Answer
See below
Work Step by Step
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$[s^2(Y)-sY(0)-y'(0)]_4[sY(s)-y(0)]+3Y(s)=e^{-2s}\\
s(Y)(s^2+4s+3)-s-3=e^{-2s}$
which implies that:
$Y(s)=\frac{e^{-2s}}{(s+1)(s+3)}+\frac{s}{(s+1)(s+3)}+\frac{3}{(s+1)(s+3)}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{e^{-2s}}{(s+1)(s+3)}+\frac{s}{(s+1)(s+3)}+\frac{3}{(s+1)(s+3)}]\\
=\frac{e^{-2s}}{2}(\frac{1}{s+3}-\frac{1}{s+3})+\frac{1}{s+1}$
We finally obtain
$y(t)=\frac{1}{2}[e^{-(t-2)-e^{-3(t-2)}}]u_2(t)+e^{-t}$
