Answer
$y=\pm \sqrt {\dfrac{2x^2 \ln x -x^2+C}{2}}$
Work Step by Step
We are given that $\dfrac{1}{x} \dfrac{dy}{dx}=\dfrac{1}{y} \ln x$
We will separate the variables to obtain:
$y \ dy=x \ln x \ dx$
Integrate to obtain:
$\int y \ dy=\int x \ln x \ dx$
This implies that $\dfrac{y^2}{2}=\dfrac{x^2}{2} \ln x- \dfrac{x^2}{4}+C$
Therefore, we have: $y=\pm \sqrt {\dfrac{2x^2 \ln x -x^2+C}{2}}$
