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We discuss in detail the particular solution for the first-order and the sec-
ond-order differential equations because these represent, as previously
shown in Section 4.7, important cases in circuit analysis.

Example 6.5
Find the particular solution to the first-order differential equation:

dy
a + by = Acos(ω t) (6.54)
dt

Solution: We guess that the particular solution of this ODE is a sinusoidal of
the form:

y t ( ) = Bcos(ω t − φ ) = B[cos( )cos(φ ω t) sin( )sin(+ φ ω t)]
partic.
(6.55)
= B cos(ω t) + B sin(ω t)
c s
Our task now is to find B and B that would force Eq. (6.55) to be the solution
c
s
of Eq. (6.54). Therefore, we substitute this trial solution in the differential
equation and require that, separately, the coefficients of sin(ωt) and cos(ωt)
terms match on both sides of the resulting equation. These requirements are
necessary for the trial solution to be valid at all times. The resulting condi-
tions are

aω Ab
B = B B = (6.56)
s c c 2 2 2
b a ω + b

from which we can also deduce the polar form of the solution, giving:

B = A 2 tan( φ) = aω (6.57)
2
a ω 2 + b 2 b
2
Example 6.6
Find the particular solution to the second-order differential equation:

2
dy dy
a + b + cy = Acos(ω t) (6.58)
dt 2 dt
Solution: Again, take the trial particular solution to be of the form:

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