Question

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

Answer 1

We have, 
\[y dx + \left\{ x \log \left( \frac{y}{x} \right) \right\} dy - 2x dy = 0\]
\[ \Rightarrow \left\{ 2x - x \log \left( \frac{y}{x} \right) \right\} dy = y dx\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{2x - x \log \left( \frac{y}{x} \right)}\]
This is a homogenoeus differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{vx}{2x - x \log v}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v}{2 - \log v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v}{2 - \log v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v - 2v + v \log v}{2 - \log v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v \log v - v}{2 - \log v}\]
\[ \Rightarrow \frac{2 - \log v}{v \log v - v}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{2 - \log v}{v \log v - v}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{1 - \left( \log v - 1 \right)}{v\left( \log v - 1 \right)}dv = \int\frac{1}{x}dx\]
\[\text{ Putting }\log v - 1 = t\]
\[ \Rightarrow \frac{1}{v}dv = dt\]
\[ \therefore \int\frac{1 - t}{t}dt = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\left( \frac{1}{t} - 1 \right)dt = \int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| t \right| - t = \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| \log v - 1 \right| - \left( \log v - 1 \right) = \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| \log v - 1 \right| - \log v = \log \left| x \right| + \log C_1 ...........\left(\text{where, }\log C_1 = \log C - 1 \right)\]
\[ \Rightarrow \log \left| \frac{\log v - 1}{v} \right| = \log \left| C_1 x \right|\]
\[ \Rightarrow \frac{\log v - 1}{v} = C_1 x\]
\[ \Rightarrow \log v - 1 = C_1 xv\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[\log \frac{y}{x} - 1 = C_1 x \times \frac{y}{x}\]
\[ \Rightarrow \log \frac{y}{x} - 1 = C_1 y\]
\[\text{ Hence, }\log \frac{y}{x} - 1 = C_1 y\text{ is the required solution }.\]