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1.7 Modeling Problems Using First-Order Linear Differential Equations 57
ForProblems33–38,useadifferentialequationsolvertode- 35. ⋄ The initial-value problem in Problem 17.
terminethesolutiontoeachoftheinitial-valueproblemsand 36. ⋄ The initial-value problem in Problem 18.
sketch the corresponding solution curve.
33. ⋄ The initial-value problem in Problem 15. 37. ⋄ The initial-value problem in Problem 19.
34. ⋄ The initial-value problem in Problem 16. 38. ⋄ The initial-value problem in Problem 20.
1.7 Modeling Problems Using First-Order Linear Differential Equations
There are many examples of applied problems whose mathematical formulation leads
to a rst-order linear differential equation. In this section we analyze two in detail.
Mixing Problems
StatementoftheProblem:ConsiderthesituationdepictedinFigure1.7.1.Atankinitially
contains V liters of a solution in which is dissolved A grams of a certain chemical. A
0 0
solution containing c1 grams/liter of the same chemical ows into the tank at a constant
rateofr liters/minute,andthemixtureowsoutataconstantrateofr liters/minute.We
1 2
assumethatthemixtureiskeptuniformbystirring.Thenatanytimet theconcentration
of chemical in the tank, c (t), is the same throughout the tank and is given by
2
c = A(t), (1.7.1)
2 V(t)
where V(t)denotes the volume of solution in the tank at time t and A(t) denotes the
amount of chemical in the tank at time t.
Solution of concentration c grams/liter
1
flows in at a rate of r liters/minute
1
A(t) amount of chemical in the tank at time t
V(t) volume of solution in the tank at time t
c (t) A(t)/V(t) concentration of chemical in the tank at time t
2
Solution of concentration
c grams/liter flows out at
2
a rate of r2 liters/minute
Figure 1.7.1: A mixing problem.
Mathematical Formulation: The two functions in the problem are V(t)and A(t).In
order to determine how they change with time, we rst consider their change during a
short time interval,
t minutes. In time
t, r
t liters of solution ow into the tank,
1
whereasr
t litersowout.Thusduringthetimeinterval
t,thechangeinthevolume
2
of solution in the tank is
V =r
tr
t =(r r )
t. (1.7.2)
1 2 1 2
Since the concentration of chemical in the inow is c grams/liter (assumed constant),
1
it follows that in the time interval
t the amount of chemical that ows into the tank is
c r
t. Similarly, the amount of chemical that ows out in this same time interval is
1 1
approximately6 c r
t. Thus, the total change in the amount of chemical in the tank
2 2
6This is only an approximation, since c2 is not constant over the time interval
t. The approximation will
becomemoreaccurate as
t → 0.
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58 CHAPTER1 First-Order Differential Equations
during the time interval
t, denoted by
A, is approximately
A≈c r
tc r
t =(c r c r )
t. (1.7.3)
1 1 2 2 1 1 2 2
Dividing Equations (1.7.2) and (1.7.3) by
t yields
V =r r and
A ≈c r c r ,
t 1 2
t 1 1 2 2
respectively. These equations describe the rates of change of V and A over the short, but
nite, time interval
t. In order to determine the instantaneous rates of change of V and
A, we take the limit as
t → 0 to obtain
dV =r r (1.7.4)
dt 1 2
and
dA=c r Ar , (1.7.5)
dt 1 1 V 2
where we have substituted for c from Equation (1.7.1). Since r and r are constants,
2 1 2
wecanintegrate Equation (1.7.4) directly, obtaining
V(t)=(r r )t +V ,
1 2 0
where V is an integration constant. Substituting for V into Equation (1.7.5) and rear-
0
ranging terms yields the linear equation for A(t) :
dA r
+ 2 A=c r . (1.7.6)
dt (r r )t +V 1 1
1 2 0
This differential equation can be solved, subject to the initial condition A(0) = A0,to
determine the behavior of A(t).
Remark The reader need not memorize Equation (1.7.6), since it is better to derive
it for each specic example.
Example 1.7.1 Atankcontains8L(liters) of water in which is dissolved 32 g (grams) of chemical. A
solution containing 2 g/L of the chemical ows into the tank at a rate of 4 L/min, and
the well-stirred mixture ows out at a rate of 2 L/min.
