linear differential equations - cengage · application to electric circuits let’s consider the...
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LINEAR DIFFERENTIAL EQUATIONS
A first-order linear differential equation is one that can be put into the form
where and are continuous functions on a given interval. This type of equation occursfrequently in various sciences, as we will see.
An example of a linear equation is because, for , it can be writtenin the form
Notice that this differential equation is not separable because it’s impossible to factor theexpression for as a function of x times a function of y. But we can still solve the equa-tion by noticing, by the Product Rule, that
and so we can rewrite the equation as
If we now integrate both sides of this equation, we get
or
If we had been given the differential equation in the form of Equation 2, we would havehad to take the preliminary step of multiplying each side of the equation by x.
It turns out that every first-order linear differential equation can be solved in a similarfashion by multiplying both sides of Equation 1 by a suitable function called an integrating factor. We try to find so that the left side of Equation 1, when multiplied by
, becomes the derivative of the product :
If we can find such a function , then Equation 1 becomes
Integrating both sides, we would have
so the solution would be
To find such an , we expand Equation 3 and cancel terms:
I�x�P�x� � I��x�
I�x�y� � I�x�P�x�y � �I�x�y�� � I��x�y � I�x�y�
I
y�x� �1
I�x� �y I�x�Q�x� dx � C�4
I�x�y � y I�x�Q�x� dx � C
�I�x�y�� � I�x�Q�x�
I
I�x��y� � P�x�y� � �I�x�y��3
I�x�yI�x�I
I�x�
y � x �C
xxy � x 2 � C
�xy�� � 2x
xy� � y � �xy��
y�
y� �1
x y � 22
x � 0xy� � y � 2x
QP
dy
dx� P�x�y � Q�x�1
1
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This is a separable differential equation for , which we solve as follows:
where . We are looking for a particular integrating factor, not the most generalone, so we take A � 1 and use
Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where is given by Equation 5. Instead of memorizing this formula, however, we just rememberthe form of the integrating factor.
To solve the linear differential equation , multiply both sides bythe integrating factor and integrate both sides.
EXAMPLE 1 Solve the differential equation .
SOLUTION The given equation is linear since it has the form of Equation 1 withand . An integrating factor is
Multiplying both sides of the differential equation by , we get
or
Integrating both sides, we have
EXAMPLE 2 Find the solution of the initial-value problem
SOLUTION We must first divide both sides by the coefficient of to put the differentialequation into standard form:
The integrating factor is
I�x� � e x �1�x� dx � e ln x � x
x � 0y� �1
x y �
1
x 26
y�
y�1� � 2x � 0x 2y� � xy � 1
y � 2 � Ce�x 3
ex 3y � y 6x 2ex 3 dx � 2ex 3
� C
d
dx �ex 3
y� � 6x 2ex 3
ex 3 dy
dx� 3x 2ex 3
y � 6x 2ex 3
ex 3
I�x� � e x 3x 2 dx � ex3
Q�x� � 6x 2P�x� � 3x 2
dy
dx� 3x 2 y � 6x 2
I�x� � e x P�x� dxy� � P�x�y � Q�x�
I
I�x� � e x P�x� dx5
A � �eC
I � Ae x P�x� dx
ln � I � � y P�x� dx
y dI
I� y P�x� dx
I
2 ■ L INEAR D I F FERENT IAL EQUAT IONS
FIGURE 1
6
_3
_1.5 1.8
C=2
C=1
C=_2
C=_1
C=0
■ ■ Figure 1 shows the graphs of several mem-bers of the family of solutions in Example 1.Notice that they all approach as .x l �2
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Multiplication of Equation 6 by gives
Then
and so
Since , we have
Therefore, the solution to the initial-value problem is
EXAMPLE 3 Solve .
SOLUTION The given equation is in the standard form for a linear equation. Multiplying bythe integrating factor
we get
or
Therefore
Recall from Section 6.4 that can’t be expressed in terms of elementary functions.Nonetheless, it’s a perfectly good function and we can leave the answer as
Another way of writing the solution is
(Any number can be chosen for the lower limit of integration.)
APPLICATION TO ELECTRIC CIRCUITS
Let’s consider the simple electric circuit shown in Figure 4: An electromotive force (usuallya battery or generator) produces a voltage of volts (V) and a current of amperes(A) at time . The circuit also contains a resistor with a resistance of ohms ( ) and aninductor with an inductance of henries (H).
