Solve the differential equation $dy = \cos x(2 - y{\mathop{\rm cosec}\nolimits} x)dx$ given that $y = 2,$ when $x = \frac{\pi }{2}$.

Given differential equation,

$dy = \cos x(2 - y{\mathop{\rm cosec}\nolimits} x)dx$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \cos x(2 - y{\mathop{\rm cosec}\nolimits} x)$

$\Rightarrow$ $\frac{{dy}}{{dx}} = 2\cos x - y{\mathop{\rm cosec}\nolimits} x \cdot \cos x$

$\Rightarrow$ $\frac{{dy}}{{dx}} = 2\cos x - y\cot x$

$\Rightarrow$ $\frac{{dy}}{{dx}} + y\cot x = 2\cos x$

which is a linear differential equation.
On comparing it with

$\frac{{dy}}{{dx}} + Py = Q,$

we get
$P = \cot x,Q = 2\cos x$

${\rm{IF}} = {e^{\int P dx}} = {e^{\int {\cot } xdx}} = {e^{\log \sin x}} = \sin x$

The general solution is
$y \cdot \sin x = \int 2 \cos x \cdot \sin xdx + C$

$\Rightarrow$ $y \cdot \sin x = \int {\sin } 2xdx + C$

$\Rightarrow$ $y \cdot \sin x = - \frac{{\cos 2x}}{2} + C$

…..(i)
When $x = \frac{\pi }{2}$
and $y = 2,$ then

$2 \cdot \sin \frac{\pi }{2} = - \frac{{\cos \left( {2 \times \frac{\pi }{2}} \right)}}{2} + C$

$\Rightarrow$ $2 \cdot 1 = + \frac{1}{2} + C$

$\Rightarrow$ $2 - \frac{1}{2} = C \Rightarrow \frac{{4 - 1}}{2} = C$

$\Rightarrow$ $\therefore C = \frac{3}{2}$

On substituting the value of $C$ in Eq.

(i), we get
$y\sin x = - \frac{1}{2}\cos 2x + \frac{3}{2}$