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 :
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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