There are several papers giving the functional equation for the various derivatives of Zeta(s) as a complex function over the whole plane - the one for the first derivative has been used for many years and is the basis of many papers - but that is not what you asked. On the critical line, the argument of the zeta function obeys a very simple differential equation. See Eq (6.1) of my paper:Research Exploring Riemann's Functional Equation - Final version

Thanks for your answer , your paper mixed me with the gaven comment of Terry Tao where he showed that zeta function can't obey any algebraic differential equation and also the other gaven answers there in MO , but it's good to find a nice paper as you have which gives a rigorous answer to my question !!!

My answer to your question was submitted in a hurry and I should have added the fact that the sum of the arguments of the zeta function on opposite sides of the critical line also satisfies a differential equation. See eq. 3.17 of my paper Article Further Exploration of Riemann's Functional Equation -v2

Further to this question, I have recently come across the paper "A NOTE ON FUNCTIONAL INDEPENDENCE OF SOME ZETA-FUNCTIONS" by R. Kacinskaite, which can be found easily enough on the internet. It was published in Annales Univ. Sci. Budapest., Sect. Comp. 41 (2013) 5-13. Reference 14 of that paper points to the 1973 proof by Voronin (in Russian) that says that zeta(s) doesn't obey a general differential equation of particular form. The first section of Kacinskaite's paper also gives a reading list pointing to later results extending Voronin's proof to more general types of differential equations.

Interesting question Rafik Zeraoulia .

The Riemann Zeta function is the solution of the differential equation (11) in the following paper. You find h(s) and g(s) from the paper, both functions are finite when Rs(s) is larger than 1/2. This function is not a known function in the sense that it would appear in some earlier context. It is an ordinary first order differential equation that directly follows by derivating the product expression of zeta, i.e., take ln zeta and derivate the product form.

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Indeed Jormakka is correct - he has proven that Zeta(s) obeys a differential equation, but it is not an equation that will shake the mathematical world. See attached file.

To Michael Milgram. Let us wait and see what the journal referees say.

To Michael Milgram. I looked at the paper and Lemma 3 says D=d/ds, so I do not know what you are arguing. Convergence of the part g(s) comes in the following way. You take the absolutely convergent product form of zeta for Rs(s)>1. As Re exp(s)=exp(x) for s=x+iy, you derivate this series with respect to x and get a convergent series. Two times this series majors the absolute value series of g(g). In a similar way you conclude it to u(s,z). Convergence of h(s) is the result of Lemma 3 and it is also correct. Then you say that the differential function is zeta'= zeta'. Of course, you can put exactly every differential equation in this form. For instance, if y'=y+1, then you say, but y+1=y', so this only means y'= y', not interesting. But here you are wrong, y'=y+1 has a solution and it is a nontrivial differential equation. The same is with the differential equation of zeta. The important issue in Lemma 3 is that h(s) is finite for x>1/2. For x>1 it is clearly finite, so for that area we always have a differential equation for zeta. Such existence of a differential equation for zeta for x>1 does not contradict any results. If for you it seems to contradict, then you have not understood the results.