If \( y=\log [x+\sqrt{x^2+1}],\) prove that \( (x^2+1)\large\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0.\)

Answer 1

Toolbox:

• $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$

Step 1:

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

$\Rightarrow \large\frac{dy}{dx}=\bigg[\large\frac{1}{x+\sqrt{x^2+1}}.$$(1+\large\frac{1}{2\sqrt{x^2+1}.2x}\bigg]$

$\Rightarrow \large\frac{1}{x+\sqrt{x^2+1}}\bigg[\large\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\bigg]$

$\Rightarrow \large\frac{1}{\sqrt{x^2+1}}$

Step 2:

$\large\frac{d^2y}{dx^2}=\large\frac{0-1.1/2\sqrt{x^2+1}.2x}{(\sqrt{x^2+1)^2}}$

$\Rightarrow \large\frac{-x}{(x^2+1)\sqrt{x^2+1}}$

$\Rightarrow (x^2+1)\large\frac{d^2y}{dx^2}=\frac{-x}{\sqrt{x^2+1}}$

$\therefore (x^2+1)\large\frac{d^2y}{dx^2}$$+x\large\frac{dy}{dx}=\frac{-x}{\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1}}$

$\Rightarrow 0$

Hence proved.