Solve ${x^2}\frac{{dy}}{{dx}} = {x^2} + xy + {y^2}$.

Given that, ${x^2}\frac{{dy}}{{dx}} = {x^2} + xy + {y^2}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = 1 + \frac{y}{x} + \frac{{{y^2}}}{{{x^2}}}$

………(i)
Let $f(x,y) = 1 + \frac{y}{x} + \frac{{{y^2}}}{{{x^2}}}$

$f(\lambda x,\lambda y) = 1 + \frac{{\lambda y}}{{\lambda x}} + \frac{{{\lambda ^2}{y^2}}}{{{\lambda ^2}{x^2}}}$

$f(\lambda x,\lambda y) = {\lambda ^0}\left( {1 + \frac{y}{x} + \frac{{{y^2}}}{{{x^2}}}} \right)$

$= {\lambda ^0}f(x,y)$

which is homogeneous expression of degree 0.
Put $y = vx \Rightarrow \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

On substituting these values in Eq.(i),

we get
$\left( {v + x\frac{{dv}}{{dx}}} \right) = 1 + v + {v^2}$

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

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

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

On integrating both sides,

we get
${\tan ^{ - 1}}v = \log |x| + C$
$\Rightarrow$ ${\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = \log |x| + C$