Differential Equations — Class 12 Maths Solution

Given, differential equation is

${y^2}dx + \left( {{x^2} - xy + {y^2}} \right)dy = 0$

$\Rightarrow$ ${y^2}dx = - \left( {{x^2} - xy + {y^2}} \right)dy$

$\Rightarrow$ ${y^2}\frac{{dx}}{{dy}} = - \left( {{x^2} - xy + {y^2}} \right)$

$\Rightarrow$ $\frac{{dx}}{{dy}} = - \left( {\frac{{{x^2}}}{{{y^2}}} - \frac{x}{y} + 1} \right)$
……(i)

which is a homogeneous differential equation.
Put $\frac{x}{y} = v$ or $x = vy$

$\Rightarrow$ $\frac{{dx}}{{dy}} = v + y\frac{{dv}}{{dy}}$

On substituting these values in Eq. (i),

we get $v + y\frac{{dv}}{{dy}} = - \left[ {{v^2} - v + 1} \right]$

$\Rightarrow$ $y\frac{{dv}}{{dy}} = - {v^2} + v - 1 - v$

$\Rightarrow$ $y\frac{{dv}}{{dy}} = - {v^2} - 1 \Rightarrow \frac{{dv}}{{{v^2} + 1}} = - \frac{{dy}}{y}$

On integrating both sides,

we get
${\tan ^{ - 1}}(v) = - \log y + C$

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