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:
$[s^2(Y)-sY(0)-y'(0)]-3(sY(s)-y(0))+2Y(s)=e^{-s}\\
s(Y)(s^2-3s+2)-s+3=e^{-s}$
which implies that:
$Y(s)=\frac{e^{-s}}{s^2-3s+2}+\frac{s}{s^2-3s+2}-\frac{3}{s^2-3s+2}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{e^{-s}}{s^2-3s+2}+\frac{s}{s^2-3s+2}-\frac{3}{s^2-3s+2}]\\
=L^{-1}[e^{-s}(\frac{1}{s-2}-\frac{1}{s-1})-\frac{1}{s-2}+\frac{2}{s-1}]$
Since $L^{-1}\{e^{-s}(\frac{1}{s-2}-\frac{1}{s-1})\}=u_1(t)[e^{2(t-1)}e^{t-1}]$ we finally obtain
$y(t)=u_1(t)[e^{2(t-1)}e^{t-1}]-e^{2t}+2e^t$
