<s>[INST] Need help solving this differential equation $$p^3 - 2xyp + 4y^2 = 0$$
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
$$(xy^2)′=y^2+2xyy′$$
 [/INST] 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.
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