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y t ( ) = Bcos(ω t − φ ) = B[cos( )cos(φ ω t) sin( )sin(+ φ ω t)]
partic.
(6.59)
= B cos(ω t) + B sin(ω t)
c s
Repeating the same steps as in Example 6.5, we find:

bω ( c − aω )
2
B = A B = A (6.60)
s 22 2 2 c 22 2 2
( c − aω ) + ω b ( c − aω ) + ω b

B = A 2 tan( )φ = bω (6.61)
2
b
)
( c − aω 22 + ω 2 2 c − aω 2
6.5.3 Applications to Circuit Analysis

An important application of the above forms for the particular solutions is in
circuit analysis with inductors, resistors, and capacitors as elements. We
describe later a more efficient analytical method (phasor representation) for
solving this kind of problem; however, we believe that it is important that
you also become familiar with the present technique.

6.5.3.1 RC Circuit
Referring to the RC circuit shown in Figure 4.4, we derived the differential
equation that the potential difference across the capacitor must satisfy;
namely:

dV C
RC + V = V cos(ω t) (6.62)
0
C
dt
This is a first-order differential equation, the particular solution of which is
given in Example 6.5 if we were to identify the coefficients in the ODE as fol-
lows: a = RC, b = 1, A = V .
0

6.5.3.2 RLC Circuit
Referring to the circuit, shown in Figure 4.5, the voltage across the capacitor
satisfies the following ODE:

2
dV c dV C
LC + RC + V = V cos(ω t) (6.63)
C
0
dt 2 dt
This equation can be identified with that given in Example 6.6 if the ODE
coefficients are specified as follows: a = LC, b = RC, c = 1, A = V .
0
© 2001 by CRC Press LLC