$({x^2} - {y^2})dx + 2xydy = 0$

: $({x^2} - {y^2})dx + 2xydy = 0$
which can be written as $\cfrac{{dy}}{{dx}} = \cfrac{{{y^2} - {x^2}}}{{2xy}} = \cfrac{{{{\left( {\cfrac{y}{x}} \right)}^2} - 1}}{{2\left( {\cfrac{y}{x}} \right)}}$

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

Putting values of $y$ and $\cfrac{{dy}}{{dx}}$ in (1),

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

$\Rightarrow x\cfrac{{dv}}{{dx}} = - \cfrac{{1 + {v^2}}}{{2v}} \Rightarrow \cfrac{{2vdv}}{{{v^2} + 1}} = - \cfrac{{dv}}{{dx}}$
…(2)
Integrating (2) both sides,

we get
$\log |{v^2} + 1| = - \log |x| + C = \;\log |({v^2} + 1)x| = {C_1}$

$\Rightarrow \log \left| {\cfrac{{{y^2} + {x^2}}}{x}} \right| = {C_1}$

$\Rightarrow \left| {\cfrac{{{y^2} + {x^2}}}{x}} \right| = {e^{{C_1}}} \Rightarrow \cfrac{{{x^2} + {y^2}}}{x} = \pm {e^{{C_1}}} = C$

(say)
$\Rightarrow {x^2} + {y^2} = Cx$

which is the required general solution.