Answer
$${y}=c_{1}\exp^{-2x}+c_{2}\exp^{-x}$$
Work Step by Step
$$y''+3y'+2y=0$$Let $y-e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2+3{\lambda}+2=0$$
$$(\lambda+2)(\lambda+1)=0$$
$$\lambda_{1,2}=-2,-1$$
$$\therefore{y}=c_{1}\exp^{-2x}+c_{2}\exp^{-x}$$