33
7 DIFFERENTIAL EQUATIONS 7.1 Modeling with Differential Equations 1. = 2 3 + 2 0 = 2 3 2 2 . To show that is a solution of the differential equation, we will substitute the expressions for and 0 in the left-hand side of the equation and show that the left-hand side is equal to the right hand side. LHS = 0 +2 = 2 3 2 2 +2 2 3 + 2 = 2 3 2 2 + 4 3 +2 2 = 6 3 =2 = RHS 2. = cos = ( sin ) + cos (1) 1= sin cos 1. LHS = = ( sin cos 1) = 2 sin cos = 2 sin + = RHS, so is a solution of the differential equation. Also ()= cos = (1) = =0, so the initial condition is satised. 3. (a) = 0 = 00 = 2 . Substituting these expressions into the differential equation 2 00 + 0 =0, we get 2 2 + =0 (2 2 + 1) =0 (2 1)( + 1) = 0 [since is never zero] = 1 2 or 1. (b) Let 1 = 1 2 and 2 = 1, so we need to show that every member of the family of functions = 2 + is a solution of the differential equation 2 00 + 0 =0. = 2 + 0 = 1 2 2 00 = 1 4 2 + . LHS = 2 00 + 0 =2 1 4 2 + + 1 2 2 ( 2 + ) = 1 2 2 +2 + 1 2 2 2 = 1 2 + 1 2 2 + (2 ) = 0 = RHS 4. (a) = cos 0 = sin 00 = 2 cos . Substituting these expressions into the differential equation 4 00 = 25, we get 4( 2 cos )= 25(cos ) (25 4 2 ) cos =0 [for all ] 25 4 2 =0 2 = 25 4 = ± 5 2 . (b) = sin + cos 0 = cos sin 00 = 2 sin 2 cos . The given differential equation 4 00 = 25 is equivalent to 4 00 + 25 =0. Thus, LHS = 4 00 + 25 = 4( 2 sin 2 cos ) + 25( sin + cos ) = 4 2 sin 4 2 cos + 25 sin + 25 cos = (25 4 2 ) sin + (25 4 2 ) cos =0 since 2 = 25 4 . 593 NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.

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Page 1: 7 DIFFERENTIAL EQUATIONS - Pepperdine Universitydt.pepperdine.edu/courses/151sp17/7.1-7.4.pdf7 DIFFERENTIAL EQUATIONS 7.1 ModelingwithDifferentialEquations 1. =2 3 0 +0 =2 0 2 3 0

7 DIFFERENTIAL EQUATIONS7.1 Modeling with Differential Equations

1. = 230� + 0�2� � 0 = 2

30� � 20�2�. To show that is a solution of the differential equation, we will substitute the

expressions for and 0 in the left-hand side of the equation and show that the left-hand side is equal to the right hand side.

LHS= 0 + 2 = 230� � 20�2� + 2� 2

30� + 0�2�

�= 2

30� � 20�2� + 4

30� + 20�2�

= 630� = 20� = RHS

2. = �� cos �� � � ���� = ��(� sin �) + cos �(�1)� 1 = � sin �� cos �� 1.

LHS= ��

��= �(� sin �� cos �� 1) = �2 sin �� � cos �� �

= �2 sin �+ = RHS,

so is a solution of the differential equation. Also (�) = �� cos� � � = ��(�1)� � = � � � = 0, so the initialcondition is satis�ed.

3. (a) = 0�� � 0 = �0�� � 00 = �20��. Substituting these expressions into the differential equation

200 + 0 � = 0, we get 2�20�� + �0�� � 0�� = 0 � (2�2 + � � 1)0�� = 0 �(2� � 1)(� + 1) = 0 [since 0�� is never zero] � � = 1

2 or �1.

(b) Let �1 = 12

and �2 = �1, so we need to show that every member of the family of functions = �0��2 + 0�� is a

solution of the differential equation 200 + 0 � = 0.

= �0��2 + 0�� � 0 = 12�0��2 � 0�� � 00 = 1

4�0��2 + 0��.

LHS = 200 + 0 � = 2�14�0��2 + 0��

�+�12�0��2 � 0��

�� (�0��2 + 0��)

= 12�0

��2 + 20�� + 12�0

��2 � 0�� � �0��2 � 0��

=�12�+ 1

2�� �

�0��2 + (2� � )0��

= 0 = RHS

4. (a) = cos �� � 0 = �� sin �� � 00 = ��2 cos ��. Substituting these expressions into the differential equation

400 = �25, we get 4(��2 cos ��) = �25(cos ��) � (25� 4�2) cos �� = 0 [for all �] � 25� 4�2 = 0 ��2 = 25

4� � = ± 5

2.

(b) = � sin ��+� cos �� � 0 = �� cos ����� sin�� � 00 = ���2 sin �����2 cos ��.

The given differential equation 400 = �25 is equivalent to 400 + 25 = 0. Thus,

LHS = 400 + 25 = 4(���2 sin�����2 cos ��) + 25(� sin ��+� cos ��)

= �4��2 sin ��� 4��2 cos ��+ 25� sin ��+ 25� cos ��= (25� 4�2)� sin ��+ (25� 4�2)� cos ��= 0 since �2 = 25

4.

593

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594 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

5. (a) = sin� � 0 = cos� � 00 = � sin�.

LHS = 00 + = � sin�+ sin� = 0 6= sin�, so = sin� is not a solution of the differential equation.

(b) = cos� � 0 = � sin� � 00 = � cos�.

LHS = 00 + = � cos�+ cos� = 0 6= sin�, so = cos� is not a solution of the differential equation.

(c) = 12� sin� � 0 = 1

2(� cos�+ sin�) � 00 = 1

2(�� sin�+ cos�+ cos�).

LHS = 00 + = 12(�� sin�+ 2cos�) + 1

2� sin� = cos� 6= sin�, so = 1

2� sin� is not a solution of the

differential equation.

(d) = �12� cos� � 0 = � 1

2 (�� sin�+ cos�) � 00 = � 12 (�� cos�� sin�� sin�).

LHS = 00 + = � 12(�� cos�� 2 sin�) + �� 1

2� cos�

�= sin� = RHS, so = � 1

2� cos� is a solution of the

differential equation.

6. (a) = ln�+�

�� 0 =

� · (1��)� (ln�+�)

�2=1� ln�� �

�2.

LHS = �20 + � = �2 · 1� ln���

�2+ � · ln�+ �

= 1� ln�� � + ln�+� = 1 = RHS, so is a solution of the differential equation.

(b) A few notes about the graph of = (ln�+�)��:

(1) There is a vertical asymptote of � = 0.

(2) There is a horizontal asymptote of = 0.

(3) = 0 � ln�+ � = 0 � � = 0�- ,so there is an �-intercept at 0�- .

(4) 0 = 0 � ln� = 1� � � � = 01�- ,so there is a local maximum at � = 01�- .

(c) (1) = 2 � 2 =ln 1 +�

1� 2 = �, so the solution is = ln�+ 2

�[shown in part (b)].

(d) (2) = 1 � 1 =ln 2 +�

2� 2 + ln 2 +� � � = 2� ln 2, so the solution is = ln�+ 2� ln 2

[shown in part (b)].

7. (a) Since the derivative 0 = �2 is always negative (or 0 if = 0), the function must be decreasing (or equal to 0) on any

interval on which it is de�ned.

(b) = 1

�+ �� 0 = � 1

(�+ �)2. LHS = 0 = � 1

(�+�)2= �

�1

�+�

�2= �2 = RHS

(c) = 0 is a solution of 0 = �2 that is not a member of the family in part (b).

(d) If (�) = 1

�+�, then (0) = 1

0 +�=1

�. Since (0) = 0�5, 1

�=1

2� � = 2, so = 1

�+ 2.

8. (a) If � is close to 0, then �3 is close to 0, and hence, 0 is close to 0. Thus, the graph of must have a tangent line that is

nearly horizontal. If � is large, then �3 is large, and the graph of must have a tangent line that is nearly vertical.

(In both cases, we assume reasonable values for .)

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SECTION 7.1 MODELING WITH DIFFERENTIAL EQUATIONS ¤ 595

(b) = (+� �2)�1�2 � 0 = �(+� �2)�3�2. RHS = �3 = �[�+� �2

��1�2]3 = �(+� �2)�3�2 = 0 = LHS

(c) When � is close to 0, 0 is also close to 0.

As � gets larger, so does |0|.

(d) (0) = (+� 0)�1�2 = 1��+ and (0) = 2 �

�+ = 1

2� + = 1

4, so =

�14� �2

��1�2.

9. (a) ����

= 1�2�

�1� �

4200

�. Now ��

�� 0 � 1� �

4200 0 [assuming that � 0] � �

4200� 1 �

� � 4200 � the population is increasing for 0 � � � 4200.

(b) ����

� 0 � � 4200

(c) ����

= 0 � � = 4200 or � = 0

10. (a) = � � 0 = 0, so �

��= 4 � 63 + 52 � 0 = �4 � 6�3 + 5�2 � �2

��2 � 6� + 5� = 0 �

�2(� � 1)(� � 5) = 0 � � = 0, 1, or 5

(b) is increasing � �

�� 0 � 2( � 1)( � 5) 0 � � (��� 0) (0� 1) (5��)

(c) is decreasing � �

��� 0 � � (1� 5)

11. (a) This function is increasing and also decreasing. But ���� = 0�( � 1)2 � 0 for all �, implying that the graph of the

solution of the differential equation cannot be decreasing on any interval.

(b) When = 1, ���� = 0, but the graph does not have a horizontal tangent line.

12. The graph for this exercise is shown in the �gure at the right.

A. 0 = 1+ � 1 for points in the �rst quadrant, but we can

see that 0 � 0 for some points in the �rst quadrant.

B. 0 = �2� = 0 when � = 0, but we can see that 0 0 for � = 0.

Thus, equations A and B are incorrect, so the correct equation is C.

C. 0 = 1� 2� seems reasonable since:(1) When � = 0, 0 could be 1.

(2) When � � 0, 0 could be greater than 1.

(3) Solving 0 = 1� 2� for gives us = 1� 0

2�. If 0 takes on small negative values, then as � �, 0+,

as shown in the �gure.

13. (a) 0 = 1 + �2 + 2 � 1 and 0 � as � �. The only curve satisfying these conditions is labeled III.

(b) 0 = �0��2��2 0 if � 0 and 0 � 0 if � � 0. The only curve with negative tangent slopes when � � 0 and positivetangent slopes when � 0 is labeled I.

NOT FOR SALE SECTION 7 1N MODELING WITH DIFF

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596 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

(c) 0 = 1

1 + 0�2+�2 0 and 0 0 as � �. The only curve satisfying these conditions is labeled IV.

(d) 0 = sin(�) cos(�) = 0 if = 0, which is the solution graph labeled II.

14. (a) The coffee cools most quickly as soon as it is removed from the heat source. The rate of cooling decreases toward 0 since

the coffee approaches room temperature.

(b) ���= �( �.), where � is a proportionality constant, is the

temperature of the coffee, and . is the room temperature. The initial

condition is (0) = 95�C. The answer and the model support each

other because as approaches ., ���� approaches 0, so the model

seems appropriate.

