I am studying ODE systems myself and have an example in the book with the following ODE system\begin{equation*}
\begin{cases}
\dot x=x,
\\
\dot y = x+2y.
\end{cases}
\end{equation*}
The equilibrium is in $x=0, y=0$. Than there is a following explanation after finding characteristic equation roots "on $(x,y)$ surface we a have a family of phase trajectories of parabolic type, adjoining the equilibrium state, and 5 special trajectories: equilibrium state and four semi-line, adjoining the equilibrium state. All the trajectories are adjoining to equilibrium state with $t \rightarrow \infty$"
After that they find eigenvectors and say the following "on $(x,y)$ surface we plot $x+y=0, x=0$ lines, directed towards eigenvectors, after that the curvilinear trajectories, tangenting the $x+y=0$ in $(0,0)$"
There is a sketch they provide
My questions are, if you managed to understand my point :)
- Why trajectories have that type as on the sketch?
- From where the $x+y=0$ come from?
- Why the line on a sketch consists of three trajectories?
If you are refering to why they are in that certain direction, try putting in some values to find the direction of $(\dot{x}, \dot{y})^T$. For example, $x=1, y=1$ gives $(\dot{x}, \dot{y})^T = (1, 3)^T$. So at the point $(1,1)$, the direction of the derivative should be in $(1,3)^T$, that is away from the origin. You can continue this for more points and find that the trajectories always move away from the origin. This of course connects to the fact that $(0,0)$ is a unstable fix-point, however inserting some points can always be good to get a feeling.
This line is most likely drawn in the direction of one of the eigenvectors. One of the eigenvectors is $(1, -1)$, which when extended to a line becomes the line $x+y = 0$.
You could of course draw any amount of trajectories, the number 3 is arbitrary. However, drawing the eigenvectors is usually a good idea as they tell you a lot about the behaviour of your system.
Hope this clarifies somethings for you!