dydx
� �0.2y
79. (a)
(b)
When 75% has changed:
t �ln�1�4�ln�4�5� � 6.2 hours
14
� et ln�4�5�
5 � 20et ln�4�5�
y � 20et ln�4�5�
y�1� � 16 � 20ek ⇒ k � ln 1620
� ln�45�
y�0� � 20 ⇒ C � 20
y � ekt�C1 � Cekt
ln y � kt � C1
�dyy
� �k dt
dydt
� ky
Section 6.3 Differential Equations: Separation of Variables 551
85.
Since when it follows that Therefore, we have
A �Pr
�ert � 1�.
C � P.t � 0,A � 0
A �1r�Cert � P�
rA � P � Cert
1r ln�rA � P� � t � C1
� 1
rA � P dA � � dt 86. (a)
(b)
� $3,243,606.35
A �250,000
0.05�e0.05�10� � 1�
� $583,098.01
�100,000
0.06�e0.06�5� � 1�
A �Pr
�ert � 1�
87. From Exercise 85,
Since when and wehave
� $11,068,161.12.
��120,000,000��0.0725�
e�0.0725��8� � 1
P �Ar
er t � 1
r � 0.0725,t � 8A � 120,000,000
A �Pr
�ert � 1�.
88. From Exercise 85,
which implies that
Since and you have
t �1
0.08 ln�800,000�0.08�
75,000� 1� � 7.71 years.
r � 0.08,P � 75,000,A � 800,000,
t �1r ln�Ar
P� 1�.
rt � ln�ArP
� 1�
ert �ArP
� 1
ert � 1 �ArP
A �Pr
�ert � 1�,
89.
(a)
(c) Using a computer algebra system or separation ofvariables, the general solution is
Using the initial condition you obtain
Thus, y � 5000e�2.3026e�0.02t.
500 � 5000e�C ⇒ C � ln 10 � 2.3026.
y�0� � 500,
y � 5000e�Ce�k t� 5000e�Ce�0.02t
.
00 300
5000
dydt
� 0.02y ln�5000y �
(b) As
(d)
The graph is concave upward on and concavedownward on �41.7, ��.
�0, 41.7�
0 3000
5000
y → L � 5000.t →�,
552 Chapter 6 Differential Equations
90.
(a)
(b) As
(c) Using a computer algebra system or separation ofvariables, the general solution is
Using the initial condition you obtain
Thus,
(d)
(e) The graph is concave upward on and concave downward on �16.7, ��.
�0, 16.7�
00 100
1200
y � 1000e�2.3026e�0.05t.
100 � 1000e�C ⇒ C � ln 10 � 2.3026.
y�0� � 100,
y � 1000e�Ce�k t� 1000e�Ce�0.05t
.
y → L � 1000.t →�,
00 100
1200
dydt
� 0.05y ln�1000y � 91. A differential equation can be solved by separation of
variables if it can be written in the form
To solve a separable equation, rewrite as
and integrate both sides.
M�x� dx � �N�y� dy
M�x� � N�y� dydx
� 0.
93. Two families of curves are mutually orthogonal if eachcurve in the first family intersects each curve in thesecond family at right angles.
92. where M and N arehomogeneous functions of the same degree. See Example 4.
M�x, y� dx � N�x, y� dy � 0,
94. (a)
Initial conditions:
when and when
Particular solution:
or v � 20�1 � e�0.2877t�v � 20�1 � eln�3�4�t � � 20�1 � �34�
t
�
C � 20, k � �ln�34�
t � 1.v � 5t � 0,W � 20, v � 0
v � W � Ce�kt
�ln�W � v� � k t � C1
� dvW � v
� �k dt
dvdt
� k�W � v� (b)
Since and we have
s � 20t � 69.5�e�0.2877t � 1�.
s�0� � 0, C � �69.5
� 20t � 3.4761e�0.2877t � C
s � �20�1 � e�0.2877t � dt
95. False. is separable, but
is not a solution.y � 0
dydx
�xy
96. True
dydx
� �x � 2��y � 1�
97. False
� t 2f �x, y�
f �t x, t y� � t 2x 2 � t 2xy � 2
Section 6.4 The Logistic Equation 553
98. True
�y2 � x2
x2 � y2 � �1
�x2 � y2 � 2x2
x2 � y2 � 2y2
�2Kx � 2x2
2Cy � 2y2
x
C � y�
K � xy
�Kx � x2
Cy � y 2
dydx
�K � x
y dydx
�x
C � y
x2 � y2 � 2Kxx 2 � y 2 � 2Cy
99. Product Rule
Need so avoid
Hence, there exists g and interval as long as12 � �a, b�.
�a, b�,
g�x� � Cex�2x � 1�1�2
ln�g�x�� � x �12
ln�2x � 1� � C1
g�
g�
f�f� � f
�2xex2
�2x � 1�ex2 � 1 �1
2x � 1
x �12.
f � f� � ex2� 2xex2
� �1 � 2x�ex2� 0,
g� �f�
f � f�g � 0
� f � f��g� � gf� � 0
fg� � gf� � f�g�
1.
Since it matches (c) or (d).
Since (d) approaches its horizontal asymptote slower than(c), it matches (d).
y�0� � 6,
y �12
1 � e�x 2.
Since it matches (a).y�0� �124
� 3,
y �12
1 � 3e�x
3.
Since it matches (b).y�0� �12
3�2� 8,
y �12
1 � �1�2�e�x 4.
Since it matches (c) or (d).
Since y approaches faster for (c), it matches (c).L � 12
y�0� � 6,
y �12
1 � e�2x
Section 6.4 The Logistic Equation
5.
y�0� �4
1 � 1� 2
� 2y�1 �y4�
� 2y�1 �4
4�1 � e�2t��
� 2y�1 �1
1 � e�2t�
� 2� 41 � e�2t�� e�2t
1 � e�2t� y� � �4�1 � e�2t��2��2e�2t�
y � 4�1 � e�2t��1; L � 4, k � 2, b � 1 6.
y�0� �5
1 � 3�
54
� 4y�1 �y5�
� 4y�1 �5
5�1 � 3e�4t��
� 4y�1 �1
1 � 3e�4t�
� 4� 51 � 3e�4t�� 3e�4t
1 � 3e�4t� y� � �5�1 � 3e�4t��2��12e�4t�
y � 5�1 � 3e�4t��1; L � 5, k � 4, b � 3
554 Chapter 6 Differential Equations
7.
y�0� �12
1 � 6�
127
� y�1 �y
12�
� y�1 �12
12�1 � 6e�t��
� y�1 �1
1 � 6e�t�
� � 121 � 6e�t�� 6e�t
1 � 6e�t� y� � �12�1 � 6e�t��2��6e�t�
y � 12�1 � 6e�t��1; L � 12, k � 1, b � 6 8.
y�0� �14
1 � 5�
73
� 3y�1 �y
14�
� 3y�1 �14
14�1 � 5e�3t��
� 3y�1 �1
1 � 5e�3t�
� 3� 141 � 5e�3t�� 5e�3t
1 � 5e�3t� y� � �14�1 � 5e�3t��2��15e�3t�
y � 14�1 � 5e�3t��1; L � 14, k � 3, b � 5
9.
(a) (b)
(c)
(d)
(e) P�0� � 60dPdt
� 0.75P�1 �P
1500�,
t �ln 240.75
� 4.2374
�0.75t � ln� 124� � �ln 24
e�0.75t �1
24
1 � 24e�0.75t � 2
750 �1500
1 � 24e�0.75t
P�0� �1500
1 � 24� 60
L � 1500k � 0.75
P�t� �1500
1 � 24e�0.75t 10.
(a) (b)
(c)
(d)
(e) P�0� � 125dPdt
� 0.2P�1 �P
5000�,
t �ln 390.2
� 18.3178
�0.2t � ln� 139� � �ln 39
e�0.2t �1
39
1 � 39e�0.2t � 2
2500 �5000
1 � 39e�0.2t
P�0� �5000
1 � 39� 125
L � 5000k � 0.2
P�t� �5000
1 � 39e�0.2t
11.
(a) (b)
(c)
(d)
(e)dPdt
� 0.8P�1 �P
6000�
t �ln 4999
0.8� 10.65 years
�0.8t � ln� 14999� � �ln 4999
e�0.8t �1
4999
1 � 4999e�0.8t � 2
3000 �6000
1 � 4999e�0.8t
P�0� �6000
1 � 4999�
65
L � 6000k � 0.8
P�t� �6000
1 � 4999e�0.8t 12.
(a) (b)
(c)
(d)
(e)dPdt
� 0.2P�1 �P
1000�
t �ln 80.2
� 10.40 years
�0.2t � ln�18� � �ln 8
e�0.2t �18
1 � 8e�0.2t � 2
500 �1000
1 � 8e�0.2t
P�0� �10001 � 8
�1000
9
L � 1000k � 0.2
P�t� �1000
1 � 8e�0.2t
14.
(a) (b)
(c)
(d) is a maximum for
(see Exercise 13).
P �2502
� 125dPdt
4 8 12 16 20
50
100
150
200
250
300
x
y
L � 250k � 0.5
dPdt
� 0.5P�1 �P
250� 15.
(a) (b)
(c)
(d) (same argument as in Exercise 13)P �250
2� 125
100−20
300
x
y
L � 250k � 0.1 �1
10
� 0.1P�1 �P
250�
dPdt
� 0.1P � 0.0004P2 � 0.1P�1 � 0.004P�
Section 6.4 The Logistic Equation 555
13.
(a)
(b)
(c)
1 2 3 4 5
20
40
60
80
100
120
t
P
L � 100
k � 3
dPdt
� 3P�1 �P
100�(d)
for and by the first Derivative Test, this is a maximum.
�Note: P � 50 �L2
�1002 �
P � 50,d2Pdt2 � 0
� 9P�1 �P
100��1 �2P100�
� 9P�1 �P
100��1 �P
100�
P100�
� 3�3P�1 �P
100��1 �P
100� �3P100�3P�1 �
P100�
d2Pdt2 � 3P��1 �
P100� � 3P��P�
100 �
16.
(a) (c)
(b)
(d) is a maximum for
(see Exercise 13).
