Question

If $y={\sin }^{-1}x$, then prove that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=0$.

Answer 1

Given

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

Diff. w.r.t $x$ on both

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

Again Diff. w.r.t $x$ on both sides,

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

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

$=\frac{-1}{2}(1-x^2{)}^{-3/2}.(-2x)$

$=\frac{x}{(1-x^2{)}^{-3/2}}$

$=\frac{x}{(1-x^2)\sqrt{1-x^2}}$

So, $(1-x^2)\frac{d^{2}y}{d{x}^2}=x\frac{1}{\sqrt{1-x^2}}$

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

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