System of ODE - Solution I have a system of ODE to solve
$$ A_{5 \times 5}\ddot{q}(t)_{5 \times 1}+  B_{5 \times 5}\dot{q}(t)_{5 \times 1}+ C_{5 \times 1} =0\tag 1$$
Given Data


*

*$A,B,C$ are constants.We know what is $q(0),q'(0)$  

*$\dot{q}$ means derivative of vector $q$ w.r.t $t$


Question(2 Nos)


*

*How can we solve this ODE in closed form?.  

*How can we solve this ODE by numerical methods?.


NB:: Method using reducing order can also be considered
 A: Let $v = \dot{q}$, so your system is
$$ A \dot{v} + B v + C = 0 $$
I suppose you're given the initial conditions $v(0)$ (not just $q(0)$).  Of course once $v(t)$ is found, integration (with the initial conditions $q(0)$)
gives you $q(t)$.
If $A$ is invertible, your system is equivalent to 
$$\dot{v} + A^{-1} B v + A^{-1} C = 0$$
If $A$ is not invertible, use Gauss-Jordan elimination.  Possibly re-ordering the variables, you get an invertible matrix $E$ such that $EA$ is of the block form
$$ \pmatrix{I & A_{2}\cr 0 & 0\cr}$$
Multiplying your system by $E$, and writing $v = \pmatrix{x\cr y\cr}$,  the system becomes
$$ \pmatrix{I & A_2 \cr 0 & 0\cr} \pmatrix{\dot{x}\cr \dot{y}\cr} + \pmatrix{B_1 & B_2\cr B_3 & B_4\cr} \pmatrix{x\cr y\cr} + \pmatrix{C_1\cr C_2\cr} = 0 $$
i.e.
$$ \eqalign{\dot{x} + A_2 \dot{y} + B_1 x + B_2 y + C_1 &= 0\cr
B_3 x + B_4 y + C_2 &= 0\cr} \tag{1}$$ 
For simplicity, let's suppose $B_4$ is invertible, so we can solve
the second equation of (1) as 
$$ y = -B_4^{-1} B_3 x - B_4^{-1} C_2$$
This must be true for the initial conditions $x(0), y(0)$, and then we must have
$$ \dot{y} = - B_4^{-1} B_3 \dot{x}$$
Substitute these into the first equation of (1), and you get an equation for $\dot{x}$ and $x$:
$$ (I - A_2 B_4^{-1} B_3) \dot{x} + (B_1 - B_2 B_4^{-1} B_3) x + C_1 - B_2 B_4^{-1} C_2 = 0 $$
Of course it's possible that $I - A_2 B_4^{-1} B_3$ is not invertible, in which case you go through this process again...
