Question

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

Answer 1

We've $y=cos\left(mco{s}^{-1}x\right)\Rightarrow \frac{dy}{dx}=-sin\left(mco{s}^{-1}x\right)\times m\times (-\frac{1}{\sqrt{1-x^2}})$
$\Rightarrow \sqrt{1-x^2}\frac{dy}{dx}=msin\left(mco{s}^{-1}x\right)$
On squaring both sides, we get: $(1-x^2){\left(\frac{dy}{dx}\right)}^2=m^{2}si{n}^2\left(mco{s}^{-1}x\right)=m^2\{1-co{s}^2(mco{s}^{-1}x\left)\right\}$
$\Rightarrow (1-x^2){\left(\frac{dy}{dx}\right)}^2=m^2\{1-y^2\}$
On differentiating again w.r.t.x: $(1-x^2)2\left(\frac{dy}{dx}\right)\times \frac{d^{2}y}{d{x}^2}-2x{\left(\frac{dy}{dx}\right)}^2=m^2(-2y\frac{dy}{dx})$
$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=m^2(-y)\therefore (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+m^{2}y=0.$
Alternative: We've $y=cos\left(mco{s}^{-1}x\right)\Rightarrow \frac{dy}{dx}=-sin\left(mco{s}^{-1}x\right)\times m\times (-\frac{1}{\sqrt{1-x^2}})$
$\Rightarrow \sqrt{1-x^2}\frac{dy}{dx}=msin\left(mco{s}^{-1}x\right)$
On differentiating again w.r.t. $x:\sqrt{1-x^2}\frac{d^{2}y}{d{x}^2}-\frac{2x}{2\sqrt{1-x^2}}\times \frac{dy}{dx}=mcos\left(mco{s}^{-1}x\right)\times \frac{-m}{\sqrt{1-x^2}}$
$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=-m^{2}y\therefore (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+m^{2}y=0$.