Solution to $\frac{d^n y}{dx^n} = y$ for integer $n$ In short, my question is about the solution of $\frac{d^n y}{dx^n} = y$ for positive integer $n$, where $y$ is real. For example, if $n = 2$, the solution is $y = c_1e^x+c_2e^{-x}$.
The work that I and my old friend, WolframAlpha, made on this is such: Assume that a solution is proportional to $e^{\lambda x}$, with $\lambda$ a constant. Substituting that into the original equation finds $$\frac{d^n}{dx^n} (e^{\lambda x}) - e^{\lambda x} = 0 \to \lambda^n e^{\lambda x}- e^{\lambda x} = 0$$
This can be factored as $$(\lambda^n -1)e^{\lambda x} = 0$$
Since $e^{\lambda x}$ cannot equal $0$, $\lambda^n -1 = 0$. Solving this for $\lambda$ is equivalent to finding the group of the $n$th roots of unity, so $$\lambda_k = e^{2k\pi i/n}, k \in \{1, 2, ..., n\}$$
The overall solution is then $$y = \sum_{k=1}^{n} c_k e^{\lambda_k x}$$
It seems like we are done. However, this can in fact be written somewhat differently (I'm cautious about calling it simplified), as the sums of $e^{px}\cos(qx)$ and $e^{px}\sin(qx)$. For example, if $n = 3$, $$y = c_1 e^x+c_2 e^{-x/2} \cos\left( \frac{x\sqrt{3}}{2}\right) + c_3 e^{-x/2} \sin\left( \frac{x\sqrt{3}}{2}\right)$$
How can I get from the first representation of $y$ to the second one for higher $n$?
For anyone interested, the Mathematica code for finding $y$ is 
DSolve[D[y[x], {x, 5}] == y[x], y[x], x]

 A: Recall the definition of $e^z$ where $z \in \Bbb{C}$. When $x, y \in \Bbb{R}$, we have
$$e^{x + iy} = e^x \operatorname{cis}(y) = e^x(\cos(y) + i \sin(y)).$$
The $n$th roots of unity can therefore be expressed by
$$\lambda_k = e^{2\pi i k / n} = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right),$$
and thus your fundamental solutions are of the form
\begin{align*}
f_k(x) &= \exp(\lambda_k x) \\
&= \exp\left(\cos\left(\frac{2\pi k}{n}\right)x + i \sin\left(\frac{2\pi k}{n}\right)x\right) \\
&= \exp\left(\cos\left(\frac{2\pi k}{n}\right)x\right) \cdot\left(\cos\left(\sin\left(\frac{2 \pi k}{n}\right) x\right) + i \sin\left(\sin\left(\frac{2 \pi k}{n}\right)x\right)\right).
\end{align*}
All your solutions to this ODE are linear combinations of the above functions. Note that these are complex-valued functions, even with real inputs, but every real-valued function will have to be a complex linear combination of these functions. In particular, using the evenness of $\sin$ and the oddness of $\cos$, it's not hard to verify that
$$\frac{f_k(x) + f_{-k}(x)}{2} = \exp\left(\cos\left(\frac{2\pi k}{n}\right)x\right) \cdot \cos\left(\sin\left(\frac{2 \pi k}{n}\right) x\right),$$
and that
$$\frac{f_k(x) - f_{-k}(x)}{2i} = \exp\left(\cos\left(\frac{2\pi k}{n}\right)x\right) \cdot \sin\left(\sin\left(\frac{2 \pi k}{n}\right) x\right).$$
Provided $\sin(2 \pi k / n) \neq 0$ (i.e. the root of unity is not real), we can replace $f_k$ and $f_{-k}$ by the above functions, and get another set of fundamental solutions to the ODE. This is the fundamental solutions that you've run across.
A: The general solution is:
$$y=c_0e^x+\sum_{k=1}^{\operatorname{floor}\left(\frac{n}{2}\right)}c_{2k-1}e^{\cos\left(\frac{2\pi}{n}\cdot k\right)\cdot x}\cos\left(\sin\left(\frac{2\pi}{n}\cdot k\right)x\right)+c_{2k}e^{\cos\left(\frac{2\pi}{n}\cdot k\right)\cdot x}\sin\left(\sin\left(\frac{2\pi}{n}\cdot k\right)x\right)$$
