Question

If $x=\sin t,y=\sin kt$,then show that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-ky=0$

Answer 1

Given:$x=\sin t$

$\Rightarrow 1-x^2=1-{\sin }^{2}t={\cos }^{2}t$

$y=\sin kt\Rightarrow \frac{dy}{dt}=-k\cos kt$

$x=\sin t\Rightarrow \frac{dx}{dt}=\cos t$

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-k\cos kt}{\cos t}=-k$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(-k)=0$

Substituting the above in $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-ky$

$={\cos }^{2}t\left(0\right)-x\times -k-ky$

$=0+kx-ky$

$=k\sin t-k\sin kt=0$

Hence proved.