Question

(x + 2y) dx − (2x − y) dy = 0

Answer 1

\[\left( x + 2y \right)dx - \left( 2x - y \right) dy = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x + 2y}{2x - y}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x + 2vx}{2x - vx}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + 2v}{2 - v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + v^2}{2 - v}\]
\[ \Rightarrow \frac{2 - v}{1 + v^2}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{2 - v}{1 + v^2}dv = \int\frac{1}{x}dx . . . . . (1)\]
\[ \Rightarrow \int\frac{2}{1 + v^2}dv - \int\frac{v}{1 + v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{2}{1 + v^2}dv - \frac{1}{2}\int\frac{2v}{1 + v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow 2 \tan^{- 1} v - \frac{1}{2}\log \left| 1 + v^2 \right| = \log \left| x \right| + \log C\]
\[ \Rightarrow 2 \tan^{- 1} v = \log \left| x \right| + \log C + \log \left| \left( 1 + v^2 \right)^\frac{1}{2} \right|\]
\[ \Rightarrow 2 \tan^{- 1} v = \log \left| Cx\sqrt{1 + v^2} \right|\]
\[ \Rightarrow \left| Cx\sqrt{1 + v^2} \right| = e^{2 \tan^{- 1} v} \]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \left| Cx\sqrt{1 + \left( \frac{y}{x} \right)^2} \right| = e^{2 \tan^{- 1} \left( \frac{y}{x} \right)} \]
\[ \Rightarrow C\sqrt{x^2 + y^2} = e^{2 \tan^{- 1} \left( \frac{y}{x} \right)} \]
\[\text{ Hence, }\sqrt{x^2 + y^2} = K e^{- 2 \tan^{- 1} \frac{y}{x}}\text{ is the required solution }.\]