Question

\[\frac{dy}{dx} = \frac{y - x}{y + x}\]

Answer 1

We have, 
\[\frac{dy}{dx} = \frac{y - x}{y + x}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{vx - x}{vx + x}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{x\left( v - 1 \right)}{x\left( v + 1 \right)}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v - 1}{v + 1} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v - 1 - v^2 - v}{v + 1}\]
\[ \Rightarrow x\frac{dv}{dx} = - \frac{v^2 + 1}{v + 1}\]
\[ \Rightarrow \frac{v + 1}{v^2 + 1}dv = - \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{v + 1}{v^2 + 1}dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{v}{v^2 + 1}dv + \int\frac{1}{v^2 + 1}dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \frac{1}{2}\int\frac{2v}{v^2 + 1}dv + \int\frac{1}{v^2 + 1}dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \frac{1}{2}\log \left| v^2 + 1 \right| + \tan^{- 1} v = - \log \left| x \right| + C\]
\[ \Rightarrow \frac{1}{2}\log \left| v^2 + 1 \right| + \log \left| x \right| + \tan^{- 1} v = C\]
\[ \Rightarrow \log \left| v^2 + 1 \right| + 2 \log \left| x \right| + 2 \tan^{- 1} v = 2C\]
\[ \Rightarrow \log \left| v^2 + 1 \right| + \log \left| x^2 \right| + 2 \tan^{- 1} v = 2C\]
\[ \Rightarrow \log \left| \left( v^2 + 1 \right) x^2 \right| + 2 \tan^{- 1} v = 2C \]
\[\text{ Substituting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \log \left| \left( \frac{y^2}{x^2} + 1 \right) x^2 \right| + 2 \tan^{- 1} \frac{y}{x} = 2C\]
\[ \Rightarrow \log \left| \left( y^2 + x^2 \right) \right| + 2 \tan^{- 1} \frac{y}{x} = k .........\left(\text{where }k = 2C \right)\]