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