∵把原微分方程两端同乘以1/(x+y),得
dx-dy=(dx+dy)/(x+y) ==>d(x-y)=d(x+y)/(x+y)
==>d(x-y)-d(x+y)/(x+y)=0
==>d(x-y)-d[ln(x+y)]=0
==>d[x-y-ln(x+y)]=0
∴原微分方程两端同乘以1/(x+y)后就构成了全微分方程
故原微分方程得积分因子就是1/(x+y)