Answer
See below
Work Step by Step
We are given that $y'-3y=10e^{-(t-a)}\sin[2(t-a)]u_a(t)$
and $y(0)=5$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[sY(s)-y(0)]-3Y(s)=\frac{20e^{-as}}{(s+1)^2+4}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s-3)-5=\frac{20e^{-as}}{(s+1)^2+4}\\
Y(s)(s-3)=\frac{20e^{-as}}{(s+1)^2+4}+5$
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:
$Y(s)=\frac{20e^{-as}}{[(s+1)^2+4](s-3)}+\frac{5}{s-3}\\
=e^{-as}(\frac{1}{s-3}-\frac{s-1}{(s+1)^2+4}-\frac{4}{(s+1)^2+4})+\frac{5}{s-3}$
Now,$f(t)= L^{-1}[e^{-as}(\frac{1}{s-3}-\frac{s-1}{(s+1)^2+4}-\frac{4}{(s+1)^2+4})+\frac{5}{s-3}]\\
=e^{3(t-a)}u_a(t)-e^{-(t-a)}\cos 2(t-a)u_a(t)-2u_a(t)\sin 2(t-a)+5e^{3t}$
