Answer
See below
Work Step by Step
Given:
$y'-y=6\cos t$
and $y(0)=2$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[sY(x)-y(0)]-Y(s)=\frac{6s}{s^2+1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s-1)-2=\frac{6s}{s^2+1}$
That is,
$Y(s)(s-1)=2+\frac{6s}{s^2+1}=\frac{2s^2+6s+2}{s^2+1}$
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:
$Y(s)=\frac{2s^2+6s+2}{(s-1)(s^2+1)}=\frac{-3s}{s^2+1}+\frac{3}{s^2+1}+\frac{5}{s-1}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=-3\cos t +3\sin t+5e^t$
