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

Section 9.5 Linear Equations 623
The solution of the initial-value problem
in Example 2 is shown in Figure 2.
5
Then
and so
xy − y 1 dx − ln x 1 C
x
y − ln x 1 C
x
(1, 2)
0 4
Since ys1d − 2, we have
2 − ln 1 1 C
1
− C
_5
FIGURE 2
Therefore the solution to the initial-value problem is
Example 3 Solve y9 1 2xy − 1.
y − ln x 1 2
x
SOLUTION The given equation is in the standard form for a linear equation. Multiplying
by the integrating factor
e y 2x dx − e x 2
n
Even though the solutions of the differential
equation in Example 3 are
expressed in terms of an integral, they
can still be graphed by a computer
algebra system (Figure 3).
2.5
C=2
_2.5 2.5
FIGURE 3
_2.5
C=_2
we get e x 2 y9 1 2xe x 2 y − e x 2
or (e x 2 y) 9 − e x 2
Therefore
e x 2 y − y e x 2 dx 1 C
Recall from Section 7.5 that y e x 2 dx 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
y − e 2x 2 y e x 2 dx 1 Ce 2x 2
y − e 2x 2 y x
e t 2 dt 1 Ce 2x 2
(Any number can be chosen for the lower limit of integration.)
0
n
E
FIGURE 4
R
switch
L
Application to Electric Circuits
In Section 9.2 we considered the simple electric circuit shown in Figure 4: An electro -
motive force (usually a battery or generator) produces a voltage of Estd volts (V) and a
current of Istd amperes (A) at time t. The circuit also contains a resistor with a resistance
of R ohms (V) and an inductor with an inductance of L henries (H).
Ohm’s Law gives the drop in voltage due to the resistor as RI. The voltage drop due to
the inductor is LsdIydtd. One of Kirchhoff’s laws says that the sum of the voltage drops
is equal to the supplied voltage Estd. Thus we have
7
L dI
dt
1 RI − Estd
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