The solution of the differential equation{1+x(x2+y2)−−−−−−−−√}dx+{(x2+y2)−−−−−−−−√−1}ydy=0 is

[c] Rearranging the equation, we have dx−ydy+(x2+y2)−−−−−−−−√(xdx+ydy)=0 ⇒dx−ydy+12(x2+y2)−−−−−−−−√d(x2+y2)=0 On integrating, we get x−y22+12∫t√dt=c,{t=(x2+y2)−−−−−−−−√} Or x−y22+13(x2+y2)3/2=c