Answer
See below
Work Step by Step
Given:
$y''-y'-2y=10\cos t$
and $y(0)=0\\
y'(0)=-1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-[sY(s)-y(0)]-2Y(s)=\frac{10s}{s^2+1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-y-2)+1=\frac{10}{s^2+1}$
That is,
$Y(s)(s^2-y-2)=\frac{10s}{s^2+1}-1=\frac{-s^2+10s-1}{s^2+1}$
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:
$Y(s)=\frac{-s^2+10s-1}{(s^2+1)(s-2)(s+1)}=\frac{1}{s-2}+\frac{2}{s+1}-\frac{3s-1}{s^2+1}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=e^{2t}+2e^{-t}-3\cos t-\sin t$
