Solve:
$(2 x^{2} + 6 x y - y^{2}) d x + (3 x^{2} - 2 x y + y^{2}) d y = 0$

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Solution

Given integral equation $\begin{matrix} (2 x^{2} + 6 x y - y^{2}) d x + (3 x^{2} - 2 x y + y^{2}) d y = 0 \end{matrix}$
Group the given terms as follow,
$2 x^{2} d x + y^{2} d y - (y^{2} d x + 2 x y d y) + 3 x^{2} d y + 6 x y d x = 0 2 x^{2} d x + y^{2} d y - d (x y^{2}) + 3 d (x^{2} y) = 0$
Integrating, we get
$\int 2 x^{2} d x + \int y^{2} d y - \int d (x y^{2}) + \int 3 d (x^{2} y) = 0$
$\frac{2 x^{3}}{3} + \frac{y^{3}}{3} - x y^{2} + 3 x^{2} y = C ∴ 2 x^{3} + y^{3} - 3 x y^{2} + 9 x^{2} y = C^{'}$ Where $C^{'} = 3 C$

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