determine the stability of an equilibrium point(x,0).

Question

I would like to determine the stability of equilibrium point $(x,0)$ of the differential equation

$\dot x = Ax$

$ A= \bigg[ \begin{matrix} 0&0\\0&a \end{matrix} \bigg] $ and $ a >0 $

I got $ x'(t) = 0\ $ $\ y'(t) = a*y$

So, the equilibrium point is $(x,0)$.

How can I determine the stability of these equilibrium points?

Here is the phase portrait:

Any help will be appreciated!

Answer 1

We want to determine the stability of the equilibrium points of the system $\dot x = Ax$, where

$$ A= \bigg[ \begin{matrix} 0&0\\0&a \end{matrix} \bigg], ~\text{with}~a >0 $$

The critical points are where we simultaneously have $x' = y' = 0$ and we get the entire $x-$axis as

$$(x, y) = (x, 0)$$

Since this system is decoupled, we can write $$\begin{align} x'(t) &= 0 \implies x(t) = c \\ y'(t) &= ay \implies y(t) = c e^{a t} \end{align}$$

The phase portrait is

Using all of the information above, we determine that the critical point is unstable.