Answer
See below
Work Step by Step
We are given that $y''-4y=u_1(t)-u_2(t) $
and $y(0)=0\\
y'(0)=4$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(s)-sY(0)-Y'(0)]-4Y(s)=\frac{e^{-s}}{s}-\frac{e^{-2s}}{s}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-4)-4=\frac{e^{-s}}{s}-\frac{e^{-2s}}{s}\\
Y(s)(s^2-4)=\frac{e^{-s}}{s}-\frac{e^{-2s}}{s}+4$
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:
$Y(s)=\frac{e^{-s}}{s(s-1)(s+2)}-\frac{e^{-2s}}{s(s-2)(s+2)}+\frac{4}{(s-2)(s+2)}\\
=\frac{e^{-s}}{4}(\frac{-1}{s}+\frac{1}{s-2}+\frac{1}{s+2})+\frac{e^{-2s}}{4}(\frac{1}{s}-\frac{1}{s-2}-\frac{1}{s+2})+(\frac{1}{s-2}-\frac{1}{s+2})$
Now,$f(t)= L^{-1}[\frac{e^{-s}}{4}(\frac{-1}{s}+\frac{1}{s-2}+\frac{1}{s+2})+\frac{e^{-2s}}{4}(\frac{1}{s}-\frac{1}{s-2}-\frac{1}{s+2})+(\frac{1}{s-2}-\frac{1}{s+2})]\\
=\frac{-1}{4}(1-e^{2(t-1)}+e^{-2(t+1)})u_1(t)+\frac{1}{4}(1-e^{2(t-1)}-e^{-2(t-2)})u_2(t)+e^{2t}-e^{-2t}$
