(1+x)dx-(1-y)dy=(dx-dy)+(xdx+ydy)=d(x-y)+d(x^2+y^2)=0 即d(x-y)=-d(x^2+y^2) 两端积分,得x-y=-(x^2+y^2)/2+c 所以,(x^2+y^2)/2+x-y=c
