Question

If $y=\sin \left(m{\sin }^{-1}x\right)$, then $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+m^{2}y=$

Answer 1

We have,

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

On differentiating both sides w.r.t $x$, we get

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

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

On differentiating both sides w.r.t $x$, we get

$\frac{d^{2}y}{d{x}^2}=m\left[\frac{\left(\sqrt{1-x^2}\right)(-\sin \left(m\left({\sin }^{-1}x\right)\right))\times \frac{m}{\sqrt{1-x^2}}-\left(\cos \left(m\left({\sin }^{-1}x\right)\right)\right)\times \frac{1}{2\sqrt{1-x^2}}\times (0-2x)}{{\left(\sqrt{1-x^2}\right)}^2}\right]$

$\frac{d^{2}y}{d{x}^2}=m\left[\frac{(-m\sin \left(m\left({\sin }^{-1}x\right)\right))+\left(\cos \left(m\left({\sin }^{-1}x\right)\right)\right)\times \frac{x}{\sqrt{1-x^2}}}{1-x^2}\right]$

$\frac{d^{2}y}{d{x}^2}=m[\frac{-m\sin \left(m\left({\sin }^{-1}x\right)\right)}{1-x^2}+\frac{x\cos \left(m\left({\sin }^{-1}x\right)\right)}{(1-x^2)\left(\sqrt{1-x^2}\right)}]$

$\frac{d^{2}y}{d{x}^2}=m[\frac{-m\sin \left(m\left({\sin }^{-1}x\right)\right)}{1-x^2}+\frac{x\frac{dy}{dx}}{(1-x^2)}]....[\frac{dy}{dx}=\frac{m\cos \left(m\left({\sin }^{-1}x\right)\right)}{\sqrt{1-x^2}}]$

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

$(1-x^2)\frac{d^{2}y}{d{x}^2}=mx\frac{dy}{dx}-m^{2}y.......[y=\sin \left(m\left({\sin }^{-1}x\right)\right)]$

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

Hence, the value is $0$.