Affine plane curves with constant curvature Question
I want to solve this differential equation for $P : \mathbb{R} \to \mathbb{A}^2$, a plane affine curve.

$ P'''(t) = \frac{P'(t)}{t^2}$

Someone recognize this equation? Is a famous curve? Is algebraic? I think it will be a curve invariant under the action of a 1-parameter subgroup of the affine group.
Thanks in advance.
Summary
I considered again the results found in
Affine arc length
I started studying some specials curves:


*

*$\omega^1 = 0$. These are points and are invariant under the action of a 4-parameter subgroup of the affine group. For example for $x^2+y^2=0$
$$
    \begin{bmatrix}
    1 & 0 & 0 \\
    0 & a & b \\
    0 & c & d \\
    \end{bmatrix}
$$

*$\omega_1^2 = 0$. These are straight lines and are invariant under the action of a 4-parameter subgroup of the affine group. For example for $y = 0$
$$
    \begin{bmatrix}
    1 & 0 & 0 \\
    x & a & b \\
    0 & 0 & d \\
    \end{bmatrix}
$$

*$\omega_2^1  = 0$. These are parabolas and are invariant under the action of a 2-parameter subgroup of the affine group. For example for $y = x^2$
$$
    \begin{bmatrix}
    1 & 0 & 0 \\
    x & a & 0 \\
    x^2 & 2ax & a^2 \\
    \end{bmatrix}
$$

*$k = 0$. These are ellipses and hyperbolas and are invariant under the action of a 1-parameter subgroup of the affine group. For example for $x^2+y^2=1$
$$
    \begin{bmatrix}
    1 & 0 & 0 \\
    0 & \cos\theta & \sin\theta \\
    0 & -\sin\theta & \cos\theta \\
    \end{bmatrix}
$$
and for $xy=1$
$$
    \begin{bmatrix}
    1 & 0 & 0 \\
    0 & a & 0 \\
    0 & 0 & 1/a \\
    \end{bmatrix}
$$

*$k = K$. I got the differential equation $ P'''(t) = \frac{P'(t)}{t^2}$, but I don't know if it is a well-known curve and especially if it is still algebraic.
 A: I relaxed a bit the conditions about the parametrization.
I searched for P such that $P''' = \alpha P'' + \beta P'$ with $\alpha, \beta$ constant. So $\lambda^{-1} = \sqrt{\alpha + \frac{2}{9}\beta^2}$. In particular I can choose $\lambda = 1, i$. For example, if I choose $\lambda =1$, I get $\alpha = 1 -  \frac{2}{9}\beta^2$ and $\kappa = \frac{\beta}{3}$ and the equation $P''' - \beta P'' - (1 -  \frac{2}{9}\beta^2)P' = 0$ and that's easy to solve. So I have found:


*

*$d\sigma^2 = 1$


*

*$P = (e^{\frac{3k-\sqrt{k^2+4}}{2}t}, e^{\frac{3k+\sqrt{k^2+4}}{2}t})$ for $|k| \neq \frac{1}{\sqrt{2}}$

*$P = (t, e^{\pm\frac{3}{\sqrt{2}}t})$ for $|k| = \frac{1}{\sqrt{2}}$


*$d\sigma^2 = -1$


*

*$P = (e^{\frac{3}{2}kt}\cos{\sqrt{4-k^2}t}, e^{\frac{3}{2}kt}\sin{\sqrt{4-k^2}t}$) for $|k| < 2$

*$P = (e^{\pm3t}, te^{\pm3t})$ for $|k| = 2$

*$P = (e^{\frac{3k-\sqrt{k^2-4}}{2}t}, e^{\frac{3k+\sqrt{k^2-4}}{2}t})$ for $|k| > 2$
So every curve with constant affine curvature is congruent to one of


*

*$P = (t, t^{\mu})$ with $\mu \ne \frac{1}{2}, 2$ (from 1.1 and 2.3)

*$P = (t, e^{\pm t})$ (from 1.2)

*$P = (e^t, \pm te^t)$ (from 2.2)

*$P = (e^{\lambda t}\cos t, e^{\lambda t}\sin t) $ (from 2.1)


The algebraic one are only of type 1 with $\mu \in \mathbb{Q}$ (or type 4 with $\lambda = 0$)
