Stability and Homogeneous Equations? I’m having some trouble understanding the concept of stability in the context of a homogenous system $x’ = Ax$. According to my textbook, a system is stable if it has a fundamental matrix whose entries all remain bounded as $t \rightarrow + \infty$. Can someone please explain what this actually means. 
Also why is it that if A has distinct real eigenvalues, then $x’(t) = Ax(t)$ is stable iff all eigenvalues are negative. This is a theorem that I don’t quite understand either. Please keep explanations simple if at all possible.  
 A: Theorem:
A linear homogeneous system is stable if and only if each of its solutions $x(t)$ is bounded for $t \ge t_0$.
Corollary:
All solutions of a stable linear homogeneous system are simultaneously either bounded or unbounded for $t \ge t_0$.
Lets do some easy examples first.
Example 1:
$\displaystyle x' = - x + e^t \rightarrow x(t) = ce^{-t} + \frac{e^t}{2}$
In this solution, all the solutions are unbounded.
Example 2:
$\displaystyle x' = - x + e^{-t} \rightarrow x(t) = ce^{-t} + t\frac{e^{-t}}{2}$
In this solution, all the solutions are bounded.
Definition:
A solution $\Phi(t)$ is said to be stable if for every $\epsilon \gt 0$ and every $t_0 \ge 0$ there exists a $\delta > 0$) ($\delta$ depends on both $\epsilon$ and possibly $t_0$) such that whenever $|\Phi(t_0)-y_0| \lt \delta$, the solution $\Psi(t, t_0, y_0)$ exists for $t \gt t_0$ and satisfies $|\Phi(t) - \Psi(t, t_0, y_0)| \lt \epsilon$ for $t \ge t_0$.
Example 3:
The scalar equation $\displaystyle y' = -\frac{y}{1+t}$ has $y = 0$ as a solution.
By separation of variables, $\displaystyle \Psi(t, t_0, y_0) = y_0\frac{1+t_0}{1+t}$ is the solution through the point $(t_0, y_o)$. Clearly, therefore, the zero solution is stable by the definition. Note that since $\displaystyle \left|\frac{1+t_0}{1+t}\right| \le 1$ for $t \ge t_0$, both $\delta$ and $\delta_0$ are independent of $t_0$. Also, we may take $\delta = \epsilon$ to satisfy the definition.
Example 4:
Use $y_0 = 0$ and show that the zero solution of the scalar equation $y' = -\alpha y$ ($\alpha \gt 0$) is stable. Find suitable numbers $\delta$ and $\delta_0$ and show that in this case they do not depend on $t_0$. Sketch the solution in the $(t, y) space and in the phase space so you can see what is going on.
It is worth noting that even for a stable solution, the neighboring solution may behave badly for $t \lt t_0$.
Example 5:
The solution $y = 0$ of $y' = -y^3$ is asymptotically stable. By separation of variables, we find that $\displaystyle \Psi(t, t_0, y_0) = |y_0| \frac{1}{\sqrt{1 + 2y_0^{2}(t-t_0)}}$, which approaches zero as $\displaystyle t \rightarrow \infty$, but it does not exist for all $t \lt t_0$; in particular, it becomes infinite as $\displaystyle t \rightarrow t_0 - \frac{1}{y_o^{2}}$
A: This is for the second part of your question...
A good way to think about this to gain intuition (NOT to be mathematically rigorous) is to forget about the matrix nature of the system for a second, but just think about it as if it's a scalar equation, with constant coefficients.  Then we have:
$$x' = Ax$$
...for some constant $A$.  Solving, using separation of variables, we have:
$$x = ce^{At}$$
Now, what happens if $A>0$?  Well, as $t\to\infty$, $x$ gets larger and larger!  If $A<0$, we know the equation is asymptotically stable.  
An "intuitive" way to think of the eigenvalues of a matrix is that they are scalars that act like the matrix.  So, for the scalar case above, replace the conditionals on $A$ with $\text{eig} A$.  
