[tex]y''+2y'+y=2t^2-t+3e^{-2t}[/tex]
The characteristic equation is
[tex]r^2+2r+1=(r+1)^2=0\implies r=-1[/tex]
so the characteristic solution is
[tex]y_c=C_1e^{-t}+C_2te^{-t}[/tex]
For the particular solution, we can try looking for a solution of the form
[tex]y_p=at^2+bt+c+de^{-2t}[/tex]
[tex]\implies{y_p}'=2at+b-2de^{-2t}[/tex]
[tex]\implies{y_p}''=2a+4de^{-2t}[/tex]
Substituting into the ODE, we have
[tex](2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}[/tex]
[tex]at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}[/tex]
[tex]\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3[/tex]
So the general solution to the ODE is
[tex]y=y_c+y_p[/tex]
[tex]y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}[/tex]
