Question

For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}\]

Answer 1

We have,

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

\[ \Rightarrow \frac{dy}{dx} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}}\]

\[ \Rightarrow \frac{dy}{dx} = \tan^2 \frac{x}{2}\]

\[ \Rightarrow dy = \left( \tan^2 \frac{x}{2} \right)dx\]

Integrating both sides, we get

\[\int dy = \int\left( \tan^2 \frac{x}{2} \right)dx\]

\[ \Rightarrow \int dy = \int\left( \sec^2 \frac{x}{2} - 1 \right)dx\]

\[ \Rightarrow y = 2 \tan \frac{x}{2} - x + C\]