Answer
$$ y= 1+Ce^{-x}$$
Work Step by Step
Given $$y^{\prime}+y=1$$
Since the equation is linear and $p(x)= 1$ and $q(x)= 1$, 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^xy&= \int e^xdx\\
&=e^x+C
\end{align*}
Then
$$ y= 1+Ce^{-x}$$
