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