,Solution manual For Elementary Differential
Equations 12e Boyce
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, CHAPTER 1
Introduction
1.1
1.
For � > 3∕2, the slopes are negative, therefore the solutions are decreasing. For � < 3∕2, the
slopes are positive, hence the solutions are increasing. The equilibrium solution appears to be
�(�) = 3∕2, to which all other solutions converge.
2.
For � > 3∕2, the slopes are positive, therefore the solutions increase. For � < 3∕2, the slopes
are negative, therefore, the solutions decrease. As a result, � diverges from 3∕2 as � → ∞ if
�(0) 3∕2.
3.
For � > −1∕2, the slopes are negative, therefore the solutions decrease. For � < −1∕2, the
slopes are positive, therefore, the solutions increase. As a result, � → −1∕2 as � → ∞.
1
,2 CHAPTER 1 Introduction
4.
For � > −1∕2, the slopes are positive, and hence the solutions increase. For � < −1∕2, the
slopes are negative, and hence the solutions decrease. All solutions diverge away from the
equilibrium solution �(�) = −1∕2.
5. For all solutions to approach the equilibrium solution �(�) = 2∕3, we must have � ′ < 0 for
� > 2∕3, and � ′ > 0 for � < 2∕3. The required rates are satisfied by the differential equation
� ′ = 2 − 3�.
6. For solutions other than �(�) = 2 to diverge from � = 2, �(�) must be an increasing func-
tion for � > 2, and a decreasing function for � < 2. The simplest differential equation whose
solutions satisfy these criteria is � ′ = � − 2.
7.
For � = 0 and � = 4 we have � ′ = 0 and thus � = 0 and � = 4 are equilibrium solutions. For
� > 4, � ′ < 0 so if �(0) > 4 the solution approaches � = 4 from above. If 0 < �(0) < 4, then
� ′ > 0 and the solutions “grow” to � = 4 as � → ∞. For �(0) < 0 we see that � ′ < 0 and the
solutions diverge from 0.
8.
Note that � ′ = 0 for � = 0 and � = 5. The two equilibrium solutions are �(�) = 0 and �(�) = 5.
Based on the direction field, � ′ > 0 for � > 5; thus solutions with initial values greater than
5 diverge from the solution �(�) = 5. For 0 < � < 5, the slopes are negative, and hence solu-
tions with initial values between 0 and 5 all decrease toward the solution �(�) = 0. For
� < 0, the slopes are all positive; thus solutions with initial values less than 0 approach the
solution �(�) = 0.
, 1.1 3
9.
Since � ′ = � 2 , � = 0 is the only equilibrium solution and � ′ > 0 for all �. Thus � → 0 if the
initial value is negative; � diverges from 0 if the initial value is positive.
10.
Observe that � ′ = 0 for � = 0 and � = 2. The two equilibrium solutions are �(�) = 0 and
�(�) = 2. Based on the direction field, � ′ > 0 for � > 2; thus solutions with initial values
greater than 2 diverge from �(�) = 2. For 0 < � < 2, the slopes are also positive, and hence
solutions with initial values between 0 and 2 all increase toward the solution �(�) = 2. For
� < 0, the slopes are all negative; thus solutions with initial values less than 0 diverge from the
solution �(�) = 0.
11. -(�) � ′ = 2 − �.
12. From Figure 1.1.6 we can see that � = 2 is an equilibrium solution and thus (c) and (j) are
the only possible differential equations to consider. Since ��∕�� > 0 for � > 2, and ��∕�� < 0
for � < 2 we conclude that (c) is the correct answer: � ′ = � − 2.
13. -(�) � ′ = −2 − �.
14. -(�) � ′ = 2 + �.
15. From Figure 1.1.9 we can see that � = 0 and � = 3 are equilibrium solutions, so (e) and
(h) are the only possible differential equations. Furthermore, we have ��∕�� < 0 for � > 3 and
for � < 0, and ��∕�� > 0 for 0 < � < 3. This tells us that (h) is the desired differential equation:
� ′ = � (3 − �).
16. -(�) � ′ = � (� − 3).
17. (a) Let �(�) denote the amount of chemical in the pond at time �. The amount � will be
measured in grams and the time � will be measured in hours. The rate at which the chemical
is entering the pond is given by 300 gal/h ⋅ .01 g/gal = 3 g/h. The rate at which the chemical
leaves the pond is given by 300 gal/h ⋅ �∕106 g/gal = (3 × 10−4 )� g/h. Thus the differential
equation is given by ��∕�� = 3 − (3 × 10−4 )�.
