From power series to differential equations

I was wondering whether for any convergent real power series (or a Frobenius series) we can find (or prove that there exists) a corresponding differential equation that characterizes it. I am aware of Hölder's theorem. So, in effect I am looking for results in these lines but, of course , for real analytic functions. (Generally, my question reads: Can real analytic functions be characterized by differential equations?)

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Note that my question here is not restricted to algebraic differential equations. – Unknown Dec 26 '10 at 14:07
$x=f$ is the unique solution to $x'=(f'/f)x, x(0)=f(0)$ if $f(0)\ne 0$. This is of course stupid, but it shows shows that you need to make your question a bit more precise. – Felipe Voloch Dec 26 '10 at 14:53
Thanks. I reworked it exactly as you explained(even with the same variable t) before commenting but wondered why you chose to do so with the $x$'s and $f$'s in that cognitively dissonant way. – Unknown Dec 26 '10 at 17:42
Maybe an algebraist thinks $x$ is an unknown to be solved for, while an analyst thinks $x$ is the independent variable of a function. – Gerald Edgar Dec 27 '10 at 1:23
If you consider invertible functions as in oeis.org/A145271, then the autonomous diff. eqn. in that entry applies to characterize f. – Tom Copeland Feb 1 '12 at 9:37

@Gerry: not really: for every (formal!) power series $f$ there is an ODE $u'(x)=f(x)$, so there are at least as many ODEs as series. – Mariano Suárez-Alvarez Dec 26 '10 at 15:44
@Mariano, yes, but: if you propose to write the right side out as an infinite power series, then you can't actually write it down, while if you just call the right side $f$, then eventually you run out of names for power series, without running out of power series. – Gerry Myerson Dec 26 '10 at 16:00
@Mariano, so, if the question had been whether for any real number $\alpha$ we can find, or prove there exists, a linear polynomial that characterizes it, you would say, yes, $x-\alpha$, and I would say that's cheating, because you generally can't write $\alpha$. I take your point; but I suspect OP had in mind an equation that didn't make such blatant use of the given power series, but rather used only, say, rational numbers and common functions. So I think it's up to OP to clarify intent. – Gerry Myerson Dec 27 '10 at 2:40