Answer
$y_{p}=\frac{3t^{2}e^{-t}}{2}$
Work Step by Step
$$y''+2y'+y=3e^{-t}$$
Consider the homogeneous LHS. Let $y=e^{\lambda{t}}$ so that $(lny)'=\lambda$.
$$\therefore(\lambda+1)^2=0$$
$$\lambda=-1$$
$\therefore{y}=(c_{1}+c_{2})e^{-t}$ is the general solution. Let $u(t)$ be the function associated with the solution in the inhomogeneous case.
$$\therefore{y}=u(t)e^{-t},\quad{y'}=u'(t)e^{-t}-u(t)e^{-t},\quad{y''}=u''(t)e^{-t}-2u'(t)e^{-t}+u(t)e^{-t}$$
$$\therefore{y''}+2y'+y=u''e^{-t}=3e^{-t}$$
By association, $u''(t)=3$
$$\therefore{u(t)}=\int{\int{3}dt}dt=3(\frac{t^2}{2}+c_{1}t+c_{2})\equiv(c_{1}+c_{2}t)e^{-t}+\frac{3t^{2}e^{-t}}{2}\\
{\Rightarrow}y_{p}=\frac{3t^{2}e^{-t}}{2}$$
Substituting particular solution $y_{p}$ into the original equation,
$$LHS:3e^{-t}-6te^{-t}+\frac{3t^{2}e^{-t}}{2}+2(3te^{-t}-\frac{3t^{2}e^{-t}}{2})+\frac{3t^{2}e^{-t}}{2}=3e^{-t}=RHS,\quad{t}\in\mathbb{R}$$
