If $y=\sqrt{x+1}+\sqrt{x-1}$ then $(x^2-1)y_2+xy_1=$
[where $y_2=\frac{d^2y}{dx^2}$ & $y_1=\frac{dy}{dx}$]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{1}{4}y$
$y=\sqrt{x+1}+\sqrt{x-1}$
$\frac{dy}{dx}=\frac{1}{2\sqrt{x+1}}+\frac{1}{2\sqrt{x-1}}=\frac{1}{2}\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}\sqrt{x-1}}\right)$
$\frac{d^2y}{dx^2}=\frac{1}{2}\left(-\frac{1}{2}×\frac{1}{(x+1)^{3/2}}-\frac{1}{2}×\frac{1}{(x-1)^{3/2}}\right)$
$=-\frac{1}{4}\left(\frac{1}{(x+1)^{3/2}}+\frac{1}{(x-1)^{3/2}}\right)$
$(x^2-1)y_2+y_1=(x-1)(x+1)\left(-\frac{1}{4}\left[\frac{1}{(x+1)^{3/2}}+\frac{1}{(x-1)^{3/2}}\right]\right)+\frac{1}{2}\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}\sqrt{x-1}}\right)$
$=-\frac{1}{4}\left[\frac{x-1}{\sqrt{x+1}}+\frac{x+1}{\sqrt{x-1}}\right]+\frac{1}{2}\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}\sqrt{x-1}}\right)$
$=\frac{1}{4}[\sqrt{x+1}+\sqrt{x-1}]$