system of differential equations with matrix $x′ = Ax$
where $$A = \begin{pmatrix}
3&1&0\\
0&3&0\\
0&0&2
\end{pmatrix}
$$
what is the logic to solve such type of equations?
I found that $det(A-λI) = (3-λ)^2(2-λ)=0$, so the eigenvalues are 3 and 2 
 A: The logic is the same as to solve the ''scalar'' ordinary differential equation 
$
\frac{dx}{dt}=ax
$
that, with the initial condition $x(0)=x_0$ has the solution $x(t)=x_0e^{at}$.
For a ''vector'' equation $\frac{d\vec x}{dt}=A\vec x$ we have the solution $\vec x(t)=e^{At}\vec x(0)$.
Where the exponential of the matrix $A$ is defined as you can see here.
A: The full process requires these steps:


*

*Compute the characteristic polynomial and find the eigenvalues, say $\lambda, \mu, \nu$. I'll assume theses eigenvalues are distinct, so the matrix is diagonalisable in a basis of eigenvectors.

*Find a basis of eigenvectors, say $u,v,w$. Let  $P$ be the change of basis matrix from the canonical basis to this basis of eigenvectors.

*In this basis, the matrix $A$ takes the form $\;\begin{bmatrix}\lambda&0&0\\0&\mu&0\\0&0&\nu\end{bmatrix}$, hence the  exponential is
$$\exp(A't)=\begin{bmatrix}\mathrm e^{\lambda t}&0&0\\0&\mathrm e^{\mu t}&0\\0&0&\mathrm e^{\nu t}\end{bmatrix}$$

*The relation between $A$ and $A'$ is : $\;A'=P^{-1}AP$, whence $\;A=PA'P^{-1}$, and similarly
$$ \exp(At)=P\exp(A't)P^{-1}. $$


If the matrix is not diagonalisable: 


*

*You have to find a Jordan basis. In such a basis, the matrix takes the form
$$D+N=\begin{bmatrix}\lambda&0&0\\0&\mu&0\\0&0&\nu\end{bmatrix}+N,$$
where $\lambda, \mu,\nu$ are the eigenvalues of $A$ and $N$ is a nilpotent matrix. 

*Now the exponential of a nilpotent matrix is easy to compute:  if $r$ is the index of nilpotency of $N$ (it is at most the dimension of the matrix), then
$$\exp Nt=I+Nt+\frac{N^2}2t^2+\dots+\frac{N^{k-1}}{(k-1)!}t^{k-1}$$

*And finally we have 
$$\exp(At)=P\bigl(\exp(Dt)\exp(Nt)\bigr)P^{-1}.$$
