Q. If $y = \left( x + \sqrt{1 + x} \right)^{n}$, then $\left( 1 + x^{2} \right)\frac{d^{2}y}{dx^{2}} + x\ \frac{dy}{dx}$ is

A) $n^{2}y$
B) $- n^{2}y$
C) $- y$
D) $2x^{2}y$
Explanation

(a)
Given, $y = \left( x + \sqrt{1 + x^{2}} \right)^{n}$
$\Rightarrow$ $\frac{dy}{dx} = n\left\lbrack x + \sqrt{1 + x^{2}} \right\rbrack^{n - 1}\left( 1 + \frac{x}{\sqrt{x^{2} + 1}} \right)$
$= \frac{n\left\lbrack x + \sqrt{1 + x^{2}} \right\rbrack^{n}}{\sqrt{1 + x^{2}}}$
$\Rightarrow$ $\left( \frac{dy}{dx} \right)^{2}\left( 1 + x^{2} \right) = n^{2}y^{2}$
Again, differentiating, we get
$2\ \frac{dy}{dx}.\frac{d^{2}y}{dx^{2}}\left( 1 + x^{2} \right) + 2x\left( \frac{dy}{dx} \right)^{2} = 2n^{2}y\ \frac{dy}{dx}$
$\Rightarrow$ $\frac{d^{2}y}{dx^{2}}\left( 1 + x^{2} \right) + x\frac{dy}{dx} = n^{2}y$ [divide by $2\frac{dy}{dx}$]

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.