(1+y)dx-(1-x)dy=0
(1+y)dx=(1-x)dy
∫ [1/(1-x)] dx = ∫ [1/(1+y) ]dy
-ln|1-x| = ln|1+y|
ln|1-x||1-y| = c'
(1-x)(1+y) = e^(c') = C从第四步起应该是：-ln|1-x|=ln|1+y|+cln|1-x||1+y|=c(1-x)(1+y)=e^c=c对了