Question

(x² + 3xy + y²) dx − x² dy = 0

Answer 1

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