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प्रश्न
(sin x + cos x) dy + (cos x − sin x) dx = 0
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उत्तर
We have,
\[\left( \sin x + \cos x \right)dy + \left( \cos x - \sin x \right)dx = 0\]
\[ \Rightarrow dy = - \left( \frac{\cos x - \sin x}{\sin x + \cos x} \right)dx\]
Integrating both sides, we get
\[\int dy = - \int\left( \frac{\cos x - \sin x}{\sin x + \cos x} \right)dx\]
\[ \Rightarrow y = - \int\left( \frac{\cos x - \sin x}{\sin x + \cos x} \right)dx\]
\[\text{ Putting }\sin x + \cos x = t\]
\[ \Rightarrow \left( \cos x - \sin x \right) dx = dt\]
\[ \therefore y = - \int\frac{dt}{t}\]
\[ \Rightarrow y = - \log\left| t \right| + C\]
\[ \Rightarrow y = - \log\left| \sin x + \cos x \right| + C\]
\[ \Rightarrow y + \log\left| \sin x + \cos x \right| = C\]
\[\text{ Hence, }y + \log\left| \sin x + \cos x \right| =\text{ C is the solution to the given differential equation }.\]
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