Integral of Vector Equations of Motion I have a problem in my textbook, it states
"At time t=o, vectors $E=E_o$ and $B=B_o$, where the unit vectors $E_o$ and $B_o$ are fixed and orthogonal. The equations of motion are
$\frac{dE}{dt}=E_o+B\times E_o$
$\frac{dB}{dt}=B_o+E\times B_o$
Find E and B at a general time t, showing that after a long time the directions of E and B have almost interchanged."
At first instinct I want to integrate E and B w.r.t. t. However, I'm not sure how to do this if E depends on B and B depends on E, while they both depend on t. I just need help starting the question, once I have direction I'm sure I can finish as well answer the second part.
 A: [Collecting valuable insights from the comments, I hope this answer will clarify what inner and cross products to take, and why.]
To me, the key to solving this problem is the orthogonality of the initial vectors $E_0$ and $B_0$. Consider the three vectors $E_0$, $B_0$ and $(E_0 \times B_0)$. These three vectors are mutually orthogonal, so (and this is important!) the triple $\{E_0,B_0,E_0 \times B_0\}$ forms a basis of three-dimensional space. That is, you can write any vector in $\mathbb{R}^3$ in terms of its components 'along' these vectors. More precisely, for any vector $V \in \mathbb{R}^3$, we can write
$$
 V = \left[\frac{V \cdot E_0}{E_0 \cdot E_0}\right]E_0 + \left[\frac{V \cdot B_0}{B_0 \cdot B_0}\right]B_0 + \left[\frac{V \cdot (E_0\times B_0)}{(E_0\times B_0) \cdot (E_0\times B_0)}\right](E_0\times B_0).\tag{1}
$$
The inner products in the denominators are just there for consistency, such that, for example, 
\begin{align}
 E_0 &= \left[\frac{E_0 \cdot E_0}{E_0 \cdot E_0}\right]E_0 + \left[\frac{E_0 \cdot B_0}{B_0 \cdot B_0}\right]B_0 + \left[\frac{E_0 \cdot (E_0\times B_0)}{(E_0\times B_0) \cdot (E_0\times B_0)}\right](E_0\times B_0)\\
&= \left[1\right]\,E_0 + [0]\,B_0 + [0]\,(E_0\times B_0).
\end{align}
The term $\left[\frac{V \cdot E_0}{E_0 \cdot E_0}\right]E_0 $ in $(1)$ is called a projection onto $E_0$.
It turns out that this particular choice of basis $\{E_0,B_0,E_0 \times B_0\}$ simplifies the equations for the electric and magnetic field considerably. For example, looking at the component of $E$ along $E_0$, that is, projecting $E$ onto $E_0$, we get
\begin{align}
 \left[\frac{\frac{\text{d} E}{\text{d} t} \cdot E_0}{E_0 \cdot E_0}\right]E_0 &= \left[\frac{(E_0 + (B \times E_0) \cdot E_0}{E_0 \cdot E_0}\right]E_0\\
&= \left[\frac{E_0 \cdot E_0}{E_0 \cdot E_0}\right]E_0\\
&= [1] E_0,
\end{align}
so we can conclude that
$$
\frac{\text{d}}{\text{d} t} \left(E \cdot E_0 \right) = \frac{\text{d} E}{\text{d} t} \cdot E_0 = E_0 \cdot E_0 = \text{constant}.
$$
Now, going through the motions (i.e. projecting both equations onto all three basis vectors), we obtain the following simple system:
\begin{align}
 \frac{\text{d} x_1}{\text{d} t} &= c_1,\\
 \frac{\text{d} x_2}{\text{d} t} &= - z_1,\\
 \frac{\text{d} y_1}{\text{d} t} &= z_2,\\ \tag{2}
 \frac{\text{d} y_2}{\text{d} t} &= c_2,\\
 \frac{\text{d} z_1}{\text{d} t} &= - c_1 y_2,\\
 \frac{\text{d} z_2}{\text{d} t} &= c_2 x_1.
\end{align}
Here, $x_1 = E \cdot E_0, x_2 = B \cdot E_0, y_1 = E \cdot B_0, y_2 = B \cdot B_0, z_1 = E \cdot (E_0 \times B_0), z_2 = B \cdot (E_0 \times B_0)$, and $c_1 = E_0 \cdot E_0, c_2 = B_0 \cdot B_0$ are constants. System $(2)$ is very easy to solve iteratively. Note that the initial condition $E(t=0) = E_0$ translates into $x_1(t=0) = c_1$ and $x_2(t=0) = 0$, etc.
Solving the system $(2)$ and substituting the results in $(1)$, you get
\begin{align}
 E &= (1+t) E_0 + \left(\frac{1}{2}t^2 + \frac{1}{6} t^3\right) B_0 - \left(t + \frac{1}{2} t^2\right) \left(E_0 \times B_0\right),\\
 B &= \left(\frac{1}{2}t^2 + \frac{1}{6} t^3\right) E_0 + (1+t) B_0 + \left(t + \frac{1}{2} t^2\right) \left(E_0 \times B_0\right). \tag{3}
\end{align}
Now, you can clearly see what happens when $t$ gets large. The cubic $t^3$-term grows faster than the quadratic $t^2$ and linear $t$ terms, so we observe that $E \leadsto \frac{1}{6} t^3 B_0$ as $t \to \infty$, and equivalently $B \leadsto \frac{1}{6} t^3 E_0$ as $t \to \infty$. In other words, the dominant component of $E$ is the one along $B_0$, and hence $E$ tends to the direction of $B_0$ as $t \to \infty$ (and $B$ tends to the direction of $E_0$). Since $E$ and $B$ started along (being equal to, actually) $E_0$ resp. $B_0$, the fields indeed reverse direction as $t \to \infty$.