P �1600
2� 800
dPdt
L � 1600
4 8 12 16 20
400
800
1200
1600
x
yk � 0.4
dPdt
� 0.4P � 0.00025P2 � 0.4P�1 �P
1600� 17.
Solution:
y�100� �40
1 � 4e�100 � 40.0
y�5� �40
1 � 4e�5 � 38.95
y �40
1 � 4e�t
8 �40
1 � b ⇒ b � 4y�0� � 8:
y �L
1 � be�kt �40
1 � be�t
L � 40k � 1,
y�0� � 8dydt
� y�1 �y
40�,
556 Chapter 6 Differential Equations
18.
Solution:
y�100� �8
1 � �3�5�e�1.2�100� � 8.0
y�5� �8
1 � �3�5�e�1.2�5� � 7.99
y �8
1 � �3�5�e�1.2t
5 �8
1 � b ⇒ 1 � b �
85
⇒ b �35
y�0� � 5:
y �L
1 � be�kt �8
1 � be�1.2t
L � 8k � 1.2,
y�0� � 5dydt
� 1.2y�1 �y8�, 19.
Solution:
y�100� �120
1 � 14e�0.8�100� � 120.0
y�5� �120
1 � 14e�0.8�5� � 95.51
y �120
1 � 14e�0.8t
8 �120
1 � b ⇒ b � 14y�0� � 8:
y �L
1 � be�kt �120
1 � be�0.8t
L � 120k �45
� 0.8,
y�0� � 8dydt
�4y5
�y2
150�
45
y�1 �y
120�,
20.
Solution:
y�100� �240
1 � 15e��3�20��100� � 240.0
y�5� �240
1 � 15e��3�20��5� � 29.68
y �240
1 � 15e��3�20�t
15 �240
1 � b ⇒ b � 15y�0� � 15:
y �L
1 � be�kt �240
1 � be��3�20�t
L � 240k �3
20,
y�0� � 15dydt
�3y20
�y2
1600�
320
y�1 �y
240�; 21. and
Matches (c).
y�0� � 350L � 250
22. and
Matches (d).
y�0� � 100L � 100 23. and
Matches (b).
y�0� � 50L � 250 24. and
Matches (a).
y�0� � 50L � 100
25.
(a)
(0, 105)
(0, 500)
10 20 30 40 50
200
400
600
800
1000
t
y
dydt
� 0.2y�1 �y
1000�(b)
y �1000
1 � �179�21�e�0.2t
b �17921
� 8.524
1 � b �1000105
�20021
y�0� � 105 �10001 � b
00 60
1000 y �1000
1 � be�0.2t
k � 0.2, L � 1000
Section 6.4 The Logistic Equation 557
26.
(a)
2 4 6 8 10
60
120
180
240
300
t
y
dydt
� 0.9y�1 �y
200�
27.
(a)
1 2 3 4 5 6 7 8 9 10
100200300400500600700800900
1000
t
y
dydt
� 0.6y�1 �y
700�
28.
(a)
2 4 6 8 10
140
280
420
560
700
t
y
dydt
� 0.4y�1 �y
500�
29. represents the value that approaches as tends toinfinity. is the carrying capacity.L
tyL 30. No, it is not possible to determine However,and You need an initial condition to determine b.k � 0.75.
L � 2500b.
(b)
y �500
1 � �1�3�e�0.4t
b �13
1 � b �500375
�43
y�0� � 375 �500
1 � b
00 10
700 y �500
1 � be�0.4t
k � 0.4, L � 500
(b)
y �200
1 � �1�6�e�0.9t
b � �16
1 � b �200240
�56
y�0� � 240 �200
1 � b
00 10
300 y �200
1 � be�0.9t
k � 0.9, L � 200
(b)
y �700
1 � 0.3e�0.6t
b � �310
� �0.3
1 � b �700
1000�
710
y�0� � 1000 �700
1 � b
00 10
1000 y �700
1 � be�0.6t
k � 0.6, L � 700
558 Chapter 6 Differential Equations
31. (a)
y �200
1 � 7e�0.2640t
k � �12
ln�2339� �
12
ln�3923� � 0.2640
e�2k �2339
1 � 7e�2k �20039
39 �200
1 � 7e�k�2�
25 �200
1 � b ⇒ b � 7
y�0� � 25L � 200,y �L
1 � be�kt , (b) For panthers.
(c)
(d)
Using Euler’s Method, when
(e) is increasing most rapidly where corresponds to years.t � 7.37
y �2002 � 100,y
t � 5.y � 65.6
y�0� � 25dydt
� ky�1 �yL� � 0.264y�1 �
y200�,
t � 7.37 years
�0.264t � ln�17�
1 � 7e�0.264t � 2
100 �200
1 � 7e�0.264t
y � 70t � 5,
32. (a)
y �200
1 � �173�27�e�0.2812t
k � 0.2812
�2k � ln�42397439�
e�2k �42397439
1 �17327
e�2k �20043
y�2� � 43 �200
1 � �173�27�e�k�2�
27 �200
1 � b ⇒ b �
17327
y �L
1 � be�kt, L � 200, y�0� � 27 (b) For panthers.
(c)
(d)
Using Euler’s Method,
(e) is increasing most rapidly when corresponding to 6.6 years.
y �2002 � 100,y
y�5� � 77.33.
y�0� � 27dydt
� ky�1 �yL� � 0.2812y�1 �
y200�,
100 �200
1 � �173�27�e�0.2812t ⇒ t � 6.6 years
t � 5, y � 78
33. (a)
Note: y �10
1 � 9e�t ln�3�2� �10
1 � 9�3�2��t �10
1 � 9�2�3�t
y �10
1 � 9e�0.40547t
k � �12
ln�49� �
12
ln�94� � ln�3
2� � 0.4055
e�2k �49
1 � 9e�2k � 5
2 �10
1 � 9e�2k
1 �10
1 � b ⇒ b � 9
y�0� � 1L � 10,y �L
1 � be�kt , (b) For
(c)
(d)
For Euler’s Method gives grams.
(e) The weight is increasing most rapidly whencorresponding to hours.t � 5.42y � 10�2 � 5,
y � 4.09t � 5,
dydt
� ln�32�y�1 �
y10� � 0.4055y�1 �
y10�
⇒ t � 8.84 hours
⇒ �32�
t
� 36
8 �10
1 � 9e�t ln�3�2� ⇒ 72e�t ln�3�2� � 2
y � 4.58 grams.t � 5,
t 0 1 2 3 4 5
Exact 1.0 1.4286 2.0 2.7273 3.6 4.5763
Euler 1.0 1.3649 1.8428 2.4523 3.2028 4.0855
Section 6.5 First-Order Linear Differential Equations 559
34. (a)
y �1.25
1 � 0.25e�0.1792t
k � 0.1792
�10k � ln�16�
e�10k �16
1 � 0.25e�10k �125120
�2524
1.2 �1.25
1 � 0.25e�k�10�
1 �1.25
1 � b ⇒ b � 0.25
y �L
1 � be�kt, L � 1.25, y�0� � 1, y�10� � 1.2 (b)
(c) The culture will never reach 8 grams
(d)
Using Euler’s Method,
(e) is increasing most rapidly when
The corresponding value is t � �7.7.t-
y �L2
�1.25
2� 0.625.
y
y�5� � 1.1349 g.
dydt
� 0.1792y�1 �y
1.25�, y�0� � 1
�L � 1.25�.
y�5� �1.25
1 � 0.25e�0.1792�5� � 1.13 g
35. False. If then and the populationdecreases.
dy�dt < 0y > L, 36. True. If then and the populationincreases.
dy�dt > 00 < y < L,
37.
Hence, when
By the First Derivative Test, this is a maximum.
1 �2yL
� 0 ⇒ y �L2
.d 2ydt 2 � 0
� k2�1 �yL�y�1 �
2yL �
� k2�1 �yL�y��1 �
yL� �
yL�
� k2y�1 �yL�
2
� ky ��ky�1 �yL�
L � d 2ydt 2 � ky��1 �
yL� � ky��
y�
L �
dydt
� ky�1 �yL�, y�0� < L 38.
� ky�1 � y�
�k
�1 � be�kt� � �1 �1
1 � be�kt�
�k
�1 � be�kt� �1 � be�kt � 1
�1 � be�kt�
�k
�1 � be�kt� �be�kt
�1 � be�kt�
y� ��1
�1 � be�kt�2��bke�kt�
y �1
1 � be�kt
Section 6.5 First-Order Linear Differential Equations
1.
Linear
y� �1x2 y �
1x3 �ex � 1�
x3y� � xy � ex � 1 2.
Linear
y� ��1 � 2x�
ln xy � 0
�ln x�y� � �1 � 2x�y � 0
2xy � y� ln x � y
560 Chapter 6 Differential Equations
3.
Not linear, because of the -term.xy2
y� � y cos x � xy2 4.
Linear
y� � 3xy � 1
1 � y� � 3xy
1 � y�
y� 3x
5.
Integrating factor:
y � x2 � 2x �Cx
xy � x�3x � 4� dx � x3 � 2x2 � C
e�1�x� dx � eln x � x
dydx
� �1x�y � 3x � 4
7.
Integrating factor:
y � �10 � Ce x
ye�x � 10e�x dx � �10e�x � C
e�xy� � e�xy � 10e�x
e �1 dx � e�x
y� � y � 10
6.
Integrating factor:
y �34
x2 �23
x �Cx2
x2y � x2�3x � 2� dx �34
x4 �2x3
3� C
e 2�x dx � eln x2� x2
dydx
�2xy � 3x � 2
8.
Integrating factor:
y � 2 � Ce�x 2
yex2� 4xex2 dx � 2ex2
� C
e 2x dx � e x 2
y� � 2xy � 4x
9.
Integrating factor:
y � �1 � Ce sin x
� �e�sin x � C
ye�sin x � �cos x�e�sin x dx
y�e�sin x � �cos x�e�sin xy � �cos x�e�sin x
e �cos x dx � e�sin x
y� � �cos x�y � cos x
y� � �y � 1� cos x � y cos x � cos x
�y � 1� cos x dx � dy
11.
Integrating factor:
y �x3 � 3x � C
3�x � 1�
y�x � 1� � �x2 � 1� dx �13
x3 � x � C1
e �1��x�1�� dx � eln x�1 � x � 1
y� � � 1x � 1�y � x � 1
�x � 1�y� � y � x2 � 1
10.
