Answer
$${y}=-1-\exp^{-3x}$$
Work Step by Step
$$y''+3y'=0,\quad{y}(0)=-2,\quad{y'}(0)=3$$Let $y=e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2+3{\lambda}=0$$
$$\lambda(\lambda+3)=0$$
$$\lambda_{1,2}=0,-3$$
The general solution is ${y}=c_{1}+c_{2}\exp^{-3x}$. Substituting in the constraints, we obtain
$c_{1}+c_{2}=-2$ and $-3c_{2}=3\Rightarrow{c}_{2}=-1=c_{1}$.
$$\therefore{y}=-1-\exp^{-3x}$$
