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