differential equation of conics Function $y(x)$ on the plane defines a conic (strictly speaking, at most second order algebraic curve) if and only if $\frac{d^3}{dx^3} (y'')^{-2/3}=0$. What is intuition behind this? How may I see without calculations that this equation is symmetric in $x$, $y$, or, say, that it is affine invariant? Quick calculation comes from considering Wronskian of 6 functions $1,x,y,x^2,y^2,xy$ which vanishes iff $(x,y)$ lie on a conic. But it is not quite satisfactory explanation.
 A: By choosing polar coordinates x,y symmetry is automatically built in it.
May I take opportunity to mention my earlier found differential equation of all conics in polar coordinates using isometric curvature invariants? The formulation is by differentiation of the latus-rectum of any conic 
$$ \boxed{p\cdot k_{gE} = \sin^{3}\psi } \tag{1}$$ 
where $r$ is radius vector, $\psi$ is angle between $r$ and conic. $kg_E, kg_H$  elliptic and hyperbolic curvatures defined by Liouville's and my curvature formulas  respectively as: 
$$  kg_E,  kg_H= \psi^{'} \pm\frac {\sin \psi}{r} \tag{2}$$
Primes are taken on arc of conic.
$$ \psi^{''}= 3   \cot \psi \cdot \psi^{'} \cdot kgE - \frac {\cos \psi}{r}\cdot kgH.  \tag{3} $$
A conic is traced on the basis of the above, and both curvatures, elliptic and hyperbolic, are plotted.
EDIT 1:
Just after the above posting here it came to light that only elongated conics can have hyperbolic points of inflection.  The limiting condition is that the eccentricity of a ellipse conic $ \epsilon \ge \frac12 $ to have points of local hyperbolic straightness; hyperbolas and parabolas would naturally qualify for this ( $ \epsilon \gt \frac12 $ ) ... am sharing this with you hoping it stays well within the scope of DEs of conics.
EDIT2:
Another most useful form of conics ode is
$$ \boxed{  \tan \psi \cdot \psi^{''} = (\frac {\sin  \psi}{r} )^2  + 2 \psi^{'2}  = 3 \psi^{'} \cdot kgE - \frac {\sin  \psi}{r}\cdot kgH}
\tag{4} $$
EDIT 3:
My earlier work regarding intrinsic differential equation of conics:
Mathforum__sci.math_Referece
EDIT4:
The ellipse below is computed and depicted from conics differential equation (1) I had derived earlier.  Note that the its focus position is arbitrary not at origin , it is not a  Newtonian ellipse rotated about focus. 

EDIT 5
Conic sections arise by sectioning a right circular cone ($ \gamma\, = $semi vertical angle) with planes of variable inclination $\alpha.$ Differential equations of conics in their original/formative natural setting are shown below with direct comprehensive relevance in the 3-space.
Consider differential triangle  $ \Delta\,MVR  $
$$ \tan \gamma= \frac{MR}{VR}=\frac{y- r_{min}}{z-z_{min}} = \frac{r \sin \theta- r_{min}}{z-z_{min}}  = \frac{dr}{dz}\tag{5/1}$$
$ {\psi,\gamma}$ are angles meridian makes to 3D conic arc and the symmetry axis respectively. Now differentiate using cylindrical coordinates priming differentials on arc length using differential relations:
Note: The same symbol for $\psi$ is used for denoting angle between radius-arc in the previous case when it is in the flat plane and between meridian-arc in 3D. This is strictly speaking incorrect. At a later date it would be however attempted to be reconciled with a new notation. It is temporarily retained because $(r-\psi)$ relations are invariant in $(kgE,kgH)$ above definitions  valid in 2D and 3D as for example $ r \sin \psi = c $ is a geodesic invariant in either case. 
$$r'= \sin \phi \cos \psi\,;  \theta^{'}= \frac{\sin \psi}{r} \,;z'= \cos \phi \cos \psi; \tag{5/2}$$
Simplify and what we obtain appears to be a pure geometric relation between two variables $ (\psi,\theta):$
$$\tan \alpha \cos \gamma = \tan \psi \cos \theta+ \sin \gamma \sin \theta \tag{5/3} $$
This however is a first order ordinary differential equation if we bear in mind differential relations (1) which formed  relations (5/2), but it is not merely a trigonometrical relationship.
The initial condition  $ \theta =0 $ at point $P$ 
$$ \tan \psi= \tan \alpha \cos \gamma \tag{5/4}$$
It is known
$$\epsilon= \dfrac{\cos \alpha}{\cos \gamma}\tag{5/5}$$
(5/4) and (5/5) supply boundary conditions $\psi_i$ suitable to generate any choice of conic with eccentricity $\epsilon$.
$$ \tan \psi_i= \sin \alpha/\epsilon \tag{5/6} $$
Writing the conics differential equation instead of arc, in terms of $\theta$ and representing differentials with dots (instead of dashes) we have
$$ \boxed{\dot{\psi} = \cos^2 \psi ( \tan \psi . \tan \theta -\sin \gamma )} \tag{5/7}$$
The relations are integrated together and displayed below:

