Question

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

Answer 1

Here x=sint t,y=sin kt
$\therefore \frac{dx}{dt}=cost,\frac{dy}{dt}=kcoskt\Rightarrow \frac{dy}{dx}=k\frac{coskt}{cost}$
$\Rightarrow cost\frac{dy}{dx}=kcoskt\Rightarrow co{s}^{2}t{\left(\frac{dy}{dx}\right)}^2=k^{2}co{s}^{2}kt$
$\Rightarrow (1-si{n}^{2}t){\left(\frac{dy}{dx}\right)}^2=k^2(1-si{n}^{2}kt)\Rightarrow (1-x^2){\left(\frac{dy}{dx}\right)}^2=k^2(1-y^2)$

Differentiating w.r.t. x both the sides,

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