Answer
$y(t)=e^{-2t}(1-4t)+(t-4)e^{-2(t-4)}u_4(t)$
Work Step by Step
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$[s^2(Y)-sY(0)-y'(0)]+4[s(Y)-Y(0)]+4Y(s)=e^{-4s}\\
s(Y)(s^2+4s+4)=e^{-4s}+s-2$
which implies that:
$Y(s)=\frac{e^{-4s}}{(s+2)^2}+\frac{s-2}{(s+2)^2}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{e^{-4s}}{(s+2)^2}+\frac{s-2}{(s+2)^2}]$
We finally obtain
$y(t)=e^{-2t}(1-4t)+(t-4)e^{-2(t-4)}u_4(t)$
