y'=1/(x+y)令u=x+y则y'=u'-1代入方程： u'-1=1/u即u'=(1+u)/udu*u/(1+u)=dxdu[1-1/(1+u)]=dx积分：u-ln|1+u|=x+C即(x+y)-ln|1+x+y|=x+C得：y-ln|1+x+y|=C
