Question

If $y=e^{aco{s}^{-1}x}$, then
$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-a^{2}y=?$

Answer 1

$\frac{dy}{dx}=e^{a{\cos }^{-1}n}\left(\frac{-a}{\sqrt{1-x^2}}\right)$

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

$\frac{d^{2}y}{d{x}^2}=-a\left[\frac{\sqrt{1+x^2}.e^{a{\cos }^{-1}x}.\frac{-a}{\sqrt{1-x^2}}-e^{a{\cos }^{-1}x}.\frac{l^{-1}x}{\sqrt{1-x^2}}}{1-x^2}\right]$

$\frac{d^{2}y}{d{x}^2}=-a\left[\frac{\sqrt{1-x^2}{e}^{a{\cos }^{-1}x}(-q)+x{e}^{a{\cos }^{-1}x}}{(1-x^2{)}^{3/2}}\right]$

$(1-x^2)\frac{d^{2}y}{d{x}^2}=-a[-a{e}^{a{\cos }^{-1}x}+\frac{x{e}^{a{\cos }^{-1}x}}{\sqrt{1-x^2}}]$

$(1-x^2)\frac{d^{2}y}{d{x}^2}=-a[-ay+\frac{xdy}{adx}]$

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

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

Hence proved