(c)

15. (a) � increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills associated with

learning a skill. As � increases, we would expect ����� to remain positive, but decrease. This is because as time

progresses, the only points left to learn are the more dif�cult ones.

(b) ����

= �(- � � ) is always positive, so the level of performance �

is increasing. As � gets close to - , ����� gets close to 0; that is,

the performance levels off, as explained in part (a).

(c)

7.2 Direction Fields and Euler's Method

1. (a) (b) It appears that the constant functions = 0�5 and = 1�5 are

equilibrium solutions. Note that these two values of satisfy the

given differential equation 0 = � cos�.

2. (a) (b) It appears that the constant functions = 0, = 2, and = 4 are

equilibrium solutions. Note that these three values of satisfy the

given differential equation 0 = tan�12�

�.

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SECTION 7.2 DIRECTION FIELDS AND EULER’S METHOD ¤ 597

3. 0 = 2� . The slopes at each point are independent of �, so the slopes are the same along each line parallel to the �-axis.

Thus, III is the direction �eld for this equation. Note that for = 2, 0 = 0.

4. 0 = �(2� ) = 0 on the lines � = 0 and = 2. Direction �eld I satis�es these conditions.

5. 0 = �+ � 1 = 0 on the line = ��+1. Direction �eld IV satis�es this condition. Notice also that on the line = �� we

have 0 = �1, which is true in IV.

6. 0 = sin� sin = 0 on the lines � = 0 and = 0, and 0 0 for 0 � � � �, 0 � � �. Direction �eld II satis�es these

conditions.

7. (a) (0) = 1

(b) (0) = 2

(c) (0) = �1

8. (a) (0) = �1(b) (0) = 0

(c) (0) = 1

9.� 0 = 1

2

0 0 0

0 1 0�5

0 2 1

0 �3 �1�50 �2 �1

Note that for = 0, 0 = 0. The three solution curves sketched go

through (0� 0), (0� 1), and (0��1).

10.� 0 = �� + 1

�1 0 0

�1 �1 1

0 0 1

0 1 0

0 2 �10 �1 2

0 �2 3

1 0 2

1 1 1

Note that 0 = 0 for = �+ 1 and that 0 = 1 for = �. For any

constant value of �, 0 decreases as increases and 0 increases as

decreases. The three solution curves sketched go through (0� 0),

(0� 1), and (0��1).

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598 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

11.� 0 = � 2��2 �2 2

�2 2 6

2 2 �22 �2 �6

Note that 0 = 0 for any point on the line = 2�. The slopes are

positive to the left of the line and negative to the right of the line. The

solution curve in the graph passes through (1� 0).

12.� 0 = � � �2

2 3 2

�2 �3 2

±2 0 �40 0 0

2 2 0

0 = � � �2 = �( � �), so 0 = 0 for � = 0 and = �. The

slopes are positive only in the regions in quadrants I and III that are

bounded by � = 0 and = �. The solution curve in the graph passes

through (0� 1).

13.� 0 = + �

0 ±2 ±21 ±2 ±4

�3 ±2 �4

Note that 0 = (�+ 1) = 0 for any point on = 0 or on � = �1.

The slopes are positive when the factors and �+ 1 have the same

sign and negative when they have opposite signs. The solution curve

in the graph passes through (0� 1).

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SECTION 7.2 DIRECTION FIELDS AND EULER’S METHOD ¤ 599

14.� 0 = �+ 2

�2 ±1 �1�2 ±2 2

2 ±1 3

0 ±2 4

0 0 0

Note that 0 = �+ 2 = 0 only on the parabola � = �2. The

slopes are positive “outside” � = �2 and negative “inside”

� = �2. The solution curve in the graph passes through (0� 0).

15. In Maple, we can use either directionfield (in Maple’s share library) or

DEtools[DEplot] to plot the direction �eld. To plot the solution, we can

either use the initial-value option in directionfield, or actually solve the

equation.

In Mathematica, we use PlotVectorField for the direction �eld, and the

Plot[Evaluate[� � �]] construction to plot the solution, which is

= 2arctan�0�

3�3 · tan 12

�.

In Derive, use Direction_Field (in utility �le ODE_APPR) to plot the direction �eld. Then use

DSOLVE1(-xˆ2*SIN(y),1,x,y,0,1) (in utility �le ODE1) to solve the equation. Simplify each result.

16. See Exercise 15 for speci�c CAS directions. The exact solution is

=2�3� 02�

2�

02�2 + 3

17. The direction �eld is for the differential equation 0 = 3 � 4.

$ = lim���

(�) exists for�2 � + � 2;

$ = ±2 for + = ±2 and $ = 0 for �2 � + � 2.

For other values of +, $ does not exist.

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600 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

18. Note that when �() = 0 on the graph in the text, we have 0 = �() = 0; so we

get horizontal segments at = ±1, ±2. We get segments with negative slopes only

for 1 � || � 2. All other segments have positive slope. For the limiting behavior

of solutions:

• If (0) 2, then lim���

=� and lim����

= 2.

• If 1 � (0) � 2, then lim���

= 1 and lim����

= 2.

• If �1 � (0) � 1, then lim���

= 1 and lim����

= �1.

• If �2 � (0) � �1, then lim���

= �2 and lim����

= �1.

• If � �2, then lim���

= �2 and lim����

= ��.

19. (a) 0 = ! (�� ) = and (0) = 1 � �0 = 0, 0 = 1.

(i) � = 0�4 and 1 = 0 + �! (�0� 0) � 1 = 1 + 0�4 · 1 = 1�4. �1 = �0 + � = 0 + 0�4 = 0�4,

so 1 = (0�4) = 1�4.

(ii) � = 0�2 � �1 = 0�2 and �2 = 0�4, so we need to �nd 2.

1 = 0 + �! (�0� 0) = 1 + 0�20 = 1 + 0�2 · 1 = 1�2,

2 = 1 + �! (�1� 1) = 1�2 + 0�21 = 1�2 + 0�2 · 1�2 = 1�44.

(iii) � = 0�1 � �4 = 0�4, so we need to �nd 4. 1 = 0 + �! (�0� 0) = 1 + 0�10 = 1 + 0�1 · 1 = 1�1,

2 = 1 + �! (�1� 1) = 1�1 + 0�11 = 1�1 + 0�1 · 1�1 = 1�21,

3 = 2 + �! (�2� 2) = 1�21 + 0�12 = 1�21 + 0�1 · 1�21 = 1�331,

4 = 3 + �! (�3� 3) = 1�331 + 0�13 = 1�331 + 0�1 · 1�331 = 1�4641.

(b) We see that the estimates are underestimates since

they are all below the graph of = 0�.

(c) (i) For � = 0�4: (exact value)� (approximate value) = 004 � 1�4 0�0918(ii) For � = 0�2: (exact value)� (approximate value) = 004 � 1�44 0�0518

(iii) For � = 0�1: (exact value)� (approximate value) = 004 � 1�4641 0�0277

Each time the step size is halved, the error estimate also appears to be halved (approximately).

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SECTION 7.2 DIRECTION FIELDS AND EULER’S METHOD ¤ 601

20. As � increases, the slopes decrease and all of the

estimates are above the true values. Thus, all of

the estimates are overestimates.

21. � = 0�5, �0 = 1, 0 = 0, and ! (�� ) = � 2�.

Note that �1 = �0 + � = 1 + 0�5 = 1�5, �2 = 2, and �3 = 2�5.

1 = 0 + �! (�0� 0) = 0 + 0�5! (1� 0) = 0�5[0� 2(1)] = �1.

2 = 1 + �! (�1� 1) = �1 + 0�5! (1�5��1) = �1 + 0�5[�1� 2(1�5)] = �3.

3 = 2 + �! (�2� 2) = �3 + 0�5! (2��3) = �3 + 0�5[�3� 2(2)] = �6�5.

4 = 3 + �! (�3� 3) = �6�5 + 0�5! (2�5��6�5) = �6�5 + 0�5[�6�5� 2(2�5)] = �12�25.

22. � = 0�2, �0 = 0, 0 = 1, and ! (�� ) = � � �2.

Note that �1 = �0 + � = 0 + 0�2 = 0�2, �2 = 0�4, �3 = 0�6, �4 = 0�8, and �5 = 1�0.

1 = 0 + �! (�0� 0) = 1 + 0�2! (0� 1) = 1 + 0�2 (0) = 1.

2 = 1 + �! (�1� 1) = 1 + 0�2! (0�2� 1) = 1 + 0�2 (0�16) = 1�032.

3 = 2 + �! (�2� 2) = 1�032 + 0�2! (0�4� 1�032) = 1�032 + 0�2 (0�2528) = 1�08256.

4 = 3 + �! (�3� 3) = 1�08256 + 0�2! (0�6� 1�08256) = 1�08256 + 0�2 (0�289536) = 1�1404672.

5 = 4 + �! (�4� 4) = 1�1404672 + 0�2! (0�8� 1�1404672) = 1�1404672 + 0�2 (0�27237376) = 1�194941952.

Thus, (1) 1�1949.

23. � = 0�1, �0 = 0, 0 = 1, and ! (�� ) = + �.

Note that �1 = �0 + � = 0 + 0�1 = 0�1, �2 = 0�2, �3 = 0�3, and �4 = 0�4.

1 = 0 + �! (�0� 0) = 1 + 0�1! (0� 1) = 1 + 0�1[1 + (0)(1)] = 1�1.

2 = 1 + �! (�1� 1) = 1�1 + 0�1! (0�1� 1�1) = 1�1 + 0�1[1�1 + (0�1)(1�1)] = 1�221�

3 = 2 + �! (�2� 2) = 1�221 + 0�1! (0�2� 1�221) = 1�221 + 0�1[1�221 + (0�2)(1�221)] = 1�36752.

4 = 3 + �! (�3� 3) = 1�36752 + 0�1! (0�3� 1�36752) = 1�36752 + 0�1[1�36752 + (0�3)(1�36752)]

= 1�5452976.

5 = 4 + �! (�4� 4) = 1�5452976 + 0�1! (0�4� 1�5452976)

= 1�5452976 + 0�1[1�5452976 + (0�4)(1�5452976)] = 1�761639264.

Thus, (0�5) 1�7616.

24. (a) � = 0�2, �0 = 0, 0 = 0, and ! (�� ) = �+ 2.

Note that �1 = �0 + � = 0 + 0�2 = 0�2, and �2 = �1 + � = 0�4.

1 = 0 + �! (�0� 0) = 0 + 0�2! (0� 0) = 0�2 (0) = 0.

2 = 1 + �! (�1� 1) = 0 + 0�2! (0�2� 0) = 0�2 (0�2) = 0�04.

Thus, (0�4) 0�04.

SECTION 7 2ON DIRECTION FIELDS

INSTRUCTOR USE ONLY Thus,Thus, (0(0��4)4) 00��0404..

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602 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

(b) Now �1 = �0 + � = 0 + 0�1 = 0�1, �2 = 0�2, �3 = 0�3, and �4 = 0�4.

1 = 0 + �! (�0� 0) = 0 + 0�1! (0� 0) = 0�1(0) = 0.

