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Let $E(z)$ be an entire elementary function of (complex) $z$ and $N(z)$ be an entire nonelementary function of (complex) $z$.

$e^{N(z)}$$N'(z) = E(z)$

The ' means derivative with respect to $z$.

How to find such $E(z)$ and $N(z)$ ?

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The left side is the derivative of $e^N$. So $e^N=\int E+C$, $N=\log(\int E+C)$. If $\int E$ is not elementary, then neither is $N$. But making $N$ entire.... –  Gerry Myerson Oct 8 '12 at 12:11
    
@GerryMyerson : The appearance of a log does not exclude the property of entire. Also Im aware that the left side is the derivative of $e^N$. Without the requirement that N is entire it would be too simple to ask. Basicly to be honest you just restated the problem in a nonsurprising way. –  mick Oct 8 '12 at 12:43
    
I'll interpret that as your way to thank me for being the only person to take any interest in your question. You're welcome. –  Gerry Myerson Oct 8 '12 at 22:43
    
I do not know in advance how many people will take interest in my question. Thank you for your interest if thats what you want to hear. But with respect your comment did not bring us a step closer to the answer. Maybe it clarifies for someone else but for me it did nothing. No offense. –  mick Oct 11 '12 at 11:51
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