Question:
If $y=\left(\frac{2}{\pi} x-1\right) \operatorname{cosecx}$ is the solution of the
differential equation,
$\frac{d y}{d x}+p(x) y=\frac{2}{\pi} \operatorname{cosec} x, 0 function $\mathrm{p}(\mathrm{x})$ is equal to
Correct Option: 1,
Solution:
$y=\left(\frac{2 x}{\pi}-1\right) \operatorname{cosecx}$ $\ldots(1)$
$\frac{d y}{d x}=\frac{2}{\pi} \operatorname{cosecx}-\left(\frac{2 x}{\pi}-1\right) \operatorname{cosecx} \cot x$
$\frac{d y}{d x}=\frac{2 \operatorname{cosecx}}{\pi}-y \cot x$
using equation (1)
$\frac{d y}{d x}+y \cot x=\frac{2 \operatorname{cosec} x}{\pi}$
$\frac{d y}{d x}+p(x) \cdot y=\frac{2 \operatorname{cosec} x}{\pi} \quad x \in\left(0, \frac{\pi}{2}\right)$
Compare : $p(x)=\cot x$