Question

Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.

Answer 1

We have, 
\[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y\cos y}\]
\[ \Rightarrow \left( \sin y + y\cos y \right) dy = 2x\left( \log x + 1 \right) dx\]
Integrating both sides, we get
\[\int\left( \sin y + y\cos y \right) dy = \int2x\left( \log x + 1 \right) dx\]
\[ \Rightarrow \int\sin y\ dy + \int y\cos y\ dy = \int2x \log x\ dx + \int2x\ dx\]
\[ \Rightarrow - \cos y + \left[ y\int\cos y\ dy - \int\left\{ \frac{d}{dy}\left( y \right)\int\cos y\ dy \right\}dy \right] = 2\left[ \log x \int x\ dx - \int\left\{ \frac{d}{dx}\left( \log x \right)\int x\ dx \right\}dx \right] + x^2 \]
\[ \Rightarrow - \cos y + \left[ y \sin y + \cos y \right] = 2\left[ \log x \times \frac{x}{2}^2 - \frac{x^2}{4} \right] + x^2 + C\]
\[ \Rightarrow y \sin y = x^2 \log x - \frac{x^2}{2} + x^2 + C\]
\[ \Rightarrow y \sin y = x^2 \log x + \frac{x^2}{2} + C ..........(1)\]
\[\text{ Given:- }x = 1, y = 0 . \]
Substituting the values of x and y in (1), we get
\[ 0 = 0 + \frac{1}{2} + C\]
\[ \Rightarrow C = - \frac{1}{2}\]
Substituting the value of C in (1), we get
\[y \sin y = x^2 \log x + \frac{x^2}{2} - \frac{1}{2}\]
\[ \Rightarrow 2y \sin y = 2 x^2 \log x + x^2 - 1\]
\[\text{ Hence, }2y \sin y = 2 x^2 \log x + x^2 - 1\text{ is the required solution.} \]