Question

dy/dx=(1-y)/(y-x)的通解

Answer 1

（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-xdy=-ydy
==>dx/(y-1)-xdy/(y-1)^2=-ydy/(y-1)^2 (等式两端同除(y-1)^2)
==>d(x/(y-1))=[-1/(y-1)-1/(y-1)^2]dy
==>x/(y-1)=-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-1=Ce^((x-1)/(1-y))
==>y=1+Ce^((x-1)/(1-y))
∴y=1+Ce^((x-1)/(1-y))也是原方程的解
故综合（1）和（2）知,原方程的通解是y=1和y=1+Ce^((x-1)/(1-y))（y≠1）.