Question

If $x=\sin \theta ,y=\cos p\theta$, and $m(1-x^2)y_2-x{y}_1+p^{2}y=0$, where $y_2=\frac{d^{2}y}{d{x}^2}$ and $y_1=\frac{dy}{dx}$ Find $m$

Answer 1

We have, $x=\sin \theta ,y=\cos p\theta$
$\frac{dx}{d\theta }=\cos x=\sqrt{1-x^2},\frac{dy}{d\theta }=-p\sin p\theta =-p\sqrt{1-y^2}$
$\therefore \frac{dy}{dx}=y_1=\frac{-p\sqrt{1-y^2}}{\sqrt{1-x^2}}$
$\Rightarrow y_1^2=\frac{p^2(1-y^2)}{(1-x^2)}\Rightarrow y_1^2(1-x^2)=p^2(1-y^2)$
Again differentiating both side w.r.t $x$
$(1-x^2)2y_1{y}_2-2y_{1}x=-2p^2{y}_1\Rightarrow (1-x^2)y_2-x{y}_1+p^{2}y=0$