## Calculus 10th Edition

$y=ke^{\frac{\ln^2{x}}{2}}$
$y\ln{x}-xy'=0$ $y\ln{x}=xy'$ $\frac{\ln{x}}{x}=\frac{y'}{y}$ $\int\frac{dy}{y}=\int\frac{\ln{x}}{x}dx$ $\ln{y}=\frac{\ln^2{x}}{2}+C$ $e^{\ln{y}}=y=e^{\frac{\ln^2{x}}{2}+C}=e^{\frac{\ln^2{x}}{2}}e^C$ Let $k=e^C$ then $y=ke^{\frac{\ln^2{x}}{2}}$