Question
Solve the following differential equation:
$y^2dx + (x^2 – xy + y^2)dy = 0$

Answer

$y^2 dx + (x^2– xy + y^2) dy = 0$
$\Rightarrow \frac{\text{dx}}{\text{dy}} = -\frac{(\text{x}^{2} - \text{xy + y}^{2})}{\text{y}^{2}}$
$\text{put x = vy} \Rightarrow \frac{\text{dx}}{\text{dy}} = \text{v + y} \frac{\text{dv}}{\text{dy}}$
$\text{v + y} \frac{\text{dv}}{\text{dy}} = \frac{\text{(v}^{2}\text{y}^{2} - \text{y}^{2} \text{v} + \text{y}^{2})}{\text{y}^{2}}$
$\Rightarrow \frac{\text{dv}}{\text{v}^{2} + 1} = -\frac{\text{dy}}{\text{y}}$
Integrating both sides
$\tan^{-1} \text{v} = -\log \text{y + c}$
$\Rightarrow \tan^{-1} \frac{\text{x}}{\text{y}} = -\log \text{y + c}$

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