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

602 Chapter 9 Differential Equations
y
6
Figure 3 shows a direction field for the
differential equation in Example 3.
Compare it with Figure 4, in which we
use the equation y − Ae x 3 y3
to graph
solutions for several values of A. If
you use the direction field to sketch
solution curves with y-intercepts 5, 2,
1, 21, and 22, they will resemble the
curves in Figure 4.
_2
_1
6
4
2
0 1 2 x
_2
_4
_2 2
_6
_6
FIGURE 3
FIGURE 4
R
Example 4 In Section 9.2 we modeled the current Istd in the electric circuit shown in
Figure 5 by the differential equation
E
L
L dI
dt
1 RI − Estd
FIGURE 5
switch
Find an expression for the current in a circuit where the resistance is 12 V, the inductance
is 4 H, a battery gives a constant voltage of 60 V, and the switch is turned on
when t − 0. What is the limiting value of the current?
SOLUTION With L − 4, R − 12, and Estd − 60, the equation becomes
and the initial-value problem is
4 dI
dt 1 12I − 60 or dI
− 15 2 3I
dt
dI
dt
− 15 2 3I Is0d − 0
We recognize this equation as being separable, and we solve it as follows:
Figure 6 shows how the solution in
Example 4 (the current) approaches
its limiting value. Comparison with
Figure 9.2.10 shows that we were able
to draw a fairly accurate solution curve
from the direction field.
6
y=5
y
dI
15 2 3I − y dt s15 2 3I ± 0d
2 1 3 ln | 15 2 3I | − t 1 C
| 15 2 3I | − e23st1Cd
15 2 3I − 6e 23C e 23t − Ae 23t
I − 5 2 1 3 Ae23t
Since Is0d − 0, we have 5 2 1 3 A − 0, so A − 15 and the solution is
Istd − 5 2 5e 23t
0 2.5
FIGURE 6
The limiting current, in amperes, is
lim Istd − lim s5 2
t l ` t l ` 5e23t d − 5 2 5 lim e 23t − 5 2 0 − 5
t l `
n
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