Find a particular solution of the differential equation $\frac{{dy}}{{dx}} + y\cot x = 4x{\mathop{\rm cosec}\nolimits} x(x \ne 0)$,
given that $y = 0$ when $x = \frac{\pi }{2}$

The given differential equation is:
$\frac{{dy}}{{dx}} + y\cot x = 4x{\mathop{\rm cosec}\nolimits} x$

This equation is a linear differential equation of the form $\frac{{dy}}{{dx}} + py = Q$,

where $p = \cot x$ and $Q = 4x{\mathop{\rm cosec}\nolimits} x$.

Now, ${\rm{I}}{\rm{.F}} = {e^{\int {pdx} }} = {e^{\int {\cot xdx} }} = {e^{\log \left| {\sin x} \right|}} = \sin x$

The general solution of the given differential equation is given by,

$y({\rm{I}}.{\rm{F}}.) = \int {({\rm{Q}} \times {\rm{I}}.{\rm{F}}.)} dx + {\rm{C}}$

$\Rightarrow$ $y\sin x = \int {(4x{\mathop{\rm cosec}\nolimits} x \cdot \sin x)} dx + {\rm{C}}$
$\Rightarrow$ $y\sin x = 4\int x dx + {\rm{C}}$
$\Rightarrow$ $y\sin x = 4 \cdot \frac{{{x^2}}}{2} + {\rm{C}}$

$\Rightarrow$ $y\sin x = 2{x^2} + C$

….(1)
Now,$y = 0$ at $x = \frac{\pi }{2}$.

Therefore, equation (1) becomes:
$0 = 2 \times \frac{{{\pi ^2}}}{4} + {\rm{C}}$
$\Rightarrow$ ${\rm{C}} = - \frac{{{\pi ^2}}}{2}$

Substituting ${\rm{C}} = - \frac{{{\pi ^2}}}{2}$

$y\sin x = 2{x^2} - \frac{{{\pi ^2}}}{2}$

This is the required particular solution of the given differential equation.