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 $y=\log (x+\sqrt{x^2+a^2})\Rightarrow e^y=x+\sqrt{x^2+a^2}$

Differentiate both sides w.r.t x

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

$e^y\frac{dy}{dx}=\frac{e^y}{\sqrt{x^2+a^2}}\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{x^2+a^2}}$

Again differentiate on both sides w.r.t x

$\frac{d^{2}y}{d{x}^2}=\frac{-2x}{2{(x^2+a^2)}^{3/2}}$

$(x^2+a^2)\frac{d^{2}y}{d{x}^2}=\frac{-x}{\sqrt{x^2+a^2}}$

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

$\therefore (x^2+a^2)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=0$