Question
Solve the following differential equation: $e^x \tan y\ dx + (1 – e^x) \sec^2y\ dy = 0.$

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 - e^x|+\log|\tan y| $
$= \log |c|\log |\tan y| = \log|c (1-e^x)$
$\therefore \tan y = c (1 - e^x).$

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