I am having some trouble with the resolution of the following ODE:
$$y(x^2+y^2)dx + x(3x^2-5y^2)dy = 0$$
I have defined $$P(x,y) = y(x^2+y^2) = x^2y + y^3$$
and
$$Q(x,y)=x(3x^2-5y^2)=3x^3-5xy^2$$
so I can find the rotatinal as so:
$$rot = \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} = (x^2 + 3y^2) - (9x^2 - 5y^2) = -8x^2 +8y^2$$
And then I try to find a $$p = \frac{rot}{P(x,y)}= \frac{-8x^2+8y^2}{x^2y+y^3} \neq p(y)$$
or
$$p = \frac{rot}{Q(x,y)} = \frac{-8x^2+8y^2}{3x^3 - 5xy^2} \neq p(x)$$
If $p$ was such that $p = p(x)$, I would gladly find an integrating factor $\mu$ defined as:
$$\mu(x) = e^{\int{p(x)dx}} = e^{\int{\frac{\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}}{Q(x,y)}dx}}$$
and if $p$ was such that $p = p(y)$, I would find an integrating factor $\mu$ defined as:
$$\mu(y) = e^{-\int{p(y)dy}} = e^{-\int{\frac{\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}}{P(x,y)}dx}}$$
But, as seen, $p \neq p(x)$ and $p \neq p(y)$. Is my method alright? If so, how could I find a proper integrating factor for that equation? Or there is another way to solve it? Thank you for the attention.