Answer
$y(x)=c_{1}e^{\frac{x}{2}}+c_{2}e^{\frac{-x}{2}}$
Work Step by Step
The given differential equation is $4y’’-y=0$
The auxiliary equation is
$4r^{2}-1=0$
This implies
$4r^{2}=1$
Thus, $r=±\frac{1}{2}$are the roots.
The general solution is
$y(x)=c_{1}e^{\frac{x}{2}}+c_{2}e^{\frac{-x}{2}}$
