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:
$s(Y)-Y(0)+4Y=3e^{-s}\\
s(Y)(s+4)-2=3e^{-s}$
which implies that:
$Y(s)=\frac{3e^{-s}}{s+4}+\frac{2}{s+4}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}\{\frac{3e^{-s}}{s+4}\}+L^{-1}\{\frac{2}{s+4}\}$
Since $L^{-1}\{\frac{e^{-s}}{s+4}\}=u_1(t)e^{-4(t-1)}$ we finally obtain
$y(t)=3u_1(t)e^{-4(t-1)}+2e^{-4t}$
