How to sketch the phase portrait near the critical point at the origin. A linear system and its general solution.
$dx/dt$ = $6x - 2y$
$dy/dt$ = $4x + 2y$
It has a general solution of this:
$$\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = A\begin{bmatrix} cos(2t) \\ cos(2t)+sin(2t) \end{bmatrix} e^{4t} + B\begin{bmatrix} sin(2t) \\ sin(2t) - cos(2t) \end{bmatrix}e^{4t}$$ 
Sketch the phase portrait near the critical point at the origin.
Discuss the type and stability of the critical point. 
I don't know how to approach this since I'm used to drawing in the 2 eigenvectors and figuring out if they face in or out against the origin. Then I would draw the orbits. How do I do this one?
 A: Here is a phase portrait for this system.



*

*What do you notice about the direction fields?

*What are your eigenvalues telling you about stability?

*What is happening to the solution curves as $t$ is increasing?


Update
Lets find the critical point, that is, where $x'$ and $y'$ are simultaneouly equal to zero.
We have:


*

*$6x - 2y = 0$

*$4x + 2y = 0$


This leads to a single critical point at $(0,0)$.
Next, the eigenvalues for this system are:
$$\lambda_{1,2} = 4 \pm 2i$$
Since we have complex conjugate, with positive real part, these are unstable spirals and that is what the phase portrait is telling us. Here is a handy stabilty reference classifying eigenvalues.
So, we need to learn to do three things, find eigenvalues, find critical points and draw phase portraits.
Next, look at your $x(t)$ and $y(t)$ result. What do you notice as $t$ increases? Now, what if you plot $x(t)$ versus $y(t)$? You get the phase portrait showing that as $t$ increases, you get directions leaving the critical point off to infinity and this is unstable.
