$$\text{Equation}$$
How to solve the equation $$ \begin{align*} \sum\limits_{k = 0}^{\infty}\left[ a_{k} \cdot \frac{\operatorname{d}^{2}y\left( x + b_{k} \right)}{\operatorname{d}x^{2}} + c_{k} \cdot \frac{\operatorname{d}y\left( x + d_{k} \right)}{\operatorname{d}x} + e_{k} \cdot y\left( x + f_{k} \right) \right] &= 0 \tag{1} \label{eq: 1}\\ \end{align*} $$
where $m$ and $n$ are whole number and every $a$, $b$, $c$, $d$, $e$ and $f$ are complex constants but there is one or more non-zero $a$ ($\left\{ m,\, n \right\} \in \mathbb{N} \cup \left\{ 0 \right\} \wedge \forall i \in \mathbb{N}:\,\left\{ a_{i},\, b_{i},\, c_{i},\, d_{i},\, e_{i},\, f_{i} \mid \exists i\left( a_{i} \ne 0 \right) \right\} \in \mathbb{C}$)?
Or alternatively: $$ \begin{align*} \sum\limits_{k = 0}^{m}\left[ a_{k} \cdot \frac{\operatorname{d}^{2}y\left( x + b_{k} \right)}{\operatorname{d}x^{2}} \right] + \sum\limits_{k = 0}^{n}\left[ c_{k} \cdot \frac{\operatorname{d}y\left( x + d_{k} \right)}{\operatorname{d}x} + e_{k} \cdot y\left( x + f_{k} \right) \right] &= 0\\ \end{align*} $$
$$\text{Context}$$
I'm very interested in differential equations of all kinds. Out of interest I was wondering how to solve DDEs of certain forms, scince their solutions can be quite interesting
This is also not a useless form of equation which cannot find application in reality, e.g. in mechanical engineering (a bit more complex) e.g. in describing stuff like:
An efficient transfer of energy from a self-excited oscillator to a resonant load is possible when the wave returns from the load to the oscillator with a favorable phase, provided that the phase shifter is optimally tuned. In many cases this efficiency is impeded by the fact that the frequency band of the resonance load is much narrower than the band of free self-excitedoscillations. - Source
$$\text{My Work And Idias}$$
Solving a homogeneous first-order linear DDE with constant coefficients and constnt delay is usually quite easy, even if we often cannot avoid the Lambert W-Function.
There is actually a good way to solve certain DDEs of this type with just one (some times two) delay(s), namely the direct two-point block method but $\eqref{eq: 1}$ can have way more delays. At least that's how I interpreted the derivation. There are also variable multistep methods (these often work with higher integer order DDEs as well) but I don't know of any such method for multiple delays.
My ideas are:
We could reduce $\eqref{eq: 1}$ to its "characteristic equation" (which is no longer a polynomial) and try to solve this new equation $\eqref{eq: 2}$ using a special function, like maybe a generalized version of the Lambert W-Function ($y\left( x + p \right) := e^{\lambda \cdot \left( x + p \right)}$): $$ \begin{align*} \sum\limits_{k = 0}^{\infty}\left[ a_{k} \cdot \lambda^{2} \cdot e^{b_{k} \cdot \lambda} + c_{k} \cdot \lambda \cdot e^{d_{k} \cdot \lambda} + e_{k} \cdot e^{f_{k} \cdot \lambda} \right] &= 0 \tag{2} \label{eq: 2}\\ \end{align*} $$ However, I don't know of any such function. But I have a question here about solving equations of a simple form, which might be useful: How to solve $a \cdot x^{2} + b \cdot x + c + \exp\left( d \cdot x + e \right) = 0$ for $x$ where $x \in \mathbb{C} \cup \{ \hat{\infty} \}$?
One could perhaps also derive a solution method analogous to the two-point block method aka a generalization of the two-point block method. But I'm not sure if that could work and I think that if it works then the derivation will be quite complicated e.g. in terms of notation.
Iff $\forall i \in \mathbb{Z}:\, e_{i} = 0$, then we could integrate the equation from some constant $x_{-1}$ to $x$ or from $x$ to some constant $x_{+1}$ (so that there is no integration constant) to get a first-order linear DDE with constant coefficients and constnt delay, which could be solved with the characteristic equation $\eqref{eq: 3.1}$ mentioned earlier. Alternatively, one could perhaps find a particular solution to first-order DDE $\eqref{eq: 3.2}$, which then also gives a more general solution: $$ \begin{align*} \sum\limits_{k = 0}^{\infty}\left[ a_{k} \cdot \lambda \cdot e^{b_{k} \cdot \lambda} + c_{k} \cdot e^{d_{k} \cdot \lambda} \right] &= 0 \tag{3.1} \label{eq: 3.1}\\ \sum\limits_{k = 0}^{\infty}\left[ a_{k} \cdot \lambda \cdot e^{b_{k} \cdot \lambda} + c_{k} \cdot e^{d_{k} \cdot \lambda} \right] &= \rm{C} \tag{3.2} \label{eq: 3.2}\\ \end{align*} $$
