Solve the differential equation is
$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{2\mathrm{xy}}{1+{\mathrm{x}}^2}={\mathrm{x}}^2+2$

OR

Solve the differential equation $\mathrm{dy}=\cos \mathrm{x}(2-\mathrm{ycosec} \mathrm{x})\mathrm{dx}, \mathrm{given} \mathrm{that} \mathrm{y}=2$, when $\mathrm{x}=\mathrm{\pi }/2$

Given differential equation is
$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{y}={\mathrm{x}}^2+2 .....\left(\mathrm{i}\right)$
This is a linear differential equation of the form $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$
$\mathrm{with} \mathrm{P}=\frac{-2\mathrm{x}}{1+{\mathrm{x}}^2}$and $\mathrm{Q}={\mathrm{x}}^2+2$
$\therefore \mathrm{IF}={\mathrm{e}}^{\int \frac{-2\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}}={\mathrm{e}}^{-\int \frac{2\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}}={\mathrm{e}}^{-\log \left({\mathrm{x}}^2+1\right)}=\frac{1}{{\mathrm{x}}^2+1}$
$solution is y\cdot ( \mathrm{IF} )=\int \mathrm{Q}\cdot ( \mathrm{IF} )\mathrm{dx}+\mathrm{C},\mathrm{we} \mathrm{get}$
$\begin{array}{l}\therefore \mathrm{y}\cdot \frac{1}{\left({\mathrm{x}}^2+1\right)}=\int \left({\mathrm{x}}^2+2\right)\cdot \frac{1}{\left({\mathrm{x}}^2+1\right)}\mathrm{dx}+\mathrm{C}\\ \frac{\mathrm{y}}{{\mathrm{x}}^2+1}=\int \frac{\left({\mathrm{x}}^2+1\right)+1}{{\mathrm{x}}^2+1}\mathrm{dx}+\mathrm{C}\\ \frac{\mathrm{y}}{{\mathrm{x}}^2+1}=\int 1\mathrm{dx}+\int \frac{1}{{\mathrm{x}}^2+1}\mathrm{dx}+\mathrm{C}\\ \frac{\mathrm{y}}{{\mathrm{x}}^2+1}=\mathrm{x}+{\tan }^{-1} \mathrm{x}+\mathrm{C}\\ \Rightarrow \mathrm{y}=\mathrm{x}\left({\mathrm{x}}^2+1\right)+{\tan }^{-1} \mathrm{x}\cdot \left({\mathrm{x}}^2+1\right)+\mathrm{C}\left({\mathrm{x}}^2+1\right)\end{array}$

OR

Given differential equation is
$\begin{array}{l}\mathrm{dy}=\cos \mathrm{x}(2-\mathrm{ycosec} \mathrm{x})\mathrm{dx}\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos \mathrm{x}(2-\mathrm{ycosec} \mathrm{x})=2\cos \mathrm{x}-\mathrm{ycosec} \mathrm{xcos} \mathrm{x}\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2\cos \mathrm{x}-\mathrm{ycot} \mathrm{x}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ycot} \mathrm{x}=2\cos \mathrm{x}\end{array}$
which is a linear differential equation of the form of
$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$
Here, $\mathrm{P}=\cot \mathrm{x}\text{ and }\mathrm{Q}=2\cos \mathrm{x}$
Now, $\mathbf{I}\mathbf{.}\mathbf{F}={\mathrm{e}}^{\int \mathrm{Fdx}}={\mathrm{e}}^{\int \cot \mathrm{xdx}}={\mathrm{e}}^{\log \sin \mathrm{x}}=\sin \mathrm{x}$
and the required solution is given by
$\begin{array}{l}\mathrm{y}\cdot \mathrm{IF}=\int (\mathrm{Q}\cdot \mathrm{IF})\mathrm{dx}+\mathrm{C}\\ \mathrm{y}\cdot \sin \mathrm{x}=\int 2\cos \mathrm{x}\cdot \sin \mathrm{xdx}+\mathrm{C}\\ \Rightarrow \mathrm{y}\cdot \sin \mathrm{x}=\int \sin 2\mathrm{xdx}+\mathrm{C}[\because \sin 2\mathrm{\theta }=2\sin \mathrm{\theta cos} \mathrm{\theta }]\\ \Rightarrow \mathrm{y}\cdot \sin \mathrm{x}=-\frac{\cos 2\mathrm{x}}{2}+\mathrm{C} ......\left(\mathrm{i}\right)\end{array}$
Also, given y=2, when $\mathrm{x}=\frac{\mathrm{\pi }}{2},$, therefore we have
$\begin{array}{l}2\cdot \sin \frac{\mathrm{\pi }}{2}=-\frac{\cos \left(2\times \frac{\mathrm{\pi }}{2}\right)}{2}+\mathrm{C}\\ \Rightarrow 2\cdot 1=+\frac{1}{2}+\mathrm{C}\Rightarrow 2-\frac{1}{2}=\mathrm{C}\\ \Rightarrow \frac{4-1}{2}=\mathrm{C}\Rightarrow \mathrm{C}=\frac{3}{2}\\ \end{array}$
On putting the value of C in Eq. (i), we get
$\mathrm{ysin} \mathrm{x}=-\frac{1}{2}\cos 2\mathrm{x}+\frac{3}{2}$