What is the solution of the equation $xyp^2 + (3x^2 - 2y^2)p - 6xy=0$, where $p = \frac{dx}{dy}$

Question

What is the solution of the equation $xyp^2 + (3x^2 - 2y^2)p - 6xy=0$, where $p = \frac{dx}{dy}$

I was trying to solve it by dividing the whole equation by $xy$ and then integrate it

$\frac{dy}{dx}[\frac{dy}{dx} + (\frac{3x}{y} - \frac{2y}{x})] = 6$, but still this equation is non separable.

Please tell me how to solve it.

Answer 1

Using quadratic formula, we have $$p=\frac{2y^2 - 3x^2\pm\sqrt{(3x^2-2y^2)^2+24x^2y^2}}{2xy}$$ $$=\frac{2y^2 - 3x^2\pm(3x^2+2y^2)}{2xy}. $$

That is: $$\frac{dy}{dx}=p=\frac{2y}{x} \text{ or } \frac{-3x}{y}. $$ In both cases, we can easily solve the ODE.

Answer 2

Let $y=\sqrt z$ which makes the equation to be $$\left(z'+6 x\right) \left(x z'-4 z\right)=0$$

Answer 3

For $xy\ne0$, using Vieta $$p^2+\left(\frac{3x}y-\frac{2y}x\right)p+6=0,$$

factors as

$$\left(p-\frac{3x}y\right)\left(p-\frac{2y}x\right)=0$$

which is easy.

$y=0$ is also a solution.