Question

2xy dx + (x² + 2y²) dy = 0

Answer 1

\[2xy dx + \left( x^2 + 2 y^2 \right) dy = 0\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{2xy}{x^2 + 2 y^2}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = - \frac{2v x^2}{x^2 + 2 v^2 x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = - \frac{2v}{1 + 2 v^2}\]
\[ \Rightarrow x\frac{dv}{dx} = - \frac{2v}{1 + 2 v^2} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{- 3v - 2 v^3}{1 + 2 v^2}\]
\[ \Rightarrow \frac{1 + 2 v^2}{3v + 2 v^3}dv = - \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{1 + 2 v^2}{3v + 2 v^3}dv = - \int\frac{1}{x}dx\]
\[\text{ Substituting }3v + 2 v^3 = t,\text{ we get }\]
\[3\left( 1 + 2 v^2 \right) dv = dt\]
\[ \therefore \frac{1}{3}\int\frac{dt}{t}dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow \frac{1}{3}\log \left| t \right| = - \log \left| x \right| + \log C\]
\[ \Rightarrow \frac{1}{3}\log \left| 3v + 2 v^3 \right| = - \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| 3v + 2 v^3 \right| = - 3 \log \left| x \right| + 3 \log C\]
\[ \Rightarrow \log \left| \left( 3v + 2 v^3 \right) \times x^3 \right| = \log C^3 \]
\[ \Rightarrow \left( 3v + 2 v^3 \right) \times x^3 = C^3 \]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \left[ \left( 3 \times \frac{y}{x} + 2 \times \frac{y^3}{x^3} \right) \times x^3 \right] = C^3 \]
\[ \Rightarrow 3y x^2 + 2 y^3 = C_1 \]
\[\text{ Hence, }3y x^2 + 2 y^3 = C_1\text{ is the required solution } .\]