Differential equation involving cross product. I have the differential equation $$f'=c \times f$$ for $f: \mathbb{R} \to \mathbb{R}^3$ and constants $c \in \mathbb{R}^3$. 
How can I solve something like this?
 A: Note that
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}|f(t)|^2
&=2f(t)\cdot\left(c\times f(t)\right)\\
&=0\tag1
\end{align}
$$
so we have that $|f(t)|$ is constant. Thus, $f(t)$ lives on a sphere of radius $|f(0)|$.
Furthermore,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}c\cdot f(t)
&=c\cdot\left(c\times f(t)\right)\\
&=0\tag2
\end{align}
$$
so we have that $c\cdot f(t)$ is constant. That is, $f(t)$ lives on the plane perpendicular to $c$ containing $f(0)$.
The speed is also constant:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}|f'(t)|^2
&=\frac{\mathrm{d}}{\mathrm{d}t}|c\times f(t)|^2\\
&=2\color{#C00}{(c\times f(t))}\cdot c\times\color{#C00}{(c\times f(t))}\\[4pt]
&=0\tag3
\end{align}
$$
Since the circumference of the circle is $\frac{2\pi}{|c|}|c\times f(0)|$ and the speed is $|c\times f(0)|$, the period is $\frac{2\pi}{|c|}$.
Therefore, the solution can be gotten geometrically since $f(t)$ circles the projection of $f(0)$ onto $c$ at a constant distance with a period $\frac{2\pi}{|c|}$.

$$f(t)=c\frac{c\cdot f(0)}{c\cdot c}+\left(f(0)-c\frac{c\cdot f(0)}{c\cdot c}\right)\cos\left(|c|\,t\right)+\frac{c}{|c|}\times f(0)\sin\left(|c|\,t\right)\tag4$$

A: Is it the typical cross product you are talking about? ($n=3$) If so, then compute $c \times f$ directly and try figuring a constant matrix $A$ for which your differential equation becomes $f' = Af (\equiv c\times f)$. Then you can solve this new equation by means of the eigendecomposition of $A$, if the latter has any. (Spoiler alert: $A$ is skew-symmetric and has zeroes over the main diagonal.)
A: In this case it helps, a little to increase the order of the ODE, as
$$
\ddot f=c\times \dot f=c\times(c\times f) = \langle c,f\rangle\, c -\|c\|^2f
$$
which means that perpendicular to $c$ it is an oscillation equation, in the direction of $c$ it is constant. Taking the next derivative confirms that as
$$
\dddot f+\|c\|^2\dot f = \langle c,c\times f\rangle\, c=0
$$
It has the general solution form
$$
f(t)=A\cos(\|c\|\,t)+B\sin(\|c\|\,t)+C
$$
where the constant vectors are bound by the original first order equation, $$
\begin{cases}A+C&=f(0),\\ \|c\|A&=-c×B,\\\|c\|B&=c×A,\\ c×C&=0\end{cases}
\implies
\begin{cases}
B&=\frac1{\|c\|}c×f(0)\\
A&=\frac1{\|c\|^2}c×(c×f(0))=f(0)-\frac1{\|c\|^2}⟨c,f(0)⟩\,c\\
C&=\frac1{\|c\|^2}⟨c,f(0)⟩\,c
\end{cases}
$$
This all combines to give the Rodrigues formula
$$
f(t)=\exp(t(c×))f(0)
=f(0)\cos(\|c\|\,t)+\frac{c×f(0)}{\|c\|}\sin(\|c\|\,t)+\frac{\langle c,f(0)\rangle\,c}{\|c\|^2} (1-\cos(\|c\|\,t)).
$$
