Q:

Can a space state system be solved using seperation of variable?

There are 4 matrices with following sizes. X is nx1, A is nxn, U is mx1 and B is nxm. As we know, a state space equation is like the following.
$$\dot X(t) = A \cdot X(t) + B \cdot U(t)$$
It means, we have n output state variable, m input state variable. Of course, here the system is linear. And the system is time invariant if A and B matrices are constant(i.e not depentent on time). Such a system can be solved using laplace transform and I won't talk about it. However, now I want to solve when A and B matrices are time dependent i.e system is time variant. Therefore laplace transform is very hard to operate. From now on, assume A and B are time dependent. In short, I am looking for a general solution of linear state space system for both time variant and invariant of course because time variant systems are more general than time independent ones.
$$\dot X(t) = A(t) \cdot X(t) + B(t) \cdot U(t)$$
By the way, when $t=0$, $x_i=0$ and $u_j=0$ for all $i\le n$ and $j\le m$ (Initial conditions).
For a minute, assume that all matrices are not matrix, only one variable.
$$\dot x(t) = a(t) \cdot x(t) + b(t) \cdot u(t)$$
Multiply both sides with $\mu(t) $ and arrange the equation.
$$\mu(t)\cdot \dot x(t) - \mu(t)\cdot a(t) \cdot x(t) = \mu(t)\cdot b(t) \cdot u(t)$$
And if $$\frac{d\mu(t)}{dt}=-\mu(t)\cdot a(t) $$ is satisfied, the equation will become like the following. 
$$\mu(t)\cdot \dot x(t) + \dot \mu(t) \cdot x(t) = \mu(t)\cdot b(t) \cdot u(t)$$
which yields,
$$ \frac{d}{dt}(\mu(t) \cdot  x(t))  = \mu(t)\cdot b(t) \cdot u(t)$$
$$ x(t) = \mu^{-1}(t)\int_0^t \mu(t)\cdot b(t) \cdot u(t) \cdot dt$$
Above, $\mu^{-1}(t)$ is reverse of $\mu(t)$ in multiplication i.e  $\mu^{-1}(t) = {1 \over \mu(t)}$
For matrix form, the same thing can be written. Here $M(t)$ matrix corresponds to $\mu(t)$ and $M(t)$ is nxn matrix.
$$M(t)\cdot \dot X(t) + \dot M(t) \cdot X(t) = M(t)\cdot B(t) \cdot U(t)$$
which yields,
$$ \frac{d}{dt}(M(t) \cdot  X(t))  = M(t)\cdot B(t) \cdot U(t)$$
$$ X(t) = M^{-1}(t)\int_0^t M(t)\cdot B(t) \cdot U(t) \cdot dt$$
For non matrix form, the solution for $\mu(t)$ is like the following.
$$\frac{d\mu(t)}{dt}=-\mu(t)\cdot a(t) $$
$$ \mu(t) =  k \cdot e^{-\int_0^t a(t) \cdot dt}$$
Constant k is not important because it will cancelled out when the solution is calculated.
But in matrix form, in order to find $M(t)$ following equation has to be solved.
$$\frac{dM(t)}{dt}=-M(t)\cdot A(t) $$
I tried this,
$$ M^{-1}(t) \cdot dM(t) = - A(t) \cdot dt $$
where $M^{-1}(t)$ is inverse of $M(t)$ . 
And I came to a deadlock here. How can I find $M(t)$ as a general solution. Is it possible ? If it is, how ? Is it worth to find it ? Thanks a lot from now :)

A:

-The linear time varying (LTI) systems have not been solved analytically in general. They are very difficult to solve even for second order systems. 
-For a homogeneous second order LTI system, if you already have one solution then you can find a second independent solution using  the method of Reduction of Order. For more details, see section 4.2 in "differential equations with boundary value problem" for Dennis Zill 7th ed.
-If you you want to add a contribution, you can work on the homogeneous systems alone. Because the particular solution can be given by the Variation of Parameters.
-It will be very appealing if you can find a general method to solve second order LTV systems. 