(b) The equilibrium solution occurs when �′ = 0, or � = 104 grams. Since �′ > 0 for � < 104
g and �′ < 0 for � > 104 g, all solutions approach the equilibrium solution independent of the
amount present at � = 0.
(c) Let �(�) denote the amount of chemical in the pond at time �. From part (a) the
function �(�) satisfies the differential equation ��∕�� = 3 − (3 × 10−4 )�. Thus in terms of
the concentration �(�) = �(�)∕106 , ��∕�� = (1∕106 )(��∕��) = (1∕106 )(3 − (3 × 10−4 )�) = (3 ×
10−6 ) − (10−6 )(3 × 10−4 )� = (3 × 10−6 ) − (3 × 10−4 )�.
,4 CHAPTER 1 Introduction
18. The surface area of a spherical raindrop of radius � is given by � = 4��2 . The volume of a
spherical raindrop is given by � = 4��3 ∕3. Therefore, we see that the surface area � = �� 2∕3
for some constant �. If the raindrop evaporates at a rate proportional to its surface area, then
��∕�� = −�� 2∕3 for some � > 0.
19. The difference between the temperature of the object and the ambient temperature
is � − 70 (� in ◦ F). Since the object is cooling when � > 70, and the rate constant is
� = 0.05 min−1 , the governing differential equation for the temperature of the object is
��∕�� = −.05 (� − 70).
20. (a) Let �(�) be the total amount of the drug (in milligrams) in the patient’s body at any
given time � (hr). The drug enters the body at a constant rate of 500 mg/hr. The rate at which
the drug leaves the bloodstream is given by 0.4 �(�). Hence the accumulation rate of the drug
is described by the differential equation ��∕�� = 500 − 0.4 � (mg/hr).
(b)
Based on the direction field, the amount of drug in the bloodstream approaches the equilib-
rium level of 1250 mg (within a few hours).
21. (a) Following the discussion in the text, the differential equation is �(��∕��) =
�� − � � 2 , or equivalently, ��∕�� = � − �� 2 ∕�.
√ a long time, ��∕�� ≈ 0. Hence the object attains a terminal velocity given by
(b) After
�∞ = ��∕� .
2
(c) Using the relation � �∞ = ��, the required drag coefficient is � = 2∕49 kg/s.
(d)
22.
All solutions become asymptotic to the line � = � − 3 as � → ∞.
, 1.2 5
23.
The solutions approach −∞, 0, or ∞ depending upon the value of �(0).
24.
The solutions approach −∞, ∞, or oscillate depending upon the value of �(0).
25.
For all �(0), there is a number �� that depends on the value of �(0) such that �(�� ) = 0 and the
solution does not exist for � > �� .
1.2
1. (a) The differential equation can be rewritten as
��
= ��.
5−�
Integrating both sides of this equation results in − ln |5 − �| = � + �1 , or equivalently, 5 − � =
� �−� . Applying the initial condition �(0) = �0 results in the specification of the constant as
� = 5 − �0 . Hence the solution is �(�) = 5 + (�0 − 5)�−� .
,6 CHAPTER 1 Introduction
All solutions appear to converge to the equilibrium solution �(�) = 5.
(b) The differential equation can be rewritten as
��
= ��.
5 − 2�
Integrating both sides of this equation results in −(1∕2) ln |5 − 2�| = � + �1 , or equivalently,
5 − 2� = � �−2� . Applying the initial condition �(0) = �0 results in the specification of the
constant as � = 5 − 2�0 . Hence �(�) = 5∕2 + (�0 − 5∕2)�−2� .
All solutions appear to converge to the equilibrium solution �(�) = 5∕2, at a faster rate than in
part (a).
(c) Rewrite the differential equation as
��
= ��.
10 − 2�
Integrating both sides of this equation results in −(1∕2) ln |10 − 2�| = � + �1 , or equivalently,
5 − � = � �−2� . Applying the initial condition �(0) = �0 results in the specification of the
constant as � = 5 − �0 . Hence �(�) = 5 + (�0 − 5)�−2� .
All solutions appear to converge to the equilibrium solution �(�) = 5, at a faster rate than in
part (a), and at the same rate as in part (b).
2. (a) The differential equation can be rewritten as
��
= ��.