Integrating factor:
y � 1 � Ce�cos x
yecos x � �sinxecos x dx � ecos x � C
e�sin x dx � ecos x
y� � �sin x�y � �sin x
��y � 1� sin x� dx � dy � 0
12.
Integrating factor:
y �16
e3x � Ce�3x
ye3x � e3xe3x dx � e6x dx �16
e6x � C
e 3 dx � e3x
y� � 3y � e3x
Section 6.5 First-Order Linear Differential Equations 561
14.
Integrating factor:
y �12
�sin x � cos x� � Cex
ye�x � e�x cos x dx �12
e�x��cos x � sin x� � C
e�1 dx � e�x
y� � y � cos x13.
Integrating factor:
y � �x � C�e x3
ye�x3� ex3e�x3 dx � dx � x � C
e� 3x2 dx � e�x3
y� � 3x2y � e x3
15. (a), (c)
−6
−2
6
6
x−4
−3
4
5
y (b)
y �12
ex �12
e�x �12
�ex � e�x�
yex �12
e2x �12
y�0� � 1 ⇒ 1 �12
� C ⇒ C �12
yex �12
e2x � C
�yex� � e2x dx
exy� � exy � e2x
dydx
� y � ex Integrating factor: edx � e x
dydx
� ex � y
16. (a), (c)
4
−4
−4
4
x
y
4
4
−4
−4
(b)
y �1x ��
12
cos x2 �12�
0 �1
����12
cos � � C� ⇒ C � �12
y �1x��
12
cos x2 � C�
yx � x sin x2 dx � �12
cos x2 � C
y�x � y � x sin x2
u�x� � e �1�x� dx � eln x � x
y� �1x
y � sin x2, P�x� �1x, Q�x� � sin x2
18.
Integrating factor:
Initial condition:
Particular solution: y � e1�x2�3x2 � 12x2 �
y�1� � e, C � 3
y � e1�x2�Cx2 � 12x2 �
ye�1�x2�
1x3 dx � �
12x2 � C1
e�2�x3� dx � e��1�x2�
y� � � 2x3�y �
1x3 e1�x2
x3y� � 2y � e1�x 217.
Integrating factor:
Initial condition:
Particular solution: y � 1 � 4e�tan x
y�0� � 5, C � 4
y � 1 � Ce�tan x
yetan x � sec2 xetan x dx � etan x � C
e sec2 x dx � etan x
y� � �sec2 x�y � sec2 x
y� cos2 x � y � 1 � 0
562 Chapter 6 Differential Equations
20.
Integrating factor:
Initial condition:
Particular solution:
y � 1 �3
sec x � tan x� 1 �
3 cos x1 � sin x
y�0� � 4, 4 � 1 �C
1 � 0, C � 3
y � 1 �C
sec x � tan x
� sec x � tan x � C
y�sec x � tan x� � �sec x � tan x� sec x dx
e sec x dx � eln sec x� tan x � sec x � tan x
y� � y sec x � sec x19.
Integrating factor:
Initial condition:
Particular solution: y � sin x � �x � 1� cos x
y�0� � 1, 1 � C
y � sin x � x cos x � C cos x
y sec x � sec x�sec x � cos x� dx � tan x � x � C
e tan x dx � eln sec x � sec x
y� � y tan x � sec x � cos x
22.
Integrating factor:
Separation of variables:
Initial condition:
Particular solution: y � 2ex�x2
y�1� � 2, 2 � C
y � Cex�x 2
yC1 � ex�x2
ln y � ln C1 � x � x2
1y dy � �1 � 2x� dx
y � Cex�x2
yex 2�x � C
e�2x�1� dx � ex2�x
y� � �2x � 1�y � 021.
Integrating factor:
Separation of variables:
Initial condition:
Particular solution: xy � 4
y�2� � 2, C � 4
xy � C
ln xy � ln C
ln y � �ln x � ln C
1y dy � �
1x dx
dydx
� �yx
e�1�x� dx � eln x � x
y� � �1x�y � 0
23.
y � �2 � x ln x � 12x
y�1� � 10 � �2 � C ⇒ C � 12
� �2 � x ln x � Cx
� x �ln x ��2x
� C�
y � x �1 �2x�
1x dx � x �1
x�
2x2� dx
u�x� � e��1�x� dx �1x
dydx
�1x
y � 1 �2x Linear
dydx
�x � y � 2
x�
yx
� 1 �2x
x dy � �x � y � 2� dx 24.
y �x3
5� x �
175�x
y�4� � 2 �645
� 4 � 2C ⇒ C � �175
�x3
5� x � C�x
� x1�2�x5�2
5� x1�2 � C�
y � x1�2�x2
2�
12�
1x1�2 dx � x1�2�x3�2
2�
x�1�2
2 � dx
u�x� � e��1�2x� dx �1
x1�2
dydx
�12x
y �x2
2�
12
, Linear
2xy� � y � x3 � x
Section 6.5 First-Order Linear Differential Equations 563
25.
1y2 � Ce2x3
�13
y�2 �13
� Ce2x3
y�2e�2x3�
13
e�2x3� C
y�2e�2x3� � 2x2e�2x3 dx
y�2e��2�3x2 dx � ��2�x2e��2�3x2 dx dx
n � 3, Q � x2, P � 3x2
y� � 3x2y � x2y3 26.
y2 � 1 � Ce�x2
y2e x2� 2xex2 dx � e x2
� C
e 2x dx � ex2P � x,Q � x,n � �1,
y� � xy � xy�1
31. (a)
(c)
−5
−4 4
5
−5
−4 4
5 (b)
Integrating factor:
(2, 8): 8 �82
� 2C ⇒ C � 2 ⇒ y �x3
2� 2x �
12
x�x2 � 4�
��2, 4�: 4 ��82
� 2C ⇒ C � �4 ⇒ y �x3
2� 4x �
12
x�x2 � 8�
y �x3
2� Cx
�1x
y� � x dx �x2
2� C
1x
y� �1x2 y � x
e�1�x dx � e�ln x �1x
dydx
�1x
y � x2
27.
y �1
Cx � x2
1y
� �x2 � Cx
y�1x�1 � �x�x�1� dx � �x � C
e��1�x� dx � e�ln x � x�1
n � 2, Q � x, P � x�1
y� � �1x�y � xy2
29.
y2�3 � 2e x � Ce2x�3
y2�3e��2�3�x � 2e�1�3�x � C
y2�3e��2�3�x � 23
e xe��2�3�x dx � 23
e�1�3�x dx
e��2�3� dx � e��2�3�x
y� � y � e x 3�y, n �13
, Q � e x, P � �1
28.
y ��x5�2 � C�2
25x
�15
x5�2 � C1 �x5�2 � C
5
y1�2x1�2 � 12
x1�2�x� dx
e�1�2��1�x� dx � e�1�2� ln x � �x
n �12
, Q � x, P � x�1
y� � �1x�y � x�y
30.
y2 � �23
ex � Ce4x
y2e�4x � 2e�4xex dx � �23
e�3x � C
e2��2� dx � e�4x
n � �1, Q � ex, P � �2
y� � 2y � exy�1
yy� � 2y2 � ex
564 Chapter 6 Differential Equations
32. (a)
(c)
−4
−2
4
6
−4
−2
4
6 (b)
Integrating factor:
�0, �12�: �
12
�14
� C ⇒ C � �34
⇒ y �14
�34
e�x 4
�0, 72�: 7
2�
14
� C ⇒ C �134
⇒ y �14
�134
e�x 4
y �14
� Ce�x4
ye x4� x3e x4 dx �
14
e x4� C
y�e x4� 4x3ye x4
� x3e x4
e 4x3 dx � e x4
y� � 4x3y � x3
33. (a)
(c)
−2
−3
6
3
−2
−3
6
3 (b)
Integrating factor:
y � �2 cot x � �2 cos 3 � sin 3� csc x
�3, �1�: �1 � �2 cot 3 � C csc 3 ⇒ C �2 cot 3 � 1
csc 3� 2 cos 3 � sin 3
y � �2 cot x � �sin 1 � 2 cos 1� csc x
(1, 1�: 1 � �2 cot 1 � C csc 1 ⇒ C �1 � 2 cot 1
csc 1� sin 1 � 2 cos 1
y � �2 cot x � C csc x
y sin x � 2 sin x dx � �2 cos x � C
y� sin x � �cos x�y � 2 sin x
e cot x dx � eln sin x � sin x
y� � �cot x�y � 2
34. (a)
(c)
−4
−2
4
6
−4
−2
4
6 (b)
Bernoulli equation, letting you obtain and The solution is:
y �2
1 � e x 2
�0, 1�: 1 �2
1 � 2C ⇒ 1 � 2C � 2 ⇒ C �
12
y �2
1 � �e x 2�3��
6
3 � e x 2
�0, 3�: 3 �2
1 � 2C ⇒ 1 � 2C �
23
⇒ C � �16
y �2
1 � 2Ce x2
1y
�12
� Ce x 2�
1 � 2Ce x2
2
y�1e�x2�
12
e�x2� C
��1�xe�x2 dx �
12e�x2.
e�2x dx � e�x2z � y1�2 � y�1,n � 2
y� � 2xy � xy2
Section 6.5 First-Order Linear Differential Equations 565
35. (a)
(b)
Let then the integrating factor is
(c) limt→�
Q �qk
Q �qk
� �Q0 �qk�e�kt
Q0 �qk
� C ⇒ C � Q0 �qk
When t � 0: Q � Q0
Q � e�kt qekt dt � e�kt�qk
ekt � C� �qk
� Ce�kt
u�t� � ekt.Q�t� � q,P�t� � k,
Q� � kQ � q
dQdt
� q � kQ, q constant
36. (a)
(c) For
For
N � 40 � 30.5685e�0.0188t
C � �30ek � �30.5685�30 � Ce�k ⇒
k �1
19 ln�10
7 � � 0.0188ln�107 � � 19k ⇒
3021
�e�k
e�20k � e19k
�21 � Ce�20k19 � 40 � Ce�20k ⇒
t � 20, N � 19:
�30 � Ce�k10 � 40 � Ce�k ⇒
t � 1, N � 10:
dNdt
� k�40 � N�
37. Let Q be the number of pounds of concentrate in the solution at any time t. Since the number of gallons of solution in the tank at any time t is and since the tank loses gallons of solution per minute, it must lose concentrate at the rate
The solution gains concentrate at the rate Therefore, the net rate of change is
ordQdt
�r2Q
v0 � �r1 � r2�t� q1r1.
dQdt
� q1r1 � � Qv0 � �r1 � r2�t�r2
r1q1.
