Solve the differential equation $\;\;$ $y \; \log y \; dx - x \; dy = 0$
The given differential equation is $\;\;$ $y \; \log y \; dx - x \; dy = 0$
i.e. $y \; \log y \; dx = x \; dy$
i.e. $\dfrac{dx}{x} = \dfrac{dy}{y \; \log y}$ $\;\;\; \cdots \; (1)$
Integrating equation $(1)$ gives
$\displaystyle \int \dfrac{dx}{x} = \int \dfrac{dy}{y \; \log y}$ $\;\;\; \cdots \; (2)$
Now, $\displaystyle \int \dfrac{dx}{x} = \log x + \log c_1$ $\;\;\; \cdots \; (3)$
Consider $\displaystyle \int \dfrac{dy}{y \; \log y}$ $\;\;\; \cdots \; (4)$
Let $\log y = t$ $\;\;\; \cdots \; (4a)$
Differentiating equation $(4a)$ gives
$\dfrac{dy}{y} = dt$ $\;\;\; \cdots \; (4b)$
$\therefore$ $\;$ In view of equations $(4a)$ and $(4b)$, equation $(4)$ becomes
$\begin{aligned}
\int \dfrac{dy}{y \; \log y} & = \int \dfrac{dt}{t} \\\\
& = \log t + \log c_2 \\\\
& = \log \left|\log y\right| + \log c_2 \;\;\; \cdots \; (5) \hspace{2em} \left[\text{By equation } (4a)\right]
\end{aligned}$
$\therefore$ $\;$ In view of equations $(3)$ and $(5)$, equation $(2)$ becomes
$\log x + \log c_1 = \log \left|\log y\right| + \log c_2$
i.e. $\log x = \log \left|\log y\right| + \log c_2 - \log c_1$
i.e. $\log x = \log \left|\log y\right| + \log \left|\dfrac{c_2}{c_1}\right|$
i.e. $\log x = \log \left|\log y\right| + \log c$ $\;$ where $c = \dfrac{c_2}{c_1}$
i.e. $\log x = \log \left|c \; \log y\right|$
i.e. $x = c \; \log y$