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X D Y D X + Cot Y = 0

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प्रश्न

\[x\frac{dy}{dx} + \cot y = 0\]
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उत्तर

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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अध्याय 21: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 21 Differential Equations
Exercise 22.07 | Q 13 | पृष्ठ ५५

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