118 Chapter 1 Fourier Series and Integrals

̇y(t)=

∑∞

n= 1

−nAnsin(nt)+nBncos(nt),

̈y(t)=

∑∞

n= 1

−n^2 Ancos(nt)−n^2 Bnsin(nt).

Then the differential equation can be written in the form

βA 0 +

∑∞

n= 1

(

−n^2 An+αnBn+βAn

)

cos(nt)

+

∑∞

n= 1

(

−n^2 Bn−αnAn+βBn

)

sin(nt)=a 0 +

∑∞

n= 1

ancos(nt)+bnsin(nt).

TheA’s andB’s are now determined by matching coefficients

βA 0 =a 0 ,

(

β−n^2

)

An+αnBn=an,

−αnAn+

(

β−n^2

)

Bn=bn.

When these equations are solved for theA’s andB’s, we find

An=(β−n

(^2) )an−αnbn
, Bn=(β−n
(^2) )bn+αnan
,
where

(

β−n^2

) 2

+α^2 n^2.
Now, given the functionf,thea’s andb’s can be determined, thus giving the
A’s andB’s. The functiony(t)represented by the series found is the periodic
part of the response. Depending on the initial conditions, there may also be a
transient response, which dies out astincreases.

Example.
Consider the differential equation

y ̈+ 0. 4 y ̇+ 1. 04 y=r(t).

Ifr(t)=sin(nt), the corresponding particular solution is

y(t)=−^0.^4 ncos(nt)+(^1.^04 −n

(^2) )sin(nt)
( 1. 04 −n^2 )^2 +( 0. 4 n)^2

.