Question

If $y=\log [x+\sqrt{x^2+a^2}]$, then show that $(x^2+a^2)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=0$

Answer 1

Given, $y=\log [x+\sqrt{x^2+a^2}]$
Differentiate both sides with respect to $x$
$\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}[1+\frac{1\left(2x\right)}{2\sqrt{x^2+a^2}}]$
$\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}\left[\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}\right]$
$\frac{dy}{dx}=\frac{1}{\sqrt{x^2+a^2}}$ .....(1)
Again differentiate both sides with respect to $x$
$\frac{d^{2}y}{d{x}^2}=\frac{-1}{2}{(x^2+a^2)}^{\frac{-3}{2}}\left(2x\right)$
$\frac{d^{2}y}{d{x}^2}=\frac{-x}{(x^2+a^2)\sqrt{x^2+a^2}}$
$(x^2+a^2)\frac{d^{2}y}{d{x}^2}=\frac{-x}{\sqrt{x^2+a^2}}$
$(x^2+a^2)\frac{d^{2}y}{d{x}^2}=-x\frac{dy}{dx}$ .....[From equation 1]
$(x^2+a^2)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=0$
Hence, proved.