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In a recent MO question the author asks: Could the Riemann zeta function be a solution for a known differential equation?

I don't understand why the answer isn't a trivial yes? For example, $y' = -\sum log(n)/n^s $ is an equation with y being the zeta function. It's easy to come up with many trivial examples of this. What am I misunderstanding?

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    The differential equation should not contain any nontrivial functions of $s$ except via $y$ and its derivatives. – 2017-01-22
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    Here is a link to the MO post the OP is (probably) talking about, for reference: http://mathoverflow.net/questions/260196/could-the-riemann-zeta-function-be-a-solution-for-a-known-differential-equation – 2017-01-22
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    Yes, it's really easy to make more. – 2017-01-22
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    For example the $\Gamma$ function is a solution for a very basic functional equation $\Gamma(s+1) = s \Gamma(s)$. Quite similarly, the Riemann zeta function is a solution for $\zeta(s) = \chi(s) \zeta(1-s)$ where $\chi(s) = \pi^{s-1} 2^s \sin(\pi s/2) \Gamma(1-s)$. So those functions obeys to some functional equations, but we don't know any similar **differential** equations for them. – 2017-01-22

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The zeta function is trivially a solution to many differential equations

You are correct that we can construct differential equations for which $\zeta(s)$ is a solution in this manner. But what you have. . .

$y'(s) = \displaystyle \sum_{n=1} ^\infty \frac{\log(n)}{n^s}$

. . . is quite a complicated equation. Much more complicated that, for example, a finite linear equation like. . .

$A_n(s)y^{(n)} + A_{n-1}(s)y^{(n-1)}+ \ldots + A_{1}(s)y' + A_{0}(s)y(s) = 0$

. . . where $A_i(s)$ are some functions. The paper points out that $\zeta(s)$ does not solve any finite linear equation. Then it asks how complicated we need to make an equation in order for $\zeta(s)$ to solve it. Does it need to be as complicated as yours or can we do better?

The answer given in the paper is that $\zeta(s)$ solves an infinite linear equation of the form

$ \ldots + A_n(s)y^{(n)} + A_{n-1}(s)y^{(n-1)}+ \ldots + A_{1}(s)y' + A_{0}(s)y(s) = 0$.

This equation is simpler than yours in that there are no functions of $s$ other than $y(s)$ and its derivatives. However it is more complicated in the sense that it includes the higher derivatives of $y(s)$.

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    $(s-1) \zeta(s)$ solves such an infinite differential equation, as any entire function ? – 2017-01-22