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:
$[s^2(Y)-sY(0)-y'(0)]-3(sY(s)-4Y(s)=e^{-3s}\\
s(Y)(s^2-4)-1=e^{-3s}$
which implies that:
$Y(s)=\frac{e^{-3s}}{(s-2)(s+2)}+\frac{1}{(s-2)(s+2)}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{e^{-3s}}{(s-2)(s+2)}+\frac{1}{(s-2)(s+2)}]\\
=L^{-1}[\frac{e^{-3s}}{4}(\frac{1}{s-2}-\frac{1}{s+2})+\frac{1}{4}(\frac{1}{s-2}-\frac{1}{s+2})]$
Since $L^{-1}\{e^{-3s}(\frac{1}{s-2}-\frac{1}{s+2})\}=u_3(t)[e^{2(t-3)}-e^{-2(t-3)}]$ we finally obtain
$y(t)=\frac{u_3(t)}{4}[e^{2(t-3)}-e^{-2(t-3)}]+\frac{1}{4}(e^{2t}-e^{-2t})$
