If $y=\sqrt{x+1}+\sqrt{x-1}$ then $(x^2-1)y_2+xy_1=$

[where $y_2=\frac{d^2y}{dx^2}$ & $y_1=\frac{dy}{dx}$]

$\frac{1}{4}y$

The correct answer is Option (4) → $\frac{1}{4}y$

$y=\sqrt{x+1}+\sqrt{x-1}$

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

$\frac{d^2y}{dx^2}=\frac{1}{2}\left(-\frac{1}{2}×\frac{1}{(x+1)^{3/2}}-\frac{1}{2}×\frac{1}{(x-1)^{3/2}}\right)$

$=-\frac{1}{4}\left(\frac{1}{(x+1)^{3/2}}+\frac{1}{(x-1)^{3/2}}\right)$

$(x^2-1)y_2+y_1=(x-1)(x+1)\left(-\frac{1}{4}\left[\frac{1}{(x+1)^{3/2}}+\frac{1}{(x-1)^{3/2}}\right]\right)+\frac{1}{2}\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}\sqrt{x-1}}\right)$

$=-\frac{1}{4}\left[\frac{x-1}{\sqrt{x+1}}+\frac{x+1}{\sqrt{x-1}}\right]+\frac{1}{2}\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}\sqrt{x-1}}\right)$

$=\frac{1}{4}[\sqrt{x+1}+\sqrt{x-1}]$