Question

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

Answer 1

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

\[ - \int\frac{\left( 1 + v \right)}{2 v^2} dv = \int\frac{1}{x}dx\]

\[ \Rightarrow - \frac{1}{2}\int\frac{1}{v^2}dv - \frac{1}{2}\int\frac{1}{v}dv = \int\frac{1}{x}dx\]

\[ \Rightarrow \frac{1}{2v} - \frac{1}{2}\log v = \log x + \log C\]

\[ \Rightarrow \frac{x}{y} - \log \frac{y}{x} = 2\log x + 2\log C\]

\[ \Rightarrow \frac{x}{y} - \log \frac{y}{x} = \log C^2 x^2 \]

\[ \Rightarrow \frac{x}{y} = \log C^2 x^2 + \log \frac{y}{x}\]

\[ \Rightarrow \frac{x}{y} = \log C^2 xy\]

\[ \Rightarrow e^\frac{x}{y} = C^2 xy\]

\[ \Rightarrow e^\frac{x}{y} = k\ xy,\text{ where }k = C^2 \]