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प्रश्न
For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}\]
उत्तर
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
\[\int dy = \int\left( \tan^2 \frac{x}{2} \right)dx\]
\[ \Rightarrow \int dy = \int\left( \sec^2 \frac{x}{2} - 1 \right)dx\]
\[ \Rightarrow y = 2 \tan \frac{x}{2} - x + C\]
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