Ohm’s Law gives the drop in voltage due to the resistor as . The voltage drop due tothe inductor is . One of Kirchhoff’s laws says that the sum of the voltage drops isequal to the supplied voltage . Thus, we have
which is a first-order linear differential equation. The solution gives the current at time .tI
L dI
dt� RI � E�t�7
E�t�L�dI�dt�
RIL
�RtI�t�E�t�
y � e�x 2 y
x
0 e t 2 dt � Ce�x 2
y � e�x 2 y ex 2 dx � Ce�x 2
x ex 2 dx
ex 2y � y ex 2
dx � C
(ex 2y)� � ex 2
ex 2y� � 2xex 2
y � ex 2
e x 2x dx � ex 2
y� � 2xy � 1
y �ln x � 2
x
2 �ln 1 � C
1� C
y�1� � 2
y �ln x � C
x
xy � y 1
x dx � ln x � C
�xy�� �1
xorxy� � y �
1
x
x
L INEAR D I F FERENT IAL EQUAT IONS ■ 3
■ ■ The solution of the initial-value problem inExample 2 is shown in Figure 2.
FIGURE 2
(1, 2)
5
_5
0 4
■ ■ Even though the solutions of the differentialequation in Example 3 are expressed in terms ofan integral, they can still be graphed by a com-puter algebra system (Figure 3).
FIGURE 3
C=2
C=_2
2.5
_2.5
_2.5 2.5
FIGURE 4
R
E
switch
L
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4 ■ L INEAR D I F FERENT IAL EQUAT IONS
EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is and theinductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closedwhen so the current starts with , find (a) , (b) the current after 1 s, and(c) the limiting value of the current.
SOLUTION(a) If we put , , and in Equation 7, we obtain the initial-valueproblem
or
Multiplying by the integrating factor , we get
Since , we have , so and
(b) After 1 second the current is
(c)
EXAMPLE 5 Suppose that the resistance and inductance remain as in Example 4 but, instead of the battery, we use a generator that produces a variable voltage of
volts. Find .
SOLUTION This time the differential equation becomes
The same integrating factor gives
Using Formula 98 in the Table of Integrals, we have
I � 5101 �sin 30t � 10 cos 30t� � Ce�3t
e 3tI � y 15e 3t sin 30t dt � 15 e 3t
909 �3 sin 30t � 30 cos 30t� � C
d
dt �e 3tI� � e 3t
dI
dt� 3e 3tI � 15e 3t sin 30t
e 3t
dI
dt� 3I � 15 sin 30tor4
dI
dt� 12I � 60 sin 30t
I�t�E�t� � 60 sin 30t
� 5 � 0 � 5
� 5 � 5 limt l �
e�3t
limt l �
I�t� � limt l �
5�1 � e�3t �
I�1� � 5�1 � e�3 � � 4.75 A
I�t� � 5�1 � e�3t �
C � �55 � C � 0I�0� � 0
I�t� � 5 � Ce�3t
e 3tI � y 15e 3t dt � 5e 3t � C
d
dt �e 3tI� � 15e 3t
e 3t dI
dt� 3e 3tI � 15e 3t
e x 3 dt � e 3t
I�0� � 0 dI
dt� 3I � 15
I�0� � 0 4 dI
dt� 12I � 60
E�t� � 60R � 12L � 4
I�t�I�0� � 0t � 0
12 �
■ ■ The differential equation in Example 4 isboth linear and separable, so an alternativemethod is to solve it as a separable equation. If we replace the battery by a generator, how-ever, we get an equation that is linear but notseparable (Example 5).
■ ■ Figure 6 shows the graph of the currentwhen the battery is replaced by a generator.
FIGURE 6
2
_2
2.50
FIGURE 5
6
0 2.5
y=5
■ ■ Figure 5 shows how the current in Example4 approaches its limiting value.
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L INEAR D I F FERENT IAL EQUAT IONS ■ 5
Since , we get
so I�t� � 5101 �sin 30t � 10 cos 30t� �
50101 e�3t
�50
101 � C � 0
I�0� � 0
EXERCISES
1–4 Determine whether the differential equation is linear.
1. 2.
3. 4.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
5–14 Solve the differential equation.
5. 6.
7. 8.
9. 10.
11.
12. ,
13. ,
14.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
15–20 Solve the initial-value problem.
15. ,
16. , ,
17. ,
18. , ,
19. ,
20. , ,
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
; 21–22 Solve the differential equation and use a graphing cal-culator or computer to graph several members of the family ofsolutions. How does the solution curve change as varies?
