Answer
$$y(x)=\frac{1}{5} e^{2 x}-\frac{6}{5} e^{-3 x}$$
Work Step by Step
Given $$y^{\prime}+3 y=e^{2 x}, \quad y(0)=-1$$
This a linear equation with $p(x) =3$ and $q(x) =e^{2x}$; then
\begin{align*}
\mu(x) &= e^{\int p(x)dx}\\
&=e^{\int3 dx}\\
&= e^{3x}
\end{align*}
Hence, the general solution is
\begin{align*}
\mu(x) y &=\int \mu (x)xq(x)dx\\
e^{3x} y&=\int e^{3x} e^{2x}dx \\
&=\int e^{5x}dx\\
&=\frac{1}{5}e^{5x}+C
\end{align*}
Then $$ y= \frac{1}{5}e^{2x}+C e^{-3x}$$
Since $ y(0)= -1 $, then $C= \frac{-6}{5}$, hence
$$y(x)=\frac{1}{5} e^{2 x}-\frac{6}{5} e^{-3 x}$$
