Is there a way to solve this general linear ODE: $$\sum_{k=0}^n y^{(k)}=0$$
For the first few $n$ here are the solutions:
$$\begin{array}{c|c} n & y \\ \hline 0 & 0 \\ 1 & c_1 e^x \\ 2 & c_1 e^{x/2} \sin \left( \frac{\sqrt 3}2 x\right)+c_2 e^{x/2} \cos \left( \frac{\sqrt 3}2 x\right)\\ 3 & c_1e^{-x}+c_2 \sin x + c_3 \cos x\\ 4 & c_1 e^{- \left(1+\sqrt5\right) x/4} \sin\left(\sqrt{\frac{5-\sqrt5}8} x\right)+c_2e^{- \left(1-\sqrt5\right) x/4} \cos\left(\sqrt{\frac{5+\sqrt5}8} x\right)+c_3 e^{- \left(1-\sqrt5\right) x/4} \sin\left(\sqrt{\frac{5+\sqrt5}8} x\right)+c_4 e^{- \left(1+\sqrt5\right) x/4} \cos\left(\sqrt{\frac{5-\sqrt5}8} x\right) \\ \vdots & \vdots \end{array}$$
My attempt:
We have $$\mathscr{L}\left\{\sum_{k=0}^ny^{(k)}\right\}=0\\
\sum_{k=0}^n \left( s^k \bar y-\sum_{i=0}^{k-1} s^i c_{i+1}\right)=0\\
\bar y = \frac{\displaystyle \sum_{k=0}^n \sum_{i=0}^{k-1}c_{i+1} s^i}{\displaystyle\sum_{k=0}^ns^k}\\
y = \mathscr L^{-1}\left\{\frac{\displaystyle \sum_{k=0}^n \sum_{i=0}^{k-1}c_{i+1} s^i}{\displaystyle\sum_{k=0}^ns^k}\right\}$$
I do not know how to evaluate that Laplace expression.
Changing the summation index of the top sum and realising that the bottom is a geometric series we get that $$y=\mathscr L ^{-1}\left \{ \sum^n_{k=1}(n+1-k)\frac{(1-s)s^{k-1}}{1-s^{n+1}}\right \}$$
So I am just left with the evaluation of $$\mathscr L ^{-1}\left \{ \frac{(1-s)s^{k-1}}{1-s^{n+1}}\right \}$$
Mathematica cannot seems to evaluate this Laplace expression. Let $*$ define the convolution operator such that: $\mathscr L \left \{ f*g\right \}=\bar f \bar g$.
Therefore $$\mathscr L ^{-1}\left \{ \frac{(1-s)s^{k-1}}{1-s^{n+1}}\right \}=\left[x^{-k} \left ( \frac1{\Gamma(1-k)}-\frac1{x\Gamma(-k)}\right )\right]*\mathscr L^{-1}\left\{\frac1{1-s^{n+1}}\right\}$$ Now I am really stuck.
If $y(x) = e^{ax}$, $y^{(k)}(x) =a^k e^{ax} $, so $\sum_{k=0}^n a^k = 0$ so, multiplying by $a-1$, $a^{n+1}-1 = 0 $.
Therefore, $e^{ax}$ is a particular solution for all $a$ such that $a^{n+1} = 1$ with $a \ne 1$.