Question

Solve $e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$.

Answer 1

$e^{x}tanydx+(1-e^x)se{c}^{2}ydy=0$

$e^{x}tanydx=(e^x-1)se{c}^{2}ydy$

$\frac{e^x}{e^x-1}dx=\frac{se{c}^{2}y}{tany}dy$

Integrating on both sides, we get,

$log|e^x-1|=log|tany|+logc$

$log|e^x-1|=logc\left(tany\right)$

$c.tany=e^x-1$

$tany=k(e^x-1)$ is the general solution.