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Question
\[\frac{dy}{dx} = \frac{y\left( x - y \right)}{x\left( x + y \right)}\]
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Solution
We have,
\[\frac{dy}{dx} = \frac{y\left( x - y \right)}{x\left( x + y \right)}\]
Putting `y = vx,` we get
\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[ \therefore v + x\frac{dv}{dx} = \frac{vx\left( x - vx \right)}{x\left( x + vx \right)}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v\left( 1 - v \right)}{1 + v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v\left( 1 - v \right)}{1 + v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v - v^2 - v - v^2}{1 + v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{- 2 v^2}{1 + v}\]
\[ \Rightarrow - \frac{\left( 1 + v \right)}{2 v^2} dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[ - \int\frac{\left( 1 + v \right)}{2 v^2} dv = \int\frac{1}{x}dx\]
\[ \Rightarrow - \frac{1}{2}\int\frac{1}{v^2}dv - \frac{1}{2}\int\frac{1}{v}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \frac{1}{2v} - \frac{1}{2}\log v = \log x + \log C\]
\[ \Rightarrow \frac{x}{y} - \log \frac{y}{x} = 2\log x + 2\log C\]
\[ \Rightarrow \frac{x}{y} - \log \frac{y}{x} = \log C^2 x^2 \]
\[ \Rightarrow \frac{x}{y} = \log C^2 x^2 + \log \frac{y}{x}\]
\[ \Rightarrow \frac{x}{y} = \log C^2 xy\]
\[ \Rightarrow e^\frac{x}{y} = C^2 xy\]
\[ \Rightarrow e^\frac{x}{y} = k\ xy,\text{ where }k = C^2 \]