� Qv0 � �r1 � r2�t�r2.
r2
v0 � �r1 � r2�t
38. From Exercise 37, and using
dQdt
�rQv0
� q1r.
r1 � r2 � r,
(b)
Integrating factor:
N � 40 � Ce�kt
Nekt � 40kekt dt � 40ekt � C
ekt
N� � kN � 40k
566 Chapter 6 Differential Equations
41. (a) The volume of the solution in the tank is given by Therefore, at minutes.
(b)
Integrating factor:
Initial condition:
Particular solution:
Q�50� � 100 � 505�2�100��3�2 � 100 �25�2
� 82.32 lbs
Q � �50 � t� � 505�2�50 � t��3�2
Q�0� � 0 � 50 � C�50�3�2�, C � �505�2
Q � �50 � t� � C�50 � t��3�2
Q�50 � t�3�2 � 2.5�50 � t�3�2 dt � �50 � t�5�2 � C
e �3��100�2t�� dt � �50 � t�3�2
Q�0� � q0 � 0, q1 � 0.5, v0 � 100, r1 � 5, r2 � 3, Q� �3
100 � 2tQ � 2.5
Q� �r2Q
v0 � �r1 � r2�t � q1r1
t � 50100 � �5 � 3�t � 200v0 � �r1 � r2�t.
39. (a)
Initial condition:
Particular solution: Q � 25e��1�20�t
Q�0� � 25, C � 25
Q � Ce��1�20�t
ln Q � �1
20t � ln C1
1Q
dQ � �1
20 dt
r1 � 10, r2 � 10, Q� �1
20Q � 0
Q�0� � q0 � 25, q1 � 0, v0 � 200,
Q� �r2Q
v0 � �r1 � r2�t � q1r1
40. (a)
Integrating factor:
Q � 8 � 17e��1�20�t
Q�0� � 25 � 8 � C ⇒ C � 17
Q � 8 � Ce��1�20�t
Qe�1�20�t � 0.4e�1�20�t dt � 8e�1�20�t � C
e�1�20�t
r1 � 10, r2 � 10, Q� �1
20Q � 0.4
Q�0� � q0 � 25, q1 � 0.04, v0 � 200,
Q� �r2Q
v0 � �r1 � r2�t � q1r1
(b)
(c) limt→�
25e��1�20�t � 0
t � �20 ln�35� � 10.2 min
ln�35� � �
120
t
15 � 25e��1�20�t
(b)
(c) limt→�
Q�t� � 8 lbs
ln� 717� � �
120
t ⇒ t � �20 ln� 717� � 17.75 min
7 � 17e��1�20�t
15 � 8 � 17e��1�20�t
Section 6.5 First-Order Linear Differential Equations 567
43.
implies that
Using a graphing utility, and
As The graph of is shownbelow.
50
−200
0 40
vv → �159.47 ft�sec.t →�,
v � �159.47�1 � e�0.2007t�.
k � 0.050165,
�101 ��8k
�1 � e�5k��1�4��.m ��8g
�14
v�5� � �101,mg � �8,g � �32,
v �mgk
�1 � e�kt�m�, solution
dvdt
�kvm
� g 44.
The graph of is shown below.
when t � 36.33 sec.s�t� � 0
0 40
−500
6000
s�t�
s�t� � �159.47t � 794.57e�0.2007t � 5794.57
s�0� � 5000 � �794.57 � C ⇒ C � 5794.57
� �159.47t � 794.57e�0.2007t � C
� �159.47�1 � e�0.2007t� dt
s�t� � v�t� dt
42. (a) The volume of the solution is given by minutes.
(b)
Integrating factor is as in #41.
When (double the answer to #41)t � 50, Q � 200 � 2�50�5�2�100��3�2 � 164.64 lbs
Q � 2�50 � t� � 2�50�5�2�50 � t��3�2
Q�0� � 0: 0 � 100 � C�50��3�2 ⇒ C � �100�50�3�2 � �2�50�5�2
Q � 2�50 � t� � C�50 � t��3�2
Q�50 � t�3�2 � 5�50 � t�3�2 dt � 2�50 � t�5�2 � C
�50 � t�3�2,
Q� �3Q
100 � 2t� 5
Q�0� � q0 � 0, q1 � 1, v0 � 100, r1 � 5, r2 � 3
Q� �r2Q
v0 � �r1 � r2�t � q1r1
v0 � �r1 � r2�t � 100 � �5 � 3�t � 200 ⇒ t � 50
45.
Integrating factor:
I �E0
R� Ce�Rt�L
I eRt�L � E0
LeRt�L dt �
E0
ReRt�L � C
e�R�L� dt � eRt�L
I� �RL
I �E0
LL
dIdt
� RI � E0, 46.
t �ln�0.1��150
� 0.0154 sec
�150t � ln�0.1�
e�150t � 0.1
0.9 � 1 � e�150t
�0.90�15
� 0.18 �15
�1 � e�150t�
limt→�
I �15
amp
I �15
�15
e�150t
�0� �120600
� C ⇒ C � �15
I �E0
R� Ce�Rt�L
I�0� � 0, E0 � 120 volts, R � 600 ohms, L � 4 henrys
568 Chapter 6 Differential Equations
47. Standard form
Integrating factoru�x� � eP�x� dx
dydx
� P�x�y � Q�x�
48. Standard form
Let Multiplying by produces
Linear z� � �1 � n�P�x�z � �1 � n�Q�x�.
�1 � n�y�n y� � �1 � n�P�x�y1�n � �1 � n�Q�x�
�1 � n�y�n�n � 0, 1�.z � y1�n
y� � P�x�y � Q�x�yn
50.
Matches d.
y � Ce2x
ln y � 2x � C1
dyy
� 2 dx
y� � 2y � 049.
Matches c.
y � x2 � C
dy � 2x dx
y� � 2x � 0
51.
Matches a.
y � Cex2
ln y � x2 � C1
dyy
� 2x dx
y� � 2xy � 0 52.
Matches b.
y � �12
� Cex2
2y � 1 � C2ex2
12
ln�2y � 1� �12
x2 � C1
dy
2y � 1� x dx
y� � 2xy � x
53.
Separation of variables:
2ex � e�2y � C
ex � �12
e�2y � C1
ex dx � e�2y dy
e2xey dx � exe�y dy
e2x�y dx � ex�y dy � 0 54.
Separation of variables:
3x2 � 6x � 2y3 � 6y2 � C
12
x2 � x �13
y3 � y2 � C1
�x � 1� dx � �y2 � 2y� dy
�x � 1� dx � �y2 � 2y� dy � 0
55.
Separation of variables:
y � Ce�sin x � 1
ln�y � 1� � �sin x � ln C
sin x � �ln�y � 1� � ln C
cos x dx � �1
y � 1 dy
�y cos x � cos x� dx � dy � 0 56.
Separation of variables:
y � sin�x2 � C�
arcsin y � x2 � C
1
�1 � y2 dy � 2x dx
y� � 2x�1 � y2
Section 6.5 First-Order Linear Differential Equations 569
59.
Integrating factor:
y �ex
x2�x � 1� �Cx2
yx2 � x21x
ex dx � ex�x � 1� � C
e�2�x� dx � eln x2� x2
Linear: y� � �2x�y �
1x
ex
�2y � ex� dx � x dy � 0 60.
Homogeneous:
y �x
C � ln x
ln x � �1v
� C
1x dx �
1v2 dv
v2 dx � x dv � 0
�v2x2 � vx2� dx � x2�v dx � x dv� � 0
y � vx, dy � v dx � x dv
�y2 � xy� dx � x2 dy � 0
57.
Homogeneous:
x3y2 � x4y � C
x5�v2 � v� � C
ln x5 � ln v2 � v � ln C
5x dx � �2v � 1
v2 � v� dv � 0
�3v2x2 � 4vx2� dx � �2vx2 � x2��v dx � x dv� � 0
y � vx, dy � v dx � x dv
�3y2 � 4xy� dx � �2xy � x2� dy � 0 58.
Linear:
Integrating factor:
y � x�ln x � C�
y1x
� 1x dx � ln x � C
e��1�x� dx � eln x�1 �1x
y� �1x
y � 1
�x � y� dx � x dy � 0
61.
Bernoulli:
x4y4 � 2x2 � C
y4x4 � 4�x�3��x4� dx � 2x2 � C
e �4�x� dx � eln x 4� x 4
n � �3, Q � x�3, P � x�1,
y� � �1x�y � x�3y�3
�x2y4 � 1� dx � x3y3 dy � 0
63.
Integrating factor:
y �125
x2 �Cx3
yx3 � 12x 4 dx �125
x5 � C
y�x3 �3x
x3y � 12x�x3� � 12x 4
e�3�x� dx � e3 ln x � x3
y� �3x
y � 12x
x dydx
� �3y � 12x2
3�y � 4x2� dx � �x dy
62.
Homogeneous:
y3�x � y� � C
y4�v � 1� � C
ln v � 1 � �ln y4 � ln C
1
v � 1 dv � �
4y dy
y�v dy � y dv� � �3vy � 4y� dy � 0
x � vy, dx � v dy � y dv
y dx � �3x � 4y� dy � 0
64.
Separation of variables:
ln�x2 � 1� � y2 � 2ey � C
12
ln�x2 � 1� � �12
y2 � ey � C1
x
x2 � 1 dx � ��y � ey� dy
x dx � �y � ey��x2 � 1� dy � 0
65. False. The equation contains �y. 66. True. is linear.y� � �x � ex�y � 0
Section 6.6 Predator-Prey Differential Equations
570 Chapter 6 Differential Equations
1.
when and
� �57.14, 14.0�.
� �4007
, 14�
�x, y� � �mn
, ab� � � 0.4
0.007,
0.70.05�
�x, y� � �0, 0�x� � y� � 0
dydt
� �my � nxy � �0.4y � 0.007xy
dxdt
� ax � bxy � 0.7x � 0.05xy 2.
when and
� �30, 125�.
