## Calculus 10th Edition

$y=\frac{1}{8}(\ln{x^2})^2+2$
$2xy'-\ln{x^2}=0$ $y'=\frac{\ln{x^2}}{2x}$ $\frac{dy}{dx}=\frac{\ln{x^2}}{2x}$ $dy=\frac{\ln{x^2}}{2x}dx$ $\int{dy}=\int{\frac{\ln{x^2}}{2x}dx}$ Let $u=\ln{x^2}$ and $du=\frac{2x}{x^2}=\frac{2}{x}$. $y=\frac{1}{4}\int{\frac{2\ln{x^2}}{x}dx}$ $y=\frac{1}{4}\int{udu}$ $y=\frac{1}{8}u^2+C$ $y=\frac{1}{8}(\ln{x^2})^2+C$ Using the initial condition $y(1)=2$: $2=\frac{1}{8}(\ln{1^2})^2+C$ $2=C$ $y=\frac{1}{8}(\ln{x^2})^2+2$