If $ y = \left(\frac{2x}{\pi }-1\right)$ cosec x is the solution of the differential equation $\frac{dy}{dx} + p(x) y =\frac{2}{\pi }cosec \, x, 0 < x < \frac{\pi }{2}$ then the function P(x) is equal to

cot x

The correct answer is option (3) : cot x

We have, $y = \left(\frac{2x}{\pi}-1\right) cosec\, x$

$⇒\frac{dy}{dx}=\frac{2}{\pi } cose x\, x - \left(\frac{2x}{\pi } - 1\right) cosec \, x cot \, x $

$⇒\frac{dy}{dx} = \frac{2}{ \pi } cose c x- y cot x $

$⇒\frac{dy}{dx} + y\, cot \, x=\frac{2}{\pi } cosec\, x $

Comparing this with $\frac{dy}{dx} + p(x) y =\frac{2}{\pi} cosec \, x $, we obtain

$p(x) = cot (x)$.