if, $y=\ln\left(\frac{x}{1-x}\right)$ prove $x^3\frac{d^2y}{dx^2}= \left(x\frac{dy}{dx}-y\right)^2$

Question

If $y=\ln\left(\frac{x}{1-x}\right)$, prove

$x^3\frac{d^2y}{dx^2}= \left(x\frac{dy}{dx}-y\right)^2$

My Try

I was able to differentiate this expression twice without a truoble, but I'm having difficulty in obtaining 'y' in the second derivative expression. Please help.

Answer 1

This is an exercise showing your given function satisfies the given differential equation!

According to your equation, we need $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$: $$\frac{dy}{dx}=\left(\ln\left(\frac{x}{1-x}\right)\right)^{'}=\frac{1}{\frac{x}{1-x}}\cdot\left(\frac{x}{1-x}\right)^{' }=\frac{1-x}{x}\cdot\frac{(1-x)-x(-1)}{(1-x)^2}=\frac{1}{x(1-x)}=\frac{1}{x}-\frac{1}{x-1}$$ $$\frac{d^2y}{dx^2}=\left(\frac{1}{x}-\frac{1}{x-1}\right)^{'}=-\frac{1}{x^2}+\frac{1}{(x-1)^2}$$

Plugging all of this into $x^3\cdot\frac{d^2y}{dx^2}=\left(x\frac{dy}{dx}-y\right)^2$: $$x^3\left(-\frac{1}{x^2}+\frac{1}{(x-1)^2}\right)=^?\left(x\left(\frac{1}{x}-\frac{1}{x-1}\right)-\ln\left(\frac{x}{1-x}\right)\right)^2$$ $$-x+\frac{x^3}{(x-1)^2}=^?\left(1-\frac{x}{x-1}-\ln\left(\frac{x}{1-x}\right)\right)^2$$

However the above relation can't be true since a rational function and polynomial are on the left while a logarithm is on the RHS.

Answer 2

I do not think the statement to be proven is true. We clearly have $$y=\ln x-\ln(1-x),$$ and so $$y'=\frac 1x+\frac{1}{1-x},\,\,\,\,y''=-\frac{1}{x^2}+\frac{1}{(1-x)^2}.$$ Let $x=1/2$. Then it is easy to see that the LHS is $$x^3 y''=0$$ but the RHS is $$(xy'-y)^2=\left(\frac 12\times 4-0\right)^2=4.$$