Answer
${y_{p}}=-\frac{2te^{-t}}{3}$
Work Step by Step
$$y''-y-2y=2e^-t$$
Consider the homogeneous LHS. Let $y=e^{\lambda{t}}$ so that $(lny)'=\lambda$.
$$\therefore{\lambda}^2-\lambda-2=0$$
$$(\lambda+1)(\lambda-2)=0$$
$$\lambda_{1,2}=-1,2$$
$\therefore{y}=c_{1}e^{-t}+c_{2}e^{2t}$ is the general solution. Let $u(t)$ and $v(t)$ be the external functions associated with the solutions in the inhomogeneous case.
$$\therefore{y}=u(t)e^{-t}+v(t)e^{2t},\quad{y'}=u'(t)e^{-t}+v'(t)e^{2t}-u(t)e^{-t}+2v(t)e^{2t}$$
To avoid over-complicating the inhomogeneous equation with variables, let us set up expression $y'$ in such a way that it would be most apparent that the homogeneous solutions $e^{-t}$ and $e^{2t}$ are twice differentiable.
$$\therefore{u'(t)}e^{-t}+v'(t)e^{2t}=0(\Rightarrow\frac{v'(t)}{u'(t)}=-e^{-3t})$$
$$\therefore{y'}=-u(t)e^{-t}+2v(t)e^{2t}\Rightarrow{y''}=-u'(t)e^{-t}+2v'(t)e^{2t}+u(t)e^{-t}+4v(t)e^{2t}$$
Substituting this into the original equation,
$$-u'(t)e^{-t}+2v'(t)e^{2t}+e^{-t}(u(t)+u(t)-2u(t))+e^{2t}(4v(t)-2v(t)-2v(t))=2e^{-t}$$
$$-u'(t)e^{-t}+2v'(t)e^{2t}=2e^{-t}$$
$$\therefore(\begin{matrix} e^{-t} & e^{2t}\\-e^{-t} & 2e^{2t}\end{matrix})(\begin{matrix}u'(t)\\v'(t)\end{matrix})=(\begin{matrix}0\\2e^{-t}\end{matrix})$$
$$(\begin{matrix}u'(t)\\v'(t)\end{matrix})=-\frac{1}{3e^{t}}(\begin{matrix} 2e^{2t} & -e^{2t}\\e^{-t} & e^{-t}\end{matrix})(\begin{matrix}0\\2e^{-t}\end{matrix})=(\begin{matrix}-\frac{2}{3}\\\frac{2}{3}e^{-3t}\end{matrix})$$
$$\therefore{u(t)}=-\int{\frac{2}{3}}dt=-\frac{2t}{3}+c_{1},\quad{v(t)}=\int{\frac{2e^{-3t}}{3}}dt=-\frac{2e^{-3t}}{9}+c_{2}$$
$$\therefore{y}=c_{1}e^{-t}+c_{2}e^{2t}-\frac{2te^{-t}}{3}-\frac{2e^{-t}}{9}=c_{1}e^{-t}+c_{2}e^{2t}-\frac{2e^{-t}}{3}(t-\frac{1}{3})\\
{\equiv}c_{1}e^{-t}+c_{2}e^{2t}-\frac{2te^{-t}}{3}\Rightarrow{y_{p}}=-\frac{2te^{-t}}{3}\quad(\because{t}\in\mathbb{R})$$
Substituting particular solution $y_{p}$ into the original equation,
$$LHS:(\frac{4e^{-t}}{3}-\frac{2te^{-t}}{3}-(-\frac{2e^{-t}}{3}+\frac{2te^{-t}}{3})+\frac{4te^{-t}}{3}=2e^{-t}=RHS,\quad{t}\in\mathbb{R}$$
(Note #1: Following the typical procedure for the method of undetermined coefficients (i.e. substituting $y=ke^{-t}$ and comparing coefficients) here will cause the LHS=0. In such a case, follow the instructions of the question by plugging the particular solution into the equation first and observe how the coefficients interact.)
