Question

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

Answer 1

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