$\text{ If }\mathrm{x}=\text{ sint and }\mathrm{y}=\text{ sinpt then the value of }\left(1-x^2\right)\frac{dy}{d{x}^2}-x\frac{dy}{dx}+p^{2}y=$

$\begin{array}{l}\frac{dx}{dt}=\cos t\\ \frac{dy}{dt}=p\cos pt\Rightarrow \frac{dy}{dx}=\frac{p\cos pt}{\cos t}=\frac{p\sqrt{1-{\sin }^{2}pt}}{\sqrt{1-{\sin }^{2}t}}\\ \frac{dy}{dx}=\frac{p\sqrt{1-y^2}}{\sqrt{1-x^2}}\\ {\left(\frac{dy}{dx}\right)}^2\left(1-x^2\right)=p^2\left(1-y^2\right)\\ \text{ Diff }\end{array}$

$\begin{array}{l}2\frac{dy}{dx}\frac{d^{2}y}{d{x}^2}\left(1-x^2\right)+{\left(\frac{dy}{dx}\right)}^2(-2x)=p^2(-2y)\frac{dy}{dx}\\ \frac{d^{2}y}{d{x}^2}\left(1-x^2\right)-x\frac{dy}{dx}=-p^{2}y\\ \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+p^{2}y=0\end{array}$