x^2 y" - x y' + y = 2 log x+ x^2 log x + x^2 + 1/x
You can also let $\displaystyle u = x y' - y$ so $\displaystyle u' = x y'' $ so your equation becomes
$\displaystyle xu' - u = 2 \log x+ x^2 \log x + x^2 + 1/x$
a first order linear equation. Once you solve it you then solve $\displaystyle xy' - y = u$ which also a first order linear equation.