�x, y� � �mn
, ab� � � 0.3
0.01,
0.50.004�
�x, y� � �0, 0�x� � y� � 0
dydt
� �my � nxy � �0.3y � 0.01xy
dxdt
� ax � bxy � 0.5x � 0.004xy
3.
when and
� �55.56, 50.0�.
� �5009
, 50�
�x, y� � �mn
, ab� � � 0.5
0.009,
0.30.006�
�x, y� � �0, 0�x� � y� � 0
dydt
� �my � nxy � �0.5y � 0.009xy
dxdt
� ax � bxy � 0.3x � 0.006xy 4.
when and
� �60, 30�.
�x, y� � �mn
, ab� � � 0.6
0.01,
0.60.02�
�x, y� � �0, 0�x� � y� � 0
dydt
� �my � nxy � �0.6y � 0.01xy
dxdt
� ax � bxy � 0.6x � 0.02xy
5.
80 160 240 320 400
10
20
30
40
x
y
(150, 30)
6.
10
2
4
6
8
4 8 12 16 20x
y
(15, 3)
7. (a) The initial conditions areand
(b)
(40, 20)
20 40 60 80 100
20
40
60
80
100
x
y
y�0� � 20.x�0� � 40
8. (a) The initial conditions areand
(b)
20
40
60
80
100
20 40 60 80 100x
y
(60, 10)
y�0� � 10.x�0� � 609. Critical points are
and
� �50, 20�.
� � 0.30.006
, 0.8
0.04�
�x, y� � �mn
, ab�
�x, y� � �0, 0� 10.
00 36
x(t)
y(t)
150
Section 6.6 Predator-Prey Differential Equations 571
11.
00 150
50 12.
(55, 10)00 150
50 13. Critical points are and
� �10,000, 1250�.
� � 0.40.00004
, 0.1
0.00008�
�x, y� � �mn
, ab�
�x, y� � �0, 0�
14.
00 240
25,000
x(t)
y(t)
15.
00 25,000
5,000 16.
00 25,000
5,000
(4000, 1000)
17. Using and you obtain the constantsolutions and
The slope field is the same, but the solution curve reducesto a single point at
00 150
50
(50, 20)
�50, 20�.
00 36
60
x
y
y � 20.x � 50y�0� � 20,x�0� � 50 18. Using and you obtain the
constant solutions and
The slope field is the same, but the solution curve reducesto a single point at
20,00000
2,500
(10,000, 1,250)
�10,000, 1250�.
15,000
00 240
x(t)
y(t)
y � 1250.x � 10,000y�0� � 1250,x�0� � 10,000
19.
when
�x, y� � �an � mcbn � cp
, bm � apbn � cp � � �1
3,
13�.
�x, y� � �ab
, 0� � �12
, 0�
�x, y� � �0, mn � � �0,
12�
�x, y� � �0, 0�
x� � y� � 0
dydt
� y � 2y2 � xy
dxdt
� x � 2x2 � xy 20.
when
�x, y� � �an � mcbn � cp
, bm � apbn � cp � � �3
3,
33� � �1, 1�.
�x, y� � �ab
, 0� � �2, 0�
�x, y� � �0, mn � � �0,
54�
�x, y� � �0, 0�
x� � y� � 0
dydt
� 5y � 4y2 � xy
dxdt
� 2x � x2 � xy
572 Chapter 6 Differential Equations
21.
when
�x, y� � �an � mcbn � cp
, bm � apbn � cp � � �0.03
0.17,
0.010.17� � � 3
17,
117�.
�x, y� � �ab
, 0� � �0.10.4
, 0� � �14
, 0�
�x, y� � �0, mn � � �0,
0.10.8� � �0,
18�
�x, y� � �0, 0�
x� � y� � 0
dydt
� 0.1y � 0.8y2 � 0.3xy
dxdt
� 0.1x � 0.4x2 � 0.5xy 22.
when
� �0.0210.1
, 0.0020.1 � � � 21
100,
150�.
�x, y� � �an � mcbn � cp
, bm � apbn � cp �
�x, y� � �ab
, 0� � �0.050.2
, 0� � �14
, 0�
�x, y� � �0, mn � � �0,
0.060.9 � � �0,
115�
�x, y� � �0, 0�
x� � y� � 0
dydt
� 0.06y � 0.9y2 � 0.2xy
dxdt
� 0.05x � 0.2x2 � 0.4xy
23.
Four critical points:
�an � mcbn � cp
, bm � apbn � cp � � �0.45
0.23,
0.040.23� � �45
23,
423�
�ab
, 0� � �0.80.4
, 0� � �2, 0�
�0, mn � � �0,
0.30.6� � �0,
12�
�0, 0�
a � 0.8, b � 0.4, c � 0.1, m � 0.3, n � 0.6, p � 0.1 24.
Both species survive.
00 36
10
x(t)y(t)
25.
Four critical points:
�an � mcbn � cp
, bm � apbn � cp � � � 0.18
�0.76,
�0.68�0.76� � ��
938
, 1719�
�ab
, 0� � �0.80.4
, 0� � �2, 0�
�0, mn � � �0,
0.30.6� � �0,
12�
�0, 0�
a � 0.8, b � 0.4, c � 1, m � 0.3, n � 0.6, p � 1 26.
One species (the trout) becomes extinct.
00 36
10
x(t)
y(t)
27. Assuming the initial conditions are the critical points
you obtain constant solutions.
00 36
3
x
y
�x�0�, y�0�� � �4523, 4
23�28. Assuming the initial conditions are the critical points
you obtain constant solutions.
00 36
2
x(t)
y(t)
�x�0�, y�0�� � �0, 12�
Section 6.6 Predator-Prey Differential Equations 573
29. Solve the equations
to obtain the critical points
and
The solutions will be constant forthese initial conditions.
�mn
, ab�.�0, 0�
dydt
� �my � nxy � 0
dxdt
� ax � bxy � 0
30. As in Exercise 29, using any of the four critical points as initialconditions will yield constant solutions.
31. False
32. False 33. False. The predator-prey equationsare a special case of the competingspecies equations.
34. False
35. (a) If then
which is a logistic equation.
(b)
is a critical point. If then andis a critical point. If then
Thus, and
The third critical point is �60, 16�.
0.4�1 �60
100� � 0.01y ⇒ y � 16.
x �0.3
0.005� 60
�0.3 � 0.005x � 0.
0.4�1 �x
100� � 0.01y
x, y � 0,�100, 0�x � 100y � 0,�0, 0�
dydt
� �0.3y � 0.005xy
dxdt
� 0.4x�1 �x
100� � 0.01xy
dxdt
� ax�1 �xL�,
y � 0, (c)
(d)
(e)
00 100
80
00 100
80
00 72
100
y
x
36. (b)
Critical numbers:
(c)
00 72
160
x(t)
y(t)
�0, 0�, �60, 40�
dydt
� �my � nx � �0.3y � 0.005xy
dxdt
� ax � bxy � 0.4x � 0.01xy (d)
(e)
00 140
100
(40, 80)
00 140
100
574 Chapter 6 Differential Equations
Review Exercises for Chapter 6
1.
Not a solution
x2y� � 3y � x2�3x2� � 3�x3� � 3�x4 � x3� � 6x3
y� � 3x2y � x3,
3.
y � ��2x2 � 5� dx �2x3
3� 5x � C
dydx
� 2x2 � 5 4.
y � ��x3 � 2x� dx �x4
4� x2 � C
dydx
� x3 � 2x
5.
y � � cos 2x dx �12
sin 2x � C
dydx
� cos 2x 6.
y � � 2 sin x dx � �2 cos x � C
dydx
� 2 sin x
2.
Not a solution
y��� � 8y � �16 cos 2x � 8�2 sin 2x� � 0
y��� � �16 cos 2x
y� � �8 sin 2x
y� � 4 cos 2x
y � 2 sin 2x
8.
y � �3e�x�3 dx � �9e�x�3 � C
dydx
� 3e�x�37.
Let
�4
15�x � 7�3�2�3x � 14� � C
�45
�x � 7�5�2 �283
�x � 7�3�2 � C
�45
u5�2 �283
u3�2 � C
y � �2�u � 7�u1�2 du
x � u � 7.du � dx,u � x � 7,
y � �2x�x � 7 dx
dydx
� 2x�x � 7
9.dydx
�2xy
x 0 2 4 8
y 2 0 4 4 6 8
Undef. 0 1 24�3�4dy�dx
�2�4
10.dydx
� x sin ��y4
x 0 2 4 8
y 2 0 4 4 6 8
0 0 0 0�4�4dy�dx
�2�4
Review Exercises for Chapter 6 575
11.
x
y
1
−3
2
(−1, 1)
��1, 1�y� � �x � 2, 12.
−4 4
−4
4
y
x
�0, 2�y� � 2x2 � x, 13.
−4 4
−4
4
x
y
(0, 3)
�0, 3�y� �14 x2 �
13 x,
14.
−4 4
−4
4
y
x
�2, 1�y� � y � 3x, 15.
4
−4
4
x
y
(0, 1)
−4
�0, 1�y� �xy
x2 � 4, 16.
5
−7
1
y
x
�0, �2�y� �y
x2 � 1,
17.
� 6x �x2
2� C
y � ��6 � x� dx
dydx
� 6 � x 18.
y � �6 � Cex
y � 6 � ex �C1 � Cex
ln y � 6 � x � C1
� dyy � 6
� �dx
dydx
� y � 6 19.
y � �3 �1
x � C
3 � y ��1
x � C
��3 � y��1 � x � C
��3 � y��2 dy � �dx
dydx
� �3 � y�2
20.
y � �2x � C�2
y1�2 � 2x � C
2y1�2 � 4x � C1
�y�1�2 dy � �4 dx
dydx
� 4�y 21.
y � Cex�2 � x��2 �Cex
�2 � x�2
lny � x � 2 ln2 � x � C1
1y dy � �1 �
22 � x dx
1y dy �
x2 � x
dx
�2 � x� dydx
� xy
�2 � x�y� � xy � 0
576 Chapter 6 Differential Equations
22.
y � Cxex
lny � x � lnx � C1
�dyy
� � x � 1x
dx
x dydx
� �x � 1�y
xy� � �x � 1�y � 0 23.
y �34
e�ln �20�3��5�t �34
e0.379t
k �15
ln�203
203
� e5k
5 �34
ek�5��5, 5�:
34
� C�0, 34:
y � Cekt
24.
