If $y = log (\sqrt{x+1}-\sqrt{x-1}), $ then $(x^2-1) \frac{d^2y}{dx^2}+x\frac{dy}{dx}$ is equal to :

$0$

The correct answer is Option (4) → $0$

$y = \log (\sqrt{x+1}-\sqrt{x-1})$

$\frac{dy}{dx}=\left(\frac{1}{\sqrt{x+1}-\sqrt{x-1}}\right)×\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x-1}}$

$\frac{dy}{dx}=\left(\frac{1}{\sqrt{x+1}-\sqrt{x-1}}\right)-\frac{1}{4(x^2-1)}$

$(x^2-1)\frac{dy}{dx}=-\frac{1}{4}\left(\frac{1}{\sqrt{x+1}-\sqrt{x-1}}\right)$

$(x^2-1)\frac{d^2y}{dx^2}=-x\frac{dy}{dx}$

$⇒(x^2-1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0$