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The ODE system (see below), where $F$ is a given function together the algebraic condition (SYM) imply that $$\boxed{y(t)=k-x(t)} \tag{*}$$ for some $k$ constant.

The result that $y$ is a translation of $x$ reminds me of Lie symmetries of ODEs. The condition (SYM) feels like, smells like, tastes like a symmetry condition.

But symmetry methods involve changing the coordinates of the ODE to canonical coordinates and here we are not doing this? Or are we? My question is: Is there a way of interpreting (*) as arising from some change of variables as in the symmetry analysis of ODEs? If so, how? Sorry if this is all too obvious.

\begin{align} &F(x,x)\,x^\prime(t)-F(x,y)\,y^\prime(t)-1=0 \tag{ODE_1}\\ \\ &F(y,x)\,x^\prime(t)-F(y,y)\,y^\prime(t)-1=0\tag{ODE_2}\\ \\ &F(x(t),x(t))+F(x(t),y(t))=F(y(t),x(t))+F(y(t),y(t)) \tag{SYM}. \end{align}

We also are assuming the ODE system is not under-determined, that is we can be solve it for $x^\prime$ and $y^\prime$. For almost all $x$ and $y$ we require: $$ \det \pmatrix{ F(x,x) & F(x,y) \\ F(x,y) & F(y,y)}\neq 0 $$

Notice also SYM does not hold for any $x,y$ but only for $x(t)$ and $y(t)$ that solve the ODE system.

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  • $\begingroup$ Of course, one line of algebra allows to show that (ODE_1)+(ODE_2)+(SYM) implies (*). Is this what you are after? $\endgroup$ – Did Nov 27 '14 at 8:13
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    $\begingroup$ @Did: Thanks for your comment but no this was not what I am after. I stated this fact in the first paragraph. My question is if SYM is related to a Lie point symmetry of the system (OD1+ODE2): en.wikipedia.org/wiki/Lie_point_symmetry#Types_of_symmetries $\endgroup$ – Sergio Parreiras Dec 1 '14 at 16:38

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