Question

Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .

Answer 1

We have,

\[y^2 dx + \left( x^2 - xy + y^2 \right) dy = 0\]

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

This is homogeneous differential equation.
Putting

\[y = vx \text { and} \frac{dy}{dx} = v + x\frac{dv}{dx}, \text { we get }\]

\[v + x\frac{dv}{dx} = \frac{- v^2 x^2}{x^2 - v x^2 + v^2 x^2}\]

\[ \Rightarrow v + x\frac{dv}{dx} = \frac{- v^2}{1 - v + v^2}\]

\[ \Rightarrow x\frac{dv}{dx} = \frac{- v^2}{1 - v + v^2} - v\]

\[\Rightarrow x\frac{dv}{dx} = \frac{- v - v^3}{1 - v + v^2}\]

\[ \Rightarrow \frac{1 - v + v^2}{v + v^3}dv = - \frac{1}{x}dx\]

\[ \Rightarrow \frac{1 + v^2 - v}{v\left( 1 + v^2 \right)}dv = - \frac{1}{x}dx\]

Integrating both sides, we have

\[\int\frac{1 + v^2 - v}{v\left( 1 + v^2 \right)}dv = - \int\frac{1}{x}dx\]

\[ \Rightarrow \int\frac{1 + v^2}{v\left( 1 + v^2 \right)}dv - \int\frac{v}{v\left( 1 + v^2 \right)}dv = - \int\frac{1}{x}dx\]

\[ \Rightarrow \int\frac{1}{v}dv - \int\frac{1}{1 + v^2}dv = - \int\frac{1}{x}dx\]

\[ \Rightarrow \log\left| v \right| - \tan^{- 1} v = - \log\left| x \right| + \log C\]

\[\Rightarrow \log \left| \frac{vx}{C} \right| = \tan^{- 1} v\]

\[ \Rightarrow \left| \frac{vx}{C} \right| = e^{\tan^{- 1}} v \]

\[\text { Putting } v = \frac{y}{x}, \text { we get }\]

\[ \Rightarrow \left| y \right| = C e^{\tan^{- 1}} v\]

Hence, 

\[\left| y \right| = C e^{\tan^{- 1} } v\]  is a required solution.