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For the Following Differential Equation, Find the General Solution:- D Y D X = 1 − Cos X 1 + Cos X

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Question

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

Sum
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Solution

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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Chapter 21: Differential Equations - Revision Exercise [Page 146]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Revision Exercise | Q 64.1 | Page 146

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