Question

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

Answer 1

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