�−5
, 1.2 7
Integrating both sides of this equation results in ln |� − 5| = � + �1 , or equivalently, � − 5 = � �� .
Applying the initial condition �(0) = �0 results in the specification of the constant as � = �0 − 5.
Hence the solution is �(�) = 5 + (�0 − 5)�� .
All solutions appear to diverge from the equilibrium solution �(�) = 5.
(b) Rewrite the differential equation as
��
= ��.
2� − 5
Integrating both sides of this equation results in (1∕2) ln |2� − 5| = � + �1 , or equivalently,
2� − 5 = � �2� . Applying the initial condition �(0) = �0 results in the specification of the
constant as � = 2�0 − 5. So the solution is �(�) = (�0 − 5∕2)�2� + 5∕2.
All solutions appear to diverge from the equilibrium solution �(�) = 5∕2.
(c) The differential equation can be rewritten as
��
= ��.
2� − 10
Integrating both sides of this equation results in (1∕2) ln |2� − 10| = � + �1 , or equivalently,
� − 5 = � �2� . Applying the initial condition �(0) = �0 results in the specification of the con-
stant as � = �0 − 5. Hence the solution is �(�) = 5 + (�0 − 5)�2� .
All solutions appear to diverge from the equilibrium solution �(�) = 5.
, 8 CHAPTER 1 Introduction
3. (a) Rewrite the differential equation as
��
= �� ,
� − ��
which is valid for � � ∕�. Integrating both sides results in −(1∕�) ln |� − ��| = � + �1 , or
equivalently, � − �� = � �−�� . Hence the general solution is �(�) = (� − � �−�� )∕�. Note that if
� = �∕�, then ��∕�� = 0, and �(�) = �∕� is an equilibrium solution.
(b)
(c) (i) As � increases, the equilibrium solution gets closer to �(�) = 0, from above. The con-
vergence rate of all solutions is �. As � increases, the solutions converge to the equilibrium
solution quicker.
(ii) As � increases, then the equilibrium solution �(�) = �∕� also becomes larger. In this case,
the convergence rate remains the same.
(iii) If � and � both increase but �∕� =constant, then the equilibrium solution �(�) = �∕�
remains the same, but the convergence rate of all solutions increases.
4. (a) The equilibrium solution satisfies the differential equation ��� ∕�� = 0. Setting
��� − � = 0, we obtain �� (�) = �∕�.
(b) Since ��∕�� = ��∕��, it follows that ��∕�� = �(� + �� ) − � = � �.
5. (a) Rewrite Eq.(ii) as (1∕�)��∕�� = � and thus ln |�| = �� + �, or �1 = ���� .
(b) If � = �1 (�) + �, then ��∕�� = ��1 ∕��. Substituting both these into Eq.(i) we get ��1 ∕�� =
�(�1 + �) − �. Since ��1 ∕�� = ��1 , this leaves �� − � = 0 and thus � = �∕�. Hence � = �1 (�) +
�∕� is the solution to Eq(i).
(c) Substitution of �1 = ���� shows this is the same as that given in Eq.(17).
6. (a) Consider the simpler equation ��1 ∕�� = −��1 . As in the previous solutions, rewrite
the equation as (1∕�1 )��1 = −� ��. Integrating both sides results in �1 (�) = � �−�� .
(b) Now set �(�) = �1 (�) + �, and substitute into the original differential equation. We find
that −��1 + 0 = −�(�1 + �) + �. That is, −�� + � = 0, and hence � = �∕�.
(c) The general solution of the differential equation is �(�) = � �−�� + �∕�. This is exactly the
form given by Eq.(17) in the text. Invoking an initial condition �(0) = �0 , the solution may also
be expressed as �(�) = �∕� + (�0 − �∕�)�−�� .
7. (a) The general solution is �(�) = 900 + � ��∕2 , that is, �(�) = 900 + (�0 − 900)��∕2 . With
�0 = 850, the specific solution becomes �(�) = 900 − 50��∕2 . This solution is a decreasing expo-
nential, and hence the time of extinction is equal to the number of months it takes, say �� , for
the population to reach zero. Solving 900 − 50��� ∕2 = 0, we find that �� = 2 ln(900∕50) ≈ 5.78
months.
(b) The solution, �(�) = 900 + (�0 − 900)��∕2 , is a decreasing exponential as long as �0 < 900.
Hence 900 + (�0 − 900)��� ∕2 = 0 has only one root, given by
900
�� = 2 ln ( ) months.