2 = 1 + �! (�1� 1) = 0 + 0�1! (0�1� 0) = 0�1(0�1) = 0�01.

3 = 2 + �! (�2� 2) = 0�01 + 0�1! (0�2� 0�01) = 0�01 + 0�1(0�2001) = 0�03001.

4 = 3 + �! (�3� 3) = 0�03001 + 0�1! (0�3� 0�03001) = 0�03001 + 0�1(0�3009006001) = 0�06010006001.

Thus, (0�4) 0�06.

25. (a) ����+ 3�2 = 6�2 � 0 = 6�2 � 3�2. Store this expression in Y1 and use the following simple program to

evaluate (1) for each part, using H = � = 1 and N = 1 for part (i), H = 0�1 and N = 10 for part (ii), and so forth.

� H: 0 X: 3 Y:

For(I, 1, N): Y+ H×Y1 Y: X+H X:

End(loop):

Display Y. [To see all iterations, include this statement in the loop.]

(i) H = 1, N = 1 � (1) = 3

(ii) H = 0�1, N = 10 � (1) 2�3928

(iii) H = 0�01, N = 100 � (1) 2�3701

(iv) H = 0�001� N = 1000 � (1) 2�3681

(b) = 2 + 0��3 � 0 = �3�20��3

LHS = 0 + 3�2 = �3�20��3 + 3�2�2 + 0��3

�= �3�20��3 + 6�2 + 3�20��3 = 6�2 = RHS

(0) = 2 + 0�0 = 2 + 1 = 3

(c) The exact value of (1) is 2 + 0�13= 2 + 0�1.

(i) For � = 1: (exact value)� (approximate value) = 2 + 0�1 � 3 �0�6321(ii) For � = 0�1: (exact value)� (approximate value) = 2 + 0�1 � 2�3928 �0�0249

(iii) For � = 0�01: (exact value)� (approximate value) = 2 + 0�1 � 2�3701 �0�0022

(iv) For � = 0�001: (exact value)� (approximate value) = 2 + 0�1 � 2�3681 �0�0002

In (ii)–(iv), it seems that when the step size is divided by 10, the error estimate is also divided by 10 (approximately).

26. (a) We use the program from the solution to Exercise 25

with Y1 = �3 � 3, H = 0�01, and N = 2�0001

= 200.

With (�0� 0) = (0� 1), we get (2) 1�9000.

(b)

Notice from the graph that (2) 1�9, which serves as

a check on our calculation in part (a).

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SECTION 7.2 DIRECTION FIELDS AND EULER’S METHOD ¤ 603

27. (a) . �/

��+1

�/ = '(�) becomes 5/0 +

1

0�05/ = 60

or /0 + 4/ = 12.

(b) From the graph, it appears that the limiting value of the

charge / is about 3.

(c) If /0 = 0, then 4/ = 12 � / = 3 is an

equilibrium solution.

(d)

(e) /0 + 4/ = 12 � /0 = 12� 4/. Now /(0) = 0, so �0 = 0 and /0 = 0.

/1 = /0 + �! (�0� /0) = 0 + 0�1(12� 4 · 0) = 1�2/2 = /1 + �! (�1� /1) = 1�2 + 0�1(12� 4 · 1�2) = 1�92/3 = /2 + �! (�2� /2) = 1�92 + 0�1(12� 4 · 1�92) = 2�352/4 = /3 + �! (�3� /3) = 2�352 + 0�1(12� 4 · 2�352) = 2�6112/5 = /4 + �! (�4� /4) = 2�6112 + 0�1(12� 4 · 2�6112) = 2�76672

Thus, /5 = /(0�5) 2�77 C.

28. (a) From Exercise 7.1.14, we have ���� = �( �.). We are given that . = 20�C and ���� = �1�C�min when

= 70�C. Thus, �1 = �(70� 20) � � = � 150

and the differential equation becomes ���� = � 150( � 20).

(b) The limiting value of the temperature is 20�C;that is, the temperature of the room.

(c) From part (a), ���� = � 150( � 20). With �0 = 0, 0 = 95, and � = 2 min, we get

1 = 0 + �! (�0� 0) = 95 + 2� 1

50(95� 20) = 92

2 = 1 + �! (�1� 1) = 92 + 2� 1

50 (92� 20)= 89�12

3 = 2 + �! (�2� 2) = 89�12 + 2� 1

50(89�12� 20) = 86�3552

4 = 3 + �! (�3� 3) = 86�3552 + 2� 1

50 (86�3552� 20)= 83�700992

5 = 4 + �! (�4� 4) = 83�700992 + 2� 1

50 (83�700992� 20)= 81�15295232

Thus, (10) 81�15�C.

NOT FOR SALE SECTION 7 2ON DIRECTION FIELDS

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604 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

7.3 Separable Equations

1. ���

= �2 � �

2= ��� [ 6= 0] �

5�2 � =

5��� � ��1 = 1

2�2 + � �

1

= �1

2�2 � � � =

1

� 12�

2 ��=

2

B � �2, where B = �2�. = 0 is also a solution.

2. ���

= �0�� � �

0��= ��� �

50� � =

5��� � 0� = 1

2�2 +� � = ln

�12�

2 + ��

3. (�2 + 1)0 = � � �

��=

�2 + 1� �

=

���

�2 + 1[ 6= 0] �

5�

=

5���

�2 + 1�

ln || = 12ln(�2 + 1) + � [� = �2 + 1, �� = 2���] = ln(�2 + 1)1�2 + ln 0- = ln

�0-��2 + 1

� �

|| = 0-��2 + 1 � = B

��2 + 1, where B = ±0- is a constant. (In our derivation, B was nonzero, but we can

restore the excluded case = 0 by allowing B to be zero.)

4. (2 + �2) 0 = 1 � 2(1 + �)�

��= 1 � 2 � =

1

1 + ��� �

52 � =

51

1 + ��� �

133 = ln |1 + �|+ � � 3 = 3 ln |1 + �|+ 3� � = 3

�3 ln |1 + �|+B, where B = 3�.

5. ( + sin ) 0 = �+ �3 � ( + sin )�

��= �+ �3 �

5( + sin ) � =

5(�+ �3) �� �

12

2 � cos = 12�

2 + 14�

4 +�. We cannot solve explicitly for .

6. ����

=1 +

��

1 +��

��1 +

����� =

�1 +

����� � 4

(1 + �1�2) �� =4(1 + �1�2) �� �

�+ 23�3�2 = � + 2

3�3�2 + �

7. ���=

�0�

�1 + 2

� �1 + 2 � = �0� �� � 4

�1 + 2 � =

4�0� �� � 1

3

�1 + 2

�3�2= �0� � 0� + �

[where the �rst integral is evaluated by substitution and the second by parts] � 1 + 2 = [3(�0� � 0� + �)]2�3 �

= ±�[3(�0� � 0� +�)]2�3 � 1

8. ���=

0� sin2 �

sec ��

0�� =

sin2 �

sec ��� � 4

0�� � =4sin2 � cos � ��. Integrating the left side by parts with

� = , �) = 0�� � and the right side by the substitution � = sin �, we obtain �0�� � 0�� = 13sin3 � + �. We cannot

solve explicitly for .

9. ����= 2 + 2�+ �+ �� � ��

��= (1 + �)(2 + �) �

5��

1 + �=

5(2 + �)�� [� 6= �1] �

ln |1 + �| = 12�2 + 2�+ � � |1 + �| = 0�

2�2+ 2�+- = B0�2�2+ 2�, where B = 0- � 1 + � = ±B0�

2�2+ 2� �

� = �1±B0�2�2+ 2� where B 0. � = �1 is also a solution, so � = �1 +�0�

2�2+ 2�, where � is an arbitrary constant.

NOT FOR SALE NTIAL EQUATIONSTIO

INSTRUCTOR USE ONLY 1±B0 where B 00. � 11 is also a solution, sois also a solution, so � 1 ++�0 , where � is an arbitrary constant.an arbitrary constant.

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SECTION 7.3 SEPARABLE EQUATIONS ¤ 605

10. �8��+ 0�+ ) = 0 � �8

��= �0�0) � 4

0�) �8 = � 40� �� � �0�) = �0� + � � 0�) = 0� � � �

1

0)= 0� � � � 0) =

1

0� � �� 8 = ln

�1

0� ��

�� 8 = � ln�0� ��

11. ���

=�

� � = ��� � 4

� =4��� � 1

22 = 1

2�2 + �. (0) = �3 �

12(�3)2 = 1

2(0)2 +� � � = 9

2, so 1

22 = 1

2�2 + 9

2� 2 = �2 + 9 � = ���2 + 9 since (0) = �3 � 0.

12. ���

=ln�

�, (1) = 2.

5 � =

5ln�

��� � 1

22 = 1

2(ln�)2 +�.

Now (1) = 2 � 12(2)2 = 1

2(ln 1)2 + � � 2 = �, so 1

22 = 1

2(ln�)2 + 2 �

2 = (ln�)2 + 4 � = ±�(ln�)2 + 4. Since (1) = 2, we have =�(ln�)2 + 4.

13. ����=2�+ sec2 �

2�, �(0) = �5.

42��� =

4 �2�+ sec2 �

��� � �2 = �2 + tan �+ �,

where [�(0)]2 = 02 + tan 0 + � � � = (�5)2 = 25. Therefore, �2 = �2 + tan �+ 25, so � = ±��2 + tan �+ 25.

Since �(0) = �5, we must have � = ���2 + tan �+ 25.

14. 0 = � sin�

+ 1, (0) = 1. + 1

��= � sin� �

5 �1 +

1

�� =

5� sin��� �

+ ln || = �� cos�+ sin�+ � [use parts with � = �, �) = sin���]. Now (0) = 1 �1 + 0 = 0 + 0 +� � � = 1, so + ln || = �� cos�+ sin�+ 1. We cannot solve explicitly for .

15. � ln� = �1 +

�3 + 2

�0, (1) = 1.

4� ln��� =

4 � +

�3 + 2

�� � 1

2�2 ln�� 4

12���

[use parts with � = ln�, �) = ���] = 122 + 1

3(3 + 2)3�2 � 1

2�2 ln�� 1

4�2 + � = 1

22 + 1

3(3 + 2)3�2.

Now (1) = 1 � 0 � 14+ � = 1

2+ 1

3(4)3�2 � � = 1

2+ 8

3+ 1

4= 41

12, so

12�2 ln�� 1

4�2 + 41

12= 1

22 + 1

3(3 + 2)3�2. We do not solve explicitly for .

16. ����

=��� � ���

�� =

�� �� � 4

��1�2 �� =4�1�2 �� � 2� 1�2 = 2

3�3�2 +�.

� (1) = 2 � 2�2 = 2

3+ � � � = 2

�2� 2

3, so 2� 1�2 = 2

3�3�2 + 2

�2� 2

3� �

� = 13�3�2 +

�2� 1

3�

� =�13�3�2 +

�2� 1

3

�2.

