Answer
$$y = \frac{1}{n^{2}+1}(n \cos x+ \sin x)+Ce^{-nx}.$$
Work Step by Step
Given $$y^{\prime}+n y=\cos x$$
This is a linear equation with $p(x)=n$ and $q(x) =\cos x$
Since
\begin{align*}
\mu(x)&=e^{\int p(x)dx}\\
&=e^{\int ndx}\\
&=e^{nx}
\end{align*}
Then
\begin{align*}
\mu(x)y&=\int \mu(x)q(x)dx\\
e^{nx} y&=\int e^{nx} \cos xdx
\end{align*}
Use $$ \int e^{a u} \cos b u d u=\frac{e^{a u}}{a^{2}+b^{2}}(a \cos b u+b \sin b u)+C$$
Then
\begin{align*}
e^{nx} y&= \frac{e^{nx}}{n^{2}+1}(n \cos x+ \sin x)+C\\
y&= \frac{1}{n^{2}+1}(n \cos x+ \sin x)+Ce^{-nx}.
\end{align*}
