Solution to Differential Equation $\sum_{i = 0}^n a_i f^{(i)} (x) = 0$ Let $p(x) = \sum_{i =0}^n a_i x^i$ be a polynomial over $\mathbb{C}$. I am interested in solutions to the differential equation $a_0 y + a_1 y' + a_2 y'' + ... + a_n y^{(n)} = 0$, where $y : \mathbb{C} \rightarrow \mathbb{C}$ is holomorphic. I realize this is a common one but I don't know the name of it.
It seems we can do this: say $D : H \rightarrow H$ from holomorphic functions on $\mathbb{C}$ to holomorphic functions on $\mathbb{C}$ is the differentiation operator, and $I$ is the identity. Factor the polynomial $\sum_{i =0}^n a_i x^i$ as $\prod_{i = 1}^n (z - b_i)$ (without loss of generality, we can take this to be monic). Then $(D - b_n I) \circ (D - b_{n-1} I) \circ \cdots \circ (D - b_0 I) (y) = 0$. 
Perhaps $y$ is then a $\mathbb{C}$-linear sum of elements $y_i$ such that $D(y_i) = b_i I (y_i)$, i.e. elements of the form $e^{b_i z}$.
 A: You can actually give a better description of the solutions with the “kernel lemma”, which is: given $P_1, \ldots P_d$ pairwise coprime polynomials, then the kernel of $P_1(u) \circ \ldots \circ P_d(u)=(P_1 \ldots P_d)(u)$ is the direct sum of the kernels of the $P_i(u)$, and the projections are polynomials in $u$ themselves (see  https://fr.m.wikipedia.org/wiki/Lemme_des_noyaux for a proof, unfortunately in French — the main argument is for $d=2$, then you use induction). 
So if $\beta_1, \ldots, \beta_r$ are the distinct roots of your polynomial, with multiplicities $m_1, \ldots, m_r$, the solutions to your differential equation are exactly the $f=f_1+\ldots+f_r$, where $f_k$ is holomorphic and $(D-\beta_kI)^{m_k}f_k=0$. 
Note that for all holomorphic $g$ and scalar $\alpha$, $(D-\alpha I)(e^{\alpha z}g(z))=e^{\alpha z}g’(z)$.
As a consequence, the solutions to your differential equation are exactly the $f(z)=f_1(z)e^{\beta_1 z}+\ldots+f_r(z)e^{\beta_r z}$, where $D^{m_k}f_k=0$, ie $f_k$ is a polynomial with degree $< m_k$. 
