## Calculus 8th Edition

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