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

Section 17.3 Applications of Second-Order Differential Equations 1173
A differential equation for the current can be obtained by differentiating Equation 7
with respect to t and remembering that I − dQydt:
L d 2 I
dt 2
1 R dI
dt 1 1 C I − E9std
ExamplE 3 Find the charge and current at time t in the circuit of Figure 7 if R − 40 V,
L − 1 H, C − 16 3 10 24 F, Estd − 100 cos 10t, and the initial charge and current are
both 0.
SOLUTION With the given values of L, R, C, and Estd, Equation 7 becomes
8
d 2 Q
dt 2
1 40 dQ
dt
1 625Q − 100 cos 10t
The auxiliary equation is r 2 1 40r 1 625 − 0 with roots
r −
240 6 s2900
2
so the solution of the complementary equation is
− 220 6 15i
Q c std − e 220t sc 1 cos 15t 1 c 2 sin 15td
For the method of undetermined coefficients we try the particular solution
Q p std − A cos 10t 1 B sin 10t
Then
Q p 9std − 210A sin 10t 1 10B cos 10t
Q p 0std − 2100A cos 10t 2 100B sin 10t
Substituting into Equation 8, we have
s2100A cos 10t 2 100B sin 10td 1 40s210A sin 10t 1 10B cos 10td
1 625sA cos 10t 1 B sin 10td − 100 cos 10t
or
s525A 1 400Bd cos 10t 1 s2400A 1 525Bd sin 10t − 100 cos 10t
Equating coefficients, we have
525A 1 400B − 100 or 21A 1 16B − 4
2400A 1 525B − 0 or 216A 1 21B − 0
The solution of this system is A − 84
697
64
and B − 697 , so a particular solution is
Q p std − 1
697 s84 cos 10t 1 64 sin 10td
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