This is a strange spiral, which is from the differential equations as follows
$$
\left\{\begin{aligned}
\dot x-\dot y=\ddot x\sqrt{\dot x^2+\dot y^2} \\
\dot x+\dot y=\ddot y\sqrt{\dot x^2+\dot y^2}
\end{aligned}\right.
$$
I don't know how to solve the equations. However, the strange spiral has an interesting property, which is the difference in its radius of curvature between two points on it is proportional to its arc-length between these two points.
This property is from a problem of physics. We consider a particle with mass $m$, and the velocity of it is $1$ along with the x-axis. There are two forces whose magnitudes are constant on the particle. One is always along with the velocity of the particle, and the other one is always perpendicular to the velocity of the particle. And we assume that the cross product of these two forces is constant.
If we set $m=1$ and $F_1=F_2=1$, we can easily get the differential equations above. I have calculated it by numerical integration as the picture shows.
Actually, I think it is like a spiral, but I don't know what kind of spiral it is. And we can calculate its arc between two points
$$
s(t)-s(t_0)=\int^{t}_{t_0}v(t)\text dt=v(t_0)(t-t_0)+\frac{(t-t_0)^2}2
$$
If we set $t_0=0$ and $v(t_0)=0$, we will get $s(t)\propto t^2$. Actually we cannot do this because $v(0)=0$ isn't what we want, but we can make an approximation that $t\gg t_0$ and $t=F_1t/m\gg v(t_0)$ which can make $s(t)\propto t^2$ approximately correct. Notice that the magnitude of $F_2$ is constant and it is always perpendicular to the velocity, therefore
$$
\frac{F_2}m=a_\perp=\frac{v^2}r=\text{const}
$$
where $r$ is the radius of curvature. Also, we can get that $v^2\propto t^2$, which means
$$
\frac{s(t)}{r(t)}=\text{const}
$$
Is it really a spiral? And what more properties does it have? Or are these differential equations solvable?
Best Answer
As is long for a comment I leave some ideas, not an answer.
Calling $u=\dot x, v=\dot y$ we have the system
$$ \cases{ \dot u = \frac{u-v}{\sqrt{u^2+v^2}}\\ \dot v = \frac{u+v}{\sqrt{u^2+v^2}} }\ \ \ \ \ \ \ (1) $$
now making the variables change
$$ \cases{ u = r\cos\theta\\ v = r\sin\theta } $$
the system $(1)$ is transformed into
$$ \dot r = 1\\ \dot\theta = \frac 1r $$
solved this ODE we arrive at the solution curves
$$ r = c_0 e^{\theta}\ \ \ \text{or}\ \ \ \sqrt{\dot x^2+\dot y^2} = c_0 e^{\arctan\left(\frac{\dot y}{\dot x}\right)} $$