Answer
$${y}=\frac{\exp^{-3x}+5\exp^{-x}}{2}$$
Work Step by Step
$$y''+4y'+3y=0,\quad{y}(0)=2,\quad{y'}(0)=-1$$Let $y-e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2+4{\lambda}+3=0$$
$$(\lambda+3)(\lambda+1)=0$$
$$\lambda_{1,2}=-3,-1$$
The general solution is ${y}=c_{1}\exp^{-3x}+c_{2}\exp^{-x}$. Substituting in the constraints, we obtain
$c_{1}+c_{2}=2$ and $-3c_{1}-c_{2}=-1\rightarrow-3(2-{c}_{2})-c_{2}=-{1}\rightarrow-6+2{c}_{2}=-1\rightarrow{c}_{2}=\frac{5}{2}\rightarrow{c}_{1}=2-\frac{5}{2}=-\frac{1}{2}$.
$$\therefore{y}=\frac{\exp^{-3x}+5\exp^{-x}}{2}$$
