(1 + y) dx + (x - 1) dy = 0
(1 + y) dx = - (x - 1) dy
- dx/(x - 1) = dy/(1 + y)
-ln(x - 1) + C = ln(1 + y)
1 + y = e^[-ln(x - 1) + C]
y = C/(x - 1) - 1
化简得dx/(1-X)=dy/(1+Y)两边同时积分得-ln(1-X)=ln(1+Y)得到Y=X/(1-X)
y=x/(1 - x) + C[1]/(1 - x)
ck
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