Solution of a variable-coefficient ODE system A system of ODE's is defined as:
$$\frac{du}{dx}  =Au$$
where $u$ is a vector and $A$ is the coefficient matrix. 
As we know, the solution is obtained by solving the eigenvalue problem:
$$det(A-rI)  =0$$
where $I$ is the identity matrix.
Here is the question:
If $A$ were dependent on the independent parameter $x$, how would i get the solution? Would "the eigenvalue solution" still valid in this case?
Best wishes..
 A: Let us asume that the initial condition is $u(x_0) = u_0$. What you could try to do in general is to note that if you integrate, you get
$$u(x) = u_0 + \int \limits _0 ^x A(x_1) u(x_1) \space \Bbb d x_1 = \\
u_0 + \int \limits _0 ^x A(x_1) \Big( u_0 + \int \limits _0 ^{x_1} A(x_2) u(x_2) \Bbb d x_2 \Big) \Bbb d x_1 = \\
u_0 + \int \limits _0 ^x A(x_1) u_0 \Bbb d x_1 + \int \limits _0 ^x \int \limits _0 ^{x_1} A(x_1) A(x_2) u(x_2) \space \Bbb d x_2 \Bbb d x_1 = \dots =\\
u_0 + \sum \limits _{n=0} ^\infty \int \limits _0 ^x \int \limits _0 ^{x_1} \dots \int \limits _0 ^{x_n} A(x_1) A(x_2) \dots A(x_n) u_0 \space \Bbb d x_n \dots \Bbb d x_2 \Bbb d x_1 .$$
Now, if $A(s)$ commutes with $A(t)$ for $s, t \in [x_0, x]$ then the above sum can be simplified to
$$u_0 + \sum \limits _{n=1} ^\infty \frac 1 {n!} \int \limits _0 ^x \int \limits _0 ^x \dots \int \limits _0 ^x A(x_1) A(x_2) \dots A(x_n) \space \Bbb d x_n \dots \Bbb d x_2 \Bbb d x_1 = \Bbb e ^{\int \limits _0 ^x A(y) \space \Bbb d y} u_0 .$$
If $A$ has various other convenient properties, smart tricks might help you say more about the solution, but in the general case do not hope for a closed-form one.
