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