Question
Solve the following differential equation:
ex tan y dx + (1 – ex) sec2y dy = 0.

Answer

Given differential equation can be written as 

$\frac{\text{e}^{\text{x}}}{\text{1-e}^{\text{x}}}\text{ dx}+\frac{\sec^{2}\text{y}}{\tan\text{y}}\text{ dy}=0$

Integrating to get - log |1 - ex |+log|tan y| = log |c|

log |tan y| = log|c (1-ex)

$\therefore$ tan y = c (1 - ex).

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