21. , 22.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
23. A Bernoulli differential equation (named after JamesBernoulli) is of the form
dy
dx� P�x�y � Q�x�y n
y� � �cos x�y � cos xx � 0xy� � y � x cos x
C
x � 0y�1� � 0x dy
dx�
y
x � 1� x
y��� � 0xy� � y � x 2 sin x
y�4� � 20x � 02xy� � y � 6x
v�0� � 5dv
dt� 2tv � 3t 2e t 2
y�1� � 0t � 0t dy
dt� 2y � t 3
y�0� � 2y� � x � y
t ln t dr
dt� r � te t
t � 0�1 � t� du
dt� u � 1 � t
���2 x ��2dy
dx� x sin 2x � y tan x
dy
dx� 2xy � x 2
1 � xy � xy�xy� � y � sx
x 2y� � 2xy � cos 2 xxy� � 2y � x 2
y� � x � 5yy� � 2y � 2e x
y� � cos y � tan xxy� � ln x � x 2y � 0
y � sin x � x 3y�y� � e xy � x 2y 2
Observe that, if or , the Bernoulli equation is linear. For other values of , show that the substitution trans-forms the Bernoulli equation into the linear equation
24–26 Use the method of Exercise 23 to solve the differentialequation.
24. 25.
26.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
27. In the circuit shown in Figure 4, a battery supplies a constantvoltage of 40 V, the inductance is 2 H, the resistance is ,and .(a) Find .(b) Find the current after s.
28. In the circuit shown in Figure 4, a generator supplies a voltageof volts, the inductance is H, the resistanceis , and A.(a) Find .(b) Find the current after s.
; (c) Use a graphing device to draw the graph of the currentfunction.
29. The figure shows a circuit containing an electromotive force,a capacitor with a capacitance of farads (F), and a resistorwith a resistance of ohms ( ). The voltage drop across the
capacitor is , where is the charge (in coulombs), so inthis case Kirchhoff’s Law gives
But , so we have
Suppose the resistance is , the capacitance is F, a battery gives a constant voltage of 60 V, and the initial chargeis C. Find the charge and the current at time .tQ�0� � 0
0.055 �
R dQ
dt�
1
C Q � E�t�
I � dQ�dt
RI �Q
C� E�t�
QQ�C
C
E R
�RC
0.1I�t�
I�0� � 120 �1E�t� � 40 sin 60t
0.1I�t�
I�0� � 010 �
y� � y � xy 3
y� �2
x y �
y 3
x 2xy� � y � �xy 2
du
dx� �1 � n�P�x�u � �1 � n�Q�x�
u � y 1�nn1n � 0
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30. In the circuit of Exercise 29, , , ,and . Find the charge and the current at time .
31. Psychologists interested in learning theory study learningcurves. A learning curve is the graph of a function , theperformance of someone learning a skill as a function of thetraining time . The derivative represents the rate atwhich performance improves.(a) When do you think increases most rapidly? What
happens to as increases? Explain.(b) If is the maximum level of performance of which the
learner is capable, explain why the differential equation
is a reasonable model for learning.(c) Solve the differential equation in part (b) as a linear differ-
ential equation and use your solution to graph the learningcurve.
32. Two new workers were hired for an assembly line. Jim pro-cessed 25 units during the first hour and 45 units during thesecond hour. Mark processed 35 units during the first hour and50 units the second hour. Using the model of Exercise 31 andassuming that , estimate the maximum number ofunits per hour that each worker is capable of processing.
33. In Section 7.6 we looked at mixing problems in which the volume of fluid remained constant and saw that such problemsgive rise to separable equations. (See Example 5 in thatsection.) If the rates of flow into and out of the system are dif-ferent, then the volume is not constant and the resulting differ-ential equation is linear but not separable.
A tank contains 100 L of water. A solution with a saltconcentration of is added at a rate of . The solution is kept mixed and is drained from the tank at a rate
5 L�min0.4 kg�L
P�0� � 0
k a positive constantdP
dt� k�M � P�
MtdP�dtP
dP�dtt
P�t�
tE�t� � 10 sin 60tQ�0� � 0C � 0.01 FR � 2 � of . If is the amount of salt (in kilograms) after
minutes, show that satisfies the differential equation
Solve this equation and find the concentration after 20 minutes.
34. A tank with a capacity of 400 L is full of a mixture of waterand chlorine with a concentration of 0.05 g of chlorine per liter.In order to reduce the concentration of chlorine, fresh water ispumped into the tank at a rate of . The mixture is keptstirred and is pumped out at a rate of . Find the amountof chlorine in the tank as a function of time.
35. An object with mass is dropped from rest and we assumethat the air resistance is proportional to the speed of the object.If is the distance dropped after seconds, then the speed is
and the acceleration is . If is the accelerationdue to gravity, then the downward force on the object is
, where is a positive constant, and Newton’s SecondLaw gives
(a) Solve this as a linear equation to show that
(b) What is the limiting velocity?(c) Find the distance the object has fallen after seconds.
