Differential Equations — Class 12 Maths Solution

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

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

…(1)
Since, R.H.S. is of the form , so it is a homogeneous function of degree zero.

Therefore, equation (1) is a homogeneous differential equation.

To solve this, we put
$\Rightarrow \cfrac{{dy}}{{dx}} = v + x\cfrac{{dv}}{{dx}}$

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

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

On integration,

we get
$\log |x| = \int {\cfrac{1}{{1 - 2{v^2}}}} dv$
$= \log |x| = \cfrac{1}{2}\int {\cfrac{{dv}}{{{{\left( {\cfrac{1}{{\sqrt 2 }}} \right)}^2} - {v^2}}}}$

$\Rightarrow \log |x| = \cfrac{1}{2}.\cfrac{1}{{2 \cdot \cfrac{1}{{\sqrt 2 }}}}\log \left| {\cfrac{{\cfrac{1}{{\sqrt 2 }} + v}}{{\cfrac{1}{{\sqrt 2 }} - v}}} \right| + C$

$\Rightarrow \log |x| = \cfrac{1}{{2\sqrt 2 }}\log \left| {\cfrac{{1 + \sqrt 2 v}}{{1 - \sqrt 2 v}}} \right| + C$

$\Rightarrow \log |x| = \cfrac{1}{{2\sqrt 2 }}\log \left| {\cfrac{{1 + \sqrt 2 \cdot \cfrac{y}{x}}}{{1 - \sqrt 2 \cdot \cfrac{y}{x}}}} \right| + C$

$\Rightarrow \log |x| = \cfrac{1}{{2\sqrt 2 }}\log \left| {\cfrac{{x + \sqrt 2 y}}{{x - \sqrt 2 y}}} \right| + C$

, where $C$ is an arbitrary constant.