Answer
$$y= e^{-x}\sin (e^x)+ce^{-x}$$
Work Step by Step
Given $$y'+y=\cos(e^x)$$
Since
\begin{align*}
\mu(x)&=e^{\int dx}\\
&=e^{x }
\end{align*}
Then
\begin{align*}
y\mu(x)&=\int \mu(x) q(x)dx\\
e^{x }y&=\int e^{x} \cos(e^x)dx\\
&= \sin (e^x)+c
\end{align*}
Hence $$y= e^{-x}\sin (e^x)+ce^{-x}$$
