Question

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

Answer 1

$y=\cos \left(m{\cos }^{-1}x\right)$

$\Rightarrow \frac{dy}{dx}=-\sin \left(m{\cos }^{-1}x\right)\frac{d}{dx}\left(m{\cos }^{-1}x\right)$

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

$\Rightarrow \frac{dy}{dx}=\sqrt{1-y^2}\frac{m}{\sqrt{1-x^2}}$ where $y=\cos \left(m{\cos }^{-1}x\right)$

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

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

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

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

$\Rightarrow (1-x^2)2\frac{dy}{dx}\times \frac{d^{2}y}{d{x}^2}-2x{\left(\frac{dy}{dx}\right)}^2=-m^2\times 2y\frac{dy}{dx}$ by dividing both sides by $2\frac{dy}{dx}$

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

Hence proved.