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)* *c _{2}=0, *

*v)* neither *c _{1}* nor

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

**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)* *c _{2}=0, *

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 *a _{1}* and

*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

*iii) *: The amplitude of * x* increases and we thus have an

**
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 *x _{1}=x-2* and

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 *(a _{0},b_{0})*. By making the change of variables