x(1+y^2)dx+y(1+x^2)dy=0x(1+y^2)dx=-y(1+x^2)dyxdx/(1+x^2)=-ydy/(1+y^2)dx^2/2(1+x^2)=-dy^2/(1+y^2)d(1+x^2)/2(1+x^2)=-d(1+y^2)/(1+y^2)ln(1+x^2) / 2 =-ln(1+y^2) /2 +Cln(1+x^2) =-ln(1+y^2) +2C 所以通解是ln(1+x^2) +ln(1+y^2) +C=0