Solve the following differential equation:

e^x tan y dx + ( 1 - e^x ) sec² y dy  = 0

The given differential equation is:

e^x tan y dx + ( 1 - e^x ) sec² y dy  = 0

$\Rightarrow$ e^x tan y dx =  - ( 1 - e^x ) sec² y dy

$\Rightarrow$  e^x tan y dx = ( e^x  - 1 ) sec² y dy

$\Rightarrow \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}} - 1} \mathrm{dx} = \frac{{\sec }^2 \mathrm{y}}{\tan {\displaystyle }{\displaystyle \mathrm{y}}} \mathrm{dy}$

On integrating on both sides,  we get

$$\begin{gathered} \int \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}} - 1} \mathrm{dx} =\int \frac{{\sec }^2 \mathrm{y}}{\tan \mathrm{y}} \mathrm{dy} ..........\left(\mathrm{i}\right) \\ \mathrm{Let} {\mathrm{I}}_{1 }=\int \frac{{\sec }^2 \mathrm{y}}{\tan \mathrm{y}} \mathrm{dy} \\ \mathrm{Put} \tan \mathrm{y} = \mathrm{t} \\ \Rightarrow {\sec }^2 \mathrm{y} \mathrm{dy} = \mathrm{dt} \\ \therefore \int \frac{{\sec }^2 \mathrm{y}}{\tan \mathrm{y}} \mathrm{dy} = \int \frac{\mathrm{dt}}{\mathrm{t}} = \log \left| \mathrm{t} \right| = \log \tan \mathrm{y} .........\left(\mathrm{ii}\right) \\ \mathrm{Let} {\mathrm{I}}_{2 = }\int \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}} - 1} \mathrm{dx} \end{gathered}$$

Put   e^x  - 1 = u

$\therefore$ e^x  dx = du

$\int \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}} - 1} \mathrm{dx} = \int \frac{\mathrm{du}}{\mathrm{u}}$

                    = log u

                    = log  ( e^x  - 1 )                        ............(iii)

From (i),  (ii),  and  (iii),  we get

log tan y = log  ( e^x  - 1 ) + log C

$\Rightarrow$ log tan y = log  C ( e^x  - 1 ) 

$\Rightarrow$ tan y  = C ( e^x  - 1 ) 

The solution of the given differential equation is  tan y = C ( e^x  - 1 ).