Answer
See below
Work Step by Step
Given:
$y''-y=6\cos t$
and $y(0)=0\\
y'(0)=4$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-Y(s)=\frac{6s}{s^2+1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-1)-4=\frac{6s}{s^2+1}=$
That is,
$Y(s)(s^2-1)=\frac{6s}{s^2+1}+4=\frac{4s^2+6s+4}{s^2+1}$
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:
$Y(s)=\frac{4s^2+6s+4}{(s^2+1)(s-1)(s+1)}=\frac{7}{2(s-1)}+\frac{3}{2(s+1)}-\frac{s+2}{s^2+1}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=\frac{7}{2}e^{t}+\frac{3}{2}e^{-t}-\cos t-2\sin t$
