∵(2x+xy^2)dx+(2y+x^2*y)dy=0
==>(xy^2dx+x^2*ydy)+(2xdx+2ydy)=0
==>d(x^2*y^2)+2d(x^2+y^2)=0
==>x^2*y^2+2(x^2+y^2)=C (C是常数)
∴原方程的通解是x^2*y^2+2(x^2+y^2)=C.