$y''+2y'+y=2(t-3)U(t-3)$ given $y(0)=2$ and $y'(0)=1$
I did the transformation and obtained $Y=e^{3s}(\frac{1}{s^2}-\frac{2}{s}-\frac{1}{s^2}+\frac{2}{s+1})+(\frac{3}{(s+1)^2}+\frac{2}{(s+1)})$
This way my result is ${y=t - \frac{t}{e^{3} e^{t}} + \frac{3 t}{e^{t}} + 1 - \frac{1}{e^{3} e^{t}} + \frac{2}{e^{t}}}$
wich is different from the solutions result $y=e^{-t } (2+3t)+2(t-5+(t-1)e^{t-3})U(t-3)$
Can you find what is wrong?
let $Y(s) = L(y) = \int_0^\infty y(t)e^{-st}dt$ be the laplace transform of $y.$ then, $$L(y^\prime) = sY - y(0) = sY - 2 , L(y^{\prime \prime} = s(sY - 2) - 1=s^2Y-2s -1$$ i am assuming that $U(t) = 1$ if $t \ge 0$ and $0$ otherwise. $$\int_0^\infty (t-3)U(t-3)e^{-st} \ dt = \int_3^\infty(t-3)e^{-st}dt= \dfrac{e^{-3s}}{s^2}$$
transforming the equation we get $$s^2Y - 2s - 1 +2(sY-2) + Y = \dfrac{2e^{-3s}}{s^2}$$ solving for $Y$ you get $$Y =\dfrac{2e^{-3s}}{s^2(s+1)^2} +\dfrac{2s+5}{(s+1)^2} =2e^{-3s}\left(\dfrac{1}{s^2} - \dfrac{2}{s}+ \dfrac{1}{(s+1)^2} +\dfrac{2}{(s+1)}\right) +\dfrac{3}{(s+1)^2} + \dfrac{2}{s+1}$$
taking the inverse transformation $$y(t) = \left((t-3) - 2 + ((t-3)+2)e^{-t+3} \right) U(t-3)+\left(3t+2 \right)e^{-t} $$