Question

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

Answer 1

$y={\sin }^{-1}x$
$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$ ..(1)
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right)=\frac{-1}{{\left(\sqrt{1-x^2}\right)}^2}\times \frac{d}{dx}\left(\sqrt{1-x^2}\right)=\frac{-1}{1-x^2}\times \frac{1}{2\sqrt{1-x^2}}\times \frac{d}{dx}(1-x^2)$
$\frac{d^{2}y}{d{x}^2}=\frac{x}{{(1-x^2)}^{\frac{3}{2}}}$..(2)
Now substituting value from (1) and (2) in,
$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}$ we get,
$(1-x^2)\times \frac{x}{{(1-x^2)}^{\frac{3}{2}}}-x\times \frac{1}{\sqrt{1-x^2}}$
$\frac{x}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}$
$0=$RHS
Hence proved