Question

求微分方程y"+2y'/(1-y）=0的通解

Answer 1

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