If y=2πx-1cosecx is the solution of the differential equation, dydx+p(x)y=2πcosecx,0<x<π2, then the function p(x) is equal to :

y=2xπ-1cosecxdydx=2xπ-1-cosecxcotx+2πcosecx=-ycotx+2πcosecx substituting 'y'⇒dydx+cotx·y=2πcosecx∴p(x)=cotx

If y=2πx-1cosecx is the solution of the differential equation, dydx+p(x)y=2πcosecx,0<x<π2, then the function p(x) is equal to :

y=2xπ-1cosecxdydx=2xπ-1-cosecxcotx+2πcosecx=-ycotx+2πcosecx substituting 'y'⇒dydx+cotx·y=2πcosecx∴p(x)=cotx

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$If y = (\frac{2}{π} x - 1) \cos e c x is the solution of the differential equation,$
$\frac{d y}{d x} + p (x) y = \frac{2}{π} cosec x, 0 < x < \frac{π}{2}, then the function p (x) is equal to :$

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$\cos e c x$

$c o t x$

$s e c x$

$\tan x$

answer is A.

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Detailed Solution

$\begin{array}{l} y = (\frac{2 x}{π} - 1) \cos ec x \\ \frac{d y}{d x} = (\frac{2 x}{π} - 1) (- cosec x \cot x) + \frac{2}{π} \cos e c x \\ = - y \cot x + \frac{2}{π} \cos e c x s u b s t i t u t i n g' y' \\ \Rightarrow \frac{d y}{d x} + \cot x \cdot y = \frac{2}{π} \cos e c x \\ ∴ p (x) = \cot x \end{array}$