1. Determine the amount of chemical in the tank after 20 minutes.
2. Whatis the concentration of chemical in the tank at that time?
Solution: Wearegiven
r =4L/min,r=2L/min,c=2g/L,V(0)=8L, and A(0)=32g.
1 2 1
For parts 1 and 2, we must nd A(20) and A(20)/V(20), respectively. Now,
V =r
tr
t
1 2
implies that
dV =2.
dt
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1.7 Modeling Problems Using First-Order Linear Differential Equations 59
Integrating this equation and imposing the initial condition that V(0) = 8 yields
V(t)=2(t +4). (1.7.7)
Further,
A≈c r
tc r
t
1 1 2 2
implies that
dA=82c .
dt 2
That is, since c = A/V,
2
dA=82A.
dt V
Substituting for V from (1.7.7), we must solve
dA+ 1 A=8. (1.7.8)
dt t +4
This rst-order linear equation has integrating factor
I = e1/(t+4)dt = t + 4.
Consequently (1.7.8) can be written in the equivalent form
d [(t + 4)A]=8(t +4),
dt
which can be integrated directly to obtain
2
(t + 4)A = 4(t +4) +c.
Hence
1 2
A(t) = t +4[4(t +4) +c].
Imposing the given initial condition A(0) = 32 g implies that c = 64. Consequently
A(t) = 4 [(t +4)2 +16].
t +4
Setting t = 20 gives us the values for parts 1 and 2:
1. We have
A(20) = 1[(24)2 +16]=296 g.
6 3
2. Furthermore, using (1.7.7),
A(20) = 1 · 296 = 37 g/L.
V(20) 48 3 18
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60 CHAPTER1 First-Order Differential Equations
Electric Circuits
An important application of differential equations arises from the analysis of simple
electric circuits. The most basic electric circuit is obtained by connecting the ends of
a wire to the terminals of a battery or generator. This causes a ow of charge, q(t),
measuredincoulombs(C),throughthewire,therebyproducingacurrent,i(t),measured
in amperes (A), dened to be the rate of change of charge. Thus,
i(t) = dq. (1.7.9)
dt
In practice a circuit will contain several components that oppose the ow of charge. As
current passes through these components, work has to be done, and the loss of energy is
described by the resulting voltage drop across each component. For the circuits that we
will consider, the behavior of the current in the circuit is governed by Kirchoff’s second
law, which can be stated as follows.
Kirchoff’s Second Law: The sum of the voltage drops around a closed circuit is zero.
In order to apply this law we need to know the relationship between the current
passing through each component in the circuit and the resulting voltage drop. The com-
ponentsofinteresttousareresistors,capacitors,andinductors.Webrieydescribeeach
of these next.
1. Resistors: A resistor is a component that, owing to its constituency, directly resists
the ow of charge through it. According to Ohm’s law, the voltage drop,
VR,
betweentheendsofaresistorisdirectlyproportionaltothecurrentthatispassing
through it. This is expressed mathematically as
V =iR (1.7.10)
R
where the constant of proportionality, R, is called the resistance of the resistor.
Theunits of resistance are ohms ( ).
2. Capacitors: A capacitor can be thought of as a component that stores charge and
therebyopposesthepassageofcurrent.Ifq(t)denotesthechargeonthecapacitor
at time t, then the drop in voltage,
V , as current passes through it is directly
C
proportional to q(t). It is usual to express this law in the form
V = 1q, (1.7.11)
C C
where the constant C is called the capacitance of the capacitor. The units of
capacitance are farads (F).
3. Inductors: The third component that is of interest to us is an inductor. This can be
consideredasacomponentthatopposesanychangeinthecurrentowingthrough
it. Thedropinvoltageascurrentpassesthroughaninductorisdirectlyproportional
to the rate at which the current is changing. We write this as
V =Ldi, (1.7.12)
L dt
where the constant L is called the inductance of the inductor, measured in units
of henrys (H).
4. EMF:Thenalcomponentinourcircuitswillbeasourceofvoltagethatproduces
an electromotive force (EMF), driving the charge through the circuit. As current
passes through the voltage source, there is a voltage gain, which we denote by
E(t)volts (that is, a voltage drop of E(t)volts).
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