Question

$If\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}=y,showthat(1-x^2)\frac{d^{2}y}{d{x}^2}-3x\frac{dy}{dx}-y=0$

Answer 1

Given,

${\sin }^{-1}x=y\sqrt{1-x^2}$

Differentiating on both sides, we get,

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

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

Again differentiating on both sides, we get,

$\frac{d}{dx}\left[\right(1-x^2)\frac{dy}{dx}-xy]=\frac{d}{dx}\left(1\right)$

$(1-x^2)\frac{dy}{dx}-2x\frac{dy}{dx}-x\frac{dy}{dx}-y=0$

$(1-x^2)\frac{dy}{dx}-3x\frac{dy}{dx}-y=0$

Hence proved.