James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

624 Chapter 9 Differential Equations
which is a first-order linear differential equation. The solution gives the current I at
time t.
Example 4 Suppose that in the simple circuit of Figure 4 the resistance is 12 V and
the inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is
closed when t − 0 so the current starts with Is0d − 0, find (a) Istd, (b) the current after
1 second, and (c) the limiting value of the current.
The differential equation in Exam ple
4 is both linear and separable, so an
alternative method is to solve it as a
separable equation (Example 9.3.4). If
we replace the battery by a generator,
however, we get an equation that is linear
but not separable (Example 5).
SOLUtioN
(a) If we put L − 4, R − 12, and Estd − 60 in Equation 7, we obtain the initial-value
problem
4 dI 1 12I − 60 Is0d − 0
dt
or
dI
dt
1 3I − 15 Is0d − 0
Multiplying by the integrating factor e y 3 dt − e 3t , we get
3t
dI
e
dt 1 3e 3t I − 15e 3t
d
dt se 3t Id − 15e 3t
Figure 5 shows how the current in
Example 4 approaches its limiting
value.
6
y=5
e 3t I − y 15e 3t dt − 5e 3t 1 C
Istd − 5 1 Ce 23t
Since Is0d − 0, we have 5 1 C − 0, so C − 25 and
Istd − 5s1 2 e 23t d
(b) After 1 second the current is
Is1d − 5s1 2 e 23 d < 4.75 A
0 2.5
FIGURE 5
Figure 6 shows the graph of the current
when the battery is replaced by a
generator.
2
(c) The limiting value of the current is given by
lim Istd − lim 5s1 2
t l ` t l ` e23t d − 5 2 5 lim
t l ` e23t − 5 2 0 − 5
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
Estd − 60 sin 30t volts. Find Istd.
SOLUTION This time the differential equation becomes
n
0
_2
FIGURE 6
2.5
4 dI
dt 1 12I − 60 sin 30t or dI
1 3I − 15 sin 30t
dt
The same integrating factor e 3t gives
d
dt se 3t 3t
dI
Id − e
dt 1 3e 3t I − 15e 3t sin 30t
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