用a／b＝In a－In b证明计算

Since a/b = ln a - ln b
So y = ln ((1+x)/(1-x)) = ln (1+x) - ln (1-x)
dy/dx = d[ln(1+x)]/dx - d[ln(1-x)]/dx
= 1/(1+x) * d(1+x)/dx = 1/(1-x) * d(1-x)/dx
= 1/(1+x) - 1/(1-x) * (-1)
= 1/(1+x) + 1/(1-x) = (1-x+1+x)/((1+x)(1-x)) = 2/(1-x^2)

Let 1-x^2 = u
d^2y/dx^2 = d[2/(1-x^2)]/dx =d(2/u)/dx = (-1)*2*u^(-2) * du/dx
=(-2) * [1/(1-x^2)^2] * (-2x) = 4x/(1-x^2)^2

Therefore,
[2x/(1-x^2)] dy/dx = [2x(1-x^2)] * [2/(1-x^2)] = 4x/(1-x^2)^2 = d^2y/dx^2

Yeah!!!!

令y'=dy/dx=p,
则y''=d^2y／dx^2=dp/dx
∴（2x／1－x^2）dy/dx－d^2y／dx^2
=p·（2x／1－x^2）-dp/dx=0;
分离变量:dp/p=dx/（2x／1－x^2）
然后接着解微分方程就行了
我没时间算了