求微分方程y"+2/（1-y）*（y'）^2=0的通解

Question

求微分方程y"+2/（1-y）*（y'）^2=0的通解

Answer 1

令p=y'
则y"=pdp/dy
代入方程：pdp/dy+2/(1-y)*p^2=0
dp/p=2dy/(y-1)
积分：ln|p|=2ln|y-1|+C
得：p=C1(y-1)^2
dy/(y-1)^2=C1dx
积分;-1/(y-1)=C1x+C2
故y=1-1/(C1x+C2)