General systems of linear differential equations with variable coefficients I am wondering what can be said in general about the fundamental matrix of a system of linear differential equations. For simplicity, let $A(t)\in\mathbb{C}^{n\times n}$ be a time dependent matrix, smooth on some interval $I$, say $[0,t_I]$.
Is is true that there is no closed form formula for the fundamental matrix $M(t)$ of the equation $A(t)v(t)=\dot{v}(t)$, with $v(t)$ a time dependent vector ? What is the best I can hope for when solving such systems in general (suppose $A(t)$ does not commute with itself at different times) ? 
In other words, if I have a given $A(t)$, which happens not to commute with iself at different time, should I hope to obtain an analytical expression for $M(t)$? Which approach would you try if say $A(t)$ was at least sparse?
 A: Assuming that $A(t)$ is analytic at $t=0$, getting a power series solution $M(t)$ at $t=0$ of $M'(t) = A(t)M(t)$ is trivial, of course.  You just write $A(t) = A_0 + A_1 t + \cdots$ and $M(t) = I_n + M_1 t + \cdots $ and then recursively solve, for $k>0$
$$
k\ M_k = A_{k-1} + A_{k-2}M_1 + \cdots + A_0 M_{k-1}\ ,
$$
so $M_1 = A_0$, $M_2 = \tfrac12(A_1+A_0^2)$, $M_3 = \tfrac16(2A_2 + 2A_1A_0 + A_0A_1+A_0^3)$, etc.  If $A(t)$ is sparse, then computing the various products of the $A_i$ might not be hard; it depends on the matrix.
One case in which one can integrate the fundamental equation by elementary means is when $A$ takes values in a solvable Lie subalgebra $\frak{s}$ of $\frak{gl}(n,\mathbb{R})$.  In this case, by a theorem of Lie and Engel, $M(t)$ can be computed from $A(t)$ by a sequence of quadratures.  (The abelian case is, of course, the simplest case of this.)
Meanwhile, it is a classical theorem that, when $n\ge 2$, there is no sequence of algebraic operations and quadratures that will produce $M(t)$ starting from the general $A(t)$.
More generally, if you add to your allowable operations, an 'integration procedure' $IP_{\frak{g}}$ for each simple Lie algebra $\frak{g}$ that will solve the equation $M'(t) = M(t)A(t)$ for any curve $A:\mathbb{R}\to\frak{g}$ for the corresponding $M:\mathbb{R}\to G$, then the equation $M'(t) = M(t)A(t)$ can be solved for any $A$ taking values in any Lie algebra $\frak{a}$ by a sequence of applications of the $IP_{\frak{g}}$ for the simple $\frak{g}$ that appear in the Levi decomposition of $\frak{a}$, followed by a sequence of quadratures.  
Look in any good book on differential algebra for a discussion of this, as well as the independence of the operations $IP_{\frak{g}}$ for different simple Lie algebras.  The case ${\frak{g}} = {\frak{sl}}(2,\mathbb{R})$ is, of course, the theory of the Riccati equation.
By the way, this result about solvability was one of the primary motivations for wanting to classify the simple Lie algebras.  In fact, all of the classical techniques (such as variation of parameters, getting the general solution of the Riccati equation by quadrature from a single solution, etc.), are special cases of general techniques for reducing or solving equations by the use of group actions, and this was the original motivation for studying Lie group actions and homogeneous spaces of Lie groups.
A: Forget about finding a closed analytical expression for the fundamental matrix. Think about the simple case of a second order scalar equation such as the Airy equation $\ddot x=tx.$
There is nevertheless a very remarkable explicit expression  for the determinant of the fundamental matrix $M(t)$ (i.e. the solution of $\dot M=AM, M(0)=Id$ ). In fact, we have
$$
\det M=\exp\int_0^t\text{trace } A(s) ds.
$$
Very remarkable indeed since the matrix $M$ is not known explicitly but its determinant has a simple closed expression: to prove the above formula is not difficult, since you just have to verify
$$
\frac d{dt} \det M=(\text{trace } A)  \det M,
$$
which is a direct consequence of multilinearity of 
$$\det M=
C_1\wedge \dots\wedge C_n,
$$
 where the $C_j$ are the columns of the matrix $M$. This (non-commutative) exterior product behaves simply with respect to differentiation and the derivative of the column $C_j$ can be found using the ODE satisfied by $M$.
A: Try by hand a piecewise constant matrix, even with  say 4 pieces on [0,1]. You can find a closed form expression, using maple or mathematica, or your pen, by what good will it do? The expression, while closed, will be completely useless. Of course you can take a $C^\infty$ approximation of it, and it will be as close as you wish to it. 
Why do like closed forms? It is


*

*Because you can compute a solution with arbitrary precision ?  Numerical methods for ODEs do that for you.

*Because you want to understand how it behaves ? Special functions (or combination of special functions) often have intricate behaviors, you will
find out more either numerically or by using the ODE directly to prove things.

*Because you will know that the solution exists / is unique? That's what Peano/ Cauchy do for you...

