Question

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

Answer 1

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