36. If we ignore air resistance, we can conclude that heavierobjects fall no faster than lighter objects. But if we take airresistance into account, our conclusion changes. Use theexpression for the velocity of a falling object in Exercise 35(a)to find and show that heavier objects do fall faster thanlighter ones.
dv�dm
t
v �mt
c �1 � e�ct�m �
m dv
dt� mt � cv
cmt � cv
ta � v��t�v � s��t�ts�t�
m
10 L�s4 L�s
dy
dt� 2 �
3y
100 � 2t
yty�t�3 L�min
6 ■ L INEAR D I F FERENT IAL EQUAT IONSTh
omso
n Br
ooks
-Col
e co
pyrig
ht 2
007
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L INEAR D I F FERENT IAL EQUAT IONS ■ 7
ANSWERS
1. No 3. Yes 5.
7. 9.
11.
13. 15.
17. 19.
21.
C=2
C=1C=0.2
C=_1C=_2
6
_4
0 10
y � sin x � �cos x��x � C�x
y � �x cos x � xv � t 3e t 2� 5e t 2
y � �x � 1 � 3e xu � �t 2 � 2t � 2C���2�t � 1�y � 1
2 x � Ce�x 2�
12 e�x 2
x e x 2 dx
y � 23 sx � C�xy � x 2 ln � x � � Cx 2
y � 23 e x � Ce�2x
25.
27. (a) (b)
29.
31. (a) At the beginning; stays positive, but decreases(c)
33. ; 0.2275 kg�L
35. (b) (c) �mt�c��t � �m�c�e�ct�m � m 2t�c 2mt�c
y � 25 �100 � 2t� � 40,000�100 � 2t��3�2
0
M
P(t)
t
P(0)
P�t� � M � Ce�kt
Q�t� � 3�1 � e�4 t �, I�t� � 12e�4 t
4 � 4e�1�2 � 1.57 AI�t� � 4 � 4e�5t
y � ��Cx 4 � 2��5x��1�2
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8 ■ L INEAR D I F FERENT IAL EQUAT IONS
SOLUTIONS
1. y0 + exy = x2y2 is not linear since it cannot be put into the standard linear form (1), y0 + P (x) y = Q(x).
2. y + sinx = x3y0 ⇒ x3y0 − y = sinx ⇒ y0 + − 1
x3y =
sinx
x3. This equation is in the standard linear
form (1), so it is linear.
3. xy0 + lnx− x2y = 0 ⇒ xy0 − x2y = − lnx ⇒ y0 + (−x) y = − lnxx, which is in the standard linear
form (1), so this equation is linear.
4. y0 + cos y = tanx is not linear since it cannot be put into the standard linear form (1), y0 + P (x) y = Q(x).
[cos y is not of the form P (x)y.]
5. Comparing the given equation, y0 + 2y = 2ex, with the general form, y0 + P (x)y = Q(x), we see that P (x) = 2
and the integrating factor is I(x) = e P (x)dx = e 2 dx = e2x. Multiplying the differential equation by I(x) gives
e2xy0 + 2e2xy = 2e3x ⇒ e2xy0= 2e3x ⇒ e2xy = 2e3x dx ⇒ e2xy = 2
3e3x + C ⇒
y = 23ex +Ce−2x.
6. y0 = x+ 5y ⇒ y0 − 5y = x. I(x) = e P (x)dx = e (−5)dx = e−5x. Multiplying the differential
equation by I(x) gives e−5xy0 − 5e−5xy = xe−5x ⇒ e−5xy0= xe−5x ⇒
e−5xy = xe−5x dx = − 15xe−5x − 1
25e−5x +C [by parts] ⇒ y = − 1
5x− 1
25+Ce5x
7. xy0 − 2y = x2 [divide by x] ⇒ y0 + − 2x
y = x (∗).
I(x) = e P (x) dx = e (−2/x) dx = e−2 ln|x| = eln|x|−2= eln(1/x
2) = 1/x2. Multiplying the differential equation
(∗) by I(x) gives 1x2
y0 − 2
x3y =
1
x⇒ 1
x2y
0=1
x⇒ 1
x2y = ln |x|+ C ⇒
y = x2(ln |x|+C ) = x2 ln |x|+Cx2.
8. x2y0 + 2xy = cos2 x ⇒ y0 +2
xy =
cos2 x
x2. I(x) = e P (x) dx = e 2/x dx = e2 ln|x| = eln(x
2) = x2.
