Can anyone provide any clues/approach to solving for the time dependent vectors ${\bf E}, {\bf B}$ from a problem given in Mathematical Methods for Physics and Engineering: A Comprehensive Guide by Riley et al?
At time $t = 0$ the vectors ${\bf E}, {\bf B}$ take the values ${\bf E}_0, {\bf B}_0$ respectively, where ${\bf E}_0, {\bf B}_0$ are unit vectors that are orthogonal.
The coupled differential equations linking ${\bf E},{\bf B}$ are $$\frac{d {\bf E}}{d t} = {\bf E}_0 + {\bf B} \times {\bf E}_0 \,\,\,\, (1)$$ $$\frac{d {\bf B}}{d t} = {\bf B}_0 + {\bf E} \times {\bf B}_0 \,\,\,\, (2).$$ My attempt was to take the time derivative of $(1)$ and use $(2)$ to eliminate $d {\bf B}/ dt$ but this seemed to be of no help.
Also I'm not clear where I'd make use of the orthogonality property. At time $t = 0$ the orthogonality property would give ${\bf E}(t = 0) \bullet {\bf B}(t = 0) = 0 $ and I guess that the property of orthogonality is preserved at later times. Indeed it is easy to show $$\frac{d {\bf E} \bullet {\bf B}}{d t} = {\bf B} \bullet {\bf E}_0 + {\bf E} \bullet {\bf B}_0$$ which vanishes at time $t=0$, but is it obvious that this vanishes for all $t$?
I also note that at $t=0$ $$ \left. \frac{d {\bf E}}{d t}\right|_{t=0} = {\bf E}_0 + {\bf B}_0 \times {\bf E}_0,$$ $$\left. \frac{d {\bf B}}{d t} \right|_{t=0} = {\bf B}_0 + {\bf E}_0 \times {\bf B}_0,$$ so that the increments to ${\bf E}$ and ${\bf B}$ are in the planes formed by the vectors ${\bf E}_0 + {\bf B}_0 \times {\bf E}_0$ and ${\bf B}_0 - {\bf B}_0 \times {\bf E}_0$ respectively.