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

572 Chapter 19 Ordinary differential equations I
Solving Equations (19.17) and (19.18) simultaneously we find α = 21
85 andβ=−18 85 ,
so thatthe particular integral is
y P
(x) = 21
85 cosx−18 85 sinx
Engineeringapplication19.7
AnLC circuitwithsinusoidalinput
The differential equation governing the flow of current in a seriesLC circuit when
subject toanapplied voltage v(t) =V 0
sinωt is
L d2 i
dt + 1 2 C i = ωV 0cos ωt
Derive this equation and then obtain its general solution.
Solution
Kirchhoff’s voltage law and the component laws give
L di
dt + 1 ∫
idt =V
C 0
sinωt
To avoid processes of differentiation and integration in the same equation let us differentiate
thisequation w.r.t.t. This yields
L d2 i
dt + 1 2 C i = ωV 0cos ωt
as required.
The homogeneous equation isL d2 i
dt + i 2 C = 0. Lettingi = ekt we find the auxiliary
equation isLk 2 + 1 C =0sothatk=± √ j . Therefore, using the result of
LC
Example 19.22, with α = 0 and β = 1 √
LC
, the complementary function is
t t
i=Acos√ +Bsin√ LC LC
Tofindaparticularintegral,tryi =Ecos ωt+F sin ωt,whereE andF areconstants.
We find
di
= −ωEsinωt + ωFcosωt
dt
d 2 i
dt 2 = −ω2 Ecosωt −ω 2 Fsinωt
Substitution into the inhomogeneous equation yields
L(−ω 2 Ecosωt − ω 2 Fsinωt) + 1 C (Ecosωt +Fsinωt) =V 0 ωcosωt
Equating coefficients of sinωt gives
−ω 2 LF + F C = 0