Multiplying by I(x) gives us our original equation back. You may have noticed this immediately, since P (x) is
the derivative of the coefficient of y0. We rewrite it as (x2y)0 = cos2 x. Thus,
x2y = cos2 xdx = 12(1 + cos 2x) dx = 1
2x+ 1
4sin 2x+ C ⇒ y =
1
2x+
1
4x2sin 2x+
C
x2or
y =1
2x+
1
2x2sinx cosx+
C
x2.
9. Since P (x) is the derivative of the coefficient of y0 [P (x) = 1 and the coefficient is x], we can write the differential
equation xy0 + y =√x in the easily integrable form (xy)0 =
√x ⇒ xy = 2
3x3/2 +C ⇒ y = 2
3
√x+C/x.
10. y0 − y = 1/x [x 6= 0], so I(x) = e (−1)dx = e−x. Multiplying the differential equation by I(x) gives
e−xy0 − e−xy = e−x/x ⇒ e−xy0= e−x/x ⇒ y = ex e−x/x dx+ C .
11. I(x) = e 2x dx = ex2
. Multiplying the differential equation y0 + 2xy = x2 by I(x) gives
ex2y0 + 2xex
2y = x2ex
2 ⇒ ex2y
0= x2ex
2. Thus
y = e−x2
x2ex2
dx+C = e−x2 1
2xex2 − 1
2ex2 dx+C = 1
2x+ Ce−x2 − e−x
2 12e
x2 dx.
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L INEAR D I F FERENT IAL EQUAT IONS ■ 9
12. I(x) = e − tan x dx = eln|cos x| = cosx (since −π2< x < π
2). Multiplying the differential equation by I(x) gives
y0 cosx− y tanx cosx = x cosx sin 2x ⇒ (y cosx)0 = x cosx sin 2x. So
y =1
cosxx cosx sin 2xdx+C =
1
cosx2x cos2 x sinxdx+ C
=1
cosx
−2x cos3 x3
+2
3sinx− sin3 x
3+ C =
−2x cos2 x3
+C
cosx+ 2 tanx
3− sin2 x9
13. (1 + t)du
dt+ u = 1 + t, t > 0 [divide by 1 + t] ⇒ du
dt+
1
1 + tu = 1 (∗), which has the form
u0 + P (t)u = Q(t). The integrating factor is I(t) = e P (t) dt = e [1/(1+t)] dt = eln(1+t) = 1 + t.
Multiplying (∗) by I(t) gives us our original equation back. We rewrite it as [(1 + t)u]0 = 1 + t. Thus,
(1 + t)u = (1 + t) dt = t+ 12t2 + C ⇒ u =
t+ 12t2 + C
1 + tor u = t2 + 2t+ 2C
2 (t+ 1).
14. t ln tdr
dt+ r = tet ⇒ dr
dt+
1
t ln tr =
et
ln t. I(t) = e dt/(t ln t) = eln(ln t) = ln t. Multiplying by ln t gives
ln tdr
dt+1
tr = et ⇒ [(ln t)r]0 = et ⇒ (ln t)r = et +C ⇒ r =
et +C
ln t.
15. y0 = x+ y ⇒ y0 + (−1)y = x. I(x) = e (−1) dx = e−x.Multiplying by e−x gives e−xy0 − e−xy = xe−x
⇒ (e−xy)0 = xe−x ⇒ e−xy = xe−x dx = −xe−x − e−x + C [integration by parts with u = x,
dv = e−x dx] ⇒ y = −x− 1 + Cex. y(0) = 2 ⇒ −1 + C = 2 ⇒ C = 3, so y = −x− 1 + 3ex.
16. tdy
dt+2y = t3, t > 0, y (1) = 0. Divide by t to get dy
dt+2
ty = t2, which is linear. I(t) = e (2/t)dt = e2 ln t = t2.
Multiplying by t2 gives t2 dydt+ 2ty = t4 ⇒ t2y
0= t4 ⇒ t2y = 1
5t5 +C ⇒ y =
t3
5+
C
t2. Thus,
0 = y(1) = 15+ C ⇒ C = − 1
5, so y = t3
5− 1
5t2.
17.dv
dt− 2tv = 3t2et2 , v (0) = 5. I(t) = e (−2t)dt = e−t
2
. Multiply the differential equation by I(t) to get
e−t2 dv
dt− 2te−t2v = 3t2 ⇒ e−t
2v
0= 3t2 ⇒ e−t
2v = 3t2 dt = t3 +C ⇒ v = t3et
2+ Cet
2.
5 = v(0) = 0 · 1 +C · 1 = C, so v = t3et2
+ 5et2
.
