The correct option is B |x2−y2|=k|x|, where k is a positive constant.
(x2+y2)dx−2xydy=0 is the given equation. It can be re-written as dydx=x2+y22xy.
Now x2+y22xy is a homogeneous function of degree zero. So it is a homogeneous differential equation.
To solve, put y = vx.
So dydx=v+xdvdx(Chain rule of differentiation)
Now we have two variables v and x. All the terms including v and x can be written separately. So we can use variable separable here onwards.
Integrating (1) we get
−log|(1−v2)|−log|x|=−log|c||x(1−v2)|=|c|=k (k > 0)
⇒∣∣∣x(x2−y2)x2∣∣∣=k.⇒|x2−y2|=k|x|. which is the required solution.