Definition: Let $I \subset \mathbb{R}$ be an interval, $d \in \mathbb{N}$ and let $A : I \to \mathbb{R}^{d×d}$ be a matrix-valued function and $b : I \to \mathbb{R}^d$ be a vector-valued function.
Then
$$y'(t) = A(t)y(t) + b(t), t \in I$$
is called a linear ODE or a system of linear ODEs.
Then one can prove that the space of solutions to this equation/system is a vector space of dimension $d$.
To solve such a system we need the solution to the related homogenous system ($b(t)=0$) and then we can use the method of variation of constants. To do so we need a fundamental matrix, i.e. a continuously differentiable matrix valued function whose columns form a basis for the solution space to the ODE.
After going through all of the general theory my lecture notes simply mention that it is in general not possible to find an explicit formula for the fundamental matrix and that it is only available for special cases.
Now my questions are:
- What does it mean that there is no explicit formula? The Picard theorem at least guarantees that there is a unique solution, but of course we do not know much more.
- In the chapter on systems of linear ODEs with constant coefficents the matrix exponential function basically pops out of nowhere and it is shown that $Z(t)=e^{tA}$ is a fundamental matrix in this case. I'd like some intuition how one comes up with this. Can it somehow be seen from the Picard iteration? In the one-dimensional case $Z(t)=e^A(t)$ is a $1 \times 1$ fundamental matrix as can be seen by comparing the solution to the general case with arbitrary $d$. Does it have anything to do with that?
I know these questions are probably very broad, so please note that I am looking for some intuition here. The proof that this is indeed a fundamental matrix is given in the notes.
Thanks a lot!
Edit:
Lioville's Theorem (as specified in my notes):
Let $I \subset R$ be an interval, $d \in \mathbb{N}$ and let $A: I \to \mathbb{R}^{d×d}$ be a continuous function. Further let $z_1,z_2,..,z_d: I \to \mathbb{R}$ be solutions to $y'(t)=A(t)y(t)$. If $Z(t)$ is the matrix with columns $z_1,z_2,..,z_d$, then $\omega(t)=det(Z(t)), t \in I$ is continuously differentiable on $I$ and
$$\omega(t)=\omega(t_0) exp(\int_{t_0}^{t} trace(A(s)) ds).$$
It depends on what you mean by "explicit". Usually it's used to mean something like: involving only basic arithmetic operations, powers, roots, exponential functions (which also leads to trigonometric and hyperbolic functions), and the inverses of any of the resulting functions. But then again, others might count the $\Gamma$-function as explicit, even though it's not expressible through the ones I've listed (as far as I know, at least). Basically, it's not a well-defined notion.
But anyway, however you define an explicit expression, let $f:\mathbb R\to\mathbb R$ be a differentiable function which has no explicit form and has no roots. Then it is a solution to $y'-\frac{f'}{f}y=0$, so that's a linear ODE whose solution can't be expressed explicitly.