$ydx + x\log \left( {\cfrac{y}{x}} \right)dy - 2xdy = 0$

: $ydx + x\log \left( {\cfrac{y}{x}} \right)dy - 2xdy = 0$

$\Rightarrow \cfrac{y}{x} + \log \left( {\cfrac{y}{x}} \right)\cfrac{{dy}}{{dx}} - 2\cfrac{{dy}}{{dx}} = 0 \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{y/x}}{{2 - \log (y/x)}}$

…(1)
Since R.H.S. is of the form $g(y/x)$, so it is a homogeneous function of degree zero.

Therefore, equation (1) is a homogeneous differential equation.

To solve this, we put $y = vx$

$\Rightarrow \cfrac{{dy}}{{dx}} = v + x\cfrac{{dv}}{{dx}}$, then (1) becomes $v + x\cfrac{{dv}}{{dx}} = \cfrac{v}{{2 - \log v}}$

$\Rightarrow x\cfrac{{dv}}{{dx}} = \cfrac{v}{{2 - \log v}} - v \Rightarrow \cfrac{{(2 - \log v)dv}}{{v\log v - v}} = \cfrac{{dx}}{x}$

$\Rightarrow \cfrac{{1 - (\log v - 1)}}{{v(\log v - 1)}}d\nu = \cfrac{{dx}}{x}$

…(2)
Integrating (2) both sides,

we get
$\int {\left( {\cfrac{1}{{v(\log v - 1)}} - \cfrac{1}{v}} \right)} dv = \int {\cfrac{{dx}}{x}}$

$\Rightarrow \int {\cfrac{{1/v}}{{\log v - 1}}} dv - \int {\cfrac{1}{v}} dv = \int {\cfrac{{dx}}{x}}$

$\Rightarrow \log |\log v - 1| - \log |v| = \log |x| + {C_1}$

$\Rightarrow \log \left| {\cfrac{{\log v - 1}}{{vx}}} \right| = {C_1} \Rightarrow \left| {\cfrac{{\log v - 1}}{{vx}}} \right| = {e^{{C_1}}}$

$\Rightarrow \cfrac{{\log v - 1}}{{vx}} = \pm {e^{{C_1}}} = C$
(say)
$\Rightarrow \log \left( {\cfrac{y}{x}} \right) - 1 = Cy$

which is the required general solution.