Answer
See below
Work Step by Step
We are given that $y'+2y=f(t)$
and $y(0)=0$
where $f(t)=(1-t)[1-u_1(t)]\\
=(1-t)+u_1(t)(t-1)$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[sY(s)-y(0)]+2Y(s)=\frac{1}{s}-\frac{1}{s^2}+\frac{e^{-s}}{s^2}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s+2)+0=\frac{1}{s}-\frac{1}{s^2}+\frac{e^{-s}}{s^2}$
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:
$Y(s)=\frac{1}{s(s+2)}-\frac{1}{s^2(s+2)}+\frac{e^{-s}}{s^2(s+2)}\\
=\frac{1}{2}(\frac{1}{s}-\frac{1}{s+2})+e^{-s}(-\frac{1}{4s}+\frac{1}{2s^2}+\frac{1}{4(s+2)})+\frac{1}{4s}-\frac{1}{2s^2}-\frac{1}{4(s+2)}$
Now,$f(t)= L^{-1}[\frac{1}{2}(\frac{1}{s}-\frac{1}{s+2})+e^{-s}(-\frac{1}{4s}+\frac{1}{2s^2}+\frac{1}{4(s+2)})+\frac{1}{4s}-\frac{1}{2s^2}-\frac{1}{4(s+2)}]\\
=\frac{1}{2}(1-e^{-2t})+[-\frac{1}{4}+\frac{1}{2}(t-1)+\frac{1}{4}e^{-2(t-1)}]+\frac{1}{4}-\frac{t}{2}-\frac{e^{-2t}}{4}$
