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

1174 Chapter 17 Second-Order Differential Equations
and the general solution is
Qstd − Q c std 1 Q p std
− e 220t sc 1 cos 15t 1 c 2 sin 15td 1 4
697 s21 cos 10t 1 16 sin 10td
Imposing the initial condition Qs0d − 0, we get
Qs0d − c 1 1 84
697 − 0 c 1 − 2 84
697
To impose the other initial condition, we first differentiate to find the current:
I − dQ
dt
− e 220t fs220c 1 1 15c 2 d cos 15t 1 s215c 1 2 20c 2 d sin 15tg
0.2
Q p
1 40
697 s221 sin 10t 1 16 cos 10td
Is0d − 220c 1 1 15c 2 1 640
697 − 0 c 2 − 2 464
2091
Thus the formula for the charge is
Qstd − F 4 e220t
s263 cos 15t 2 116 sin 15td 1 s21 cos 10t 1 16 sin 10tdG
697 3
and the expression for the current is
Istd − 1
2091 fe220t s21920 cos 15t 1 13,060 sin 15td 1 120s221 sin 10t 1 16 cos 10tdg
■
Note 1 In Example 3 the solution for Qstd consists of two parts. Since e 220t l 0 as
t l ` and both cos 15t and sin 15t are bounded functions,
Q
0 1.2
_0.2
FIGURE 8
5 m d 2 x
dt 1 c dx 1 kx − Fstd
2 dt
7 L d 2 Q
dt 2
1 R dQ
dt
1 1 C Q − Estd
Q c std − 4
2091 e220t s263 cos 15t 2 116 sin 15td l 0 as t l `
So, for large values of t,
Qstd < Q p std − 4
697 s21 cos 10t 1 16 sin 10td
and, for this reason, Q p std is called the steady state solution. Figure 8 shows how the
graph of the steady state solution compares with the graph of Q in this case.
Note 2 Comparing Equations 5 and 7, we see that mathematically they are identical.
This suggests the analogies given in the following chart between physical situations
that, at first glance, are very different.
Spring system
Electric circuit
x displacement Q charge
dxydt velocity I − dQydt current
m mass L inductance
c damping constant R resistance
k spring constant 1yC elastance
Fstd external force Estd electromotive force
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