$({x^2} + xy)dy = ({x^2} + {y^2})dx$

: $\cfrac{{dy}}{{dx}} = \cfrac{{{x^2} + {y^2}}}{{{x^2} + xy}}$ or $\cfrac{{dy}}{{dx}} = \cfrac{{{x^2}\left( {1 + {{\left( {\cfrac{y}{x}} \right)}^2}} \right)}}{{{x^2}\left( {1 + \cfrac{y}{x}} \right)}} = \cfrac{{1 + {{\left( {\cfrac{y}{x}} \right)}^2}}}{{1 + \cfrac{y}{x}}}$

…(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 $y = vx$

…(2)
Differentiating (2) w.r.t. $x$,

we get $\cfrac{{dy}}{{dx}} = v + x\cfrac{{dv}}{{dx}}$

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

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

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

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

[Integrating both sides]
$\Rightarrow \int {\left( {1 + \cfrac{2}{{v - 1}}} \right)dv} = - \log |x| + {C_1}$

$\Rightarrow v + 2\log |v - 1| = - \log |x| + {C_1}$

$\Rightarrow \log |v - 1{|^2} + \log |x| = - v + {C_1}$

$\Rightarrow \log |x{(v - 1)^2}| = - v + {C_1}$

$\Rightarrow \left| {x{{\left( {\cfrac{y}{x} - 1} \right)}^2}} \right| = {e^{ - y/x + {C_1}}} \Rightarrow \cfrac{{{{(y - x)}^2}}}{x} = \pm {e^{{C_1}}}{e^{ - y/x}}$

$\Rightarrow {(y - x)^2} = \pm x{e^{{C_1}}}{e^{ - y/x}} \Rightarrow {(y - x)^2} = Cx{e^{ - y/x}}$

where $C = \pm {e^{{C_1}}}$ is an arbitrary constant.