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:
$s(Y)-Y(0)+Y=e^{-5s}\\
s(Y)-3+Y=e^{-5s}$
which implies that:
$Y(s)=\frac{3}{s+1}+\frac{e^{-5s}}{s+1}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}\{\frac{3}{s+1}\}+L^{-1}\{\frac{e^{-5s}}{s+1}\}$
Since $L^{-1}\{\frac{e^{-5s}}{s+1}\}=u_5(t)e^{-(t-5)}$ we finally obtain
$y(t)=3e^{-t}+u_5(t)e^{-(t-5)}$
