Is there a general form for $y^{(n)}=y$? For $y' =y$ we get $y = c_1 e^x$.
For $y'' = y$ we get $y = c_1 e^x + c_2 e^{-x}$.
Is there a general solution for the $n\text{th}$ derivative of y?
 A: We look for solutions of the type $e^{rx}$.  Thus, we must have $r^n=1$.  So, the general solution can be written
$$y(x)=\sum_{m=1}^n c_me^{r_mx}$$
where $r_m=e^{i2\pi m/n}$ for $m=1,\dots,n$.
A: Consider the characteristic equation formed by this ODE.  Given $y^{(n)} = y$, the characteristic equation is $z^n - 1 =0$, which means we are looking for the $n$-th roots of unity.
We can determine these roots by substituting $z = e^{i\theta}$ and finding $n$ such that:
$$e^{i\theta n} = 1$$
This is true whenever $\theta n$ is a multiple of $2\pi$.  Thus, we have $n$ options for $\theta$:
$$\theta = 0, \frac{2\pi}{n}, \frac{4\pi}{n}, \dots, \frac{2\pi(n-1)}{n}$$
So, the solutions to the characteristic are the $n$th roots of unity:
$$z_0 = e^{0},\; z_1 = e^{2\pi i/n},\; z_2 = e^{4\pi i/n},\dots,\; z_{n-1} =  e^{2\pi(n-1) i/n}$$
Returning to the method of characteristic equations, the solution must have the form:
\begin{align*}
  y &= \sum_{k=0}^{n-1} c_k e^{z_k x}\\
    &= \sum_{k=0}^{n-1} c_k \exp\left(xe^{i 2\pi k/n}\right)
\end{align*}
A: Going off of the other answer (Dr MV's), here's what it looks like without complex numbers:
Note that $r_m = e^{i2 \pi m/n} = \cos(2 \pi m/n) + i \sin (2 \pi m/n) := a_m + ib_m$.  By reorganizing the sum, we find
$$
y = \sum_{m=1}^{\lceil n/2-1 \rceil} e^{a_m\,x}[c_{1,m} \cos(b_m x) + c_{2,m} \sin(b_m x)] + d_1 e^x + \overbrace{d_2 e^{-x}}^{\text{ if $n$ is even}}
$$