900 − �0
Equations 12e Boyce
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We have all what you need with best price
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, CHAPTER 1
Introduction
1.1
1.
For � > 3∕2, the slopes are negative, therefore the solutions are decreasing. For � < 3∕2, the
slopes are positive, hence the solutions are increasing. The equilibrium solution appears to be
�(�) = 3∕2, to which all other solutions converge.
2.
For � > 3∕2, the slopes are positive, therefore the solutions increase. For � < 3∕2, the slopes
are negative, therefore, the solutions decrease. As a result, � diverges from 3∕2 as � → ∞ if
�(0) 3∕2.
3.
For � > −1∕2, the slopes are negative, therefore the solutions decrease. For � < −1∕2, the
slopes are positive, therefore, the solutions increase. As a result, � → −1∕2 as � → ∞.
1
,2 CHAPTER 1 Introduction
4.
For � > −1∕2, the slopes are positive, and hence the solutions increase. For � < −1∕2, the
slopes are negative, and hence the solutions decrease. All solutions diverge away from the
equilibrium solution �(�) = −1∕2.
5. For all solutions to approach the equilibrium solution �(�) = 2∕3, we must have � ′ < 0 for
� > 2∕3, and � ′ > 0 for � < 2∕3. The required rates are satisfied by the differential equation
� ′ = 2 − 3�.
6. For solutions other than �(�) = 2 to diverge from � = 2, �(�) must be an increasing func-
tion for � > 2, and a decreasing function for � < 2. The simplest differential equation whose
solutions satisfy these criteria is � ′ = � − 2.
7.
For � = 0 and � = 4 we have � ′ = 0 and thus � = 0 and � = 4 are equilibrium solutions. For
� > 4, � ′ < 0 so if �(0) > 4 the solution approaches � = 4 from above. If 0 < �(0) < 4, then
� ′ > 0 and the solutions “grow” to � = 4 as � → ∞. For �(0) < 0 we see that � ′ < 0 and the
solutions diverge from 0.
8.
Note that � ′ = 0 for � = 0 and � = 5. The two equilibrium solutions are �(�) = 0 and �(�) = 5.
Based on the direction field, � ′ > 0 for � > 5; thus solutions with initial values greater than
5 diverge from the solution �(�) = 5. For 0 < � < 5, the slopes are negative, and hence solu-
tions with initial values between 0 and 5 all decrease toward the solution �(�) = 0. For
� < 0, the slopes are all positive; thus solutions with initial values less than 0 approach the
solution �(�) = 0.
, 1.1 3
9.
Since � ′ = � 2 , � = 0 is the only equilibrium solution and � ′ > 0 for all �. Thus � → 0 if the
initial value is negative; � diverges from 0 if the initial value is positive.
10.
Observe that � ′ = 0 for � = 0 and � = 2. The two equilibrium solutions are �(�) = 0 and
�(�) = 2. Based on the direction field, � ′ > 0 for � > 2; thus solutions with initial values
greater than 2 diverge from �(�) = 2. For 0 < � < 2, the slopes are also positive, and hence
solutions with initial values between 0 and 2 all increase toward the solution �(�) = 2. For
� < 0, the slopes are all negative; thus solutions with initial values less than 0 diverge from the
solution �(�) = 0.
11. -(�) � ′ = 2 − �.
12. From Figure 1.1.6 we can see that � = 2 is an equilibrium solution and thus (c) and (j) are
the only possible differential equations to consider. Since ��∕�� > 0 for � > 2, and ��∕�� < 0
for � < 2 we conclude that (c) is the correct answer: � ′ = � − 2.
13. -(�) � ′ = −2 − �.
14. -(�) � ′ = 2 + �.
15. From Figure 1.1.9 we can see that � = 0 and � = 3 are equilibrium solutions, so (e) and
(h) are the only possible differential equations. Furthermore, we have ��∕�� < 0 for � > 3 and
for � < 0, and ��∕�� > 0 for 0 < � < 3. This tells us that (h) is the desired differential equation:
� ′ = � (3 − �).
16. -(�) � ′ = � (� − 3).
17. (a) Let �(�) denote the amount of chemical in the pond at time �. The amount � will be
measured in grams and the time � will be measured in hours. The rate at which the chemical
is entering the pond is given by 300 gal/h ⋅ .01 g/gal = 3 g/h. The rate at which the chemical
leaves the pond is given by 300 gal/h ⋅ �∕106 g/gal = (3 × 10−4 )� g/h. Thus the differential
equation is given by ��∕�� = 3 − (3 × 10−4 )�.
