Answer
See below
Work Step by Step
We are given that $y'-y=f(t)$
and $y(0)=2$
where $f(t)=t[1-u_1(t)]+e^{-(t-1)}u_1(t)\\
=t+[e^{-(t-1)}-(t-1)-1]u_1(t)$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[sY(s)-y(0)]-Y(s)=\frac{1}{s^2}+\frac{e^{-s}}{s+1}-\frac{e^{-s}}{s^2}-\frac{e^{-s}}{s}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s-1)-2=\frac{1}{s^2}+e^{-s}(\frac{1}{s+1}-\frac{1}{s^2}-\frac{1}{s})\\
Y(s)(s-1)=\frac{1}{s^2}+e^{-s}(\frac{1}{s+1}-\frac{1}{s^2}-\frac{1}{s})+2$
Thus, we have solved for the Laplace transform of $y(t)$. To find y itself, we must take the inverse Laplace transform. We first decompose the right-hand side into partial fractions to obtain:
$Y(s)=\frac{1}{s^2(s-1)}+e^{-s}[\frac{1}{(s+1)(s-1)}-\frac{1}{s^2(s-1)}-\frac{1}{s(s-1)}]+\frac{2}{s-1}\\
=-\frac{1}{s}-\frac{1}{s^2}+\frac{1}{s-1}+e^{-s}[\frac{1}{2(s-1)}-\frac{1}{2(s+1)}-(\frac{-1}{s}-\frac{1}{s^2}+\frac{1}{s-1})-(\frac{1}{s-1}-\frac{1}{s})]+\frac{2}{s-1}$
Now,$f(t)= L^{-1}[-\frac{1}{s}-\frac{1}{s^2}+\frac{1}{s-1}+e^{-s}[\frac{1}{2(s-1)}-\frac{1}{2(s+1)}-(\frac{-1}{s}-\frac{1}{s^2}+\frac{1}{s-1})-(\frac{1}{s-1}-\frac{1}{s})]+\frac{2}{s-1}]\\
=-1-t+e^t+[\frac{1}{2}(e^{t-1}-e^{-(t-1)})+(1+t-e^{-(t-1)})-(e^{t-1}-1)]+2e^t$
