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 $\frac{dy}{dx}$ using the chain rule: $\frac{dx}{dt} = \cos t$ and $\frac{dy}{dt} = p \cos pt$. S

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(A) Given $x = \sin t$ and $y = \sin pt$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dx}{dt} = \cos t$ and $\frac{dy}{dt} = p \cos pt$. So,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{p \cos pt}{\cos t}$. This implies $\cos t \frac{dy}{dx} = p \cos pt$. Differentiating both sides with respect to $x$: $\frac{d}{dx} (\cos t \frac{dy}{dx}) = \frac{d}{dx} (p \cos pt)$. Using the product rule on the left: $-\sin t \frac{dt}{dx} \frac{dy}{dx} + \cos t \frac{d^2 y}{dx^2} = -p^2 \sin pt \frac{dt}{dx}$. Since $\frac{dt}{dx} = \frac{1}{\cos t}$,we substitute: $-\sin t (\frac{1}{\cos t}) \frac{dy}{dx} + \cos t \frac{d^2 y}{dx^2} = -p^2 \sin pt (\frac{1}{\cos t})$. Multiply the entire equation by $\cos t$: $-\sin t \frac{dy}{dx} + \cos^2 t \frac{d^2 y}{dx^2} = -p^2 \sin pt$. Substitute $x = \sin t$,$\cos^2 t = 1 - x^2$,and $y = \sin pt$: $-x \frac{dy}{dx} + (1 - x^2) \frac{d^2 y}{dx^2} = -p^2 y$. Rearranging the terms: $(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + p^2 y = 0$.

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

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