18. 2xy0 + y = 6x, x > 0 ⇒ y0 +1
2xy = 3. I(x) = e 1/(2x) dx = e(1/2) lnx = eln x
1/2=√x. Multiplying
by√x gives
√x y0 +
1
2√xy = 3
√x ⇒ (
√xy)
0= 3
√x ⇒ √
x y = 3√xdx = 2x3/2 + C ⇒
y = 2x+C√x. y(4) = 20 ⇒ 8 +
C
2= 20 ⇒ C = 24, so y = 2x+ 24√
x.
19. xy0 = y + x2 sinx ⇒ y0 − 1
xy = x sinx. I(x) = e (−1/x) dx = e− ln x = eln x
−1=1
x.
Multiplying by 1xgives 1
xy0 − 1
x2y = sinx ⇒ 1
xy
0= sinx ⇒ 1
xy = − cosx+ C ⇒
y = −x cosx+Cx. y(π) = 0 ⇒ −π · (−1) +Cπ = 0 ⇒ C = −1, so y = −x cosx− x.
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10 ■ L INEAR D I F FERENT IAL EQUAT IONS
20. xdy
dx− y
x+ 1= x ⇒ y0 − y
x (x+ 1)= 1 (x > 0), so I(x) = e− 1/[x(x+1)]dx = e−(ln|x|−ln|x+1|) =
x+ 1
x.
Multiplying the differential equation by I(x) gives x+ 1x
y0 − y
x (x+ 1)
x+ 1
x=
x+ 1
x⇒
x+ 1
xy
0=
x+ 1
x. Then y = x
x+ 11 +
1
xdx+C =
x
x+ 1(x+ lnx+C). But
0 = y (1) = 12[1 + C] so C = −1 and the solution to the initial-value problem is y = x
x+ 1(x− 1 + lnx).
21. y0 +1
xy = cosx (x 6= 0), so I(x) = e (1/x)dx = eln|x| = x (for
x > 0). Multiplying the differential equation by I(x) gives
xy0 + y = x cosx ⇒ (xy)0 = x cosx. Thus,
y =1
xx cosxdx+ C =
1
x[x sinx+ cosx+ C]
= sinx+cosx
x+
C
x
The solutions are asymptotic to the y-axis (except for C = −1). In fact, for C > −1, y →∞ as x→ 0+, whereas
for C < −1, y → −∞ as x→ 0+. As x gets larger, the solutions approximate y = sinx more closely. The graphs
for larger C lie above those for smaller C. The distance between the graphs lessens as x increases.
22. I(x) = e cos x dx = esin x. Multiplying the differential equation by
I(x) gives esinxy0 + cosx · esinxy = cosx · esin x ⇒
esin xy0= cosx · esin x ⇒
y = e− sin x cosx · esin x dx+ C = 1 +Ce− sinx. The graphs for
C = −3, 0, 1, and 3 are shown. As the values of C get further fromzero the graph is stretched away from the line y = 1, which is the value
for C = 0. The graphs are all periodic in x, with a period of 2π.
23. Setting u = y1−n, dudx
= (1− n) y−ndy
dxor dy
dx=
yn
1− n
du
dx=
un/(1−n)
1− n
du
dx. Then the Bernoulli differential
equation becomes un/(1−n)
1− n
du
dx+ P (x)u1/(1−n) = Q(x)un/(1−n) or du
dx+ (1− n)P (x)u = Q(x)(1− n).
24. Here y0 + y
x= −y2, so n = 2, P (x) = 1
xand Q(x) = −1. Setting u = y−1, u satisfies u0 − 1
xu = 1. Then
I(x) = e (−1/x)dx =1
x(for x > 0) and u = x
1
xdx+ C = x(ln |x|+C). Thus, y = 1
x(C + ln |x|) .
25. y0 +2
xy =
y3
x2. Here n = 3, P (x) = 2
x,Q(x) = 1
x2and setting u = y−2, u satisfies u0 − 4u
x= − 2
x2.
Then I(x) = e (−4/x)dx = x−4 and u = x4 − 2
x6dx+ C = x4
2
5x5+C = Cx4 +
2
5x.
Thus, y = ± Cx4 +2
5x
−1/2.
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L INEAR D I F FERENT IAL EQUAT IONS ■ 11
26. Here n = 3, P (x) = 1, Q(x) = x and setting u = y−2, u satisfies u0 − 2u = −2x. Then I(x) = e (−2)dx = e−2x
and u = e2x −2xe−2x dx+C = e2x xe−2x + 12e−2x + C = x+ 1
2 + Ce2x.
So y−2 = x+ 12+ Ce2x ⇒ y = ± x+ 1
2+ Ce2x
−1/2.
