ODE Solution for Canonical State Space form

Question

Suppose I have a first order ODE

$$ X'(t) = A(t)X(t), \quad \quad X(t_0) = X_0 $$ where $X\in R^n$, and time-dependent $A(t)$ has a canonical state-space form

$$ \begin{bmatrix} 0 & 1 & & &\\ 0 & 0 & 1 & &\\ \vdots & \cdots & & &\\ a_1(t) & a_2(t) & a_3(t) & \cdots & a_n(t) \end{bmatrix} $$

I have two questions:

1. Could I express the solution as

$$ X(t) = X_0\,e^{\int^t_{t_0}A(s)ds} $$ in this particular case? Does $A(t)$ commutes with its integral?

2. If no, how about if I say the coefficients $a_1(t),a_2(t)\cdots$ are selected such that $A(t)$ has $n$ identical real negative eigenvalue $r(t)$? If again no, what could be a condition of $A(t)$ such that it could?

Answer 1

For general $A(t)$ the solution is given by the Magnus Series.

Important special cases are when $A(t), A(t')$ commute or when $A(t)$ is constant. Then the solutions are given by the Matrix Exponential \begin{align} A(t)A(t') &= A(t')A(t) &&\implies &&X(t_1) = \exp \left(\int_{t_0}^{t_1} A(t') \, dt' \right) X_0 \\ A(t) &= A_0 = \text{const} &&\implies &&X(t_1) = \exp \left( A_0 (t_1-t_0) \right) X_0\,. \end{align}

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For your $A(t)$: in general it does not commute. Thus, you have to use the magnus series which will probably not help a lot.