082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

19.4 First-order linear equations: use of an integrating factor 555
(e) Showthatthesolutionin(b)canbewrittenasthesumofthezero-stateresponse
and the zero-input response.
(f) Identifythe transient and steady-state termsinthe solution topart(e).
Solution
(a) The given differential equation can berewritten as
dv C
+ 1
dt RC v C = 1
RC v S (t)
Comparing the form of this equation with (19.4) we see that it is a first-order
linearequation,withindependentvariablet,anddependentvariable v C
,inwhich
P(t) = 1
RC andQ(t) = 1
RC v S (t).
(b) The integrating factor isgiven by
µ = e ∫ (1/RC)dt = e t/(RC)
Itfollows from the second key point on page 552 that
e t/(RC) v C
= V ∫
e t/(RC) cosωtdt
RC
Thisintegralcanbeevaluatedusingintegrationbypartstwicefollowingthetechnique
inExample 14.4. You should verify that
∫
e t/(RC) cosωtdt = R2 C 2 e t/(RC) [
cos ωt
]
ωsinωt +
R 2 C 2 ω 2 +1 RC
Then
e t/(RC) v C
= VRCet/(RC)
R 2 C 2 ω 2 +1
+constant of integration
[
ωsinωt +
cos ωt
]
+K
RC
from which
VRC
[
v C
= ωsinωt + cosωt ]
+Ke −t/(RC)
R 2 C 2 ω 2 +1 RC
Thisisthegeneralsolutionofthedifferentialequation.Applyingtheinitialcondition
gives us a value forK. Whent = 0, v C
=V 0
, so
[ ]
VRC 1
V 0
=
+K
R 2 C 2 ω 2 +1 RC
from which
Finally,
K=V 0
−
v C
=
V
R 2 C 2 ω 2 +1
VRC
[
ωsinωt+
R 2 C 2 ω 2 +1
cos ωt
] (
+ V
RC 0
−
)
V
e −t/(RC)
R 2 C 2 ω 2 +1
This is the particular solution which satisfies the given initial condition. It tells
us the voltage across the capacitor as a function of timet.
➔