solve $y''-3y'+2y=g$ whereas $g(t)=\begin{cases} 0, & t>\pi \\ \sin t,& 0\leq t\leq\pi. \end{cases}$

Question

Solve $$\left\{\begin{aligned} &y''-3y'+2y=g\\ &y(0)=y'(0)=0 \end{aligned}\right.$$ whereas $$g(t)=\begin{cases} 0, & t>\pi \\ \sin t,& 0\leq t\leq\pi. \end{cases}$$

My try:

\begin{array}{c} y''-3y'+2y=g\\ t^{2}Ly-3tLy+2Ly=\int_{0}^{\pi}\sin te^{-ts}dt\\ t^{2}Ly-3tLy+2Ly=\frac{1+e^{-\pi s}}{1+s^{2}}\\ Ly(t^{2}-3t+2)=\frac{1+e^{-\pi s}}{1+s^{2}}\\ Ly=\frac{1+e^{-\pi s}}{1+s^{2}}\frac{1}{(t-1)(t-2)}=\frac{1+e^{-\pi s}}{1+s^{2}}(\frac{1}{(t-2)}-\frac{1}{(t-1)})=(\frac{1}{1+s^{2}}+\frac{1}{1+s^{2}}e^{-\pi s})(\frac{1}{(t-2)}-\frac{1}{(t-1)})\\ (\cos t+\cos t*\delta(t-\pi))*(e^{-2t}-e^{-t}) \end{array}

which seems a bit off.

Answer 1

It should be: \begin{array}{c} y''-3y'+2y=g\\ s^{2}Ly-3sLy+2Ly=\int_{0}^{\pi}\sin te^{-ts}dt\\ s^{2}Ly-3sLy+2Ly=\frac{1+e^{-\pi s}}{1+s^{2}}\\ Ly(s^{2}-3s+2)=\frac{1+e^{-\pi s}}{1+s^{2}}\\ Ly=\frac{1+e^{-\pi s}}{(1+s^{2})(s-1)(s-2)} =(1+e^{-\pi s})\left(\frac 3{10}\frac{s}{1+s^{2}} + \frac{1}{10}\frac{1}{1+s^{2}} +\frac 15\frac{1}{s-1}-\frac 12 \frac{1}{s-2}\right) \end{array}

Let $Lz=\frac 3{10}\frac{s}{1+s^{2}} + \frac{1}{10}\frac{1}{1+s^{2}} +\frac 15\frac{1}{s-1}-\frac 12 \frac{1}{s-2}$.

Then $z=\left(\frac 3{10}\cos t+\frac 1{10}\sin t+\frac 15 e^t - \frac 12 e^{2t}\right)u(t)$

So we have $Ly=(1+e^{-\pi s})Lz = Lz + e^{-\pi s}Lz$.
Therefore $y(t)=z(t)+z(t-\pi)$.

Substitute to find: $$ y(t)=\begin{cases}0 & \text{if } t<0 \\ \frac 1{10}\Big[3\cos(t) +\sin(t) + 2e^t - 5e^{2t}\Big] & \text{if }0\le t\le\pi \\ \frac 1{10}\Big[2(e^t+e^{t-\pi}) - 5(e^{2t}+e^{2(t-\pi)})\Big] & \text{if }t>\pi\end{cases}$$