Answer
$y(t)=1+\frac{1}{4}(t^2+3t)-te^t$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}'''-3{y}''+2{y}'=0 \;\;\;\;\Rightarrow \;\;\;\; r^3e^{rt}-3r^2e^{rt}+2re^{rt}=0\\\\$
$r^3-3r^2+2r=r(r-1)(r-2)=0 $
$ \rightarrow\;\;\;\;\; r_{1}=0\;\;\;\;\;\;\;or\;\;\;\;,r_{2}=1\;\;\;\;or\;\;\;\;r_{3}=2\\\\$
$\boxed{y_{c}(t)= C_{1}+C_{2}e^{t}+C_{3}e^{2t}}$
Let; $\;\;\;\;Y(t)=At^2+Bt+Cte^t$
${Y}'=2At+B+Ce^t+Cte^t$
${Y}''=2A+2Ce^t+Cte^t$
${Y}'''=3Ce^t+Cte^t$
${Y}^{(4)}=16Acos(2t)+16Bsin(2t)$
${y}'''-3{y}''+2{y}'=t+e^t$
$3Ce^t+Cte^t-6A-6Ce^t-3Cte^t+4At+2B+2Ce^t+2Cte^t=t+e^t$
$-Ce^t+4At+(-6A+2B) =t+e^t \;\;\;\;\;\;\Rightarrow \;\;\;A=\frac{1}{4}\;\;\;,\;\;\;B=\frac{3}{4}\;\;\;\;\;,\;\;\;C=-1$
$\boxed{Y(t)=\frac{1}{4}t^2+\frac{3}{4}t-te^t}$
The general solution :
$y(t)=y_{c}(t)+Y(t)$
$y(t)=C_{1}+C_{2}e^{t}+C_{3}e^{2t}+\frac{1}{4}t^2+\frac{3}{4}t-te^t$
$y(0)=C_{1}+C_{2}+C_{3}=1 \;\;\;\;\;\; \;\;\;\;$
${y}'(0)=C_{2}+2C{3}-\frac{1}{4}=\frac{1}{4} \;\;\;\;\;\; \;\;\;\;$
${y}''(0)=C_{2}+4C_{3}-\frac{-3}{2}=\frac{-3}{2}\;\;\;\;\;\; \;\;\;\;$
$C_{1}=1\;\;\;\;\;,\;\;\;C_{2}=0\;\;\;\;,\;\;\;\;C_{3}=0\;\;\;\;\;\;\;\;$
$y(t)=1+\frac{1}{4}t^2+\frac{3}{4}t-te^t$
$y(t)=1+\frac{1}{4}(t^2+3t)-te^t$
