Answer
$${y}=\frac{2\sqrt33}{33}e^{-(\frac{1-\sqrt33}{4})x}-\frac{2\sqrt33}{33}e^{-(\frac{1+\sqrt33}{4})x}$$
Work Step by Step
$$2y''+y'-4y=0,\quad{y}(0)=0,\quad{y'}(0)=1$$Let $y=e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$$2{\lambda}^2+{\lambda}-4=0$$
$$\lambda_{1,2}=\frac{-1\pm\sqrt{1+32}}{4}=\frac{-1\pm\sqrt33}{4}$$
The general solution is ${y}=c_{1}e^{-(\frac{1-\sqrt33}{4})x}+c_{2}e^{-(\frac{1+\sqrt33}{4})}{x}$. Substituting in the constraints, we obtain
$c_{1}+c_{2}=0$ and $-(\frac{1-\sqrt33}{4})c_{1}-(\frac{1+\sqrt33}{4})c_{2}=1\Rightarrow{c}_{1}=-c_{2}\Rightarrow-\frac{\sqrt33}{2}{c}_{2}=1\Rightarrow{c}_{2}=-\frac{2\sqrt33}{33}\Rightarrow{c}_{1}=\frac{2\sqrt33}{33}$.
$$\therefore{y}=\frac{2\sqrt33}{33}e^{-(\frac{1-\sqrt33}{4})x}-\frac{2\sqrt33}{33}e^{-(\frac{1+\sqrt33}{4})x}$$
