Answer
See below
Work Step by Step
We are given that $y''-y'-2y=10e^{-(t-a)}\sin[2(t-a)]u_a(t)$
and $y(0)=2\\
y'(0)=0$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(s)-sY(0)-Y'(0)]-3Y(s)=\frac{e^{-s}}{s}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-1)-2s=\frac{e^{-s}}{s}\\
Y(s)(s^2-1)=\frac{e^{-s}}{s}+2s$
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:
$Y(s)=\frac{e^{-s}}{s(s^2-1)}+\frac{2s}{s^2-1}\\
=\frac{-e^{-s}}{s}+\frac{e^{-s}}{2(s+1)}+\frac{e^{-s}}{2(s-1)}-\frac{1}{s+1}-\frac{1}{s-1}\\\
=e^{-s}(\frac{-1}{s}+\frac{1}{2(s+1)+\frac{1}{2(s-1)}}-(\frac{1}{s+1}+\frac{1}{s-1})$
Now,$f(t)= L^{-1}[e^{-s}(\frac{-1}{s}+\frac{1}{2(s+1)+\frac{1}{2(s-1)}}-(\frac{1}{s+1}+\frac{1}{s-1})]\\
=-u_1(t)+\frac{1}{2}e^{-(t-1)}u_1(t)+\frac{1}{2}u_1(t)e^{t-1}-e^{-t}-e^t\\
=u_1(t)[-1+\frac{1}{2}e^{-(t-1)}+\frac{1}{2}e^{t-1}]-e^{-t}-e^t$
