Question

If $y=\cos e{c}^{-1}x,x>1,$ then show that $x\left(x^2-1\right)\frac{d^{2}y}{d{x}^2}+\left(2x^2-1\right)\frac{dy}{dx}=0$

Answer 1

Given:

$y=\cos e{c}^{-}1x$

Differentiate both side with respect to x.

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

Again differentiate w r t x.

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

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

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

Hance proved