Answer
$y(t)=\cos 2t+\frac{4}{3}\sin t-\frac{\sin 2t}{6}+\frac{3}{2}\sin (2t-4)$
Work Step by Step
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$[s^2(Y)-sY(0)-y'(0)]+4Y(s)=\frac{4}{s^2+1}+3e^{-2s}\\
s(Y)(s^2+4)=\frac{4}{s^2+1}+3e^{-2s}+s+1$
which implies that:
$Y(s)=\frac{4}{(s^2+1)(s^2+4)}+\frac{3e^{-2s}}{s^2+4}+\frac{s+1}{s^2+4}\\
=\frac{s}{s^2+4}+\frac{1}{s^2+4}+\frac{4}{3}(\frac{1}{s^2+1}-\frac{1}{s^2+4})+\frac{3e^{-2s}}{s^2+4}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{s}{s^2+4}+\frac{1}{s^2+4}+\frac{4}{3}(\frac{1}{s^2+1}-\frac{1}{s^2+4})+\frac{3e^{-2s}}{s^2+4}]$
We finally obtain
$y(t)=\cos 2t+\frac{4}{3}\sin t-\frac{\sin 2t}{6}+\frac{3}{2}\sin (2t-4)$
