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If y=\left ( \frac{2}{\pi }x-1 \right )\csc x is the solution of the differential equation, \frac{dy}{dx}+p(x)y=\frac{2}{\pi }\csc x,\;0<x<\frac{\pi }{2}, then the function p(x) is equal to :
Option: 1 \cot x
Option: 2 \csc x
Option: 3 \sec x
Option: 4 \tan x

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\begin{aligned} &y=\left(\frac{2 x}{\pi}-1\right) \operatorname{cosec} x\\ &\frac{d y}{d x}=\frac{2}{\pi} \operatorname{cosec} x-\left(\frac{2 x}{\pi}-1\right) \operatorname{cosec} x \cot x\\ &\frac{d y}{d x}=\frac{2 \operatorname{cossec} x}{\pi}-y \cot x \end{aligned}

\begin{aligned} &\text { using equation (1) }\\ &\frac{d y}{d x}+y \cot x=\frac{2 \operatorname{cosec} x}{\pi} \end{aligned}

\\\frac{\mathrm{dy}}{\mathrm{dx}}+{\color{Red} \mathrm{p}(\mathrm{x})} \cdot \mathrm{y}=\frac{2 \operatorname{cosec} \mathrm{x}}{\pi} \mathrm{x} \in\left(0, \frac{\pi}{2}\right) \\\\ \text {Compare : } \mathrm{p}(\mathrm{x})=\cot \mathrm{x}

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himanshu.meshram

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