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Find the general solution Differential Equations Exercise 9 point 4 Q1.

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Given,

$\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}$

$\\ \implies\frac{dy}{dx} = \frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}} = tan^2\frac{x}{2} \\ \implies dy = (sec^2\frac{x}{2} - 1)dx$

$\\ \implies \int dy = \int sec^2\frac{x}{2}dx - \int dx \\ \implies y = 2tan^{-1}\frac{x}{2} - x + C$

Posted by

HARSH KANKARIA

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Find the general solution:

Q1. $\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}$

Posted by

HARSH KANKARIA

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