Linear First Order ODE/y' + y cot x = 2 x cosec x

Theorem

$(1): \quad y' + y \cot x = 2 x \csc x$

$y = x^2 \csc x + C \csc x$

$(1)$ is in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

$\map P x = \cot x$

Thus:

$\ds \int \map P x \rd x$

$=$

$\ds \int \cot x \rd x$

$\ds $

$=$

$\ds \map \ln {\sin x}$

$\ds \leadsto \ \ $

$\ds e^{\int P \rd x}$

$=$

$\ds \sin x$

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

$\dfrac {\d} {\d x} y \sin x = 2 x$

$y \sin x = x^2 + C$

or:

$y = x^2 \csc x + C \csc x$

$\blacksquare$