Answer
$y=\frac{1}{8}(\ln{x^2})^2+2$
Work Step by Step
$2xy'-\ln{x^2}=0$
$y'=\frac{\ln{x^2}}{2x}$
$\frac{dy}{dx}=\frac{\ln{x^2}}{2x}$
$dy=\frac{\ln{x^2}}{2x}dx$
$\int{dy}=\int{\frac{\ln{x^2}}{2x}dx}$
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 $y(1)=2$:
$2=\frac{1}{8}(\ln{1^2})^2+C$
$2=C$
$y=\frac{1}{8}(\ln{x^2})^2+2$
