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

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Differential Equations

y = (\frac{2}{π} x - 1) cosec 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:

Answer: (A)

Differentiate

y = (\frac{2}{π} x - 1) cosec x

to get

\frac{d y}{d x} = \frac{2}{π} cosec x - (\frac{2}{π} x - 1) cosec x cot x

. Since the second term is

y cot x

, this rearranges to

\frac{d y}{d x} + y cot x = \frac{2}{π} cosec x

. Comparing with

\frac{d y}{d x} + p (x) y = \frac{2}{π} cosec x

identifies

p (x) = cot x

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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