Answer
See below
Work Step by Step
Given:
$y''-y=8e^{t}\sin 2t$
and $y(0)=2\\
y'(0)=-2$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-Y(s)=\frac{16}{(s-1)^2+4}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-1)-2s+2=\frac{16}{(s-1)^2+4}$
That is,
$Y(s)(s^2-1)=\frac{16}{(s-1)^2+4}+2s-2=\frac{2s^3-6s^2+14s}{(s-1)^2+4}$
Thus, we have solved for the Laplace transform of $y(t)$. To find y itself, we must take the inverse Laplace transform. We first decompose the right-hand side into partial fractions to obtain:
$Y(s)=\frac{2s^3-6s^2+14s}{[(s-1)^2+4](s-1)(s+1)}=\frac{1}{s-1}+\frac{2}{s+1}-\frac{s+1}{[(s-1)^2+4]}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=e^{t}+2e^{-t}-e^{-t}\cos 2t$
