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