27. (a) 2dIdt+ 10I = 40 or dI
dt+ 5I = 20. Then the integrating factor is e 5 dt = e5t. Multiplying the differential
equation by the integrating factor gives e5t dIdt+ 5Ie5t = 20e5t ⇒ e5tI
0= 20e5t ⇒
I(t) = e−5t 20e5t dt+ C = 4 +Ce−5t. But 0 = I(0) = 4 + C, so I(t) = 4− 4e−5t.
(b) I(0.1) = 4− 4e−0.5 ≈ 1.57 A
28. (a) dIdt+ 20I = 40 sin 60t, so the integrating factor is e20t. Multiplying the differential equation by the integrating
factor gives e20t dIdt+ 20Ie20t = 40e20t sin 60t ⇒ e20tI
0= 40e20t sin 60t ⇒
I(t) = e−20t 40e20t sin 60t dt+ C = e−20t 40e20t 14000
(20 sin 60t− 60 cos 60t) + Ce−20t
=sin 60t− 3 cos 60t
5+Ce−20t
But 1 = I(0) = − 35+C, so I(t) = sin 60t− 3 cos 60t + 8e−20t
5.
(b) I(0.1) = sin 6− 3 cos 6 + 8e−25
≈ −0.42 A (c)
29. 5dQ
dt+ 20Q = 60 withQ(0) = 0 C. Then the integrating factor is e 4 dt = e4t, and multiplying the differential
equation by the integrating factor gives e4t dQdt+ 4e4tQ = 12e4t ⇒ e4tQ
0= 12e4t ⇒
Q(t) = e−4t 12e4t dt+C = 3 + Ce−4t. But 0 = Q(0) = 3 + C soQ(t) = 3 1− e−4t is the charge at
time t and I = dQ/dt = 12e−4t is the current at time t.
30. 2dQ
dt+ 100Q = 10 sin 60t or dQ
dt+ 50Q = 5 sin 60t. Then the integrating factor is e 50 dt = e50t, and
multiplying the differential equation by the integrating factor gives e50t dQdt+ 50e50tQ = 5e50t sin 60t ⇒
e50tQ0= 5e50t sin 60t ⇒
Q(t) = e−50t 5e50t sin 60t dt+ C = e−50t 5e50t 16100
(50 sin 60t− 60 cos 60t) + Ce−50t
= 1122 (5 sin 60t− 6 cos 60t) + Ce−50t
But 0 = Q(0) = − 6122
+ C so C = 361andQ(t) = 5 sin 60t− 6 cos 60t
122+3e−50t
61is the charge at time t, while
the current is I(t) = dQ
dt=150 cos 60t+ 180 sin 60t− 150e−50t
61.
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12 ■ L INEAR D I F FERENT IAL EQUAT IONS
31. (a) P increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills
associated with learning a skill. As t increases, we would expect dP/dt to remain positive, but decrease. This is
because as time progresses, the only points left to learn are the more difficult ones.
(b) dPdt
= k(M − P ) is always positive, so the level of performance P is increasing. As P gets close toM , dP/dt
gets close to 0; that is, the performance levels off, as explained in part (a).
(c) dPdt+ kP = kM , so I(t) = e k dt = ekt. Multiplying the differential
equation by I(t) gives ekt dPdt+ kPekt = kMekt ⇒
ektP0= kMekt ⇒
P (t) = e−kt kMektdt+C =M +Ce−kt, k > 0. Furthermore,
it is reasonable to assume that 0 ≤ P (0) ≤M , so −M ≤ C ≤ 0.
32. Since P (0) = 0, we have P (t) =M(1− e−kt). If P1(t) is Jim’s learning curve, then P1(1) = 25 and P1(2) = 45.
Hence, 25 = M1(1 − e−k) and 45 = M1(1− e−2k), so 1− 25/M1 = e−k or
k = − ln 1− 25
M1= ln
M1
M1 − 25 . But 45 =M1(1− e−2k) so 45 =M1 1− M1 − 25M1
2
or
45 =50M1 − 625
M1. Thus,M1 = 125 is the maximum number of units per hour Jim is capable of processing.
Similarly, if P2(t) is Mark’s learning curve, then P2(1) = 35 and P2(2) = 50. So k = lnM2
M2 − 35 and
50 =M2 1− M2 − 35M2
2
orM2 = 61.25. Hence the maximum number of units per hour for Mark is
approximately 61. Another approach would be to use the midpoints of the intervals so that P1(0.5) = 25 and
P1(1.5) = 45. Doing so gives usM1 ≈ 52.6 andM2 ≈ 51.8.
