Question

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

Answer 1

$y=\log [x+\sqrt{x^2+a^2}]$

$y_1=\frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}(1+\frac{2x}{2\sqrt{x^2+a^2}})$

$y_1=\frac{1}{x+\sqrt{x^2+a^2}}\left(\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}\right)$

$\Rightarrow y_1\left(\sqrt{x^2+a^2}\right)=1$

Again differentiate,

$\Rightarrow y_2\left(\sqrt{x^2+a^2}\right)+\frac{\mathrm{1.2.}x.y_1}{2\sqrt{x^2+a^2}}=0$

$\Rightarrow y_2(x^2+a^2)+x{y}_1=0$