The vectorial ODE $ D\left( X \right) = AX$, with $A$ a constant coefficient matrix Let $ X(t) $ a vector function in $ R^n $, and let A be an $ n \times n $ matrix with constant coefficients. Let us use $ D(X) $ to denote the derivative with respect to $t$ of the function $X$ (this will be the derivative at each coordinate). Does anyone know any material where the general case of the equation $ D(X) = AX $ is solved?
I could not find good material for this, and I really need it.
 A: As Didier Piau said, the solution is $X(t)=e^{tA}X(0)$ for every $t\in\mathbb R$. 
Moreover, $e^{tA}$ can be computed as follows, for any $n$ by $n$ matrix $A$ with complex coefficients. 
Let $X$ be an indeterminate; let 
$$ 
p(X)=\prod_{j=1}^r\ (X-a_j)^{m_j}\ \in\mathbb C[X]
$$ 
be the minimal polynomial of $A$, where the $a_j$ are the distinct eigenvalues of $A$; and let $t$ be a real number. For each $j$ put 
$$ 
e_j(X):=e^{a_jt}\ \sum_{k=0}^{m_j-1}\ \frac{t^k}{k!}\ (X-a_j)^k, 
$$ 
and for each rational fraction fraction $f(X)$ defined at $a_j$ let $T_j(f(X))$ be the degree less than $m_j$ Taylor approximation of $f(X)$ at $X=a_j$. 
Then we have $e^{tA}=e(A)$ where $e(X)\in\mathbb C[X]$ is given by 
$$ 
e(X)=\sum_j\ T_j\left(e_j(X)\frac{(X-a_j)^{m_j}}{p(X)}\right)\ \frac{p(X)}{(X-a_j)^{m_j}}\quad. 
$$ 
Moreover $e(X)$ is the only polynomial of degree less than $\deg p(X)$ which satisfies $e^{tA}=e(A)$. 
EDIT. It's easy to see that $e(X)$ is the unique degree less than $\deg p(X)$ solution to the congruences $e(X)\equiv e_j(X)$ mod $(X-a_j)^{m_j}$. 
The above formula for $e(X)$ is essentially contained in Article 36 of Gauss's Disquisitiones Arithmeticae. A French translation is available at Internet Archive and at Google:


*

*whole book at Internet Archive, 

*Article 36 at Internet Archive, 

*whole book at Google, 

*Article 36 at Google. 
Article 36 can be paraphrased as follows. 
Let $A$ be a principal ideal domain; let $p_1$, ..., $p_r$ be distinct irreducible elements of $A$; let $m_1$, ... , $m_r$ be positive integers; and let $x_1$, ... , $x_r$ be elements of $A$. Then the congruences 
$$ 
x\equiv x_i\bmod p_j^{m_j} 
$$ 
for $j=1,\dots,r$, can be solved by putting $a:=p_1^{m_1}\cdots p_r^{m_r}$ and
$$ 
x:=\sum_{j=1}^r\ T_j\left(x_j\ \frac{p_j^{m_j}}{a}\right)\cdot\frac{a}{p_j^{m_j}}\quad, 
$$ 
where $T_j(x_j\,p_j^{m_j}/a)$ is any solution $y$ to the congruence 
$$
\frac{a}{p_j^{m_j}}\ y\equiv x_j\bmod p_j^{m_j}. 
$$ 
Such a solution exists because $p_j$ doesn't divide $a/p_j^{m_j}$. 
This applies in particular to the case $A=K[Z]$, where $K$ is a field and $Z$ an indeterminate. Then our congruences 
$$ 
x(Z)\equiv x_j(Z)\bmod p_j(Z)^{m_j}\tag{1} 
$$ 
have a unique solution $x(Z)$ satisfying 
$$ 
\deg x(Z) < \deg a(Z),\tag{2} 
$$ 
where $a(Z)$ is the product of the $p_j(Z)^{m_j}$. 
Extending $K$ if necessary, we can assume $p_j(Z)=Z-a_j$, where the $a_j$ are distinct elements of $K$. 
If we define $T_j(f(Z))$ as the degree less than $m_j$ Taylor approximation of $f(Z)$ at $Z=a_j$, where $f(Z)\in K(Z)$ is any rational fraction $f(Z)\in K(Z)$ defined at $a_j$, then 
$$ 
\sum_{j=1}^r\ T_j\left(x_j(Z)\ \frac{p_j(Z)^{m_j}}{a(Z)}\right)\cdot\frac{a(Z)}{p_j(Z)^{m_j}} 
$$ 
will be the solution satisfying $(2)$. 
Many elementary problems boil down to solving congruences of the form $(1)$. 
A: The solution is $X(t)=\mathrm e^{tA}X(0)$ for every $t\geqslant0$, very much like in the one dimensional case. The matrix $\mathrm e^{tA}$ may be defined as $\mathrm e^{tA}=\sum\limits_{n=0}^{+\infty}\dfrac{t^n}{n!}A^n$. See here and especially this paragraph.
