If $y=\left(\frac{2}{\pi} x-1\right) \operatorname{cosec} x$ is the solution of the differential equation, $\frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{p}(x) y=\frac{2}{\pi} \operatorname{cosec} x, 0
Correct Option: 1
$\because y=\left(\frac{2}{\pi} x-1\right) \operatorname{cosec} x$
$\frac{d y}{d x}=\frac{2}{\pi} \operatorname{cosec} x-\left(\frac{2}{\pi} x-1\right) \operatorname{cosec} x \cdot \cot x$
$=\operatorname{cosec} x\left[\frac{2}{\pi}-\left(\frac{2}{\pi} x-1\right) \cot x\right]$
$\Rightarrow \frac{d y}{d x}-\frac{2}{\pi} \operatorname{cosec} x=y \cot x$ ...(i)
It is given that,
$\Rightarrow \frac{d y}{d x}-\frac{2}{\pi} \operatorname{cosec} x=-y p(x)$....(ii)
By comparision of (i) and (ii), we get $p(x)=\cot x$
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