Answer
$$ y= xe^x+Ce^{x}$$
Work Step by Step
Given $$y^{\prime}-y=e^{x}$$
Since the equation is linear and $p(x)=- 1$ and $q(x)= e^x$, then
\begin{align*}
\mu &=e^{\int p(x)}dx\\
&= e^{-\int dx}\\
&=e^{-x}
\end{align*}
Hence the general solution given by
\begin{align*}
\mu(x)y&=\int \mu(x)q(x)dx\\
e^{-x}y&= \int e^{-x}e^xdx\\
&= x+C
\end{align*}
Then
$$ y= xe^x+Ce^{x}$$
