The correct option is C |y+√x2+y2|=Cx2
xdydx=y+√x2+y2⇒dydx=y+√x2+y2x⋯(i)
This is an homogeneous D.E.
Put y=vx, so that,
dydx=v+xdvdx
Then, equation (i) becomes: v+xdvdx=vx+√x2+v2x2x⇒xdvdx=√1+v2
⇒∫dv√1+v2=∫dxx
⇒log∣∣v+√1+v2∣∣=log|x|+logc⇒|v+√1+v2|=|cx|
⇒|y+√x2+y2|=Cx2;
where |c|=C