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

  1. (1) $\cot x$

  2. (2) cose

  3. (3) $\sec x$

  4. (4) $\tan x$

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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