$(x - y)dy - (x + y)dx = 0$

: which can be written as
$\cfrac{{dy}}{{dx}} = \cfrac{{x + y}}{{x - y}} = \cfrac{{1 + \left( {y/x} \right)}}{{1 - \left( {y/x} \right)}}$

…(1)
Since, R.H.S. is of the form $g\left( {y/x} \right)$,

so it is a homogeneous function of degree zero.

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

To solve this, we put $y = vx$

…(2)
Differentiating (2) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = v + x\cfrac{{dv}}{{dx}}$

Substituting the value of $y$ and $\cfrac{{dy}}{{dx}}$ in (1),

we get
$v + x\cfrac{{dv}}{{dx}} = \cfrac{{1 + v}}{{1 - v}} \Rightarrow x\cfrac{{dv}}{{dx}} = \cfrac{{1 + v}}{{1 - v}} - v$

$\Rightarrow x\cfrac{{dv}}{{dx}} = \cfrac{{1 + v - v + {v^2}}}{{1 - v}} \Rightarrow x\cfrac{{dv}}{{dx}} = \cfrac{{1 + {v^2}}}{{1 - v}}$

Integrating both sides,

we get
$\int {\cfrac{{dx}}{x} = \int {\cfrac{{1 - v}}{{1 + {v^2}}}dv \Rightarrow \int {\cfrac{{dx}}{x} = \int {\cfrac{1}{{1 + {v^2}}} - \cfrac{1}{2}\int {\cfrac{{2v}}{{1 + {v^2}}}} } dv} } }$

$\Rightarrow \log x + C = {\tan ^{ - 1}}v - \cfrac{1}{2}\log \left( {1 + {v^2}} \right)$
$\Rightarrow \log x + \cfrac{1}{2}\log \left( {1 + \cfrac{{{y^2}}}{{{x^2}}}} \right) + C = {\tan ^{ - 1}}\cfrac{y}{x}$

[From (2)]
$\Rightarrow \log x + \cfrac{1}{2}\log ({x^2} + {y^2}) - \log x + C = {\tan ^{ - 1}}\cfrac{y}{x}$

$\Rightarrow \cfrac{1}{2}\log ({x^2} + {y^2}) + C = {\tan ^{ - 1}}\cfrac{y}{x}$