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