75
CHAPTER 6 Differential 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

CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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Page 1: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 2: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 3: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 4: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 5: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 6: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 7: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 8: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 9: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 10: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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�

Page 11: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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.

Page 12: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 13: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 14: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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)

Page 15: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 16: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 17: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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�

Page 18: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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�

Page 19: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 20: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 21: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 22: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 23: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 24: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 25: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 26: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 27: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 28: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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�

Page 29: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 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

Page 30: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 31: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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�

Page 32: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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.

Page 33: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 34: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 35: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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 →�,

Page 36: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 37: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 38: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 39: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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�,

Page 40: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 41: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 42: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 43: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 44: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 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

Page 45: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 46: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 47: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 48: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 49: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 50: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 51: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 52: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 53: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 54: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 55: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 56: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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�

Page 57: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 58: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 59: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 60: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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,

Page 61: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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)

Page 62: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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)

Page 63: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 64: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 65: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 66: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 67: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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,

Page 68: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 69: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 70: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 71: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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]

Page 72: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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 �.

Page 73: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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

Page 74: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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�.

Page 75: CHAPTER 6 Differential Equations - oakwood.k12.oh.uslane_jay/Site/AP_Calculus_files/csg_ch06.pdfCHAPTER 6 Differential Equations ... Particular solutions: Point Circles x 12 1 2 y

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