Question

\[\frac{dy}{dx} = \frac{\sin x + x \cos x}{y\left( 2 \log y + 1 \right)}\]

Answer 1

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

\[ \Rightarrow 2\log y\int y\ dy - 2\int\left( \frac{d}{dy}\left( \log y \right) \times \int y\ dy \right)dy + \int y\ dy = - \cos x + x\int \cos x\ dx - \int \left[ \frac{dx}{dx} \times \int\cos x \right] dx\]

\[ \Rightarrow y^2 \log y - \int y\ dy + \int y\ dy = - \cos x + x \sin x\ dx + \cos x + C\]

\[ \Rightarrow y^2 \log y = x \sin x + C\]