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X D Y D X + Cot Y = 0

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Question

\[x\frac{dy}{dx} + \cot y = 0\]
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Solution

We have, 
\[x\frac{dy}{dx} + \cot y = 0\]
\[ \Rightarrow x\frac{dy}{dx} = - \cot y\]
\[ \Rightarrow \frac{1}{x}dx = - \frac{1}{\cot y}dy\]
\[ \Rightarrow \frac{1}{x}dx = - \tan y dy\]
Integrating both sides, we get
\[\int\frac{1}{x}dx = - \int\tan y dy\]
\[ \Rightarrow \ln \left| x \right| = - \ln\left| \sec y \right| + \ln C\]
\[ \Rightarrow \ln \left| x \right| = \ln \left| \cos y \right| + \ln C\]
\[ \Rightarrow x = C \cos y \]
\[\text{ Hence, }x = C \cos y\text{ is the required solution .} \]

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Chapter 21: Differential Equations - Exercise 22.07 [Page 55]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Exercise 22.07 | Q 13 | Page 55

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