Answer
$$y= e^{-3x}+ce^{-4x}$$
Work Step by Step
Given $$ \frac{dy}{dx}+4y=e^{-3x}$$
Since
\begin{align*}
\mu(x)&=e^{\int 4dx}\\
&=e^{4x}
\end{align*}
Then
\begin{align*}
y\mu(x)&=\int \mu(x) q(x)dx\\
e^{4x}y&=\int e^{4x}e^{-3x}dx\\
&=\int e^{ x}dx\\
&=e^x+c
\end{align*}
Hence $$y= e^{-3x}+ce^{-4x}$$