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:
$s(Y)-Y(0)-5Y=\frac{2}{s+1}+e^{-3s}\\
s(Y)(s-5)-0=\frac{2}{s+1}+e^{-3s}$
which implies that:
$Y(s)=\frac{2}{(s+1)(s-5)}+\frac{e^{-3s}}{s-5}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{2}{(s+1)(s-5)}+\frac{e^{-3s}}{s-5}]\\
=\frac{1}{3}(\frac{1}{s-5}-\frac{1}{s+1})+\frac{e^{-3s}}{s-5}$
Since $L^{-1}\{\frac{e^{-3s}}{s-5}\}=u_3(t)e^{5(t-3)}$ we finally obtain
$y(t)=\frac{1}{3}(e^{5t}-e^{-t})+u_3(t)e^{5(t-3)}$