Hence,
Finally, y �9
20e1�2 ln �10�3�t.
C �32
e�2�1�2� ln�10�3� �32 �
310 �
920
.
103
� e2k ⇒ k �12
ln �103
5 � Ce4k � �32
e�2ke4k �32
e2k�4, 5�:
32
� Ce2k ⇒ C �32
e�2k�2, 32:
y � Cekt 25.
y � 5e��t ln 30��5 � 5e�0.680t
k �15
ln� 130 �
�ln 305
16
� 5e5k�5, 16:
C � 5�0, 5�:
y � Cekt
26.
Hence,
Finally, y � 12.1586e�0.3008t.
C � 9e�1�5 ln�2�9� � 9�29
�1�5
� 12.15864.
k �15
ln�29 � �0.3008
2 � Ce6k ⇒ 2 � �9e�k�e6k � 9e5k�6, 2�:
9 � Cek ⇒ C � 9e�k�1, 9�:
y � Cekt 27.
P�35,000� � 30e��35,000 ln 2��18,000 � 7.79 inches
P�h� � 30e��h ln 2��18,000
k �ln�1�2�18,000
��ln 218,000
P�18,000� � 30e18,000k � 15
P�h� � 30ekh
P�0� � 30dPdh
� kp,
28.
When y � 5e�0.000433�600� � 3.86 g.t � 600,
k �1
1599 ln�12� � �0.000433
2.5 � 5ek�1599�
y � Cekt � 5ekt
29.
(a)
S � 30eln �1�6��t � 30�16�1�t
k � ln 16 � �1.7918
5 � 30ek
limt→�
Cek�t � C � 30
5 � Cek
S � 5 when t � 1
S � Cek�t
(b) When which is 20,965 units.
(c)
00
40
30
S � 20.9646t � 5,
Review Exercises for Chapter 6 577
30.
(a)
(b) 25,000 units � limt→�
S � 25� ⇒ k � ln�21
25� � �0.1744
⇒ ek �2125
4 � 25�1 � ek�1�� ⇒ 1 � ek �425
S � 25�1 � ekt�
31.
t �ln 2
0.015� 46.21 years
ln 2 � 0.015t
2 � e0.015t
2C � Ce0.015t
P � Ce0.015t 32. (a)
When
(b)
y � 28e0.6�0.012s, s > 50.
s � 50, y � 28 � Ce�0.012�50� ⇒ C � 28e0.6
y � Ce�0.012s
�1
0.012 ln y � s � C1
�10.012
�dyy
� �ds
dyds
� �0.012y, s > 50
33.
y �x2
2� 3 lnx � C
�dy � ��x �3x dx
dydx
�x2 � 3
x34.
y � �12
ln�1 � e�2x� � C
�dy � � e�2x
1 � e�2x dx � �12� �2e�2x
1 � e�2x dx
dydx
�e�2x
1 � e�2x
35.
y � Cex2
ex2�C1 � y
lny � x2 � C1
�1y dy � �2x dx
dydx
� 2xy
y� � 2xy � 0 36.
y � ln 1cos x � C � �lncos x � C
�C � �C1� ey �1
cos x � C
�e�y � �cos x � C1
�e�y dy � �sin x dx
dydx
� ey sin x
y� � ey sin x � 0
(c) When which is 14,545 units.
(d)
0 80
25
S � 14.545t � 5,
Speed(s) 50 55 60 65 70
Miles per gallon (y) 28 26.4 24.8 23.4 22.0
37.
Let
—CONTINUED—
y � vx, dy � x dv � v dx.
�x2 � y2� dx � 2xy dy � 0
dydx
�x2 � y2
2xy, (homogeneous differential equation)
578 Chapter 6 Differential Equations
37. —CONTINUED—
or C1 �x
x2 � y2 1 �Cx
x2 � y2
x �C
1 � v2 �C
1 � �y�x�2 �Cx2
x2 � y2
lnx � �ln1 � v2 � C1 � �ln1 � v2 � ln C
�dxx
� � 2v1 � v2 dv
�1 � v2� dx � 2xv dv
�x2 � x2v2� dx � 2x3v dv
�x2 � v2x2 � 2x2v2� dx � 2x3v dv � 0
�x2 � v2x2� dx � 2x�vx��x dv � v dx� � 0
39.
y � �2x �12x3
4 � ��4C2� � 12C2 � 8C2 ⇒ C2 �12, C1 � �2
x � 2, y� � 4: 4 � C1 � 12C2
x � 2, y � 0: 0 � 2C1 � 8C2 ⇒ C1 � �4C2
� 6C2 x3 � 3C1x � 9C2x3 � 3C1x � 3C2x3 � 0
x2y� � 3xy� � 3y � x2�6C2x� � 3x�C1 � 3C2x2� � 3�C1x � C2 x3�
y� � 6C2 x
y� � C1 � 3C2 x2
y � C1x � C2 x3
38.
Let
y �x3 � 3Cx
2C
x3 � C�3x � 2y� � 3Cx � 2Cy
x2 � C�3 � 2v� � C�3 � 2�yx
x � C2�3 � 2v�1�2
lnx �12
ln3 � 2v � C1 � ln�3 � 2v�1�2 � ln C2
�1x dx � � 1
3 � 2v dv
�3 � 2v� dx � x dv
�3x � 2vx� dx � x2 dv � 0
3�x � vx� dx � x�x dv � v dx� � 0
y � vx, dy � x dv � v dx.
3�x � y� dx � x dy � 0
dydx
�3�x � y�
x, (homogeneous differential equation)
Review Exercises for Chapter 6 579
40.
(a)
At
Note that since the object is moving downward.k < 0
v �1k�9.8 � �kv0 � 9.8�ekt�.
v0 �1k�9.8 � C3� ⇒ C3 � kv0 � 9.8t � 0,
v �1k�9.8 � C3e
kt kv � 9.8 � ekt�C2 � C3ekt
lnkv � 9.8 � kt � C2
1k lnkv � 9.8 � t � C1
� dvkv � 9.8
� �dt
dvdt
� kv � 9.8
(b)
(c)
�9.8t
k�
1k2�kv0 � 9.8��ekt � 1� � s0
s�t� �9.8t
k�
1k2�kv0 � 9.8�ekt � s0 �
1k2�kv0 � 9.8�
s�0� �1k2�kv0 � 9.8� � C ⇒ C � s0 �
1k2 �kv0 � 9.8�
�9.8t
k�
1k2�kv0 � 9.8�ekt � C
�1k�9.8t �
1k�kv0 � 9.8�ekt � C
s�t� � �1k�9.8 � �kv0 � 9.8�ekt� dt
limt→�
v�t� �9.8k
41.
4x2 � y2 � C ellipses
y2
2� �2x2 � C1
�y dy � ��4x dx
x
y
4
−4
−4 4
dydx
��4x
y42.
y �32
� Ce�2x
2y � 3 � C2e�2x
2y � 3 � C2e�2x
ln2y � 3 � �2x � 2C1
12
ln2y � 3 � �x � C1
� dy2y � 3
� ��dx
4x
y dydx
� 3 � 2y
43.
(a)
(b)
(c)
(d)
(e)dPdt
� 0.55P�1 �P
7200
t ��10.55
ln� 144 � 6.88 yrs.
e�0.55t �1
44
1 � 44e�0.55t � 2
3600 �7200
1 � 44e�0.55t
P�0� �7200
1 � 44� 160
L � 7200
k � 0.55
P�t� �7200
1 � 44e�0.55t44.
(a)
(b)
(c)
(d)
(e)dPdt
� 0.15P�1 �P
4800
t � �1
0.15 ln� 1
14 � 17.59 yrs.
14e�0.15t � 1
2400 �4800
1 � 14e�0.15t
P�0� �4800
1 � 14� 320
L � 4800
k � 0.15
P�t� �4800
1 � 14e�0.15t
580 Chapter 6 Differential Equations
45. (a)
y �20,400
1 � 16e�0.553t
k � �ln 2340
� ln 4023
� 0.553
16e�k �465
y�1� � 2000 �20,400
1 � 16e�k
y�0� � 1200 �20,4001 � b
⇒ b � 16
y �20,400
1 � be�kt
y�1� � 2000y�0� � 1200,L � 20,400,
46.
Use Euler’s Method with
Euler’s Method gives trout.y�8� � 16,170
h � 1.
y�0� � 1200dydt
� 0.553y�1 �y
20,400,
t 0 2 4 6 8
Exact 1200 3241 7414 12,915 17,117
Euler 1200 2743 5853 10,869 16,170
47.
Since when we have Thus, S � L�1 � e�kt�.C � �L.
0 � L � C ⇒t � 0,S � 0
S � L � Ce�kt
L � S � e�kt�C1
�lnL � S � kt � C1
� dS
L � S� � k dt
dSdt
� k�L � S�
48. The general solution is
(a) Since and when we have thefollowing.
Thus, the particular solution is S � 100�1 � e�0.1438t�.
k � �12 ln 34 � 0.1438
�2k � ln 34
e�2k �34
14 � 1 � e�2k
25 � 100�1 � e�2k�
t � 2,S � 25L � 100
S � L�1 � e�kt�.
(b) Since and when we have the following.
Thus, the particular solution is S � 500�1 � e�0.1054t�.
k � �ln 910 � ln 10
9 � 0.1054
�k � ln 910
e�k �910
110 � 1 � e�k
50 � 500�1 � e�k�
t � 1,S � 50L � 500
49. The differential equation is given by the following.
P �CLeLkn
1 � CeLkn �CL
e�Lkn � C
P
L � P� CeLkn
1L
�lnP � lnL � P� � kn � C1
� 1
P�L � P� dP � � k dn
dPdn
� kP�L � P�
(b)
(c) 10,000 �20,400
1 � 16e�0.553t ⇒ t � 4.94 yrs.
y�8� � 17,118 trout
Review Exercises for Chapter 6 581
50. The general solution is
(a) Since and when andwhen we have the following.
