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When given an ODE of the form $F(x, y, y', \ldots, y^{(n)}) = 0$, where $F$ is an elementary function, chances are that it has no solution of the form $y = G(x, c_1, \ldots, c_n)$, where $G$ is also elementary.

But how about the inverse? I.e. given a parametric family of functions $y = G(x, c_1, \ldots, c_n)$, where $G$ is elementary, is there always an ODE $F(x, y, y', \ldots, y^{(m)}) = 0$, where $m\geq n$, such that $G$ satisfies it?

In simple cases it is easy to produce $F$ by differentiating and getting rid of the constants. But if $G$ is complicated and has multiple repetitions of the constants, it is not evident that this will work. I was wondering if you are aware of any positive or negative result in this direction (ideally with a pointer to some corresponding algorithm like e.g. the Risch algorithm for integrals), or a simple solution I am not seeing.

*Note that it is ok for the degree of $F$ to be much larger than the number of constants, thus the solution may not be exactly $G$, but a broader family which contains it.

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Any elementary function solves an algebraic differential equation, i.e. where $F$ is a polynomial of its variables. I think you can find the result in: Eliakim Hastings Moore, "Concerning transcendentally transcendental functions", Mathematische Annalen 48, 1-2 (1896), pp. 49--74. Around page 54. It particularly means that every solution of an elementary ODE is a solution to an algebraic ODE.

As Alexandre Eremenko mentions, it may not possible to adapt the argument to the parametric case when you wish to eliminate relations in the constants that are analytic but whose inverse branches do not solve algebraic differential equations.

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  • $\begingroup$ Thanks for your answer @Loïc! I need to spend some time reading parts of the paper and trying out your suggestion. Will get back to you once I do that. $\endgroup$
    – Marios Koulakis
    Commented Sep 6, 2023 at 14:29
  • $\begingroup$ I had a look at the paper today, but it is not obvious to me how to adapt it for the parametric case. The result you are referring to at page 54 is that $((h(x))) = \mathfrak{R}_x = \mathfrak{R}[((h(x)))]$, right? Would it be possible to elaborate a bit more, or point me to a little more modern paper with the result (if there is one)? By the way, the paper though a bit hard to read, has some really nice ideas! I was not aware they were already working in those topics on the 1890s. $\endgroup$
    – Marios Koulakis
    Commented Sep 9, 2023 at 16:54
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To derive a differential equation, you differentiate $y=G(x,c)$ with respect to $x$ and then eliminate $c$. This elimination process is not always possible with elementary functions. For example $y=e^x+e^{cx}$, $y'=e^x+ce^{cx}$. How do you eliminate $c$ from these two equations, by using only elementary functions?

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  • $\begingroup$ It is a good point that one cannot always directly differentiate and solve for $c$. You can though use the information from previous derivatives to eliminate it. For example in this case, $cy - y' = ce^x + ce^{cx} - e^x - ce^{cx} = ce^x - e^x$ which implies $\frac{y' - e^x}{y - e^x} = c$. By differentiating once more, you can get a second order differential equation expressed by elementary functions. This is a simple example. Nevertheless, differentiation introduces repeated blocks which gives some hope to be able to solve after some number of differentiations in general. $\endgroup$
    – Marios Koulakis
    Commented Sep 6, 2023 at 14:20
  • $\begingroup$ Unless I'm mistaken, you actually can on your example, since $ce^{cx}=c(y-e^x)=y'-e^x$, so that you only have to differentiate the ratio $(y-e^x)/(y'-e^x)$. $\endgroup$
    – Loïc Teyssier
    Commented Sep 6, 2023 at 14:20
  • $\begingroup$ (Marios Koulakis beat me to it by just a few bred crumbles) $\endgroup$
    – Loïc Teyssier
    Commented Sep 6, 2023 at 14:21
  • $\begingroup$ Of course, this probably means, that the degree of the equation might need to be larger than the number of constants, thus formally its solution contains the family it is not exactly the family, which is ok. You have a point though, I will change a bit the question to make this clear. $\endgroup$
    – Marios Koulakis
    Commented Sep 6, 2023 at 14:30

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