Show that the given differential equation is homogeneous and solve it:
$y \, dx + x \log \left(\frac{y}{x}\right) dy - 2x \, dy = 0$

(N/A) $y \, dx + x \log \left(\frac{y}{x}\right) dy - 2x \, dy = 0$
$\Rightarrow y \, dx = [2x - x \log \left(\frac{y}{x}\right)] dy$
$\Rightarrow \frac{dy}{dx} = \frac{y}{2x - x \log \left(\frac{y}{x}\right)}$ ........... $(1)$
Let $F(x, y) = \frac{y}{2x - x \log \left(\frac{y}{x}\right)}$
$\therefore F(\lambda x, \lambda y) = \frac{\lambda y}{2(\lambda x) - (\lambda x) \log \left(\frac{\lambda y}{\lambda x}\right)} = \frac{y}{2x - x \log \left(\frac{y}{x}\right)} = \lambda^0 F(x, y)$
Since $F(\lambda x, \lambda y) = \lambda^0 F(x, y)$,the given differential equation is homogeneous.
To solve it,substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into equation $(1)$:
$v + x \frac{dv}{dx} = \frac{vx}{2x - x \log v} = \frac{v}{2 - \log v}$
$\Rightarrow x \frac{dv}{dx} = \frac{v}{2 - \log v} - v = \frac{v - 2v + v \log v}{2 - \log v} = \frac{v(\log v - 1)}{2 - \log v}$
$\Rightarrow \frac{2 - \log v}{v(\log v - 1)} dv = \frac{dx}{x}$
$\Rightarrow \left[ \frac{-( \log v - 1) + 1}{v(\log v - 1)} \right] dv = \frac{dx}{x}$
$\Rightarrow \left[ -\frac{1}{v} + \frac{1}{v(\log v - 1)} \right] dv = \frac{dx}{x}$
Integrating both sides:
$-\int \frac{1}{v} dv + \int \frac{1}{v(\log v - 1)} dv = \int \frac{1}{x} dx$
$-\log v + \log |\log v - 1| = \log |x| + C$
$\log \left| \frac{\log v - 1}{v} \right| = \log |x| + C$
Substituting $v = \frac{y}{x}$:
$\log \left| \frac{\log (y/x) - 1}{y/x} \right| = \log |x| + C$
$\frac{x}{y} [\log (y/x) - 1] = C_1 x$
$\log (y/x) - 1 = C_1 y$