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$e^{x} \tan y d x+\left(1-e^{x}\right) \sec ^{2} y d y=0$


The given differential equation is:

$e^{x} \tan y d x+\left(1-e^{x}\right) \sec ^{2} y d y=0$


$\left(1-e^{x}\right) \sec ^{2} y d y=-e^{x} \tan y d x$

Separating the variables, we get:

$\frac{\sec ^{2} y}{\tan y} d y=\frac{-e^{x}}{1-e^{x}} d x$

Integrating both sides, we get:

$\int \frac{\sec ^{2} y}{\tan y} d y=\int \frac{-e^{x}}{1-e^{x}} d x$           ...(1)

Let $\tan y=u$.

$\Rightarrow \frac{d}{d y}(\tan y)=\frac{d u}{d y}$

$\Rightarrow \sec ^{2} y=\frac{d u}{d y}$

$\Rightarrow \sec ^{2} y d y=d u$

$\therefore \int \frac{\sec ^{2} y}{\tan y} d y=\int \frac{d u}{u}=\log u=\log (\tan y)$

Now, let $1-e^{x}=t$

$\therefore \frac{d}{d x}\left(1-e^{x}\right)=\frac{d t}{d x}$

$\Rightarrow-e^{x}=\frac{d t}{d x}$

$\Rightarrow-e^{x} d x=d t$

$\Rightarrow \int \frac{-e^{x}}{1-e^{x}} d x=\int \frac{d t}{t}=\log t=\log \left(1-e^{x}\right)$

Substituting the values of $\int \frac{\sec ^{2} y}{\tan y} d y$ and $\int \frac{-e^{x}}{1-e^{x}} d x$ in equation (1), we get:

$\Rightarrow \log (\tan y)=\log \left(1-e^{x}\right)+\log \mathrm{C}$

$\Rightarrow \log (\tan y)=\log \left[\mathrm{C}\left(1-e^{x}\right)\right]$

$\Rightarrow \tan y=\mathrm{C}\left(1-e^{x}\right)$

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

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