Differential Equations — Maths STD 12 Science — Question

$\therefore\left(x^2+y^2\right) d x-2 x y \ d y=0$
$\therefore \frac{d y}{d x}=\frac{x^2+y^2}{2 x y}....(i)$
This is homogeneous differential equation
Put $y=v x$
$\therefore \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ Given equation $(i)$ becomes,
$v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 x^2 v}$
$=\frac{x^2\left(1+v^2\right)}{2 x^2 v}$
$\therefore x \frac{d v}{d x}=\frac{1+\dot{v}^2}{2 v}-v$
$\therefore x \frac{d v}{d x}=\frac{1+v^2-2 v^2}{2 v}$
$\therefore \frac{2 v}{1-v^2} d v=\frac{d x}{x}$
Integrating both sides
$\int \frac{2 v}{1-v^2} d v=\int \frac{d x}{x}$
$\therefore -\log \left(1-v^2\right)=\log x-\log c$
$\therefore \log \left[x\left(1-v^2\right)\right]=\log c$
$\therefore x\left(1-v^2\right)=c$
$\therefore x\left(1-\frac{y^2}{x^2}\right)=c$
$\therefore x\left(\frac{x^2-y^2}{x^2}\right)=c$
$\therefore x^2-y^2=c x$
This is the general solution of given $D.E.$