Solving $\displaystyle \ f^{(2n)}+f=0$

Question

I was wondering, just for fun, if it was possible to find all solutions of the equation for $x \in \mathbb{R}$ and $n \in \mathbb{N}$ given by $$ f^{\left(2n\right)}\left(x\right)+f\left(x\right)=0 $$ I know how to solve it for $n=1$, $ \ n=2$ and $n=4$. But is there a way to solve it for a particular $n$ ?

Answer 1

It's a differential equation with costant coefficients, so you can solve the equation $\lambda^{2n}+1=0$ and get that $\lambda_j=e^{-i\cdot j\pi/2n }$. Then the solution to the differential equation are $f_j=e^{\lambda_j x}$. Then note that that solutions are coniugated in pairs, so you can modify it (with trigonometric interpretation of complex numbers) to write it down as real functions.

Answer 2

Let’s be systematic. A constant solution to the equation is such that $f(x)=0$.

The equation is linear so the set of solutions is a vector subspace of the vector space of functions from $\Bbb{R}\to \Bbb{K}$ where $\Bbb{K}$ is $\Bbb{R}$ or $\Bbb{C}$.

Heuristicallly because we derive $2n$ times, one can say that this space of solutions has $2n$ degrees of freedom and one suspects it is of dimension $2n$.

Let’s look for the solutions of the form $t\to e^{\lambda t}$

Any solution of that form is such that $\lambda^{2n}+1=0$. This means the set of admissible $\lambda$ is discrete of cardinal $2n$

$$\lambda_k=e^{{i\pi\over 2n}+{k\pi\over n}}\text{ with }k\in \{0,1,\cdots,2n-1\}$$

One can check that these $2n$ solutions are independant. They generate the space of solutions so any solution is a $\Bbb{K}$- linear combination of those exponential. In the real case, we can combine them two by two that are complex conjugates and we get an $\Bbb{R}$-linear combination of trigonometric functions.