If $y=\begin{Bmatrix} log (x+\sqrt{x^2+1})\end{Bmatrix}^2$ then $(1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=$

2

The correct answer is Option (4) → 2

$y'=\frac{2\log(x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}}\left(1+\frac{x}{\sqrt{x^2+1}}\right)$

so $y'=\frac{2\log(x+\sqrt{x^2+1})}{\sqrt{x^2+1}}$

so $y'\sqrt{x^2+1}=2\log(x+\sqrt{x^2+1})$

differentiating

$y''(\sqrt{x^2+1})+\frac{xy'}{\sqrt{x^2+1}}=\frac{2}{x+\sqrt{x^2+1}}\left(1+\frac{x}{\sqrt{x^2+1}}\right)$

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