Arc length paramatrizations satisfy original system of differential equations? Say we have a system of differential equations
$$
\begin{cases}
x'''(t)+f(t)x'(t)=0\\
y'''(t)+f(t)y'(t)=0
\end{cases}
$$
on an interval $[a,b]$, along with the restriction that
$$
x'(t)^2+y'(t)^2=1
$$
on $[a,b]$.
Now, Euler proved that as long as $x(t),y(t)$ are continuous and not both zero at a point, then we can re-parametrize $x,y$ to some functions $\tilde x,\tilde y$ which have the same image as $x,y$ but are arc-length parametrized.
Now, say we have a solution $\big(x^*(t),y^*(t)\big)$ to
$$
\begin{cases}
x'''(t)+f(t)x'(t)=0\\
y'''(t)+f(t)y'(t)=0
\end{cases}
$$
that satisfies the hypotheses needed to apply the previously mentioned re-parametrization.  Say $\big(\tilde x^*(t), \tilde y^*(t)\big)$ are this re-parametrization, so they have the same image of $\big(x(t),y(t)\big)$ on $[a,b]$ and satisfy
$$
x'(t)^2+y'(t)^2=1.
$$
Is it still the case that
$$
\begin{cases}
(\tilde x^*)'''(t)+f(t)(\tilde x^*)'(t)=0\\
(\tilde y^*)'''(t)+f(t)(\tilde y^*)'(t)=0?
\end{cases}
$$
Note: I'm not necessarily looking for a proof of this, just some intuition.
 A: Solutions of that system cannot be reparametrized to have $(x')^2 + (y')^2$ constant for nonconstant $f(t)$.  Reparametrization alters the velocity vector but here $x',y'$ are the "spatial" coordinates and $x'',y''$ the velocities.  For variable positive $f(t)$ the solution curves cut across the solutions with $f=c$ for different $c$, and all of the latter are circles (for $c > 0$).
If you did reparametrize to have ${x''}^2 + {y''}^2$ constant, there would be no reason for it to satisfy the same system of differential equations, but $(x'(t),y'(t))$ would follow the same curves.
A: I agree with the view of zyx and Etienne. In the general case, i.e. for any function $f(t)$, the solutions of that system cannot be reparametrized to have 
$(x′)^2+(y′)^2=1$. Nevertheless, it is of interest to see if some particular functions $f(t)$ allow to reparametrize the solutions according to the condition $(x′)^2+(y′)^2=1$
There was a big mistake in my first answer. I am very sorry for that. The previous attachment is now remplaced by the corrected one below. In fact, the solution is much simpler than before. 
In this particular case of $c_1-c_2=n\pi$, all the solutions are circles, which could have been guessed immediately.

