Laplace Transform

y''(t) + 4y(t) = 2 cos(2t)
L[y''(t) + 4y(t)] = 2L[ cos(2t)]
[s^2+4]Y(s)-s=2[s/(s^2+4)]
Y(s)=(s^3+6s)/(s^2+4)^2
So y(t)=L^(-1)[Y(s)]
Since L^(-1)[s/(s^2+4)^2]=(1/4)(tsin2t)
L^(-1)[s^3/(s^2+4)^2]=[(1/4)(sin2t+2tcos2t)]'=(1/4)(2cos2t-4tsin2t+2cos2t)
So y(t)=(1/4)(2cos2t-4tsin2t+2cos2t)+(6/4)(tsin2t)=cos2t-tsin2t

2008-12-14 21:15:16 補充：
y(t)=(1/4)(2cos2t-4tsin2t+2cos2t)+(6/4)(tsin2t)=cos2t+(1/2)(tsin2t)