令p=y'
则y"=pdp/dy
代入原方程:
pdp/dy+2p/(1-y)=0
dp=2dy/(y-1)
积分:p=2ln|y-1|+C1
即dy/dx=2ln|y-1|+C1
得dy/(2ln|y-1|+C1)=dx
再积分:∫dy/(2ln|y-1|+C1)=x+C2
哦,谢谢,不过我好像题目弄错了,能不能在帮我解一下y"+2y'²/(1-y)=0的通解,有一步我解不出来
令p=y'则y"=pdp/dy代入原方程:pdp/dy+2p²/(1-y)=0dp/p=2dy/(y-1)积分:ln|p|=2ln|y-1|+C1得p=C1(y-1)²即dy/dx=C1(y-1)²得dy/(y-1)²=C1dx再积分:-1/(y-1)=C1x+C2故有:y=1-1/(C1x+C2)