Again consider the system
The equilibrium point obviously is (0,0). Let 
and 
be the eigenvalues to the matrix A. We have the following different possibilities:
1) If
Let
be the eigenvectors belonging to the corresponding eigenvalues and . Then the general solution of the system above is
We again have a number of possibilities:
i) c2=0, c1>0: x(t) is a curve along l1+ (away from origo for increasing t).
ii) c2=0, c1<0: x(t) is a curve along l1– (away from origo for increasing t).
iii) c1=0, c2>0: x(t) is a curve along l2+ (away from origo for increasing t).
iv) c1=0, c2<0: x(t) is a curve along l2– (away from origo for increasing t).
v) neither c1 nor c2 are zero. Then for
we have
hence
for large (in absolute value) negative t. In particular we have that
When
we have
for large positive t. Hence x(t) diverges to infinity with a slope asympotically v2 as t goes to (positive) infinity. This means that (0,0) is an unstable node. The lines defined by the eigenvectors v1 and v2 are called separatrices. The behavior is shown below in the phase portrait.
2) If
In the same way as above we realize that (0,0) is a stable node. The phase portrait looks the same but with reversed arrows.
3) If
We have the following possibilities:
i) c2=0, c1>0: x(t) is a curve along l1+ (towards origo for increasing t).
ii) c2=0, c1<0: x(t) is a curve along l1– (towards origo for increasing t).
iii) c1=0, c2>0: x(t) is a curve along l2+ (away from origo for increasing t).
iv) c1=0, c2<0: x(t) is a curve along l2– (away from origo for increasing t).
We have a saddle point. See the phase portrait below.
4) If
We have two cases:
a) We have two linearly independent eigenvectors
Then we can write the solution
where a1 and a2 are arbitrary. This means that every curve is a halfline towards origo. See the phase portrait below.
b) We only have one eigenvector
Then the solution on the form
where the vector
satisfies
For large t this means that the solution is
In this case the phase portrait will look like this.
Both these cases are examples of a stable node.
5) If
This case is analogous to case 4) above, but with reversed arrows; (0,0) is an unstable node.
6) If
By using similar arguments as in Example 9 we realize that the real solutions are
Three cases:
i) 
: Then x(t) and y(t) are periodic with period
The point (0,0) is a center.
ii) 
: The amplitude of x decreases and we thus have a stable spiral.
iii) 
: The amplitude of x increases and we thus have an unstable spiral.
Example 10: Consider the system
The coefficient matrix
has the characteristic equation
and thus the eigenvalues
This corresponds to case 3) above and we conclude that (0,0) is an unstable saddle point. The eigenvectors are obtained by solving the linear equation system:
Eigenvalue 1:
implies that
that is, the corresponding eigenvector is
Eigenvalue 2:
implies that
that is, the corresponding eigenvector is
The general solution of the system thus is
that is
The eigenvectors define the directions of the separatrices.
Example 11: Consider the system
Here we have the critical point (2,1). We make the change of variables x1=x-2 and y1=y-1 and rewrite the system as
By using the result from example 10 above we see that the point (2,1) is an unstable saddle point and that the solution to the original system is
We get the phase portrait by taking that from example 10 and move it two steps in the x-direction and one step in the y-direction.
In the same way as in Example 11 we may instead study the more general system
The equlibrium point is here (a0,b0). By making the change of variables x1=x-a0 and y1=y-b0 we can transfer the system to the ones studied above with equilibrium point (0,0).