Question

$\text{If}\text{Y}=(x+\sqrt{x^2-1}{)}^m\text{show}\text{that}$
$(x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=m^{2}y$

Answer 1

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

Differentiating w.r.t x

$\frac{dy}{dx}=m(x+\sqrt{x^2-1}{)}^{m-1}[1+\frac{1\times 2x}{2\sqrt{x^2-1}}]$

$=m(x+\sqrt{x^2-1}{)}^{m-1}\left[\frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}}\right]$

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

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

$=my$ ……….(i)

again differentiating w.r.t x

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

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

$=m\left(my\right)$ [From (i)]

$(x^2-1)\frac{d^{2}y}{d{x}^2}+\frac{xdy}{dx}=m^{2}y$.