Integrate the differential equation y2 + (x2 - xy + y2) y′ = 0

Integrate the following differential equation:

$y^2 + (x^2 - xy + y^2)y' = 0.$

First, we have

$y^2 + (x^2 - xy + y^2)y' = 0 \quad \implies \quad \frac{y^2}{x^2} + \left(1 - \frac{y}{x} + \frac{y^2}{x^2}\right) y' = 0.$

Making the substitution $v = \frac{y}{x}$ which gives us $y' = v + v'x$ we have

$\begin{align*} && \frac{y^2}{x^2} + \left( 1 - \frac{y}{x} + \frac{y^2}{x^2} \right)y' &= 0 \\[9pt] \implies && v^2 + (1 - v + v^2)(v + v'x) &= 0 \\[9pt] \implies && v + v'x &= \frac{-v^2}{1-v+v^2} \\[9pt] \implies && v'x &= \frac{-v^2 - v + v^2 - v^3}{1-v+v^2} \\[9pt] \implies && v'x &= -\frac{v(v^2+1)}{1-v+v^2}. \end{align*}$

Then, integrating both sides we have,

$\begin{align*} \implies && - \int \frac{1-v+v^2}{v(v^2+1)} \, dv &= \int \frac{1}{x} \, dx \\[9pt] \implies && - \log |v| + \arctan v &= \log |x| + C \\[9pt] \implies && - \log \left| \frac{y}{x} \right| + \arctan \frac{y}{x} &= \log |x| + C \\[9pt] \implies && -\log |y| + \arctan \frac{y}{x} &= C \\[9pt] \implies && \log |y| - \arctan \frac{y}{x} &= C \\[9pt] \implies && \log |y| + \arctan \frac{x}{y} &= C. \end{align*}$

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Integrate the differential equation y² + (x² – xy + y²) y′ = 0

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