Question

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

• A. $\cot x$
• B. $\tan x$
• C. $cosecx$
• D. $secx$

Answer 1

Answer: (A)

$y=\left(\frac{2x}{\pi }-1\right)cosecx\dots (1)$
$\frac{dy}{dx}=\frac{2}{\pi }cosecx-\left(\frac{2x}{\pi }-1\right)cosecxcotx$
$\frac{dy}{dx}=\frac{2cosecx}{\pi }-ycotx$
using equation _____ (1)
$\frac{dy}{dx}+y\cot x=\frac{2\mathrm{cosec}x}{\pi }$
$\frac{dy}{dx}+p(x)\cdot y=\frac{2\mathrm{cosec}x}{\pi }x\in \left(0,\frac{\pi }{2}\right)$
Compare : $p(x)=\cot x$