17. 0 tan� = �+ , 0 � � � ��2 � �

��=

�+

tan�� �

�+ = cot��� [�+ 6= 0] �

5�

�+ =

5cos�

sin��� � ln |�+ | = ln |sin�|+� � |�+ | = 0ln|sin �|+- = 0ln|sin�| · 0- = 0- |sin�| �

�+ = B sin�, where B = ±0- . (In our derivation, B was nonzero, but we can restore the excluded case

= �� by allowing B to be zero.) (��3) = � � �+ � = B sin��3

�� 2� = B

�3

2� B =

4��3

.

Thus, �+ =4��3sin� and so = 4��

3sin�� �.

NOT FOR SALE SECTION 7 3 S

INSTRUCTOR USE ONLY ��33

��33

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606 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

18. �$��= �$2 ln � � �$

$2= � ln � �� �

5�$

$2=

5� ln � �� � � 1

$= �� ln ��

5� ��

[by parts with � = ln �, �) = � ��] � � 1$= �� ln �� ��+ � � $ =

1

��� �� ln ���.

$(1) = �1 � �1 = 1

� � � ln 1��� � � � = 1 � � = � + 1. Thus, $ = 1

��� �� ln �� � � 1 .

19. If the slope at the point (�� ) is �, then we have �

��= � � �

= ��� [ 6= 0] �

5�

=

5��� �

ln || = 12�

2 +�. (0) = 1 � ln 1 = 0 + � � � = 0. Thus, || = 0�2�2 � = ±0�2�2, so = 0�

2�2

since (0) = 1 0. Note that = 0 is not a solution because it doesn’t satisfy the initial condition (0) = 1.

20. � 0(�) = �(�)(1� �(�)) � �

��= (1� ) � �

(1� )= �� [ 6= 0, 1] �

5�

(1� )=

5�� �

5 ��

+

1�

�� =

5�� �

5 �1

+

1

1�

�� =

5�� � ln || � ln |1� | = �+ � �

ln

����

1�

���� = �+ � �����

1�

���� = 0�+- �

1� = B0�, where B = ±0- �

= (1� )B0� = B0� � B0� � + B0� = B0� � (1 +B0�) = B0� � =B0�

1 +B0�.

�(0) = 12

� 1

2=

B

1 +B� 1 +B = 2B � B = 1, so = 0�

1 + 0�

or 1

1 + 0��

!.

Note that = 0 and = 1 are not solutions because they don’t satisfy the initial condition �(0) = 12

.

21. � = �+ � �

��(�) =

��(�+ ) � ��

��= 1 +

��, but �

��= �+ = �, so ��

��= 1 + � �

��

1 + �= �� [� 6= �1] �

5��

1 + �=

5�� � ln |1 + �| = �+ � � |1 + �| = 0�+- �

1 + � = ±0-0� � � = ±0-0� � 1 � �+ = ±0-0� � 1 � = B0� � �� 1, where B = ±0- 6= 0.

If � = �1, then �1 = �+ � = ��� 1, which is just = B0� � �� 1 with B = 0. Thus, the general solution

is = B0� � �� 1, where B � R.

22. �0 = + �0��� � 0 = ��+ 0��� � �

��= ) + 0� . Also, ) = �� � �) = � �

��= �

�)

��+ ),

so ) + 0� = ��)

��+ ) � �)

0�=

��

�[� 6= 0] �

5�)

0�=

5��

�� �0�� = ln |�|+ � �

0�� = � ln |�| �� � �) = ln(� ln |�| � �) � �� = � ln(� ln |�| ��) � = �� ln(� ln |�| � �).

23. (a) 0 = 2��1� 2 � �

��= 2�

�1� 2 � ��

1� 2= 2��� �

5��1� 2

=

52��� �

sin�1 = �2 +� for ��2 � �2 + � � �

2 .

(b) (0) = 0 � sin�1 0 = 02 +� � � = 0,

so sin�1 = �2 and = sin(�2) for ����2 � � ����2.

NOT FOR SALE NTIAL EQUATIONSTIO

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SECTION 7.3 SEPARABLE EQUATIONS ¤ 607

(c) For�1� 2 to be a real number, we must have �1 � � 1; that is, �1 � (0) � 1. Thus, the initial-value problem

0 = 2��1� 2, (0) = 2 does not have a solution.

24. 0��0 + cos� = 0 � 40�� � = � 4

cos��� � �0�� = � sin�+ �1 � = � ln(sin�+�). The solution

is periodic, with period 2�. Note that for � 1, the domain of the solution is R, but for �1 � � � 1 it is only de�ned on the

intervals where sin�+ � 0, and it is meaningless for � � �1, since then sin�+ � � 0, and the logarithm is unde�ned.

For �1 � � � 1, the solution curve consists of concave-up pieces separated by intervals on which the solution is not de�ned

(where sin�+ � � 0). For � = 1, the solution curve consists of concave-up pieces separated by vertical asymptotes at the

points where sin�+ � = 0 � sin� = �1. For � 1, the curve is continuous, and as � increases, the graph moves

downward, and the amplitude of the oscillations decreases.

25. ���

=sin�

sin , (0) =

2. So

4sin � =

4sin��� �

� cos = � cos�+� � cos = cos�� �. From the initial condition,

we need cos �2 = cos 0� � � 0 = 1� � � � = 1, so the solution is

cos = cos�� 1. Note that we cannot take cos�1 of both sides, since that would

unnecessarily restrict the solution to the case where�1 � cos�� 1 � 0 � cos�,

as cos�1 is de�ned only on [�1� 1]. Instead we plot the graph using Maple’s

plots[implicitplot] or Mathematica’s Plot[Evaluate[· · ·]].

26. ���

=���2 + 1

0�� 4

0� � =4���2 + 1 ��. We use parts on the LHS with � = , �) = 0� �, and on the RHS

we use the substitution 8 = �2 + 1, so �8 = 2���. The equation becomes 0� � 40� � = 1

2

4 �8 �8 �

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608 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

0�( � 1) = 13(�2 + 1)3�2 + �, so we see that the curves are symmetric about the -axis. Every point (�� ) in the plane lies

on one of the curves, namely the one for which � = ( � 1)0� � 13 (�

2 + 1)3�2. For example, along the -axis,

� = ( � 1)0� � 13

, so the origin lies on the curve with � = � 43

. We use Maple’s plots[implicitplot] command or

Plot[Evaluate[· · ·]] in Mathematica to plot the solution curves for various values of �.

It seems that the transitional values of � are � 43

and � 13� For � � �4

3, the graph consists of left and right branches. At

� = � 43

, the two branches become connected at the origin, and as � increases, the graph splits into top and bottom branches.

At � = � 13

, the bottom half disappears. As � increases further, the graph moves upward, but doesn’t change shape much.

27. (a) , (c) (b) 0 = 2 � �

��= 2 �

5�2 � =

5�� �

��1 = �+ � � 1

= ��� � �

=1

B � �, where B = ��. = 0 is also a solution.

NOT FOR SALE NTIAL EQUATIONSTIO

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SECTION 7.3 SEPARABLE EQUATIONS ¤ 609

28. (a) , (c) (b) 0 = � � �

��= � �

5�

=

5��� �

ln || = 12�2 +� � || = 0�

2�2+- = 0�2�20- �

= B0�2�2, where B = ±0- . Taking B = 0 gives us the

solution = 0.

29. The curves �2 + 22 = �2 form a family of ellipses with major axis on the �-axis. Differentiating gives

��(�2 + 22) =

��(�2) � 2�+ 40 = 0 � 40 = �2� � 0 =

��2

. Thus, the slope of the tangent line

at any point (�� ) on one of the ellipses is 0 = ��2

, so the orthogonal trajectories

must satisfy 0 = 2

�� �

��=2

�� �

= 2 =

��

��

5�

= 2

5��

�� ln || = 2 ln |�|+ �1 � ln || = ln |�|2 + �1 �

|| = 0ln�2+-1 � = ±�2 · 0-1 = ��2. This is a family of parabolas.

30. The curves 2 = ��3 form a family of power functions. Differentiating gives �

��(2) =

��(��3) � 20 = 3��2 �

0 =3��2

2=3(2��3)�2

2=3

2�, the slope of the tangent line at (�� ) on one of the curves. Thus, the orthogonal

trajectories must satisfy 0 = �2�3

� �

��= �2�

3�

3 � = �2��� � 43 � =

4 �2��� � 322 = ��2 + �1 �

32 = �2�2 + �2 � 2�2 + 32 = �. This is a family of ellipses.

31. The curves = ��� form a family of hyperbolas with asymptotes � = 0 and = 0. Differentiating gives

��() =

��

��

�� 0 = � �

�2� 0 = ��

�2[since = ��� � � = �] � 0 = �

�. Thus, the slope

of the tangent line at any point (�� ) on one of the hyperbolas is 0 = ���,

so the orthogonal trajectories must satisfy 0 = �� � �

��=

� = ��� � 4 � =

4��� � 1

22 = 1

2�2 + �1 �

2 = �2 + �2 � �2 � 2 = �. This is a family of hyperbolas with

asymptotes = ±�.

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610 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

32. Differentiating = �

1 + ��gives 0 = 1

(1 + ��)2, but � = ��

�, so

0 =1�

1 +��

�2 = 2

�2. Thus, the orthogonal trajectories must satisfy

0 = ��2�2 � 2 � = ��2 �� � 42 � = � 4

�2 �� �13

3 = �13�

3 + �1 � 3 = � � �3 � = 3�� � �3.

33. (�) = 2 +5 �

2

[�� �(�)] �� � 0(�) = �� �(�) [by FTC 1] � �

��= �(1� ) �

5�

1� =

5��� � � ln |1� | = 1

2�2 +�. Letting � = 2 in the original integral equation

gives us (2) = 2 + 0 = 2. Thus, � ln |1� 2| = 12(2)2 +� � 0 = 2 +� � � = �2�

Thus, � ln |1� | = 12�2 � 2 � ln |1� | = 2� 1

2�2 � |1� | = 02��2�2 �

1� = ±02��2�2 � = 1 + 02��2�2 [(2) = 2].

34. (�) = 2 +5 �

1

��

�(�), � 0 � 0(�) =

1

�(�)� �

��=

1

��

5 � =

51

��� �

122 = ln�+� [� 0]. Letting � = 1 in the original integral equation gives us (1) = 2 + 0 = 2.

Thus, 12(2)2 = ln 1 + � � � = 2. 1

22 = ln�+ 2 � 2 = 2 ln�+ 4 [ 0] � =

�2 ln�+ 4.

35. (�) = 4 +5 �

0

2��(�) �� � 0(�) = 2�

�(�) � �

��= 2�

� �

5��=

52��� �

2� = �2 + �. Letting � = 0 in the original integral equation gives us (0) = 4 + 0 = 4.

Thus, 2�4 = 02 + � � � = 4. 2� = �2 + 4 � �

= 12�2 + 2 � =

�12�2 + 2

�2.

36. (�2 + 1)� 0(�) + [�(�)]2 + 1 = 0 � (�2 + 1)�

��+ 2 + 1 = 0 � �

��=�2 � 1�2 + 1

�5

2 + 1= �

5��

�2 + 1� arctan = � arctan �+ � � arctan �+ arctan = � �

tan(arctan �+ arctan ) = tan� � tan(arctan �) + tan(arctan )

1� tan(arctan �) tan(arctan ) = tan� � �+

1� �= tan� = � �

�+ = � � �� � + �� = � � � � (1 + ��) = � � � � �(�) = =� � �

1 + ��.

