Question

\[xy \log\left( \frac{x}{y} \right) dx + \left\{ y^2 - x^2 \log\left( \frac{x}{y} \right) \right\} dy = 0\]

Answer 1

Given:-

\[xy \log\left( \frac{x}{y} \right) dx + \left\{ y^2 - x^2 \log\left( \frac{x}{y} \right) \right\}dy = 0\]

\[ \Rightarrow xy \log\left( \frac{x}{y} \right) dx = - \left\{ y^2 - x^2 log\left( \frac{x}{y} \right) \right\}dy\]
\[ \Rightarrow \frac{dx}{dy} = \frac{- \left\{ y^2 - x^2 log\left( \frac{x}{y} \right) \right\}}{xy log\left( \frac{x}{y} \right)} = \frac{x^2 \log\left( \frac{x}{y} \right) - y^2}{xy log\left( \frac{x}{y} \right)}\]
It is a homogeneous equation . 
We put x = vy
\[\frac{dx}{dy} = v + y\frac{dv}{dy}\]
\[So, v + y\frac{dv}{dy} = \frac{v^2 y^2 \log(v) - y^2}{v y^2 \log(v)}\]
\[v + y\frac{dv}{dy} = \frac{v^2 \log(v) - 1}{v \log(v)}\]
\[ \Rightarrow y\frac{dv}{dy} = \frac{v^2 log(v) - 1}{v log(v)} - v\]
\[ \Rightarrow y\frac{dv}{dy} = \frac{v^2 log(v) - 1 - v^2 log(v)}{v log(v)}\]
\[ \Rightarrow y\frac{dv}{dy} = \frac{- 1}{v log(v)}\]
\[ \Rightarrow v log(v) dv = \frac{- 1}{y}dy\]
On integrating both sides we get,
\[\int v \log(v) dv = - \int\frac{1}{y}dy\]
\[ \Rightarrow \frac{v^2}{2} \log(v) - \int\frac{v}{2} dv = - \log y + C\]
\[ \Rightarrow \frac{v^2}{2} log(v) - \frac{v^2}{4} = - \log y + C\]
\[ \Rightarrow \frac{v^2}{2}\left[ log(v) - \frac{1}{2} \right] = - \log y + C\]
\[ \Rightarrow v^2 \left[ log(v) - \frac{1}{2} \right] = - 2 log y + C\]
\[\text{ now putting back the values of v as }\frac{x}{y}\text{ we get, }\]
\[\frac{x^2}{y^2}\left[ log(v) - \frac{1}{2} \right] + \log y^2 = C\]