Answer
$=\frac{4}{25}e^{4t}+\frac{11}{25}e^{-t}-\frac{4}{5}te^{-t}$
Work Step by Step
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$[s^2(Y)-sY(0)-y'(0)]-3[s(Y)-y(0)]-4Y(s)=\frac{4}{s+1}+e^{-\frac{\pi s}{4}}\\
s(Y)(s^2-3s-4)=\frac{4}{s+1}+s-2$
which implies that:
$Y(s)=\frac{4}{(s+1)(s^2-3s-4)}+\frac{s-2}{(s^2-3s-4)(s+1)}\\
=\frac{s-2}{(s-4)(s+1)}+\frac{4}{(s-4)(s+1)^2}\\
=\frac{2}{5(s-4)}+\frac{3}{5(s+1)}+\frac{4}{25(s-4)}-\frac{4}{25(s+1)}-\frac{4}{5(s+1)^2}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{2}{5(s-4)}+\frac{3}{5(s+1)}+\frac{4}{25(s-4)}-\frac{4}{25(s+1)}-\frac{4}{5(s+1)^2}]$
We finally obtain
$y(t)=\frac{2}{5}e^{4t}+\frac{3}{5}e^{-t}+\frac{4}{25}e^{4t}-\frac{4}{25}e^{-t}-\frac{4}{5}te^{-t}\\
=\frac{4}{25}e^{4t}+\frac{11}{25}e^{-t}-\frac{4}{5}te^{-t}$