Since �(3) = 2 = � � 31 + 3�

� 2 + 6� = � � 3 � 5� = �5 � � = �1 , we have = �1� �

1 + (�1)� =�+ 1

�� 1 .

37. From Exercise 7.2.27, �/��

= 12� 4/ �5

�/

12� 4/ =

5�� � � 1

4 ln|12� 4/| = �+ � �

ln|12� 4/| = �4�� 4� � |12� 4/| = 0�4��4- � 12� 4/ = B0�4� [B = ±0�4-] �

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SECTION 7.3 SEPARABLE EQUATIONS ¤ 611

4/ = 12�B0�4� � / = 3��0�4� [� = B�4]. /(0) = 0 � 0 = 3�� � � = 3 �

/(�) = 3� 30�4�. As � �, /(�) 3� 0 = 3 (the limiting value).

38. From Exercise 7.2.28, ���= � 1

50 ( � 20) �5

� 20 =5 �� 1

50

��� � ln| � 20| = � 1

50 �+ � �

� 20 = B0���50 � (�) = B0���50 + 20. (0) = 95 � 95 = B + 20 � B = 75 �

(�) = 750���50 + 20.

39. ����

= �(- � � ) �5

��

� �-=

5(��) �� � ln|� �- | = ���+ � � |� �- | = 0���+- �

� �- = �0��� [� = ±0- ] � � =- +�0���. If we assume that performance is at level 0 when � = 0, then

� (0) = 0 � 0 =- +� � � = �- � � (�) =- �-0���. lim���

� (�) =- �- · 0 =- .

40. (a) ����= �(�� �)(� �), � 6= . Using partial fractions, 1

(�� �)(� �)=1�(� �)

�� �� 1�(� �)

� �, so

5��

(�� �)(� �)=

5� �� � 1

� �(� ln |�� �|+ ln |� �|) = ��+ � � ln

���� � �

�� �

���� = (� �)(��+ �).

The concentrations [A] = �� � and [B] = � � cannot be negative, so � �

�� �� 0 and

���� � �

�� �

���� = � �

�� �.

We now have ln�� �

�� �

�= (� �)(��+ �). Since �(0) = 0, we get ln

�= (� �)�. Hence,

ln

�� �

�� �

�= (� �)��+ ln

�� � �

�� �=

�0( ��)�� � � =

[0( ��)�� � 1]0( ��)����� 1 =

�[0( ��)�� � 1]0( ��)�� � �

molesL

.

(b) If = �, then ��

��= �(�� �)2, so

5��

(�� �)2=

5� �� and 1

�� �= ��+�. Since �(0) = 0, we get � = 1

�.

Thus, �� � =1

��+ 1��and � = �� �

���+ 1=

�2��

���+ 1

molesL

. Suppose � = [C] = ��2 when � = 20. Then

�(20) = ��2 � �

2=

20�2�

20�� + 1� 40�2� = 20�2� + � � 20�2� = � � � =

1

20�, so

� =�2��(20�)

1 + ���(20�)=

���20

1 + ��20=

��

�+ 20

molesL

.

41. (a) If � = , then ��

��= �(�� �)(� �)1�2 becomes ��

��= �(�� �)3�2 � (�� �)�3�2 �� = � �� �

4(�� �)�3�2 �� =

4� �� � 2(�� �)�1�2 = ��+ � [by substitution] � 2

��+ �=��� � �

�2

��+ �

�2= �� � � �(�) = �� 4

(��+�)2. The initial concentration of HBr is 0, so �(0) = 0 �

0 = �� 4

�2� 4

�2= � � �2 =

4

�� � = 2�

�� [� is positive since ��+ � = 2(�� �)�1�2 0].

Thus, �(�) = �� 4

(��+ 2��� )

2 .

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612 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

(b) ����= �(�� �)(� �)1�2 � ��

(�� �)�� �

= � �� �5

��

(�� �)�� �

=

5� �� (3).

From the hint, � =�� � � �2 = � � � 2��� = ���, so5��

(�� �)�� �

=

5 �2���[�� (� �2)]�

= �25

��

�� + �2= �2

5����

�� �2+ �2

17= �2

�1���

tan�1����

So (3) becomes �2���

tan�1�� ����

= ��+ �. Now �(0) = 0 � � =�2���

tan�1��

�� and we have

�2���

tan�1�� ����

= ��� 2���

tan�1��

�� � 2�

��

�tan�1

�� � tan�1

�� �

��

�= �� �

�(�) =2

����

�tan�1

�� � tan�1

�� �

��

�.

42. If & = ��

��, then �&

��=

�2�

��2. The differential equation �2�

��2+2

��

��= 0 can be written as �&

��+2

�& = 0. Thus,

�&

��=�2&�

� �&

&= �2

��� �

51

&�& =

5�2��� � ln|&| = �2 ln|�|+ �. Assuming & = ����� 0

and � 0, we have & = 0�2 ln �+- = 0ln ��20- = ��2� [� = 0-] � & =

1

�2� � ��

��=1

�2� �

�� =1

�2� �� �

5�� =

51

�2� �� � � (�) = ��

�+�.

� (1) = 15 � 15 = �� +� (1) and � (2) = 25 � 25 = �12� +� (2).

Now solve for � and �: �2(2) + (1) � �35 = ��, so � = 35 and � = 20, and � (�) = �20�� + 35.

43. (a) ����

= � � �� � ��

��= �(�� � �) �

5��

�� � �=

5��� � (1��) ln|�� � �| = ��+-1 �

ln|�� � �| = ���+-2 � |�� � �| = 0���+*2 � �� � � =-30��� � �� =-30

��� + � ��(�) = -40

��� + ���. �(0) = �0 � �0 = -4 + ��� � -4 = �0 � ��� ��(�) = (�0 � ���)0��� + ���.

(b) If �0 � ���, then �0 � ��� � 0 and the formula for �(�) shows that �(�) increases and lim���

�(�) = ���.

As � increases, the formula for �(�) shows how the role of �0 steadily diminishes as that of ��� increases.

44. (a) Use 1 billion dollars as the �-unit and 1 day as the �-unit. Initially, there is $10 billion of old currency in circulation,

so all of the $50 million returned to the banks is old. At time �, the amount of new currency is �(�) billion dollars, so

10� �(�) billion dollars of currency is old. The fraction of circulating money that is old is [10� �(�)]�10, and the amount

of old currency being returned to the banks each day is 10� �(�)

100�05 billion dollars. This amount of new currency per

day is introduced into circulation, so ��

��=10� �

10· 0�05 = 0�005(10� �) billion dollars per day.

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SECTION 7.3 SEPARABLE EQUATIONS ¤ 613

(b) ��

10� �= 0�005 �� � ���

10� �= �0�005 �� � ln(10� �) = �0�005�+ + � 10� � = �0�0005�,

where � = 0� � �(�) = 10� �0�0005�. From �(0) = 0, we get � = 10, so �(�) = 10(1� 0�0005�).

(c) The new bills make up 90% of the circulating currency when �(�) = 0�9 · 10 = 9 billion dollars.

9 = 10(1� 0�0005�) � 0�9 = 1� 0�0005� � 0�0005� = 0�1 � �0�005� = � ln 10 �� = 200 ln 10 460�517 days 1�26 years.

45. (a) Let (�) be the amount of salt (in kg) after � minutes. Then (0) = 15. The amount of liquid in the tank is 1000 L at all

times, so the concentration at time � (in minutes) is (�)�1000 kg�L and �

��= �

(�)

1000

kgL

!�10

Lmin

�= �(�)

100

kgmin

.

5�

= � 1

100

5�� � ln = � �

100+�, and (0) = 15 � ln 15 = �, so ln = ln 15� �

100.

It follows that ln�

15

�= � �

100and

15= 0���100, so = 150���100 kg.

(b) After 20 minutes, = 150�20�100 = 150�02 12�3 kg.

46. Let (�) be the amount of carbon dioxide in the room after � minutes. Then (0) = 0�0015(180) = 0�27 m3. The amount of

air in the room is 180 m3 at all times, so the percentage at time � (in mimutes) is (�)�180× 100, and the change in the

amount of carbon dioxide with respect to time is

��= (0�0005)

�2

m3

min

�� (�)

180

�2

m3

min

�= 0�001�

90=9� 1009000

m3

min

Hence,5

9� 100 =5

��

9000and � 1

100ln |9� 100| = 1

9000�+ �. Because (0) = 0�27, we have

� 1100

ln 18 = �, so � 1100

ln |9� 100| = 19000

�� 1100

ln 18 � ln|9� 100| = � 190�+ ln 18 �

ln |9� 100| = ln 0���90 + ln 18 � ln |9� 100| = ln(180���90), and |9� 100| = 180���90. Since is continuous,

(0) = 0�27, and the right-hand side is never zero, we deduce that 9� 100 is always negative. Thus, |9� 100| = 100� 9and we have 100 � 9 = 180���90 � 100 = 9 + 180���90 � = 0�09 + 0�180���90. The percentage of carbon

dioxide in the room is

�(�) =

180× 100 = 0�09 + 0�180���90

180× 100 = (0�0005 + 0�0010���90)× 100 = 0�05 + 0�10���90

In the long run, we have lim���

�(�) = 0�05 + 0�1(0) = 0�05; that is, the amount of carbon dioxide approaches 0�05% as time

goes on.

47. Let (�) be the amount of alcohol in the vat after � minutes. Then (0) = 0�04(500) = 20 gal. The amount of beer in the vat

is 500 gallons at all times, so the percentage at time � (in minutes) is (�)�500× 100, and the change in the amount of alcohol

with respect to time � is ���= rate in � rate out = 0�06

�5

galmin

�� (�)

500

�5

galmin

�= 0�3�

100=30�

100

galmin

.

Hence,5

30� =

5��

100and � ln |30� | = 1

100�+�. Because (0) = 20, we have � ln 10 = �, so

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614 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

� ln |30� | = 1100

�� ln 10 � ln |30� | = ���100 + ln 10 � ln |30� | = ln 0���100 + ln 10 �

ln |30� | = ln(100���100) � |30� | = 100���100. Since is continuous, (0) = 20, and the right-hand side is

never zero, we deduce that 30� is always positive. Thus, 30� = 100���100 � = 30� 100���100. The

percentage of alcohol is �(�) = (�)�500× 100 = (�)�5 = 6� 20���100. The percentage of alcohol after one hour is

�(60) = 6� 20�60�100 4�9.

48. (a) If (�) is the amount of salt (in kg) after � minutes, then (0) = 0 and the total amount of liquid in the tank remains

constant at 1000 L.

��=

�0�05

kgL

��5

Lmin

�+

�0�04

kgL

��10

Lmin

���(�)

1000

kgL

��15

Lmin

= 0�25 + 0�40� 0�015 = 0�65� 0�015 = 130� 3200

kgmin

Hence,5

130� 3 =5

��

200and � 1

3 ln|130� 3| = 1200 �+ �. Because (0) = 0, we have� 1

3 ln 130 = �,

so � 13ln|130� 3| = 1

200�� 1

3ln 130 � ln|130� 3| = � 3

200�+ ln 130 = ln(1300�3��200), and

|130� 3| = 1300�3��200. Since is continuous, (0) = 0, and the right-hand side is never zero, we deduce that

130� 3 is always positive. Thus, 130� 3 = 1300�3��200 and = 1303 (1� 0�3��200) kg.