(b) The equilibrium solution occurs when �′ = 0, or � = 104 grams. Since �′ > 0 for � < 104
g and �′ < 0 for � > 104 g, all solutions approach the equilibrium solution independent of the
amount present at � = 0.
(c) Let �(�) denote the amount of chemical in the pond at time �. From part (a) the
function �(�) satisfies the differential equation ��∕�� = 3 − (3 × 10−4 )�. Thus in terms of
the concentration �(�) = �(�)∕106 , ��∕�� = (1∕106 )(��∕��) = (1∕106 )(3 − (3 × 10−4 )�) = (3 ×
10−6 ) − (10−6 )(3 × 10−4 )� = (3 × 10−6 ) − (3 × 10−4 )�.
,4 CHAPTER 1 Introduction
18. The surface area of a spherical raindrop of radius � is given by � = 4��2 . The volume of a
spherical raindrop is given by � = 4��3 ∕3. Therefore, we see that the surface area � = �� 2∕3
for some constant �. If the raindrop evaporates at a rate proportional to its surface area, then
��∕�� = −�� 2∕3 for some � > 0.
19. The difference between the temperature of the object and the ambient temperature
is � − 70 (� in ◦ F). Since the object is cooling when � > 70, and the rate constant is
� = 0.05 min−1 , the governing differential equation for the temperature of the object is
��∕�� = −.05 (� − 70).
20. (a) Let �(�) be the total amount of the drug (in milligrams) in the patient’s body at any
given time � (hr). The drug enters the body at a constant rate of 500 mg/hr. The rate at which
the drug leaves the bloodstream is given by 0.4 �(�). Hence the accumulation rate of the drug
is described by the differential equation ��∕�� = 500 − 0.4 � (mg/hr).
(b)
Based on the direction field, the amount of drug in the bloodstream approaches the equilib-
rium level of 1250 mg (within a few hours).
21. (a) Following the discussion in the text, the differential equation is �(��∕��) =
�� − � � 2 , or equivalently, ��∕�� = � − �� 2 ∕�.
√ a long time, ��∕�� ≈ 0. Hence the object attains a terminal velocity given by
(b) After
�∞ = ��∕� .
2
(c) Using the relation � �∞ = ��, the required drag coefficient is � = 2∕49 kg/s.
(d)
22.
All solutions become asymptotic to the line � = � − 3 as � → ∞.
, 1.2 5
23.
The solutions approach −∞, 0, or ∞ depending upon the value of �(0).
24.
The solutions approach −∞, ∞, or oscillate depending upon the value of �(0).
25.
For all �(0), there is a number �� that depends on the value of �(0) such that �(�� ) = 0 and the
solution does not exist for � > �� .
1.2
1. (a) The differential equation can be rewritten as
��
= ��.
5−�
Integrating both sides of this equation results in − ln |5 − �| = � + �1 , or equivalently, 5 − � =
� �−� . Applying the initial condition �(0) = �0 results in the specification of the constant as
� = 5 − �0 . Hence the solution is �(�) = 5 + (�0 − 5)�−� .
,6 CHAPTER 1 Introduction
All solutions appear to converge to the equilibrium solution �(�) = 5.
(b) The differential equation can be rewritten as
��
= ��.
5 − 2�
Integrating both sides of this equation results in −(1∕2) ln |5 − 2�| = � + �1 , or equivalently,
5 − 2� = � �−2� . Applying the initial condition �(0) = �0 results in the specification of the
constant as � = 5 − 2�0 . Hence �(�) = 5∕2 + (�0 − 5∕2)�−2� .
All solutions appear to converge to the equilibrium solution �(�) = 5∕2, at a faster rate than in
part (a).
(c) Rewrite the differential equation as
��
= ��.
10 − 2�
Integrating both sides of this equation results in −(1∕2) ln |10 − 2�| = � + �1 , or equivalently,
5 − � = � �−2� . Applying the initial condition �(0) = �0 results in the specification of the
constant as � = 5 − �0 . Hence �(�) = 5 + (�0 − 5)�−2� .
All solutions appear to converge to the equilibrium solution �(�) = 5, at a faster rate than in
part (a), and at the same rate as in part (b).
2. (a) The differential equation can be rewritten as
��
= ��.
