Answer
$y=C_{1}e^{t}+C_{2}e^{-t}+C_{3}e^{2t}+\frac{1}{30}e^{4t}$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}'''-2{y}''-{y}'+2y=0 \;\;\;\;\Rightarrow \;\;\;\; r^3e^{rt}-2r^2e^{rt}-re^{rt}+2e^{rt}=0\\\\$
$r^3-2r^2-r+2=(r-1)(r+1)(r-2)=0 $
$ \rightarrow\;\;\;\;\; r_{1}=1\;\;\;\;\;\;\;or\;\;\;\;,r_{2}=-1\;\;\;\;\;\;or\;\;\;\;r_{3}=2\\\\$
$\boxed{y_{c}(t)= C_{1}e^{t}+C_{2}e^{-t}+C_{3}e^{2t}}$
$y_{1}=e^t\;\;\;\;\;,\;\;y_{2}=e^{-t}\;\;\;\;\;\;\;,\;\;\;y_{3}=e^{2t}$
$W(e^t,e^{-t},e^{2t})=\begin{vmatrix}
e^t & e^{-t} & e^{2t} \\
e^t & -e^{-t} & 2e^{2t} \\
e^t & e^{-t} & 4e^{2t}
\end{vmatrix}\;=\;-6e^t$
$W_{1}=\begin{vmatrix}
0 & e^{-t} & e^{2t} \\
0 & -e^{-t} & 2e^{2t} \\
e^{4t} & e^{-t} & 4e^{2t}
\end{vmatrix}\;=\;3e^{5t}$
$W_{2}=\begin{vmatrix}
e^t & 0 & e^{2t} \\
e^t & 0 & 2e^{2t} \\
e^t & e^{4t} & 4e^{2t}
\end{vmatrix}\;=\;-e^{7t}$
$W_{3}=\begin{vmatrix}
e^t & e^{-t} & 0 \\
e^t & -e^{-t} & 0 \\
e^t & e^{-t} & e^{4t}
\end{vmatrix}\;=\;-e^{4t}$
${u}'_{1}=\frac{w_{1}}{w}=-\frac{1}{2}e^{3t}$
$u_{1}=\int -\frac{1}{2}e^{3t}= -\frac{1}{6}e^{3t}$
${u}'_{2}=\frac{w_{2}}{w}=\frac{1}{6}e^{5t}$
$u_{2}=\int \frac{1}{6}e^{5t}= \frac{1}{30}e^{5t}$
${u}'_{3}=\frac{w_{3}}{w}=\frac{1}{3}e^{2t}$
$u_{3}=\int \frac{1}{3}e^{2t}= \frac{1}{6}e^{2t} $
$y_{p}=y_{1}u_{1}+y_{2}u_{2}+y_{3}u_{3}$
$y_{p}=\frac{1}{30}e^{4t}$
$y=y_{c}+y_{p}$
$y=C_{1}e^{t}+C_{2}e^{-t}+C_{3}e^{2t}+\frac{1}{30}e^{4t}$
