I have this problem: let $Y_1,\dots,Y_n$ be real analytic functions $\mathbb{R}^+\to\mathbb{R}^+$ such that all the $Y_1,\dots,Y_n$ and all their derivatives are algebraically independent over $\mathbb{R}$, i.e. the functions in $\{{Y_i}^{(j)}|i=1,\dots,n; j=0,1,\dots\}$ are algebraically independent. I actually don't know why such functions should exist (if you know an easy example that would be very appreciated!), however we could consider $F:=\mathbb{R}({Y_i}^{(j)}|i=1,\dots,n; j=0,1,\dots)$, which is a differential field of functions with domains certain open co-countable subsets or $\mathbb{R}^+$. Now if I have an equation $u'=v'\cdot u$, it turns out that the only solutions in $F$ are constant. It seems very reasonable, however I cannot find a reasonable proof for this fact.
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$\begingroup$ For your first question, why not $Y_i(x) = \sum_{n\geq 0} a_i(n) \frac{x^n}{n!}$ for $\{ a_i(n) \}$ characteristic functions of suitably sparse subsets $A_i$ of $\mathbb{Z}_{\geq 0}$ (e.g., $A_i$ is the set of $i$th powers of factorials)? $\endgroup$– S. Carnahan ♦Commented Sep 11, 2014 at 13:38
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$\begingroup$ Thank you very much for your help, your idea is interesting! Unfortunately I miss the tools to see its immediacy $\endgroup$– AndreCommented Sep 15, 2014 at 13:24
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