## Calculus (3rd Edition)

$$y(x)=\frac{1}{5} e^{2 x}-\frac{6}{5} e^{-3 x}$$
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 $$y(x)=\frac{1}{5} e^{2 x}-\frac{6}{5} e^{-3 x}$$