Answer
$2y''-y'-y=0$
Work Step by Step
$y=e^x+e^{-x/2}$
By the chain rule:
$y'=e^x+e^{-x/2}*\frac{d}{dx}(-\frac{x}{2})$
$y'=e^x-\frac{1}{2}e^{-x/2}$
$y''=e^x-\frac{1}{2}e^{-x/2}*\frac{d}{dx}(-\frac{x}{2})$
$y''=e^x+\frac{1}{4}e^{-x/2}$
$2y''-y'-y$
$=2(e^x+\frac{1}{4}e^{-x/2})-(e^x-\frac{1}{2}e^{-x/2})-(e^x+e^{-x/2})$
$=2e^x+\frac{1}{2}e^{-x/2}-e^x+\frac{1}{2}e^{-x/2}-e^x-e^{-x/2}$
$=0$