Find the particular solution of the differential equation $e^{x} tan y d x + (2 - e^{x}) s e c^{2} y d y = 0$ , given that $y = \frac{π}{4}$ when $x = 0$ .

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Solution

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

Integrating both sides.
$\int \frac{- e^{x}}{2 - e^{x}} d x = \int \frac{{sec}^{2} y d y}{tan y}$
$\Rightarrow log (tan y) = log (2 - e^{x}) + c$

$c = log (tan y) - log (2 - e^{x}) = log \frac{tan y}{2 - e^{x}}$

Putting $y = \frac{π}{4}$ & $x = 0$

$c = log \frac{1}{1} \Rightarrow c = 0$

$\Rightarrow log (tan y) = log (2 - e^{x})$

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