(b) After one hour, = 1303 (1� 0�3·60�200) = 130

3 (1� 0�09) 25�7 kg.

Note: As � �, (�) 1303= 43 1

3kg.

49. Assume that the raindrop begins at rest, so that )(0) = 0. ����� = �� and (�))0 = �� � �)0 + )�0 = �� �

�)0 + )(��) = �� � )0 + )� = � � �)

��= � � �) �

5�)

� � �)=

5�� �

� (1��) ln|� � �)| = �+ � � ln |� � �)| = ���� �� � � � �) = �0���. )(0) = 0 � � = �.

So �) = � � �0��� � ) = (���)(1� 0���). Since � 0, as � �, 0��� 0 and therefore, lim���

)(�) = ���.

50. (a) � �)

��= ��) � �)

)= � �

��� � ln |)| = � �

��+ �. Since )(0) = )0, ln|)0| = �. Therefore,

ln

���� ))0���� = � �

�� �

���� ))0���� = 0����� � )(�) = ±)00�����. The sign is+ when � = 0, and we assume

) is continuous, so that the sign is + for all �. Thus, )(�) = )00�����. ����� = )00

����� �

�(�) = ��)0�

0����� +�0.

From �(0) = �0, we get �0 = ��)0�

+ �0, so �0 = �0 +�)0�

and �(�) = �0 +�)0�(1� 0�����).

The distance traveled from time 0 to time � is �(�)� �0, so the total distance traveled is lim���

[�(�)� �0] =�)0�

.

Note: In �nding the limit, we use the fact that � 0 to conclude that lim���

0����� = 0.

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SECTION 7.3 SEPARABLE EQUATIONS ¤ 615

(b) � �)

��= ��)2 � �)

)2= � �

��� � �1

)= ���

�+� � 1

)=

��

���. Since )(0) = )0,

� = � 1

)0and 1

)=

��

�+1

)0. Therefore, )(�) = 1

����+ 1�)0=

�)0�)0�+�

. ��

��=

�)0�)0�+�

�(�) =�

5�)0 ��

�)0�+�=

�ln|�)0�+�|+ �0. Since �(0) = �0, we get �0 =

�ln�+ �0 �

�0 = �0 � �

�ln� � �(�) = �0 +

�(ln|�)0�+�| � ln�) = �0 +

�ln

�����)0�+�

����.We can rewrite the formulas for )(�) and �(�) as )(�) = )0

1 + (�)0��)�and �(�) = �0 +

�ln

����1 + �)0�

����.Remarks: This model of horizontal motion through a resistive medium was designed to handle the case in which )0 0.

Then the term ��)2 representing the resisting force causes the object to decelerate. The absolute value in the expression

for �(�) is unnecessary (since �, )0, and � are all positive), and lim���

�(�) =�. In other words, the object travels

in�nitely far. However, lim���

)(�) = 0. When )0 � 0, the term ��)2 increases the magnitude of the object’s negative

velocity. According to the formula for �(�), the position of the object approaches �� as � approaches ���(�)0):lim

�����(��0)�(�) = ��. Again the object travels in�nitely far, but this time the feat is accomplished in a �nite amount of

time. Notice also that lim�����(��0)

)(�) = �� when )0 � 0, showing that the speed of the object increases without limit.

51. (a) 1

$1

�$1��

= �1

$2

�$2��

� �

��(ln$1) =

��(� ln$2) �

5�

��(ln$1) �� =

5�

��(ln$�

2) �� �

ln$1 = ln$�2 +� � $1 = 0ln,

2 +- = 0ln,

2 0- � $1 = B$�

2 , where B = 0- .

(b) From part (a) with $1 = �, $2 = , and � = 0�0794, we have � = B 00794.

52. (a) ���

=1

��

5�

=1

5��

�� ln =

1

�ln�+ � [� 0, 0] �

= 0(ln �)(1��)+- = (0ln�)1��0- � = B�1�� , where B = 0- .

(b) When � = 1, we get = B�, so the relationship between � and is linear. When � �, 1�� 0, so B; that is,

the nutrient content of the consumer is constant.

53. (a) The rate of growth of the area is jointly proportional to��(�) and - ��(�); that is, the rate is proportional to the

product of those two quantities. So for some constant �, ����� = ��� (- ��). We are interested in the maximum of

the function ����� (when the tissue grows the fastest), so we differentiate, using the Chain Rule and then substituting for

����� from the differential equation:

��

���

��

�= �

�� (�1)��

��+ (- ��) · 1

2��1�2

��

��

!= 1

2���1�2

��

��[�2�+ (- ��)]

= 12���1�2

����(- ��)

�[- � 3�] = 1

2�2(- ��)(- � 3�)

This is 0 when - �� = 0 [this situation never actually occurs, since the graph of �(�) is asymptotic to the line =- ,

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616 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

as in the logistic model] and when - � 3� = 0 � �(�) =-�3. This represents a maximum by the First Derivative

Test, since �

��

���

��

�goes from positive to negative when �(�) =-�3.

(b) From the CAS, we get �(�) =-

��0

�*�� � 1

�0�*�� + 1

�2. To get � in terms of the initial area �0 and the maximum area - ,

we substitute � = 0 and � = �0 = �(0): �0 =-

�� � 1� + 1

�2� (� + 1)

��0 = (� � 1)

�- �

���0 +

��0 = �

�- ��- � �

- +��0 = �

�- � �

��0 �

�- +

��0 = �

��- ���0

�� � =

�- +

��0�

- ���0. [Notice that if �0 = 0, then � = 1.]

54. (a) According to the hint we use the Chain Rule: ��)

��= �

�)

��· ����= �)

�)

��= � ��.2

(�+.)2�

5) �) =

5 ��.2 ��

(�+.)2� )2

2=

�.2

�+.+ �. When � = 0, ) = )0, so )20

2=

�.2

0 +.+ � �

� = 12)20 � �. � 1

2)2 � 1

2)20 =

�.2

�+.� �.. Now at the top of its �ight, the rocket’s velocity will be 0, and its

height will be � = �. Solving for )0: � 12)20 =

�.2

�+.� �. � )20

2= �

� .2

.+ �+.(.+ �)

.+ �

!=

�.�

.+ ��

)0 =

�2�.�

.+ �.

(b) )� = lim���

)0 = lim���

�2�.�

.+ �= lim

���

�2�.

(.��) + 1=�2�.

(c) )� =�2 · 32 ft�s2 · 3960 mi · 5280 ft�mi 36,581 ft�s 6�93 mi�s

APPLIED PROJECT How Fast Does a Tank Drain?

1. (a) = ��2� � �

��= ��2

��

��[implicit differentiation] �

��

��=

1

��2�

��=

1

��2����2��� = 1

�22

���� 1

12

�2�2 · 32��

�= � 1

72

��

(b) ����= � 1

72

�� � ��1�2 �� = � 1

72�� � 2

�� = � 1

72�+�.

�(0) = 6 � 2�6 = 0 + � � � = 2

�6 � �(�) =

�� 1144

�+�6�2.

(c) We want to �nd � when � = 0, so we set � = 0 =�� 1

144�+

�6�2 � � = 144

�6 5 min 53 s.

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APPLIED PROJECT HOW FAST DOES A TANK DRAIN? ¤ 617

2. (a) ����= �

�� � ��1�2 �� = � �� [� 6= 0] � 2

�� = ��+� �

�(�) = 14(��+�)2. Since �(0) = 10 cm, the relation 2

��(�) = ��+�

gives us 2�10 = �. Also, �(68) = 3 cm, so 2

�3 = 68� + 2

�10 and

� = ��10��334

. Thus,

�(�) =1

4

�2�10�

�10��334

�2 10� 0�133�+ 0�00044�2.

Here is a table of values of �(�) correct to one decimal place.

� (in s) �(�) (in cm)

10 8�7

20 7�5

30 6�4

40 5�4

50 4�5

60 3�6

(b) The answers to this part are to be obtained experimentally. See the article by Tom Farmer and Fred Gass, Physical

Demonstrations in the Calculus Classroom, College Mathematics Journal 1992, pp. 146–148.

3. (�) = ��2�(�) = 100��(�) � �

��= 100� and �

��=

��

��

��= 100�

��

��.

Diameter = 2�5 inches � radius = 1�25 inches = 54· 112

foot = 548

foot. Thus, � ��

= ���2�� �

100���

��= ��� 5

48

�2�2 · 32� = �25�

288

�� � ��

��= �

��

1152� 4

��1�2 �� =4 � 1

1152 �� �

2�� = � 1

1152�+ � � �

� = � 12304

�+ � � �(�) =�� 1

2304�+ �

�2. The water pressure after � seconds is

62�5�(�) lb�ft2, so the condition that the pressure be at least 2160 lb�ft2 for 10 minutes (600 seconds) is the condition

62�5 · �(600) � 2160; that is,�� � 600

2304

�2 � 2160625

� ��� � 2596

�� � �34�56 � � � 2596+�34�56. Now �(0) = �2,

so the height of the tank should be at least�2596+�34�56

�2 37�69 ft.

4. (a) If the radius of the circular cross-section at height � is �, then the Pythagorean Theorem gives �2 = 22 � (2� �)2 since

the radius of the tank is 2 m. So �(�) = ��2 = �[4� (2� �)2] = �(4�� �2). Thus, �(�) ����= ���2�� �

�(4�� �2)��

��= ��(0�01)2�2 · 10� � (4�� �2)

��

��= �0�0001�20�.

(b) From part (a) we have (4�1�2 � �3�2) �� =��0�0001�20 ��� � 8

3�3�2 � 2

5�5�2 =

��0�0001�20 ��+ �.

�(0) = 2 � 83(2)3�2 � 2

5(2)5�2 = � � � =

�163� 8

5

��2 = 56

15

�2. To �nd out how long it will take to drain all

the water we evaluate � when � = 0: 0 =��0�0001�20 ��+ � �

� =�

0�0001�20=

56�2�15

0�0001�20=11,200

�10

3 11,806 s 3 h 17 min

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618 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

APPLIED PROJECT Which Is Faster, Going Up or Coming Down?

1. �)0 = ��) ��� � ��)

��= � (�) +��) �

5�)

�) +��=

5� 1�

�� �

1

�ln(�) +��) = � 1

��+ � [�) +�� 0]. At � = 0, ) = )0, so � = 1

�ln(�)0 +��).

Thus, 1�ln(�) +��) = � 1

��+

1

�ln(�)0 +��) � ln(�) +��) = � �

��+ ln(�)0 +��) �

�) +�� = 0�$���(�)0 +��) � �) = (�)0 +��)0�$��� ��� � )(�) =

�)0 +

��

�0�$��� � ��

�.

2. (�) =4)(�) �� =

5 �)0 +

��

�0�$��� � ��

!�� =

�)0 +

��

�0�$���

����

�� ��

��+ �.

