Question

\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]

Answer 1

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