Need help solving this differential equation

Question

$$p^3 - 2xyp + 4y^2 = 0$$ where $p = \mathrm dy / \mathrm dx$.

I don't know which type of equation it is or how to simplify it. Though there is an observation that $$(xy^2)′=y^2+2xyy′$$

Answer 1

The given equation is:

$$\tag 1 p^3 - 2xyp + 4y^2 = 0$$

Solving for $x$ yields:

$$x = \dfrac{2y}{p} + \dfrac{p^2}{2y}$$

Differentiating w.r.t. $y$ yields:

$$\dfrac{dx}{dy} = \dfrac 1p = \dfrac 2p - \dfrac{2y}{p^2}\dfrac{dp}{dy}-\dfrac {p^2}{2y^2} + \dfrac{p}{y} \dfrac{dp}{dy}$$

Simplifying yields:

$$p = 2p - 2y \dfrac{dp}{dy} - \dfrac{p^4}{2y^2} + \dfrac{p^3}{y} \dfrac{dp}{dy}$$

$$p - 2y \dfrac{dp}{dy} - \dfrac{p^4}{2y^2} + \dfrac{p^3}{y} \dfrac{dp}{dy} = 0$$

$$p \left( 1 - \dfrac{p^3}{2y^2} \right) - 2y \dfrac{dp}{dy} \left( 1 - \dfrac{p^3}{2y^2} \right) = 0$$

$$\left( 1 - \dfrac{p^3}{2y^2} \right) \left( p - 2y \dfrac{dp}{dy} \right) = 0$$

Ignoring the first term since it does not include $\dfrac{dp}{dy}$, we have:

$$p - 2y \dfrac{dp}{dy} = 0 \implies 2 \int \dfrac{dp}{p}- \int \dfrac{dy}{y} = 0 \implies 2 \ln p - \ln y = \ln c$$

This yields:

$$\tag 2 p^2 = cy$$

Now, we have to eliminate $p$ from $(1)$, yielding:

$$2xyp = p^3 + 4y^2 \implies 2xyp-p^3 = 4y^2 \implies p(2xy-p^2) = 4y^2$$

Squaring yields:

$$p^2(2xy-p^2)^2 = 16y^4$$

Substituting $(2)$ yields:

$$cy(2xy - cy)^2 = 16 y^4$$

Maybe it is possible to simplify, but I am sure you can proceed with that.