Generalizing solutions to the differential equation $y^{(n)} = y$. Question: Can you generalize to find solutions of the equation $y^{(n)} = y$?
What I have done: I know the derivative all the way to the fifth power, and what I have surmised is the further down the derivative "rabbit hole" so to speak of this differential equation, the first 2 terms are always $c_1e^x$ and $c_2e^{-x}$ with cosines and since splitting the remaining terms. But I feel that isn't the true solution, any help is much appreciated.
 A: The solutions are of the form
$$ x\mapsto\sum_{\omega\in\mathbb{U}_n}a_{\omega}e^{\omega x} $$
where $a_{\omega}\in\mathbb{R}$. Indeed the maps $x\mapsto a_{\omega}e^{\omega x}$ are linearly independent, the $\omega$'s being distinct so that $\dim{\rm span}(x\mapsto e^{\omega x})_{\omega\in\mathbb{U}_n}=n$. Furthermore, the linear map $y\mapsto(y(0),\ldots,y^{(n-1)}(0))$ from the vector space $E$ of the solutions of $y^{(n)}=y$ to $\mathbb{R}^n$ is one-to-one according to Cauchy-Lipschitz theorem, therefore $\dim E\leqslant n$. Since ${\rm span}(x\mapsto e^{\omega x})_{\omega\in\mathbb{U}_n}\subset E$ and $\dim{\rm span}(x\mapsto e^{\omega x})_{\omega\in\mathbb{U}_n}=n$, then $E={\rm span}(x\mapsto e^{\omega x})_{\omega\in\mathbb{U}_n}$.
A: Substitute $$y = ae^{bx} \tag{1}$$
$$y^{(n)} = ab^ne^{bx} \tag{2}$$
$$b^n = 1 \tag{3}$$
$$ b^n = e^{i2\pi k}, k \in \mathbb{Z} \tag{4}$$
$$ b_k = e^{\frac{i2\pi k}{n}}, k \in [0,n-1] \tag{5}$$
Since the equation is linear the solutions can be in superposition:
$$y = \sum_{k=0}^{n-1} a_k e^{b_kx}\tag{6}$$
