Solution of the equation $\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{x^2}\tan y\sin y$ is

Given the differential equation, $\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{x^2}\tan y\sin y$

Dividing both sides of the given equation by

Then, $cosecy \cot y\frac{dy}{dx}+\frac{1}{x}cosecy=\frac{1}{x^2}.$

let, $Cosec y =t ......( 1 )$

Differentiate the equation (1), we have.

$\frac{dt}{dx}=-cosecy \cot y\frac{dy}{dx}$

so, we have.

$\frac{dt}{dx}-\frac{t}{x}=-\frac{1}{x^2}$

the integration factor is:

$$\begin{gathered} \text{ I.f. }=e^{\int -\frac{1}{x}dx} \\ =e^{-\ln \left(x\right)}=\frac{1}{x}. \\ \text{ So, }\frac{1}{x}\left(t\right)=\int \frac{1}{x}\left(\frac{-1}{x^2}\right)dx. \\ \frac{t}{x}=\int -\frac{1}{x^3}dx \\ \frac{t}{x}=-\frac{x^{-3+1}}{-3+1}+C^{\text{'}} \\ \frac{t}{x}=\frac{x^2}{2}+c^{\text{'}} \\ \text{then, }\frac{t}{x}=\frac{1+c{x}^2}{2x^2} \\ \text{ Here, }2c^{\text{'}}=c \\ \text{Therfore, }\frac{cosec}{x}=\frac{1+c{x}^2}{2x^2} \\ \Rightarrow \frac{1}{\sin y}=\frac{1+c{x}^2}{2x}⟹2x=\sin y\left(1+c{x}^2\right) \end{gathered}$$

$$\begin{gathered} \text{OR, the integration factor is: } \\ \text{ I.f. }=e^{-\frac{1}{x}dx} \\ =e^{\log x^{-1}}=x^{-1} \\ \text{ so, }t{x}^{-1}=\int x^{-1}-\frac{1}{x^2}dx+C. \\ \frac{t}{x}=-\frac{x^{-2}}{-2}+C=\frac{1}{2x^2}+C \end{gathered}$$

$\Rightarrow cosecy=\frac{1}{2x}+cx$ then, we have.

$\frac{2x}{\sin y}=1+2c{x}^2\text{. }$

$2x = \sin y (1+2c{x}^2)$