Answer
See below
Work Step by Step
We are given that $y''+4y'+5y=5u_3(t)$
and $y(0)=2\\
y'(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(s)-sY(0)-Y'(0)]+4[sY(s)-y(0)]+5Y(s)=\frac{5e^{-3s}}{s}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+4s+5)-2s-9=\frac{5e^{-3s}}{s}\\
Y(s)(s^2+4s+5)=\frac{5e^{-3s}}{s}+2s+9$
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:
$Y(s)=\frac{5e^{-3s}}{s(s+1)(s+4)}+\frac{2s}{(s+1)(s+4)}+\frac{9}{(s+1)(s+4)}\\
=5e^{-3s}(\frac{1}{4s}-\frac{1}{3(s+1)}+\frac{1}{12(s+4)})+(\frac{-2}{3(s+1)}+\frac{8}{3(s+4)})+\frac{3}{s+1}-\frac{3}{s+4}$
Now,$f(t)= L^{-1}[5e^{-3s}(\frac{1}{4s}-\frac{1}{3(s+1)}+\frac{1}{12(s+4)})+(\frac{-2}{3(s+1)}+\frac{8}{3(s+4)})+\frac{3}{s+1}-\frac{3}{s+4}]\\
=5(\frac{e^{-(t-3)}}{12}+\frac{e^{-4(t-3)}}{12})u_3(t)+\frac{7}{3}e^{-t}-\frac{1}{3}e^{-4t}$
