Flow of a particle in dynamical system Suppose I have the linear system $\dot{x}=Ax$, with $A=\left[ \begin{array}{cc} -1 & 0 \\ 0 & 2\\ \end{array}\right]$. I know that the phase portrait of the linear system has a saddle in $(0,0)$, so how am I supposed to know where a particle will flow given an initial point in a small neighborhood $N_\epsilon (x_0)$ of radius $\epsilon=0.3$, given $x_0=(-3,0)$?
I know that if $x_0$ has it's second component $x_2>0$, given $x=(x_1,x_2)$, it the particle will flow upwards towards $(0,+\infty)$, and if $x_2<0$ it will go downwards to $(0,-\infty)$, but given the saddle in $(0,0)$, should the particle just stop at $(0,0)$, or bifucate itself to go to $(0,\pm\infty)$? I am quite confused to see what is the mathematical and physical explanation ...
 A: Here is phase portrait showing the phase lines for multiple points (blue trajectories), the initial point $(-3,0)$ in green and $N_\epsilon (x_0)$ of radius $\epsilon=0.3$ in red.

Some observations:


*

*Notice what happens to initial points inside the circle? If they are slightly above the y-axis, they shoot off to positive infinity, if they are slightly below the y-axis, the go to negative infinity.

*Notice what happens to the point $(-3,0)$? It stays on the horizontal axis. Why? If we write $\dfrac{y'}{x'} = \dfrac{2y}{x}$, what happens when $y$ is zero, regardless of the value of $x \ne 0$? We have a constant slope of $0$ and the trajectory stays there.

*From the previous statement, we have $\dfrac{y'}{x'} = \dfrac{2y}{x}$. What happens to the slope when $x = 0$? It is infinite (vertical).

*If we solve this system, $x' = -x, y' = 2y, x(0) = -3$, we have: 


$$x(t) = -3e^{-t}, y(t) = 0$$
What happens if you do a parametric plot of that?


*

*I would say it is safe to state that the trajectory just continues toward the origin but never quite reaches it.

