Question

If $y=\sqrt{x+1}-\sqrt{x-1}$, prove that $(x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-\frac{1}{4}y=0$

Answer 1

$y=\sqrt{x+1}-\sqrt{x-1}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x-1}}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\left(\frac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x^2-1}}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}\left(\frac{-y}{\sqrt{x^2-1}}\right)$

$\Rightarrow \sqrt{x^2-1}\frac{dy}{dx}=-\frac{1}{2}y$

Squaring both sides, we get

$\Rightarrow (x^2-1){\left(\frac{dy}{dx}\right)}^2=\frac{1}{4}{y}^2$

Differentiating both sides, w.r.t $x$ we get

$\Rightarrow 2(x^2-1)\frac{dy}{dx}\frac{d^{2}y}{d{x}^2}+2x{\left(\frac{dy}{dx}\right)}^2=\frac{1}{4}\times 2y\frac{dy}{dx}$

Divide both sides by $2\frac{dy}{dx}$ we get

$\Rightarrow (x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=\frac{1}{4}y$

$\Rightarrow (x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-\frac{1}{4}y=0$ is proved.