Differential Equations — Class 12 Maths Solution

Option d is correct

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

$\Rightarrow \cfrac{{dy}}{{dx}} = - \cfrac{{{y^2}}}{{{x^2} - xy - {y^2}}} = \cfrac{{{y^2}}}{{{y^2} + xy - {x^2}}}$

Now, $f(x,y) = \cfrac{{{y^2}}}{{{y^2} + xy - {x^2}}}$

$\therefore$ $f(\lambda x,\;\lambda y) = \cfrac{{{\lambda ^2}{y^2}}}{{{\lambda ^2}{y^2} + (\lambda x)(\lambda y) - {\lambda ^2}{y^2}}}$

$= {\lambda ^0}\left( {\cfrac{{{y^2}}}{{{y^2} + xy - {x^2}}}} \right) = {\lambda ^0}f(x,y)$

$\therefore$ $f(x,\;y)$ is homogeneous function of degree zero.

So, (1) is homogeneous differential equation.

Exercise-9.6