What is the solution of the differential equationdxdy+xy−y2=0?

[c] dxdy+xy−y2=0;dxdy+xy=y2 This is a linear differential equation of the form dxdy+P1x=Q1; Here, P=1y and Q=y2 ∴ I.F. =e∫Pdy=e∫1ydy=elogy=y So, required solution is x⋅y=∫y2⋅ydy+c;xy=∫y3dy+c xy=y44+c;4xy=y4+c