Answer
See below
Work Step by Step
We are given that $y'-2y=u_2(t)e^{t-2}$
and $y(0)=2$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[sY(s)-y(0)]-2Y(s)=\frac{e^{-2s}}{s-1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-2)-2=\frac{e^{-2s}}{s-1}\\
Y(s)(s^2-2)=\frac{e^{-2s}}{s-1}+2$
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:
$Y(s)=\frac{e^{-2s}}{(s-1)(s-2)}+\frac{2}{s-2}\\
=e^{-2s}(\frac{1}{s-2}-\frac{1}{s-1})+\frac{2}{s-2}$
Now,$f(t)= L^{-1}[e^{-2s}(\frac{1}{s-2}-\frac{1}{s-1})+\frac{2}{s-2}]\\
=e^{2(t-2)})u_2(t)+e^{t-2}u_3(t)+2e^{2t}$