�−5
, 1.2 7
Integrating both sides of this equation results in ln |� − 5| = � + �1 , or equivalently, � − 5 = � �� .
Applying the initial condition �(0) = �0 results in the specification of the constant as � = �0 − 5.
Hence the solution is �(�) = 5 + (�0 − 5)�� .
All solutions appear to diverge from the equilibrium solution �(�) = 5.
(b) Rewrite the differential equation as
��
= ��.
2� − 5
Integrating both sides of this equation results in (1∕2) ln |2� − 5| = � + �1 , or equivalently,
2� − 5 = � �2� . Applying the initial condition �(0) = �0 results in the specification of the
constant as � = 2�0 − 5. So the solution is �(�) = (�0 − 5∕2)�2� + 5∕2.
All solutions appear to diverge from the equilibrium solution �(�) = 5∕2.
(c) The differential equation can be rewritten as
��
= ��.
2� − 10
Integrating both sides of this equation results in (1∕2) ln |2� − 10| = � + �1 , or equivalently,
� − 5 = � �2� . Applying the initial condition �(0) = �0 results in the specification of the con-
stant as � = �0 − 5. Hence the solution is �(�) = 5 + (�0 − 5)�2� .
All solutions appear to diverge from the equilibrium solution �(�) = 5.
, 8 CHAPTER 1 Introduction
3. (a) Rewrite the differential equation as
��
= �� ,
� − ��
which is valid for � � ∕�. Integrating both sides results in −(1∕�) ln |� − ��| = � + �1 , or
equivalently, � − �� = � �−�� . Hence the general solution is �(�) = (� − � �−�� )∕�. Note that if
� = �∕�, then ��∕�� = 0, and �(�) = �∕� is an equilibrium solution.
(b)
(c) (i) As � increases, the equilibrium solution gets closer to �(�) = 0, from above. The con-
vergence rate of all solutions is �. As � increases, the solutions converge to the equilibrium
solution quicker.
(ii) As � increases, then the equilibrium solution �(�) = �∕� also becomes larger. In this case,
the convergence rate remains the same.
(iii) If � and � both increase but �∕� =constant, then the equilibrium solution �(�) = �∕�
remains the same, but the convergence rate of all solutions increases.
4. (a) The equilibrium solution satisfies the differential equation ��� ∕�� = 0. Setting
��� − � = 0, we obtain �� (�) = �∕�.
(b) Since ��∕�� = ��∕��, it follows that ��∕�� = �(� + �� ) − � = � �.
5. (a) Rewrite Eq.(ii) as (1∕�)��∕�� = � and thus ln |�| = �� + �, or �1 = ���� .
(b) If � = �1 (�) + �, then ��∕�� = ��1 ∕��. Substituting both these into Eq.(i) we get ��1 ∕�� =
�(�1 + �) − �. Since ��1 ∕�� = ��1 , this leaves �� − � = 0 and thus � = �∕�. Hence � = �1 (�) +
�∕� is the solution to Eq(i).
(c) Substitution of �1 = ���� shows this is the same as that given in Eq.(17).
6. (a) Consider the simpler equation ��1 ∕�� = −��1 . As in the previous solutions, rewrite
the equation as (1∕�1 )��1 = −� ��. Integrating both sides results in �1 (�) = � �−�� .
(b) Now set �(�) = �1 (�) + �, and substitute into the original differential equation. We find
that −��1 + 0 = −�(�1 + �) + �. That is, −�� + � = 0, and hence � = �∕�.
(c) The general solution of the differential equation is �(�) = � �−�� + �∕�. This is exactly the
form given by Eq.(17) in the text. Invoking an initial condition �(0) = �0 , the solution may also
be expressed as �(�) = �∕� + (�0 − �∕�)�−�� .
7. (a) The general solution is �(�) = 900 + � ��∕2 , that is, �(�) = 900 + (�0 − 900)��∕2 . With
�0 = 850, the specific solution becomes �(�) = 900 − 50��∕2 . This solution is a decreasing expo-
nential, and hence the time of extinction is equal to the number of months it takes, say �� , for
the population to reach zero. Solving 900 − 50��� ∕2 = 0, we find that �� = 2 ln(900∕50) ≈ 5.78
months.
(b) The solution, �(�) = 900 + (�0 − 900)��∕2 , is a decreasing exponential as long as �0 < 900.
Hence 900 + (�0 − 900)��� ∕2 = 0 has only one root, given by
900
�� = 2 ln ( ) months.
900 − �0
