(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|+cln|1-x||1+y|=c(1-x)(1+y)=e^c=c 求采纳 大半夜做题难
