The correct option is A tan−1(yx)−logy+c=0
y2dx+(x2−xy+y2)dy=0⇒(x2−xy+y2)dydx+y2=0
Substituting y=xv⇒dydx=v+xdvdx
(x2−x(xv)+x2v2)(v+xdvdx)+x2v=0⇒x2(xdvdx+v3+xv2dvdx+v−xvdvdx)=0⇒dvdx=−v3−vx(v2−v+1)⇒dvdx×v2−v+1−v3−v=1x
Integrating both sides, we get
∫dv(v2−v+1−v3−v)=∫1xdx⇒tan−1v−logv=logx+c⇒tan−1v−logv−logx+c=0⇒tan−1yx−logy+c=0