If y = left(frac{2}{pi} x - 1right) operatorn | vedclass

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(A) Given the solution $y = \left(\frac{2x}{\pi} - 1\right) \operatorname{cosec} x$. Differentiating both sides with respect to $x$ using the product

Vedclass · Accepted Answer

$\cot x$

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(A) Given the solution $y = \left(\frac{2x}{\pi} - 1\right) \operatorname{cosec} x$. Differentiating both sides with respect to $x$ using the product rule:$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{2x}{\pi} - 1\right) \operatorname{cosec} x + \left(\frac{2x}{\pi} - 1\right) \frac{d}{dx}(\operatorname{cosec} x)$$\frac{dy}{dx} = \frac{2}{\pi} \operatorname{cosec} x + \left(\frac{2x}{\pi} - 1\right) (-\operatorname{cosec} x \cot x)$Substituting $y = \left(\frac{2x}{\pi} - 1\right) \operatorname{cosec} x$ into the expression:$\frac{dy}{dx} = \frac{2}{\pi} \operatorname{cosec} x - y \cot x$Rearranging the terms to match the form $\frac{dy}{dx} + p(x) y = Q(x)$:$\frac{dy}{dx} + y \cot x = \frac{2}{\pi} \operatorname{cosec} x$Comparing this with the given differential equation $\frac{dy}{dx} + p(x) y = \frac{2}{\pi} \operatorname{cosec} x$,we get:$p(x) = \cot x$.

If $y = \left(\frac{2}{\pi} x - 1\right) \operatorname{cosec} x$ is the solution of the differential equation $\frac{dy}{dx} + p(x) y = \frac{2}{\pi} \operatorname{cosec} x$ for $0 < x < \frac{\pi}{2}$,then the function $p(x)$ is equal to

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