$x\cfrac{{dy}}{{dx}} + y - x + xy\cot x = 0(x \ne 0)$

: The given linear equation is $x\cfrac{{dy}}{{dx}} + y - x + xy\cot x = 0$

$\Rightarrow x\cfrac{{dy}}{{dx}} + y({\rm{l}} + x\cot x) = x \Rightarrow \cfrac{{dy}}{{dx}} + \left( {\cfrac{1}{x} + \cot x} \right)y = 1$

…(1)
which is a linear equation of the type $\cfrac{{dy}}{{dx}} + Py = Q$

Where, $P = \cfrac{1}{x} + \cot x$ and $Q = 1$
$\therefore {\rm{I}}.{\rm{F}}. = {e^{\int {Pdx} }} = {e^{\int {\left( {\cfrac{1}{x} + \cot x} \right)dx} }} = {e^{\log x}} \cdot {e^{{\rm{log}}\left| {\sin x} \right|}} = x \cdot \sin x$

$\therefore$ The solution is $y \cdot ({\rm{I}}.{\rm{F}}.) = \int Q \cdot ({\rm{I}}.{\rm{F}}.)dx + C$

$\Rightarrow y \cdot x\sin x = \int x \sin xdx + C = x( - \cos x) - \int {(1)( - \cos x)dx} + C$

(Integrating by parts)
$= - x\cos x + \int {\cos } xdx + C = - x\cos x + \sin x + C$

$\Rightarrow y = \cfrac{1}{x} - \cot x + \cfrac{C}{{x\sin x}}$,

which is the required solution.