Question

$x { \cos}^{2} (y) dx + \tan(y)dy$ = 0
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Answer 1

Step-by-step explanation:

We have,

$x \: \cos^{2} (y) \: dx + \tan(y) \: dy = 0$

$= > 1 + x \: \tan(y) \: \sec^{2} (y) \frac{dy}{dx} = 0$

$= > \tan(y) \sec^{2} (y) \frac{dy}{dx} = \frac{ - 1}{x}$

$= > \tan(y) \sec^{2} (y) \: dy = \frac{ - 1}{x} dx$

Integrating both sides,

$= > \int\tan(y) \sec^{2} (y) dy = - \int \frac{1}{x}dx$

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

$= > \tan^{2} (y) + log( {x}^{2} )+c = 0$