Medium MCQ +4 / -1 PYQ · JEE Mains 2020

If $y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$ is the solution of the differential equation,

${{dy} \over {dx}} + p\left( x \right)y = {2 \over \pi } cosec\,x$,

$0 < x < {\pi \over 2}$, then the function p(x) is equal to :

  1. A cot x Correct answer
  2. B sec x
  3. C tan x
  4. D cosec x

Solution

$y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$ <br>Differentiate w.r.t x <br><br>${{dy} \over {dx}} = {2 \over \pi }$cosec x - $\left( {{2 \over \pi }x - 1} \right)$cosec x.cot x <br><br>$\Rightarrow$ ${{dy} \over {dx}}$ + $\left( {{2 \over \pi }x - 1} \right)$cosec x.cot x = ${2 \over \pi }$cosec x <br><br>$\Rightarrow$ ${{dy} \over {dx}}$ + ycot x = ${2 \over \pi }$cosec x <br><br>Compare this differential equation with given differential equation, we get <br><br>p(x) = cot x

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree

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