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Tan Y D Y D X = Sin (X + Y) + Sin (X − Y) - Mathematics

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प्रश्न

tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y) 

 

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उत्तर

We have,
\[\tan y \frac{dy}{dx} = \sin \left( x + y \right) + \sin \left( x - y \right)\]
\[ \Rightarrow \tan y \frac{dy}{dx} = \sin x \cos y + \cos x \sin y + \sin x \cos y - \cos x \sin y\]
\[ \Rightarrow \tan y \frac{dy}{dx} = 2 \sin x\cos y\]
\[ \Rightarrow \frac{\tan y}{\cos y}dy = 2 \sin x dx\]
\[ \Rightarrow \tan y \sec y dy = 2 \sin x dx\]
Integrating both sides, we get
\[\int\tan y \sec y dy = 2\int\sin x dx\]
\[ \Rightarrow \sec y = - 2 \cos x + C\]
\[ \Rightarrow \sec y + 2 \cos x = C\]
\[\text{ Hence,} \sec y + 2 \cos x = \text{ C is the required solution .}\] 

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अध्याय 22: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 22 Differential Equations
Exercise 22.07 | Q 24 | पृष्ठ ५५

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