If x = sin(t), y = cos(pt), then (1 - x^2) fr | vedclass

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(A) Given $x = \sin(t)$, so $\frac{dx}{dt} = \cos(t)$.Given $y = \cos(pt)$, so $\frac{dy}{dt} = -p \sin(pt)$.Using the chain rule, $\frac{dy}{dx} = \f

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$-p^2y$

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(A) Given $x = \sin(t)$, so $\frac{dx}{dt} = \cos(t)$.Given $y = \cos(pt)$, so $\frac{dy}{dt} = -p \sin(pt)$.Using the chain rule, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-p \sin(pt)}{\cos(t)}$.Since $\cos(t) = \sqrt{1 - \sin^2(t)} = \sqrt{1 - x^2}$, we have $\frac{dy}{dx} = \frac{-p \sin(pt)}{\sqrt{1 - x^2}}$.Thus, $\sqrt{1 - x^2} \frac{dy}{dx} = -p \sin(pt)$.Differentiating both sides with respect to $x$:$\sqrt{1 - x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \left( \frac{-2x}{2\sqrt{1 - x^2}} \right) = -p^2 \cos(pt) \frac{dt}{dx}$.Since $\frac{dt}{dx} = \frac{1}{\cos(t)} = \frac{1}{\sqrt{1 - x^2}}$, we get:$\sqrt{1 - x^2} \frac{d^2y}{dx^2} - \frac{x}{\sqrt{1 - x^2}} \frac{dy}{dx} = -p^2 \frac{\cos(pt)}{\sqrt{1 - x^2}}$.Multiplying throughout by $\sqrt{1 - x^2}$:$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -p^2 \cos(pt) = -p^2 y$.

If $x = \sin(t)$, $y = \cos(pt)$, then $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = $

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