TITLE: Differential equation involving cross product.
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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
$$
f(t)=\exp(t(c×))f(0)
=f(0)\cos(\|c\|\,t)+\frac{c×f(0)}{\|c\|}\sin(\|c\|\,t)+\frac{\langle c,f(0)\rangle\,c}{\|c\|^2} (1-\cos(\|c\|\,t)).
$$