## Calculus 8th Edition

$y(x)=c_{1}e^{\frac{x}{2}}+c_{2}e^{\frac{-x}{2}}$
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 $y(x)=c_{1}e^{\frac{x}{2}}+c_{2}e^{\frac{-x}{2}}$