At � = 0, = 0, so � =�)0 +

��

��

�. Thus,

(�) =

�)0 +

��

��

���)0 +

��

��

�0�$��� � ���

�=

�)0 +

��

��

�1� 0�$���

�� ���

3. )(�) = 0 � ��

�=

�)0 +

��

�0�$��� � 0$��� =

�)0��

+ 1 � ��

�= ln

��)0��

+ 1

��

�1 =�

�ln

��� + �)0

��

�. With � = 1, )0 = 20, � = 1

10, and � = 9�8, we have �1 = 10 ln

�11898

� 1�86 s.

4. The �gure shows the graph of = 1180(1� 0�01�)� 98�. The zeros are

at � = 0 and �2 3�84. Thus, �1 � 0 1�86 and �2 � �1 1�98. So the

time it takes to come down is about 0�12 s longer than the time it takes to go

up; hence, going up is faster.

5. (2�1) =�)0 +

��

��

�(1� 0�2$�1��)� ��

�· 2�1

=

��)0 +��

��

�1� (0$�1��)�2

�� ��

�· 2�

�ln

��)0 +��

��

Substituting � = 0$�1�� =�)0��

+ 1 =�)0 +��

��(from Problem 3), we get

(2�1) =

�� · ��

��

�(1� ��2)� �2�

�2· 2 ln� = �2�

�2

��� 1

�� 2 ln�

�. Now � 0, � 0, �1 0 �

� = 0$�1�� 00 = 1. �(�) = �� 1

�� 2 ln� � � 0(�) = 1 +

1

�2� 2

�=

�2 � 2�+ 1�2

=(�� 1)2

�2 0

for � 1 � �(�) is increasing for � 1. Since �(1) = 0, it follows that �(�) 0 for every � 1. Therefore,

(2�1) =�2�

�2�(�) is positive, which means that the ball has not yet reached the ground at time 2�1. This tells us that the

time spent going up is always less than the time spent coming down, so ascent is faster.

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SECTION 7.4 EXPONENTIAL GROWTH AND DECAY ¤ 619

7.4 Exponential Growth and Decay

1. The relative growth rate is 1�

��

��= 0�7944, so ��

��= 0�7944� and, by Theorem 2, � (�) = � (0)007944� = 2007944�.

Thus, � (6) = 2007944(6) 234�99 or about 235 members.

2. (a) By Theorem 2, � (�) = � (0)0�� = 600��. In 20 minutes ( 13

hour), there are 120 cells, so ��13

�= 600��3 = 120 �

0��3 = 2 � ��3 = ln 2 � � = 3 ln 2 = ln�23�= ln 8.

(b) � (�) = 600(ln 8)� = 60 · 8�

(c) � (8) = 60 · 88 = 60 · 224 = 1,006,632,960

(d) ����� = �� � � 0(8) = �� (8) = (ln 8)� (8) 2�093 billion cells�h

(e) � (�) = 20,000 � 60 · 8� = 20,000 � 8� = 1000�3 � � ln 8 = ln(1000�3) �

� =ln(1000�3)

ln 8 2�79 h

3. (a) By Theorem 2, � (�) = � (0)0�� = 1000��. Now � (1) = 1000�(1) = 420 � 0� = 420100

� � = ln 4�2.

So � (�) = 1000(ln 42)� = 100(4�2)�.

(b) � (3) = 100(4�2)3 = 7408�8 7409 bacteria

(c) ����� = �� � � 0(3) = � · � (3) = (ln 4�2)�100(4�2)3� [from part (a)] 10,632 bacteria�h

(d) � (�) = 100(4�2)� = 10,000 � (4�2)� = 100 � � = (ln 100)�(ln 4�2) 3�2 hours

4. (a) (�) = (0)0�� � (2) = (0)02� = 400 and (6) = (0)06� = 25,600. Dividing these equations, we get

06��02� = 25,600�400 � 04� = 64 � 4� = ln 26 = 6 ln 2 � � = 32ln 2 1�0397, about 104% per hour.

(b) 400 = (0)02� � (0) = 400�02� � (0) = 400�03 ln 2 = 400��0ln 2

�3= 400�23 = 50.

(c) (�) = (0)0�� = 500(3�2)(ln 2)� = 50(0ln 2)(3�2)� � (�) = 50(2)15�

(d) (4�5) = 50(2)15(45) = 50(2)675 5382 bacteria

(e) ���= � =

�3

2ln 2

�(50(2)675) 5596 bacteria�h

(f ) (�) = 50,000 � 50,000 = 50(2)15� � 1000 = (2)1.5� � ln 1000 = 1.5� ln 2 �

� =ln 1000

1.5 ln 2 6.64 h

5. (a) Let the population (in millions) in the year � be � (�). Since the initial time is the year 1750, we substitute �� 1750 for � in

Theorem 2, so the exponential model gives � (�) = � (1750)0�(��1750). Then � (1800) = 980 = 7900�(1800�1750) �

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620 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

980790

= 0�(50) � ln 980790

= 50� � � = 150ln 980

790 0�0043104. So with this model, we have

� (1900) = 7900�(1900�1750) 1508 million, and � (1950) = 7900�(1950�1750) 1871 million. Both of these

estimates are much too low.

(b) In this case, the exponential model gives � (�) = � (1850)0�(��1850) � � (1900) = 1650 = 12600�(1900�1850) �

ln 16501260

= �(50) � � = 150ln 1650

1260 0�005393. So with this model, we estimate

� (1950) = 12600�(1950�1850) 2161 million. This is still too low, but closer than the estimate of � (1950) in part (a).

(c) The exponential model gives � (�) = � (1900)0�(��1900) � � (1950) = 2560 = 16500�(1950�1900) �

ln 25601650

= �(50) � � = 150ln 2560

1650 0�008785. With this model, we estimate

� (2000) = 16500�(2000�1900) 3972 million. This is much too low. The discrepancy is explained by the fact that the

world birth rate (average yearly number of births per person) is about the same as always, whereas the mortality rate

(especially the infant mortality rate) is much lower, owing mostly to advances in medical science and to the wars in the �rst

part of the twentieth century. The exponential model assumes, among other things, that the birth and mortality rates will

remain constant.

6. (a) Let � (�) be the population (in millions) in the year �. Since the initial time is the year 1951, we substitute �� 1951 for � in

Theorem 2, and �nd that the exponential model gives � (�) = � (1951)0�(��1951) �

� (1961) = 92 = 760�(1961�1951) � � = 110ln 439

361 0�0196. With this model, we estimate

� (2001) = 3610�(2001�1951) 960 million. This estimate is slightly lower than the given value, 1029 million.

(b) Substituting �� 1961 for � in Theorem 2, we �nd that the exponential model gives � (�) = � (1961)0�(��1961) �

� (1981) = 653 = 4390�(1981�1961) � � = 120ln 653

439 0�0199. With this model, we estimate

� (2001) = 4390�(2001�1961) 971 million, which is better than the estimate in part (a). The further estimates are

� (2010) = 439049� 1161 million and � (2020) = 439049� 1416 million.

(c) Both models are reasonable.

7. (a) If = [N2O5] then by Theorem 2, ���= �0�0005 � (�) = (0)0�00005� = �0�00005�.

(b) (�) = �0�00005� = 0�9� � 0�00005� = 0�9 � �0�0005� = ln 0�9 � � = �2000 ln 0�9 211 s

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SECTION 7.4 EXPONENTIAL GROWTH AND DECAY ¤ 621

8. (a) The mass remaining after � days is (�) = (0) 0�� = 500��. Since the half-life is 28 days, (28) = 50028� = 25 �

028� = 12

� 28� = ln 12

� � = �(ln 2)�28, so (�) = 500�(ln 2)��28 = 50 · 2���28.

(b) (40) = 50 · 2�40�28 18�6mg (d)

(c) (�) = 2 � 2 = 50 · 2���28 � 250= 2���28 �

(���28) ln 2 = ln 125

� � =��28 ln 1

25

�� ln 2 130 days

9. (a) If (�) is the mass (in mg) remaining after � years, then (�) = (0)0�� = 1000��.

(30) = 100030� = 12(100) � 030� = 1

2� � = �(ln 2)�30 � (�) = 1000�(ln 2)��30 = 100 · 2���30

(b) (100) = 100 · 2�100�30 9�92 mg

(c) 1000�(ln 2)��30 = 1 � �(ln 2)��30 = ln 1100 � � = �30 ln 001ln 2 199�3 years

10. (a) If (�) is the mass after � days and (0) = �, then (�) = �0��.

(1) = �0� = 0�945� � 0� = 0�945 � � = ln 0�945.

Then �0(ln 0945)� = 12� � ln 0(ln 0945)� = ln 1

2� (ln 0�945)� = ln 1

2� � = � ln 2

ln 0945 12�25 years.

(b) �0(ln 0945)� = 0�20� � (ln 0�945)� = ln 15

� � = � ln 5ln 0945

28�45 years

11. Let (�) be the level of radioactivity. Thus, (�) = (0)0��� and � is determined by using the half-life:

(5730) = 12(0) � (0)0��(5730) = 1

2(0) � 0�5730� = 1

2� �5730� = ln 1

2� � = � ln

12

5730=ln 2

5730.

If 74% of the 14C remains, then we know that (�) = 0�74(0) � 0�74 = 0��(ln 2)�5730 � ln 0�74 = � � ln 25730

� = �5730(ln 0�74)ln 2

2489 2500 years.

12. From the information given, we know that ���

= 2 � = �02� by Theorem 2. To calculate � we use the point (0� 5):

5 = �02(0) � � = 5. Thus, the equation of the curve is = 502�.

13. (a) Using Newton’s Law of Cooling, ����

= �(� � ��), we have ��

��= �(� � 75). Now let = � � 75, so

(0) = � (0)� 75 = 185� 75 = 110, so is a solution of the initial-value problem ���� = � with (0) = 110 and by

Theorem 2 we have (�) = (0)0�� = 1100��.

(30) = 110030� = 150� 75 � 030� = 75110 =

1522 � � = 1

30 ln1522 , so (�) = 1100

130

� ln( 1522 ) and

(45) = 11004530

ln( 1522 ) 62�F. Thus, � (45) 62 + 75 = 137�F.

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622 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

(b) � (�) = 100 � (�) = 25. (�) = 1100130

� ln( 1522 ) = 25 � 0130

� ln( 1522 ) = 25110

� 130� ln 15

22= ln 25

110�

� =30 ln 25

110

ln 1522

116 min.

14. Let � (�) be the temperature of the body � hours after 1:30 PM. Then � (0) = 32�5 and � (1) = 36�3. Using Newton’s Law of

Cooling, ����= �(� � ��), we have ��

��= �(� � 20). Now let = � � 20, so (0) = � (0)� 20 = 32�5� 20 = 12�5,

so is a solution to the initial value problem ���� = � with (0) = 12�5 and by Theorem 2 we have

(�) = (0)0�� = 12�50��.

(1) = 30�3� 20 � 10�3 = 12�50�(1) � 0� = 10312.5 � � = ln 103

12.5 . The murder occurred when

(�) = 37� 20 � 12.50�� = 17 � 0�� = 1712.5 � �� = ln 17

12.5 � � =�ln 17

12.5

�� ln 103

12.5 �1�588 h

�95 minutes. Thus, the murder took place about 95 minutes before 1:30 PM, or 11:55 AM.

