Question

Find the general solution of $y\log y\frac{dx}{dy}+x-\log y=0$

Answer 1

Given

$y\log y\frac{dx}{dy}+x-\log y=0$

$\frac{dx}{dy}+(\frac{1}{y\log y})x=\frac{1}{y}$

Integrating factor will be $e^{\int \frac{dy}{y\log y}}=e^{\log \log y}=\log y$

$⟹x\log y=\int \frac{\log y}{y}dy$

$⟹x\log y=\frac{1}{2}(\log y{)}^2+C$