If x = sin t and y = sin pt,then the value of | vedclass

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(A) Given $x = \sin t$ and $y = \sin pt$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = \cos t$$\frac{dy}{dt} = p \cos pt$Using the c

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(A) Given $x = \sin t$ and $y = \sin pt$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = \cos t$$\frac{dy}{dt} = p \cos pt$Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{p \cos pt}{\cos t}$.Squaring both sides:$(\frac{dy}{dx})^2 = \frac{p^2 \cos^2 pt}{\cos^2 t} = \frac{p^2(1 - \sin^2 pt)}{1 - \sin^2 t} = \frac{p^2(1 - y^2)}{1 - x^2}$.Thus,$(1 - x^2) (\frac{dy}{dx})^2 = p^2(1 - y^2)$.Differentiating both sides with respect to $x$:$(1 - x^2) \cdot 2 \frac{dy}{dx} \cdot \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 \cdot (-2x) = p^2 \cdot (-2y) \frac{dy}{dx}$.Dividing by $2 \frac{dy}{dx}$ (assuming $\frac{dy}{dx} 
eq 0$):$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -p^2y$.Rearranging the terms:$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2y = 0$.

If $x = \sin t$ and $y = \sin pt$,then the value of $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2y$ is equal to

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