Lecture to solve 2nd order differential equation in matrix form. I have an system of three differential equation coupled. I put this system to matrix form. I think it should be more easy to solve.
$$
\begin{bmatrix}
    \ddot x_1 \\ 
    \ddot x_2 \\
    \ddot x_3 \\
    \end{bmatrix}
=
    \begin{bmatrix}
    0 & a_{12} & -a_{31} \\ 
    -a_{12} & 0 & a_{23} \\
    a_{31} & -a_{23} & 0 \\
    \end{bmatrix}
\begin{bmatrix}
    \dot x_1 \\ 
    \dot x_2 \\
    \dot x_3 \\
    \end{bmatrix}+\begin{bmatrix}
    c_1 \\ 
    c_2 \\
    c_3 \\
    \end{bmatrix}
$$
All constants are positive or zero.
Could someone tell me what I should read to get the knowledge to solve this differential equation ? 
Thanks for your help.
Tof
Edit: $a_{23} -> -a_{23} $
 A: Note that the $A$ matrix is a skew symmetric matrix which represents a vectorial product operator $\vec A \times$ so the DE could read
$$
\ddot X = \vec A \times \dot X + C
$$
So we have
$$
\langle \dot X, \ddot X \rangle = \langle \dot X, C \rangle
$$
because $\langle \dot X, \vec A\times \dot X \rangle = 0$ or
$$
\frac{1}{2}\frac{d}{dt}\Vert\dot X\Vert^2 = \langle \dot X,C \rangle
$$
and after integration
$$
\Vert\dot X\Vert^2 = 2 \langle X, C \rangle + C_0
$$
which can be arranged as
$$
\big\langle X - X_0(t),C \big\rangle = 0
$$
so the movement is such that $X-X_0(t)$ is normal to $C$
etc.
NOTE
$\langle \cdot,\cdot \rangle$ represents the scalar product between two vectors. Here $X_0(t)$ represents a vector multiplied by the scalar $||\dot X||^2$ plus a constant vector. Now assuming $\Vert C \Vert > 0$
$$
\Vert\dot X\Vert^2 = \left\langle \Vert\dot X \Vert^2\frac{C}{\Vert C \Vert^2},C \right\rangle
$$
and
$$
C_0 = \langle x_0, C \rangle
$$
hence
$$
X_0 =  \Vert\dot X\Vert^2\frac{C}{\Vert C\Vert^2} + x_0
$$
A: In
$$
\ddot X=A\times \dot X+C 
$$
you can apply an orthogonal rotation 
$$
U\ddot X=\det(U)^{-1}(UA)\times (U\dot X)+UC.
$$
The rotation $U$ can be chosen so that $UA\sim e_1$. Thus one can assume w.l.o.g. that $A=ae_1$. Then the components of the system read as (set at first $V=\dot X$)
$$
\pmatrix{\dot v_1\\\dot v_2\\\dot v_3}
=a\pmatrix{0\\-v_3\\v_2}+\pmatrix{c_1\\c_2\\c_3}
$$
Integration by standard methods gives
\begin{align}
v_1&=c_1t+b_1\\
v_2&=ab_2\cos(at)-ab_3\sin(at)-\frac{c_3}a&\implies
av_3=c_2-\dot v_2&=c_2+a^2b_2\sin(at)+a^2b_3\cos(at)\\
v_3&=ab_2\sin(at)+ab_3\cos(at)+\frac{c_2}a
\end{align}
Now integrate once again to get $X$,
\begin{align}
x_1&=\frac12c_1t^2+b_1t+d_1\\
x_2&=b_2\sin(at)+b_3\cos(at)-\frac{c_3}at+d_2\\
x_3&=-b_2\cos(at)+b_3\sin(at)+\frac{c_2}at+d_3\\
\end{align}
We get as the result a superposition of an accelerated motion in direction $A=ae_1$ and of a circular and a linear motion in the plane perpendicular to $A$.
