Answer
\begin{equation}
y=\frac{t^{3}}{3} e^{2 t}+c e^{2 t}
\end{equation}
Work Step by Step
finding the integrating factor
\begin{equation}
\begin{array}{l}
y^{\prime}-2 y=t^{2} e^{2 t} \\
\mu(t)=e^{\int p(t) d t} \\
p(t)=-2 \\
\mu(t)=e^{\int-2 d t} \\
\mu(t)=e^{-2 t}
\end{array}
\end{equation}
multiplying the integrating factor on both sides
\begin{equation}
\begin{array}{c}
e^{-2 t}\left(y^{\prime}-2 y\right)=e^{-2 t} \cdot t^{2} e^{2 t} \\
e^{-2 t} y^{\prime}-2 e^{-2 t} y=t^{2} e^{2 t-2 t} \\
\left(e^{-2 t} y\right)^{\prime}=t^{2}
\end{array}
\end{equation}
integrating on both sides
\begin{equation}
\begin{array}{c}
\int\left(e^{-2 t} y\right)^{\prime} d t=\int t^{2} d t \\
e^{-2 t} y=\frac{t^{3}}{3}+c
\end{array}
\end{equation}
here, c is the integration constant. The answer is
\begin{equation}
y=\frac{t^{3}}{3} e^{2 t}+c e^{2 t}
\end{equation}
