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:
$s(Y)-Y(0)-2Y=e^{-2s}\\
s(Y)(s-2)-1=e^{-2s}$
which implies that:
$Y(s)=\frac{e^{-2s}}{s-2}+\frac{1}{s-2}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}\{\frac{e^{-2s}}{s-2}\}+L^{-1}\{\frac{1}{s-2}\}$
Since $L^{-1}\{\frac{e^{-2s}}{s-2}\}=u_2(t)e^{2(t-2)}$ we finally obtain
$y(t)=u_2(t)e^{2(t-2)}+e^{2t}$