A: I think your original 5th order differential equation and the 5th order differential equation in the references cited by Dave Renfro can be partially understood without calculation.
What matters is that the family of conics is a 5-parameter family of 1-dimensional submanifolds in the plane $P$. The dimension of the total space of the family is 6, which is the same as the dimension of the space $P^{(4)}$ of 4-jets of curves in the plane. Stated correctly, it turns out that forming 4-jets of conics defines a bijection between these 6-dimensional manifolds. Forming their 5-jets yields a section $P^{(4)}\rightarrow P^{(5)}$. Its image $E$ is the ``geometric" expression of the differential equation.
Using this line of reasoning one can show that


*

*A plane curve is a conic if and only if it makes order 5 contact with all of its (4th order) osculating conics.

*The differential equation is equivalent to the condition that the curve make order 5 contact with its osculating conics at each point.


The different versions of the differential equation are probably just different functions cutting out the same zero set $E$ in $P^{(5)}$.
A: Let me try to give a semi-conceptual answer, which is still not free of some computations. It is inspired by beautiful book "Projective Differential Geometry Old and New: From the Schwarzian Derivative to the Cohomology of Diffeomorphism Groups" by Ovsienko and Tabachnikov. 
The main idea is that there exists a correspondence between curves in $\mathbb{RP}^n$ and differential operators 
$$
A=\frac{d^{n+1}}{dx^{n+1}}+a_{n-1}(x)\frac{d^{n}}{dx^{n}}+\ldots+a_1(x)\frac{d}{dx}+a_0.
$$
Projectively equivalent curves correspond to the same operator. Various interesting invariants, like so-called projective curvature, could be obtained from the coefficients of $A$. In our case we are given a curve $\gamma$ in $\mathbb{RP}^2$ with parametric presentation
$$
[y(x):x:1],
$$
which corresponds to an operator
$$
A_{\gamma}=\frac{d^{3}}{dx^{3}}+a_{1}(x)\frac{d}{dx}+a_0.
$$
Tensor density $A_0=(a_0(x)-\frac{1}{2}a_1'(x))(dx)^3$ is invariant under parametrization of the curve and is known to vanish if and only if the curve is a conic. 
Here is a construction of operator $A_{\gamma}.$ Let's pick a lift $\Gamma(x)$ of our curve to $\mathbb{R}^3$ such that 
$$
\det (\Gamma, \Gamma',  \Gamma'')=1.
$$
Then for some functions $a_0, a_1$ we have a linear dependence $\Gamma'''+a_1 \Gamma' +a_0\Gamma.$ These functions are the coefficients of $A_{\gamma}.$
There exists a function $u(x)$ such that the lift of our curve equals to
$(u y,u x,u ).$ 
An easy computation with determinant shows that $u=(y'')^{-\frac{1}{3}}.$ 
We know that functions $u$ and $xu$ are annihilated by operator $A_\gamma.$ 
This implies that $$3u''+a_1u=0.$$ Since $A_0 u=0,$
$$
u'''+a_1u'+a_0 u=0.
$$
From these equations it is easy to see that 
$$a_0-\frac{1}{2}a_1'=0$$ if and only if $$(u^2)'''=((y'')^{-\frac{2}{3}})''' =0.$$
A: I don't know what you mean by 'without calculation'.  I don't think you'll get a simpler explanation than simply solving the equation:  If
$$
\frac{d^3\bigl((y'')^{-2/3}\bigr)}{dx^3} = 0,
$$
then
$$
y''(x) = (ax^2 +2bx + c)^{-3/2}
$$
for some constants $a$, $b$, and $c$, so, integrating twice (assuming $ac-b^2\not=0$), you get
$$
y(x) = ex + f + \frac{(ax^2 +2bx + c)^{1/2}}{(ac-b^2)}
$$
for some constants $e$ and $f$, which gives
$$
(ac-b^2)^2(y-ex-f)^2 - (ax^2 +2bx + c) = 0,
$$
which is the general equation of a conic.  (I'm sure you can handle the degenerate cases.)
Obviously, this is reversible, so what more is there to know?
Finally, you should be aware that the expression $\frac{d^3\bigl((y'')^{-2/3}\bigr)}{dx^3}$ is not, itself, an affine invariant, let alone a projective invariant:  It's not even homogeneous of degree $0$ in $y$.  The true projective invariant is the symmetric cubic form
$$
C = (y'')^{2/3}\left(\frac{d^3\bigl((y'')^{-2/3}\bigr)}{dx^3}\right)\,dx^3.
$$
