If $ y=\left(\frac{2}{\pi} x-1\right) \operatorname{cosec} x $ is the solution of the differential equation, $ \frac{d y}{d x}+p(x) y=\frac{2}{\pi} \operatorname{cosec} x, 0

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If $ y=\left(\frac{2}{\pi} x-1\right) \operatorname{cosec} x $ is the solution of the differential equation, $ \frac{d y}{d x}+p(x) y=\frac{2}{\pi} \operatorname{cosec} x, 0<x<\frac{\pi}{2} $, then the function $ p(x) $ is equal to: [JEE Main 2020, 6 Sep (Shift 2)]

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If \( y=\left(\frac{2}{\pi} x-1\right) \operatorname{cosec} x \) is the solution of the differential equation, \( \frac{…

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