33. y(0) = 0 kg. Salt is added at a rate of 0.4kgL
5Lmin
= 2kgmin
. Since solution is drained from the tank at a
rate of 3 L/min, but salt solution is added at a rate of 5 L/min, the tank, which starts out with 100 L of water,
contains (100 + 2t) L of liquid after t min. Thus, the salt concentration at time t is y(t)
100 + 2t
kgL. Salt therefore
leaves the tank at a rate of y(t)
100 + 2t
kgL
3Lmin
=3y
100 + 2t
kgmin
. Combining the rates at which salt enters
and leaves the tank, we get dydt= 2− 3y
100 + 2t. Rewriting this equation as dy
dt+
3
100 + 2ty = 2, we see that
it is linear. I(t) = exp 3 dt
100 + 2t= exp 3
2ln(100 + 2t) = (100 + 2t)3/2. Multiplying the differential
equation by I(t) gives (100 + 2t)3/2 dydt+ 3(100 + 2t)1/2y = 2(100 + 2t)3/2 ⇒
(100 + 2t)3/2y0= 2(100 + 2t)3/2 ⇒ (100 + 2t)3/2y = 2
5 (100 + 2t)5/2 +C ⇒
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L INEAR D I F FERENT IAL EQUAT IONS ■ 13
y = 25(100 + 2t) + C(100 + 2t)−3/2. Now 0 = y(0) = 2
5(100) + C · 100−3/2 = 40 + 1
1000C ⇒
C = −40,000, so y = 25(100 + 2t)− 40,000(100 + 2t)−3/2 kg. From this solution (no pun intended), we
calculate the salt concentration at time t to be C(t) = y(t)
100 + 2t=
−40,000(100 + 2t)5/2
+2
5
kgL. In particular,
C(20) =−40,0001405/2
+2
5≈ 0.2275 kg
Land y(20) = 2
5(140)− 40,000(140)−3/2 ≈ 31.85 kg.
34. Let y(t) denote the amount of chlorine in the tank at time t (in seconds). y(0) = (0.05 g/L) (400 L) = 20 g. The
amount of liquid in the tank at time t is (400− 6t) L since 4 L of water enters the tank each second and 10 L of
liquid leaves the tank each second. Thus, the concentration of chlorine at time t is y(t)
400− 6tgL. Chlorine doesn’t
enter the tank, but it leaves at a rate of y(t)
400− 6tgL
10Ls=
10 y(t)
400− 6tgs=
5 y(t)
200− 3tgs. Therefore,
dy
dt= − 5y
200− 3t ⇒ dy
y=
−5 dt200− 3t ⇒ ln y = 5
3 ln(200− 3t) +C ⇒
y = exp 53 ln(200− 3t) + C = eC(200− 3t)5/3. Now 20 = y(0) = eC · 2005/3 ⇒ eC =
20
2005/3, so
y(t) = 20(200− 3t)5/32005/3
= 20(1− 0.015t)5/3 g for 0 ≤ t ≤ 66 23s, at which time the tank is empty.
35. (a) dvdt+
c
mv = g and I(t) = e (c/m)dt = e(c/m)t, and multiplying the differential equation by I(t) gives
e(c/m)tdv
dt+
vce(c/m)t
m= ge(c/m)t ⇒ e(c/m)tv
0= ge(c/m)t. Hence,
v(t) = e−(c/m)t ge(c/m)t dt+K = mg/c+Ke−(c/m)t. But the object is dropped from rest, so v(0) = 0
andK = −mg/c. Thus, the velocity at time t is v(t) = (mg/c) 1− e−(c/m)t .
(b) limt→∞
v(t) = mg/c
(c) s(t) = v(t) dt = (mg/c) t+ (m/c)e−(c/m)t + c1 where c1 = s(0)−m2g/c2. s(0) is the initial position,
so s(0) = 0 and s(t) = (mg/c) t+ (m/c)e−(c/m)t −m2g/c2.
36. v = (mg/c)(1− e−ct/m) ⇒ dv
dm=
mg
c0− e−ct/m · ct
m2+
g
c1− e−ct/m · 1 =
−gt
me−ct/m +
g
c− g
ce−ct/m =
g
c1− e−ct/m − ct
me−ct/m ⇒
c
g
dv
dm= 1− 1 +
ct
me−ct/m = 1− 1 + ct/m
ect/m= 1− 1 +Q
eQ, where Q =
ct
m≥ 0. Since eQ > 1 +Q for all
Q > 0, it follows that dv/dm > 0 for t > 0. In other words, for all t > 0, v increases asm increases.
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