$\left\{ {x\cos \left( {\cfrac{y}{x}} \right) + y\sin \left( {\cfrac{y}{x}} \right)} \right\}ydx$ $= \left\{ {y\sin \left( {\cfrac{y}{x}} \right) - x\cos \left( {\cfrac{y}{x}} \right)} \right\}$

: $\left\{ {x\cos \left( {\cfrac{y}{x}} \right) + y\sin \left( {\cfrac{y}{x}} \right)} \right\}ydx$ $= \left\{ {y\sin \left( {\cfrac{y}{x}} \right) - x\cos \left( {\cfrac{y}{x}} \right)} \right\}$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{yx\left\{ {\cos \cfrac{y}{x} + \cfrac{y}{x}\sin \cfrac{y}{x}} \right\}}}{{xx\left\{ {\cfrac{y}{x}\sin \cfrac{y}{x} - \cos \cfrac{y}{x}} \right\}}}$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{\cfrac{y}{x}\left\{ {\cos \cfrac{y}{x} + \cfrac{y}{x}\sin \cfrac{y}{x}} \right\}}}{{\cfrac{y}{x}\sin \cfrac{y}{x} - \cos \cfrac{y}{x}}}$

… (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
$y = vx \Rightarrow \cfrac{{dy}}{{dx}} = v + x\cfrac{{dy}}{{dx}}$ ,

then the equation (1) becomes
$v + x\cfrac{{dv}}{{dx}} = \cfrac{{v(\cos v + v\sin v)}}{{v\sin v - \cos v}} \Rightarrow x\cfrac{{dv}}{{dx}} = \cfrac{{v\cos v + {v^2}\sin v}}{{v\sin v - \cos v}} - v$

$\Rightarrow x\cfrac{{dv}}{{dx}} = \cfrac{{2v\cos v}}{{v\sin v - \cos v}}$

$\Rightarrow \left( {\cfrac{{v\sin v - \cos v}}{{v\cos v}}} \right)dv = \cfrac{2}{x}dx$

$\Rightarrow \left( {\tan v - \cfrac{1}{v}} \right)dv = \cfrac{2}{x}dx$

… (2)
Integrating (2) both sides,

we get
$- \log |\cos v| - \log |v| = 2\log |x| + {C_1}$

$\Rightarrow \log |v\cos v| + 2\log |x| = - {C_1}$

$\Rightarrow \log (|v\cos v||x{|^2}) = - {C_1}$

$\Rightarrow |v\cos v||x{|^2} = {e^{ - {C_1}}} \Rightarrow {x^2}v\cos v = \pm {e^{ - {C_1}}} = C$

(say)
$\Rightarrow {x^2}\cfrac{y}{x}\cos \left( {\cfrac{y}{x}} \right) = C$

$\Rightarrow xy\cos \cfrac{y}{x} = C$ ,

which is the required general solution.