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Question
\[\frac{dy}{dx} = \frac{\sin x + x \cos x}{y\left( 2 \log y + 1 \right)}\]
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Solution
We have,
\[\frac{dy}{dx} = \frac{\sin x + x \cos x}{y\left( 2 \log y + 1 \right)}\]
\[ \Rightarrow y \left( 2 \log y + 1 \right)dy = \left( \sin x + x \cos x \right)dx\]
Integrating both sides, we get
\[ \Rightarrow 2\log y\int y\ dy - 2\int\left( \frac{d}{dy}\left( \log y \right) \times \int y\ dy \right)dy + \int y\ dy = - \cos x + x\int \cos x\ dx - \int \left[ \frac{dx}{dx} \times \int\cos x \right] dx\]
\[ \Rightarrow y^2 \log y - \int y\ dy + \int y\ dy = - \cos x + x \sin x\ dx + \cos x + C\]
\[ \Rightarrow y^2 \log y = x \sin x + C\]
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