Therefore,
P
−1
1
1 2 3 4t
P �1
1 � e�0.4337n.
0.85 �1
1 � e�4k ⇒ k � �14
ln 3
17� 0.4337
0.50 �C
C � 1 ⇒ C � 1
n � 4,P � 0.85n � 0,P � 0.50L � 1
P �CL
C � e�Lkn.
(b) Since and when and when we have the following.
Therefore,
2015105
0.6
0.8
0.4
0.2
P
t
P �4
5 � 11e�0.1887n.
0.60 ��5�11��0.80�
�5�11� � e�8k ⇒ k � �18
ln 5
33� 0.2359
0.25 �0.80CC � 1
⇒ C �5
11
n � 10,P � 0.60n � 0,P � 0.25L � 0.80
(b)
y �23
e x�2 �53
e�x �13
�2e x�2 � 5e�x�
y�0� � �1 �23
� C ⇒ C � �53
y �23
e x�2 � Ce�x
�23
e�3�2�x � C
ye x��e x�2e x dx � �e�3�2�x dx
y� � y � ex�2, Integrating factor: e�dx � ex
y� � e x�2 � y51. (a)
−1 1
1
2
3
4
2 3 4−2−3−4
y
x
52. (a)
x
y
−4 −2 2 4
−2
−4
2
4
(b)
y �15
�2 sin x � cos x� �215
e�2x
ye2x �15
e2x�2 sin x � cos x� �215
⇒ 4 ��15
� C ⇒ C �215
y�0� � 4 ⇒ 4 �15
�0 � 1� � C
ye2x �15
e2x �2 sin x � cos x� � C
�ye2x� � � e2x sin x dx
e2xy� � 2e2xy � e2x sin x
y� � 2y � sin x, Integrating factor: e� 2 dx � e2x
(c)
−9
−6
9
6
(c) 4
−4
−6 6
582 Chapter 6 Differential Equations
(b)
Integrating factor:
⇒ C �1 � cos 1
sin 1� 1.83
y�1� � 1 ⇒ 1 � �cos 1 � C sin 1
y � �cos x � C sin x
y csc x � �csc2 x dx � �cot x � C
�y csc x�� � csc2 x
csc x y� � csc x cot x y � csc2 x
e��cot x dx � e�lnsin x � csc x
dydx
� �cot x�y � csc x
dydx
� csc x � y cot x53. (a)
(c)
−4.5
−3
4.5
3
x−3 3
−3
3
y
54. (a)
(c)
−4
−4
4
4
x−2 2
−2
2
4
y
(1, 2)(b)
Integrating factor:
y � x csc x � C csc x
⇒ C � 2 sin 1 � 1 � 0.683
y�1� � 2 ⇒ 2 sin 1 � 1 � C
y sin x � x � C
�y sin x�� � 1
sin x y� � cos x y � 1
e� cot x dx � e lnsin x � sin x
dydx
� �cot x�y � csc x
dydx
� csc x � y cot x
57.
�14
xe x�4 � Ce x�4
� e�1�4�x�14
x � C
y �1
e��1�4�x � 14
ex�4e��1�4�x dx
u�x� � e���1�4� dx � e��1�4�x
Q�x� �14
e x�4P�x� � �14
,
y� �14
y �14
ex�4
4y� � ex�y � y55.
� �8 � Cex
� ex��8e�x � C�
y �1
e�x�8e�x dx
u�x� � e��dx � e�x
Q�x� � 8P�x� � �1,
y� � y � 8 56.
�13
e�x � Ce�4x
� e�4x�13
e3x � C
y �1
e4x� e�xe4x dx
u�x� � e�4 dx � e4x
Q�x� � e�xP�x� � 4,
y� � 4y � e�x
exy� � 4exy � 1
Review Exercises for Chapter 6 583
59.
�1
x � 2�x � C�
y �1
x � 2 �� 1
x � 2�x � 2� dx
u�x� � e��1�x�2� dx � elnx�2 � x � 2
Q�x� �1
x � 2P�x� �
1x � 2
,
dydx
�1
x � 2y �
1x � 2
�x � 2�y� � y � 1
60.
��x � 3�2
2�
C�x � 3�2
�1
�x � 3�2��x � 3�4
2� C
y �1
�x � 3�2 �2�x � 3��x � 3�2 dx
u�x� � e��2�x�3� dx � e2 ln�x�3� � �x � 3�2
Q�x� � 2�x � 3�P�x� �2
x � 3,
dydx
�2
x � 3y � 2�x � 3�
�x � 3�y� � 2y � 2�x � 3�2 61.
Integrating factor:
y � �1
13�3 sin 2x � 2 cos 2x� � Ce3x
�1
13e�3x��3 sin 2x � 2 cos 2x� � C
ye�3x � �e�3x sin 2x dx
e��3 dx � e�3x
y� � 3y � sin 2x
�3y � sin 2x� dx � dy � 0
62.
Integrating factor:
y � ex�1 � tan x� � C sec x
y cos x � �2ex cos x dx � ex�cos x � sin x� � C
e�� tan x dx � elncos x � cos x
dydx
� �tan x�y � 2ex
dy � �y tan x � 2ex� dx 63.
Integrating factor:
y �1
10e5x � Ce�5x
ye5x � �e10x dx �110
e10x � C
e� 5 dx � e5x
y� � 5y � e5x
58.
� �15
� Ce�5�x
�1
e5�x��15
e5�x � C
y �1
e5�x� 1x2e5�x dx
u�x� � e���5�x2�dx � e5�x
Q�x� �1x2P�x� � �
5x2,
dydx
�5yx2 �
1x2
64.
Integrating factor:
y �bx4
4 � a� Cxa
yx�a � �bx3�x�a� dx �b
4 � ax4�a � C
e���a�x� dx � e�a ln x � x�a
y� � �axy � bx3
65. Bernoulli equation
let
Linear equation
y �1
x � 1 � Cex
y�1 � x � 1 � Cex
z �1
e�x��xe�x dx � ex�xe�x � e�x � C�
u�x� � e��dx � e�x
z� � z � �x,
��y�2�y� � ��y�2�y � �x
z� � �y�2y�.z � y1�2 � y�1,n � 2,
y� � y � xy2,
584 Chapter 6 Differential Equations
66. Bernoulli equation
let
Linear equation
y �1
�1�2� � Cex2 �2
1 � C1ex2
1y
�12
� Cex2
z �1
e�x2���x�e�x2 dx � ex2�1
2e�x2
� C u�x� � e��2x dx � e�x2
z� � 2xz � �x,
��y�2�y� � 2xy��y�2� � �x
z� � �y�2y�.z � y1�2 � y�1,n � 2,
y� � 2xy � xy2, 67. Bernoulli equation
let
Linear equation
1y2 �
23x
� Cx2
z �1
x�2��2x2 �x�2� dx � x2�2x�3
3� C
u�x� � e���2�x� dx � e�2 ln x � x�2
z� �2x
z ��2x2 ,
��2y�3�y� �1x
y��2y�3� ��2x2
z� � �2y�3y�.z � y1�3 � y�2,n � 3,
y� �1x
y �y3
x2,
68.
Bernoulli equation
let
y �1
Cx � x ln x
1y
� �x ln x � Cx
z � x��1x dx � �x�ln x � C�
u�x� � e���1�x� dx �1x
z� �1x
z � �1, Linear equation
�y�2 y� �1x
y��y�2� � y2��y�2�
z� � �y�2 y�.z � y1�2 � y�1,n � 2,
y� �1x
y � y2,
xy� � y � xy2 69. Answers will vary.
Sample answer:
Solution: Let
x3 � C�x2 � y2�
� C�1 �y2
x2 x � C�1 � v2�
lnx � ln1 � v2 � C1
� dxx
� � 2v
1 � v2 dv
�1 � v2� dx � 2 xv dv
�x2 � v2x2� dx � 2x3v dv � 0
�x2 � 3v2x2� dx � 2x�vx��x dv � v dx� � 0
dy � x dv � v dx.y � vx,
�x2 � 3y2� dx � 2xy dy � 0
70. Answers will vary.
72. Answers will vary.
71. Answers will vary.
Sample answer:
y �1x2 �
1x3 �x2� dx �
1x2 �ln x � C �
u�x� � e� �2�x� dx � x2
y� �2x
y �1x3
x3y� � 2x2y � 1
73.
For this linear differential equation, we have and Therefore, the integrating factor is
and the solution is
Since when we have which implies that
A �Pr
� �A0 �Pr ert.
C � A0 � �P�r�t � 0,A � A0
A � ert��Pe�rt dt � ert�Pr
e�rt � C �Pr
� Cert.
u�x� � e��r dt � e�rt
Q�t� � �P.P�t� � �r
dAdt
� rA � �P
Review Exercises for Chapter 6 585
74.
(a)
The balance continues to increase.
(b)
The balance remains at $500,000.
(c)
The balance decreases and is depleted in years.t � �ln 6��0.10 � 17.9
A �60,0000.10
� �500,000 �60,0000.10 e0.10t � 100,000�6 � e0.10t�
2,500,000
00 30
P � 60,000
A �50,0000.10
� �500,000 �50,0000.10 e0.10t � 500,000
2,500,000
00 30
P � 50,000
A �40,0000.10
� �500,000 �40,0000.10 e0.10t � 100,000�4 � e0.10t�
2,500,000
00 30
P � 40,000
r � 0.10A0 � 500,000,
75.
t �ln�10�3�
0.14� 8.6 years
e0.14t �103
0 � 200,000�507
� �5 �507 e0.14t
A �200,000
0.14� �1,000,000 �
200,0000.14 e0.14t 76. (a)
(b) when and
(c)
00 36
80
x(t)
y(t)
�x, y� � �mn
, ab � � 0.4
0.01,
0.30.02 � �40, 15�.
�x, y� � �0, 0�x� � y� � 0
dydt
� �my � nxy � �0.4y � 0.01xy
dxdt
� ax � bxy � 0.3x � 0.02xy
77. (a)
(b) when and
(c)
00 24
50
x(t)
y(t)
�x, y� � �mn
, ab � � 0.6
0.02,
0.40.04 � �30, 10�.
