Question

求微分方程y"+(y'^2)/(1-y)=0的疑问

我求到dp/dy=-p/1-y,那个-p/1-y的负号不放进1-y行不行，为什么？

Answer 1

解：设y'=p，则y''=pdp/dy
代入原方程得pdp/dy+p²/(1-y)=0
==>dp/dy=p/(y-1)
==>dp/p=dy/(y-1)
==>ln │p│=ln│y-1│+ln│C1│ (C1是积分常数)
==>p=C1(y-1)
==>y'=C1(y-1)
==>dy/(y-1)=C1dx
==>ln│y-1│=C1x+ln│C2│ (C2是积分常数)
==>y-1=C2e^(C1x)
故原方程的通解是y=1+C2e^(C1x) （C1，C2是积分常数）。