By *analytical*, I presume that you mean *explicit*, or *in close form*. The known case so far is when $A$ and $B$ can be diagonalized in the same basis. Notice that the case where some eigenvalues come as complex conjugate pairs is subtle.

To see the importance of the condition $[A,B]=0_n$, let us make the Fourier transform of your equation:
$$\partial_t\hat f=(B-i\xi A)\hat f(t,\xi).$$
Even if you can compute the exponential of $t(B-i\xi A)$ (unlikely), it will be a complicated function of $\xi$, whose backward Fourier transform is not explicit at all. When $A$ and $B$ commutte, we have
$$\exp t(B-i\xi A)=e^{tB}e^{it\xi A},$$
and this explains that everything works well in this case. A less obvious good situation is when the commutator $[A,B]$ is a linear combination of $A$ and $B$. This argument was used (in an infinite dimensional setting) to compute explicitly the solution of the Cauchy problem for the harmonic oscillator:
$$\partial_tu-\Delta u+|x|^2u=0.$$

If you accept a less explicit formula, and if you assume that $A$ is diagonalisable with *real* eigenvalues (we say that the operator $\partial_t+A\partial_x$ is *hyperbolic*), then the Cauchy problem is well posed. You can construct the solution by using the Laplace transorm in time and Fourier transform in space. You obtain an integral formula involving the inverses $(B-i\xi A)^{-1}$.

scalarPDEs admit solutions via the method of characteristics. The same is not true for first ordersystems. (Any higher order PDE can be converted to a system of first order equations. I challenge you to solve $\triangle u = 0$ using the method of characteristics.) $\endgroup$ – Willie Wong Mar 15 '11 at 12:25