�x, y� � �0, 0�x� � y� � 0
dydt
� �my � nxy � �0.6y � 0.02xy
dxdt
� ax � bxy � 0.4x � 0.04xy 78. (a)
(b) when
(c)
00 6
4
x(t)y(t)
�x, y� � �an � mcbn � cp
, bm � apbn � cp � � 1
1�2,
1�21�2 � �2, 1�.
�x, y� � �ab
, 0 � �3, 0�,
�x, y� � �0, mn � �0, 2�,
�x, y� � �0, 0�,x� � y� � 0
dydt
� my � ny2 � pxy � 2y � y2 � 0.5xy
dxdt
� ax � bx2 � cxy � 3x � x2 � xy
586 Chapter 6 Differential Equations
79. (a)
(b) when
(c)
One species, becomes extinct.x,
00 4
15
x(t)
y(t)
�x, y� � �an � mcbn � cp
, bm � apbn � cp � � ��38
�12,
�26�12� � �19
6,
136 �.
�x, y� � �ab
, 0� � �152
, 0�,
�x, y� � �0, mn � � �0,
172 �,
�x, y� � �0, 0�,x� � y� � 0
dydt
� my � ny2 � pxy � 17y � 2y2 � 4xy
dxdt
� ax � bx2 � cxy � 15x � 2x2 � 4xy
2. Since
When Therefore,
When Therefore, and Thus,
When t � 5, y � 52e5 ln�7�13� � 20 � 22.35�.
y � 52e�ln�7�13��t � 20.k � ln 713.ek �
2852 �
713,48 � 52ek � 20,t � 1, y � 48.
C � 52.t � 0, y � 72.
y � Cekt � 20.
lny � 20 � k t � C
1y � 20
dy � k dt
dydt
� k�y � 20�,
Problem Solving for Chapter 6
1. (a)
Hence,
For T � 100, limt→T �
y � �.
y �1
�1 � 0.01t�100.
y�0� � 1: 1 �1
C100 ⇒ C � 1
y �1
�C � 0.01t�100
y0.01 �1
C � 0.01t
1
y0.01 � �0.01t � C
y�0.01
�0.01� t � C1
y�1.01 dy � dt
dydt
� y1.01 (b)
Hence,
For t → 1y �
0 �k, y →�.
y �1
� 1y �
0
� �kt�1��.
y�0� � y0 �1
C1�� ⇒ C1�� �
1y0
⇒ C � � 1y0�
�
y �1
�C � �kt�1��
y�� � ��kt � C
y��
��� kt � C1
y��1��� dy � k dt
Problem Solving for Chapter 6 587
3. (a)
Also, when And,when
Particular solution:
(c)
00
10
125
S �100
1 � 9eln�4�9�t �100
1 � 9e�0.8109t
t � 1 ⇒ k � �ln 49.S � 20t � 0 ⇒ C � 9.S � 10L � 100.
� k1S�L � S�, where k1 �kL
.
� �kL�
L1 � Ce�kt � �L �
L1 � Ce�kt�
� �kL�
L1 � Ce�kt �
C Le�kt
1 � Ce�kt
�LC ke�kt
�1 � Ce�kt �2
dSdt
� �L�1 � Ce�kt ��2��Cke�kt �
S �L
1 � Ce�kt is a solution because
dSdt
� k1S�L � S� (b)
Choosing we have:
(This is the inflection point.)
(d)
(e) Sales will decrease toward the line S � L.
t1 2 3 4
140
120
100
80
60
40
20
S
t � 2.7 months
ln�1�9�ln�4�9� � t
2 � 1 � 9eln�4�9�t
50 �100
1 � 9eln�4�9�t
S � 50,
� 0 when S � 50 or dSdt
� 0.
� k1�100 � 2S� dSdt
d2Sdt2 � k1�S��dS
dt � � �100 � S� dSdt �
dSdt
� k1S�100 � S�
4. (a)
Using the modified Euler Method, you obtain:
0 1.0
0.1 0.91
0.2 0.83805
1.0 0.73708
��
y x
h � 0.1y�0� � 1,y� � x � y, (b)
The modified Euler Method is more accurate.
1.0
0.50 1.0
[x y ][ ][0 1 ][ ][.1 .9100000000][ ][.2 .8380500000][ ][.3 .7824352500][ ][.4 .7416039013][ ][.5 .7141515307][ ][.6 .6988071353][ ][.7 .6944204575][ ][.8 .6999505140][ ][.9 .7144552152][ ][1.0 .7370819698]
[ ][.1 .9 ][ ][.2 .82 ][ ][.3 .758 ][ ][.4 .7122 ][ ][.5 .68098 ][ ][.6 .662882 ][ ][.7 .6565938 ][ ][.8 .66093442 ][ ][.9 .674840978 ][ ][1.0 .6973568802]
588 Chapter 6 Differential Equations
6.
Area of cross section
When and
The tank is completely drained when t � 1481.45 seconds � 24 minutes and 41 seconds.h � 0 ⇒
C �63�2
5��504� � �1481.45.t � 0h � 6,
h3�2
5�36h � 720� � t � C
365
h5�2 � 144h3�2 � t � C
�18h3�2 � 216h1�2� dh � dt
�12h � h2� dhdt
� �118
h1�2
�12h � h2� dhdt
� �1
144 64h
h
6 ft
x2 + (y − 6)2 = 36
x
y A�h� dhdt
� �k 2gh
A�h� � �12h � h2�,
x2 � 36 � �y � 6�2 � 12y � y2
x2 � �y � 6�2 � 36, Equation of tank
g � 32
k � � 112�
2
5. Let
Finally,
The graph of the logistics function is just a shift of the graph of the hyperbolic tangent.
�L
1 � be�kt .
�L2
2
1 � be�kt
12
L�1 � tanh�12
k�t �ln b
k ��� �L2
�1 � tanh u�
e�2u � e�k�t � �ln b�k�� � eln be�kt � be�kt
1 � tanh u � 1 �eu � e�u
eu � e�u �2
1 � e�2u
u �12
k�t �ln b
k �.
Problem Solving for Chapter 6 589
7. (a)
Hence,
At
Hence,
� 9809.1 seconds, (2 hr, 43 min, 29 sec)
h � 0 ⇒ t �6 2
0.000865
2 h � �0.000865t � 6 2.
6 2 � 4 3
1800� C � 0.000865
2 12 � �1800 C � 6 2
h � 12:t � 30�60� � 1800,
2 h � �Ct � 6 2.
�at t � 0, h � 18� 2 18 � C1
2 h � �Ct � C1
C �8k
r2h�1�2 dh ��8kr2 dt � �C dt,
r2 dhdt
� �k 64h
A�h� dhdt
� �k 2gh
9. Let the radio receiver be located at The tangent line to joins and
—CONTINUED—
xx
x0
−1 1
2
1
y x x= − 2
Transmitter
Radio
( 1, 1)−
y
�x0, 0�.��1, 1�y � x � x2�x0, 0�.
8.
h � 0 ⇒ t � 4 5�288� � 2575.95 sec
2 h ��t288
� 4 5
2 20 � C � 4 5h � 20:
2 h ��t288
� C
h�1�2 dh � �1288
dt
64 dhdt
��
368 h
A�h� dhdt
� �k 2gh
(b)
h
r
18 ft
⇒ h � 7.21 feet
t � 3600 sec ⇒ 2 h � �0.000865�3600� � 6 2
590 Chapter 6 Differential Equations
10.
(a)
(b)
(c) As and s → 184.21.Ce�0.019t → 0,t →�,
00
400
200
s � 184.21 � Ce�0.019t
0.019s � 3.5 � C3e�0.019t
3.5 � 0.019s � C3e�0.019t
ln3.5 � 0.019s � �0.019t � C2
1
0.019 ln3.5 � 0.019s � �t � C1
�ds3.5 � 0.019s
� � dt
dsdt
� 3.5 � 0.019s 11. (a)
Since when it follows that andthe function is
(b) Finally, as we have
limt→�
C � limt→�
C0e�Rt�V � 0.
t →�,
C � C0e�Rt�V.K � C0t � 0,C � C0
C � Ke�Rt�V
lnC � �RV
t � K1
dCC
� �RV
dt
9. —CONTINUED—
(a) If is the point of tangency on then
(c)
There is a vertical asymptote at which is theheight of the mountain.
h �14,
0.25 3
−2
10
x0 �4 3 � 6
6 � 3 3� 1.155.
� 6 � 3 3 � x0�6 � 3 3 � 3 � �1 � x0��6 � 3 3 �
Then 1 � 0
�1 � x0�
1 � 3 3 � 5
�1 � 1 � 3�
6 � 3 3
� 3
y � x � x2 � 3 3 � 5.
x � ��2 ± 4 � 82 � � �1 � 3
x2 � 2x � 2 � 0
x � 2x2 � 1 � 2x � x � x2 � 1
1 � 2x �y � 1x � 1
�x � x2 � 1
x � 1
y � x � x2,�x, y� (b) Now let the transmitter be located at
x0 �h 2 � h
2h � 4 � 3 2 � h� 1.
x0 � 1
h�
2 � h
2h � 4 � 3 2 � h
�2h � 4 � 3 2 � h
� 2 � h
Then, h � 0
�1 � x0�
h � �3 2 � h � h � 4��1 � ��1 � 2 � h �
� 3 2 � h � h � 4
y � x � x2
� �1 � 2 � h
x ���2 ± 4 � 4�h � 1� �
2
x2 � 2x � h � 1 � 0
x � 2x2 � 1 � 2x � x � x2 � h
1 � 2x �y � hx � 1
�x � x2 � h
x � 1
��1, h�.
Problem Solving for Chapter 6 591
12. From Exercise 11, we have
(a) For and we have
(b) For and we have
00
4
0.8
C � 0.6e�0.75t.C0 � 0.6,R � 1.5,V � 2,
00
4
0.8
C � 0.6e�0.25t.C0 � 0.6,R � 0.5,V � 2,
C � C0e�Rt�V. 13. (a)
Since when it follows that and
we have
(b) As the limit of C is Q�R.t →�,
C �QR
�1 � e�Rt�V�.
K � Qt � 0,C � 0
�1R
�Q � Ke�Rt�V�
C �1R
�Q � e�R��t�V��K1��
Q � RC � e�R��t�V��K1�
�1R
lnQ � RC �tV
� K1
1Q � RC
dC � 1V
dt