Solve $e^{x} t a n y d x + (1 - e^{x}) s e c^{2} y d y = 0$

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Solution

$e^{x} tan y d x + (1 - e^{x}) {sec}^{2} y d y = 0$
$e^{x} tan y d x = - (1 - e^{x}) {sec}^{2} y d y$
$e^{x} tan y d x = (e^{x} - 1) {sec}^{2} y d y$
$(\frac{e^{x}}{e^{x} - 1}) d x = \frac{{sec}^{2} y}{tan y} d y$
$\frac{d (e^{x} - 1)}{e^{x} - 1} = \frac{d (tan y)}{tan y}$
Integrating on both sides, we get
$\int \frac{d (e^{x} - 1)}{e^{x} - 1} = \int \frac{d (tan y)}{tan y}$
$\Rightarrow l o g | e^{x} - 1 | = l o g | t a n y | + l o g c$
$\Rightarrow l o g | e^{x} - 1 | = l o g c | tan y |$
$\Rightarrow c tan y = e^{x} - 1$ is the general solution.

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Solve extanydx+(1−ex)sec2ydy=0

extanydx+(1−ex)sec2ydy=0extanydx=−(1−ex)sec2ydyextanydx=(ex−1)sec2ydy(exex−1)dx=sec2ytanydyd(ex−1)ex−1=d(tany)tanyIntegrating on both sides, we get∫d(ex−1)ex−1=∫d(tany)tany⇒log|ex−1|=log|tany|+logc⇒log|ex−1|=logc|tany|⇒ctany=ex−1 is the general solution.

Solve $e^{x} t a n y d x + (1 - e^{x}) s e c^{2} y d y = 0$