If $y=\begin{Bmatrix}x+\sqrt{x^2+1}\end{Bmatrix}^2, $ and $(x^2+1)y_2+xy_1-ky=0, $ then k is equal to :

4

The correct answer is Option (3) → 4

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

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

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

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

$(x^2+1)y_2+xy_1-ky=2\left(2(x+\sqrt{x^2+1})^2-\frac{x(x+\sqrt{x^2+1})^2}{\sqrt{x^2+1}}\right)+\frac{2x(x+\sqrt{x^2+1})^2}{\sqrt{x^2+1}}-ky$

$(x^2+1)y_2+xy_1-ky=4(x+\sqrt{x^2+1})^2-ky=0$

$⇒k=4$