Simone
解：（1）显然，y=1是原方程的解
（2）当y≠1时，
∵dy/dx=(1-y)/(y-x)
==>(y-1)dx-xd见素度卷频乙y=-ydy
==>dx/(y-1)-xdy/(y-1)^2=-ydy/(y-1)^2
(等式两端同除(y-1察何连矛评春)^2)
==>d(x/(y-1))=[-1/(y-1)-1/(y-1)^2]dy
==>x/(y-1)=-计附客陆评防客甲ln│y-1│+1/(y-1)+ln│c│
(c是非零常数)
==>(x-1)/(y-1)=ln│c/(y-1)│
==>e^((x-1)/(y-1))=c/(y-1)
==>y析使否井脱-1=ce^((x-1)/(减减供项业病1-y))
==>y=1+ce^((x-1)/(1-y))
∴y=1+ce^((x来自-1)/(1-y)绿起甚非阳假)也是原方程的解
故综合（360问答1）和（2）知，原方程的通解是y=1和y=1+ce^((x-1盟并)/(1-y))（y≠1）。
