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

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

Vedclass · Accepted Answer

$(1 - x^2)y_2 - xy_1 + p^2y = 0$

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(D) Given $x = \sin t$ and $y = \cos pt$. First,find the derivatives with respect to $t$: $\frac{dx}{dt} = \cos t$ and $\frac{dy}{dt} = -p \sin pt$. Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-p \sin pt}{\cos t}$. Thus,$\cos t \frac{dy}{dx} = -p \sin pt$. Differentiating both sides with respect to $x$: $-\sin t + \cos t \frac{d^2y}{dx^2} = -p^2 \cos pt \frac{dt}{dx}$. Since $\frac{dt}{dx} = \frac{1}{\cos t}$,we have: $-\sin t \frac{dy}{dx} + \cos t \frac{d^2y}{dx^2} = -p^2 \cos pt \frac{1}{\cos t}$. Multiplying by $\cos t$: $-\sin t \cos t \frac{dy}{dx} + \cos^2 t \frac{d^2y}{dx^2} = -p^2 \cos pt$. Substituting $x = \sin t$,$\cos^2 t = 1 - x^2$,and $y = \cos pt$: $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2 y = 0$. Therefore,$(1 - x^2)y_2 - xy_1 + p^2y = 0$.

If $x = \sin t$ and $y = \cos pt$,then which of the following is true?

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