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प्रश्न
Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]
उत्तर
We have,
\[x\frac{dy}{dx} + \cot y = 0\]
\[ \Rightarrow x\frac{dy}{dx} = - \cot y\]
\[ \Rightarrow \tan y\ dy = - \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\tan y\ dy = - \int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| \sec y \right| = - \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left( \left| x \right| \left| \sec y \right| \right) = \log C\]
\[ \Rightarrow x \sec y = C . . . . . (1) \]
\[\text{ Given:- } x = \sqrt{2}, y = \frac{\pi}{4} . \]
Substituting the values of x and y in (1), we get
\[\sqrt{2} sec \frac{\pi}{4} = C\]
\[ \Rightarrow C = 2\]
Substituting the value of C in (1), we get
\[x \sec y = 2\]
\[ \Rightarrow x = 2 \cos y\]
\[\text{Hence, }x = 2 \cos y\text{ is the required solution . }\]
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