Answer
$$ y= \frac{1}{2}\left(\sin x+\cos x\right)+Ce^{-x}$$
Work Step by Step
Given $$y^{\prime}+ y=\cos{x}$$
This a linear equation with $p(x) =1$ and $q(x) =\cos x$, then
\begin{align*}
\mu(x) &= e^{\int p(x)dx}\\
&=e^{\int dx}\\
&= e^x
\end{align*}
Hence the general solution is
\begin{align*}
\mu(x) y &=\int \mu (x)xq(x)dx\\
e^x y&=\int e^x \cos {x}dx
\end{align*}
Use $$\int e^{a u} \cos b u d u=\frac{e^{a u}}{a^{2}+b^{2}}(a \cos b u+b \sin b u)+C $$
We get
\begin{align*}
e^x y&=\int e^x \cos {x}dx \\
&= \frac{1}{2}e^x\left(\sin x+\cos x\right)+C
\end{align*}
Hence $$ y= \frac{1}{2}\left(\sin x+\cos x\right)+Ce^{-x}$$
