Question

If x = sin t, y = sin kt, show that $\left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+k^{2}y=0\left(kcons\tan t\right)$

Answer 1

We have $x=\sin t$ and $y=\sin kt$

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

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

$\Rightarrow \cos t\frac{dy}{dx}=k\cos kt$

squaring both sides, we get

$\Rightarrow {\cos }^{2}t{\left(\frac{dy}{dx}\right)}^2=k^2{\cos }^{2}kt$

$\Rightarrow {\left(\frac{dy}{dx}\right)}^2(1-{\sin }^{2}t)=k^2(1-{\sin }^{2}kt)$

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

Differentiating both sides w.r.t $x$ we get

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

Divide both sides by $2\frac{dy}{dx}$ we get

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

Hence proved.