The correct option is A |c(x2−y2)|=|x|
(x2+y2)dx=2xy dy⇒dydx=x2+y22xy⋯(i)
This is homogeneous D.E.
Put y=vx so that
dydx=v+xdvdx.
Then, equation (i) becomes :
v+xdvdx=x2+v2x22vx2⇒xdvdx=1+v22v−v
⇒2v1−v2dv=∫dxx⇒−ln∣∣1−v2∣∣=ln|x|+ln|c|⇒1|1−v2|=|cx|⇒x2|x2−y2|=|cx|
⇒|x|=|c(x2−y2)|, which is the required solution.