15. ����= �(� � 20). Letting = � � 20, we get �

��= �, so (�) = (0)0��. (0) = � (0)� 20 = 5� 20 = �15, so

(25) = (0)025� = �15025�, and (25) = � (25)� 20 = 10� 20 = �10, so �15025� = �10 � 025� = 23

. Thus,

25� = ln�23

�and � = 1

25ln�23

�, so (�) = (0)0�� = �150(1�25) ln(2�3)�. More simply, 025� = 2

3� 0� =

�23

�1�25 �

0�� =�23

���25 � (�) = �15 · � 23

���25.

(a) � (50) = 20 + (50) = 20� 15 · � 23

�50�25= 20� 15 · � 2

3

�2= 20� 20

3= 13�3̄ �C

(b) 15 = � (�) = 20 + (�) = 20� 15 · � 23

���25 � 15 · � 23

���25= 5 � �

23

���25= 1

3�

(��25) ln�23

�= ln

�13

� � � = 25 ln�13

��ln�23

� 67�74 min.

16. ����

= �(� � 20). Let = � � 20. Then �

��= �, so (�) = (0)0��� (0) = � (0)� 20 = 95� 20 = 75,

so (�) = 750��. When � (�) = 70, ����

= �1�C�min. Equivalently, ���= �1 when (�) = 50. Thus,

�1 = �

��= �(�) = 50� and 50 = (�) = 750��. The �rst relation implies � = �1�50, so the second relation says

50 = 750���50. Thus, 0���50 = 23

� ���50 = ln� 23

� � � = �50 ln� 23

� 20�27 min.

17. (a) Let � (�) be the pressure at altitude �. Then ����� = �� � � (�) = � (0)0�� = 101�30��.

� (1000) = 101�301000� = 87�14 � 1000� = ln�87141013

� � � = 11000

ln�87141013

� �

� (�) = 101�3 01

1000� ln( 87�14101�3 ), so � (3000) = 101�303 ln(

87�14101�3 ) 64�5 kPa.

(b) � (6187) = 101�3 061871000

ln( 87�14101�3 ) 39�9 kPa

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SECTION 7.4 EXPONENTIAL GROWTH AND DECAY ¤ 623

18. (a) Using � = �0�1 +

(

��with �0 = 1000, � = 0�08, and � = 3, we have:

(i) Annually: ( = 1; � = 1000�1 + 008

1

�1·3= $1259�71

(ii) Quarterly: ( = 4; � = 1000�1 + 008

4

�4·3= $1268�24

(iii) Monthly: ( = 12; � = 1000�1 + 008

12

�12·3= $1270�24

(iv) Weekly: ( = 52 � = 1000�1 + 008

52

�52·3= $1271�01

(v) Daily: ( = 365; � = 1000�1 + 008

365

�365·3= $1271�22

(vi) Hourly: ( = 365 · 24; � = 1000�1 + 008

365 · 24�365·24·3

= $1271�25

(vii) Continuously: � = 10000(008)3 = $1271�25

(b)

�010(3) = $1349�86,

�008(3) = $1271�25, and

�006(3) = $1197�22.

19. (a) Using � = �0�1 +

(

��with �0 = 3000, � = 0�05, and � = 5, we have:

(i) Annually: ( = 1; � = 3000�1 + 005

1

�1·5= $3828�84

(ii) Semiannually: ( = 2; � = 3000�1 + 005

2

�2·5= $3840�25

(iii) Monthly: ( = 12; � = 3000�1 + 005

12

�12·5= $3850�08

(iv) Weekly: ( = 52; � = 3000�1 + 005

52

�52·5= $3851�61

(v) Daily: ( = 365; � = 3000�1 + 005

365

�365·5= $3852�01

(vi) Continuously: � = 30000(005)5 = $3852�08

(b) ����� = 0�05� and �(0) = 3000.

20. (a) �00006� = 2�0 � 0006� = 2 � 0�06� = ln 2 � � = 503ln 2 11�55, so the investment will

double in about 11�55 years.

(b) The annual interest rate in � = �0(1 + �)� is �. From part (a), we have � = �00006�. These amounts must be equal,

so (1 + �)� = 0006� � 1 + � = 0006 � � = 0006 � 1 0�0618 = 6�18%, which is the equivalent annual

interest rate.

21. (a) ����

= �� �� = ��� � �

�. Let = � � �

�, so �

��=

��

��and the differential equation becomes �

��= �.

The solution is = 00�� � � � �

�=��0 � �

�0�� � � (�) =

�+��0 � �

�0��.

(b) Since � 0, there will be an exponential expansion � �0 � �

� 0 � � � ��0.

(c) The population will be constant if �0 � �

�= 0 � � = ��0. It will decline if �0 � �

�� 0 � � ��0.

(d) �0 = 8,000,000, � = 1� ; = 0�016, � = 210,000 � � ��0 (= 128,000), so by part (c), the population was

declining.

22. (a) ���= �1+� � �1�� � = � �� � ��

�+ = ��+ �. Since (0) = 0, we have � =��0

�+ . Thus,

��

�+ = ��+��0

�+ , or �� = ��0 � +��. So � = 1

��0 � +��

=�0

1� +�0��and (�) = 0

(1� +�0��)1��

.

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624 ¤ CHAPTER 7 DIFFERENTIAL EQUATIONS

(b) (�) � as 1� +�0�� 0, that is, as � 1

+�0�. De�ne � = 1

+�0�. Then lim

��!�(�) =�.

(c) According to the data given, we have + = 0�01, (0) = 2, and (3) = 16, where the time � is given in months. Thus,

0 = 2 and 16 = (3) =0

(1� +�0� · 3)1��. Since � = 1

+�0�, we will solve for +�0�. 16 =

2

(1� 3+�0�)100�

1� 3+�0� =�18

�001= 8�001 � +�0� =

13

�1� 8�001�. Thus, doomsday occurs when

� = � =1

+�0�=

3

1� 8�001 145�77 months or 12�15 years.

APPLIED PROJECT Calculus and Baseball

1. (a) ! = �� = ��)

��, so by the Substitution Rule we have

5 �1

�0

! (�) �� =

5 �1

�0

��)

��

��� = �

5 �1

�0

�) =�)

�1�0= �)1 ��)0 = �(�1)� �(�0)

(b) (i) We have )1 = 110 mi�h = 110(5280)3600 ft�s = 161�3 ft�s, )0 = �90 mi�h = �132 ft�s, and the mass of the

baseball is � =*

�= 5�16

32 = 5512 . So the change in momentum is

�(�1)� �(�0) = �)1 ��)0 =5512

[161�3� (�132)] 2�86 slug-ft�s.

(ii) From part (a) and part (b)(i), we have4 00010

! (�) �� = �(0�001)� �(0) 2�86, so the average force over the

interval [0� 0�001] is 10001

4 00010

! (�) �� 10001

(2�86) = 2860 lb.

2. (a) % =

5 �1

�0

! (�) ��, where ! (�) = ��)

��= �

�)

��

��

��= �)

�)

��and so, by the Substitution Rule,

% =

5 �1

�0

! (�) �� =

5 �1

�0

�)�)

���� =

5 �(�1)

�(�0)

�) �) =12�)2

�1�0= 1

2�)21 � 1

2�)20

(b) From part (b)(i), 90 mi�h = 132 ft�s. Assume )0 = )(�0) = 0 and )1 = )(�1) = 132 ft�s [note that �1 is the point of

release of the baseball]. � = 5512

, so the work done is % = 12�)21 � 1

2�)20 =

12· 5512

· (132)2 85 ft-lb.

3. (a) Here we have a differential equation of the form �)��� = �), so by Theorem 7.4.2, the solution is )(�) = )(0)0��.

In this case � = � 110

and )(0) = 100 ft�s, so )(�) = 1000���10. We are interested in the time � that the ball takes to travel

280 ft, so we �nd the distance function

�(�) =4 �

0)(�) �� =

4 �

01000���10 �� = 100

��100���10

��0= �1000(0���10 � 1) = 1000(1� 0���10)

Now we set �(�) = 280 and solve for �: 280 = 1000(1� 0���10) � 1� 0���10 = 725

� 110� = ln

�1� 7

25

� � � 3�285 seconds.

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SECTION 7.5 THE LOGISTIC EQUATION ¤ 625

(b) Let � be the distance of the shortstop from home plate. We calculate the time for the ball to reach home plate as a function

of �, then differentiate with respect to � to �nd the value of � which corresponds to the minimum time. The total time that

it takes the ball to reach home is the sum of the times of the two throws, plus the relay time�12 s

�. The distance from the

�elder to the shortstop is 280� �, so to �nd the time �1 taken by the �rst throw, we solve the equation

�1(�1) = 280� � � 1� 0��1�10 =280� �

1000� �1 = �10 ln 720 + �

1000. We �nd the time �2 taken by the second

throw if the shortstop throws with velocity *, since we see that this velocity varies in the rest of the problem. We use

) = *0���10 and isolate �2 in the equation �(�2) = 10*(1� 0��2�10) = � � 0��2�10 = 1� �

10*�

�2 = �10 ln 10* � �

10*, so the total time is ��(�) =

1

2� 10

ln720 + �

1000+ ln

10* � �

10*

!.

To �nd the minimum, we differentiate: �����

= �10

1

720 + �� 1

10* � �

!, which changes from negative to positive

when 720 + � = 10* � � � � = 5* � 360. By the First Derivative Test, �� has a minimum at this distance from the

shortstop to home plate. So if the shortstop throws at * = 105 ft�s from a point � = 5(105)� 360 = 165 ft from home

plate, the minimum time is �105(165) = 12 � 10

�ln 720+165

1000 + ln 1050� 1651050

� 3�431 seconds. This is longer than the

time taken in part (a), so in this case the manager should encourage a direct throw. If * = 115 ft�s, then � = 215 ft from

home, and the minimum time is �115(215) = 12� 10�ln 720+215

1000+ ln 1150� 215

1150

� 3�242 seconds. This is less than the

time taken in part (a), so in this case, the manager should encourage a relayed throw.

(c) In general, the minimum time is ��(5* � 360) = 1

2� 10

ln360 + 5*

1000+ ln

360 + 5*

10*

!=1

2� 10 ln (* + 72)

2

400*.

We want to �nd out when this is about 3�285 seconds, the same time as the

direct throw. From the graph, we estimate that this is the case for

* 112�8 ft�s. So if the shortstop can throw the ball with this velocity,

then a relayed throw takes the same time as a direct throw.

7.5 The Logistic Equation

1. (a) ����� = 0�05� � 0�0005� 2 = 0�05� (1� 0�01� ) = 0�05� (1� ��100). Comparing to Equation 4,

����� = �� (1� ��-), we see that the carrying capacity is - = 100 and the value of � is 0�05.

(b) The slopes close to 0 occur where � is near 0 or 100. The largest slopes appear to be on the line � = 50. The solutions

are increasing for 0 � �0 � 100 and decreasing for �0 100.

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