Example 1  Solve the following IVP.

Solution

We first need the eigenvalues and eigenvectors for the matrix.

So, now that we have the eigenvalues recall that we only need to get the eigenvector for one of the eigenvalues since we can get the second eigenvector for free from the first eigenvector.

:

We need to solve the following system.

Using the first equation we get,

So, the first eigenvector is,

When finding the eigenvectors in these cases make sure that the complex number appears in the numerator of any fractions since we’ll need it in the numerator later on.  Also try to clear out any fractions by appropriately picking the constant.  This will make our life easier down the road.

Now, the second eigenvector is,

However, as we will see we won’t need this eigenvector.

The solution that we get from the first eigenvalue and eigenvector is,

So, as we can see there are complex numbers in both the exponential and vector that we will need to get rid of in order to use this as a solution.  Recall from the complex roots section of the second order differential equation chapter that we can use Euler’s formula to get the complex number out of the exponential.  Doing this gives us,

The next step is to multiply the cosines and signs into the vector.

Now combine the terms with an “i” in them and split these terms off from those terms that don’t contain an “i”.  Also factor the “i” out of this vector.

Now, it can be shown (we’ll leave the details to you) that  and  are two linearly independent solutions to the system of differential equations.  This means that we can use them to form a general solution and they are both real solutions.

So, the general solution to a system with complex roots is

where  and  are found by writing the first solution as

For our system then, the general solution is,

We now need to apply the initial condition to this to find the constants.

This leads to the following system of equations to be solved,

The actual solution is then,

Example 2  Sketch the phase portrait for the system.

Solution

When the eigenvalues of a matrix A are purely complex, as they are in this case, the trajectories of the solutions will be circles or ellipses that are centered at the origin.  The only thing that we really need to concern ourselves with here are whether they are rotating in a clockwise or counterclockwise direction.

This is easy enough to do.  Recall when we first looked at these phase portraits a couple of sections ago that if we pick a value of  and plug it into our system we will get a vector that will be tangent to the trajectory at that point and pointing in the direction that the trajectory is traveling..  So, let’s pick the following point and see what we get.

Therefore at the point (1,0) in the phase plane the trajectory will be point in a upwards direction.  The only way that this can be is if the trajectories are traveling in a counterclockwise direction.

Here is the sketch of some of the trajectories for this problem.

The equilibrium solution in the case is called a center and is stable.

Example 3  Solve the following IVP.

Solution

Let’s get the eigenvalues and eigenvectors for the matrix.

Now get the eigenvector for the first eigenvalue.

:

We need to solve the following system.

Using the second equation we get,

So, the first eigenvector is,

The solution corresponding the this eigenvalue and eigenvector is

As with the first example multiply cosines and signs into the vector and split it up.  Don’t forget about the exponential that is in the solution this time. 

The general solution to this system then,

Now apply the initial condition and find the constants.

The actual solution is then,

Example 4  Sketch the phase portrait for the system.

Solution

When the eigenvalues of a system are complex with a real part the trajectories will spiral into or out of the origin.  We can determine which one it will be by looking at the real portion.  Since the real portion will end up being the exponent of an exponential function (as we saw in the solution to this system) if the real part is positive the solution will grow very large as t increases.  Likewise, if the real part is negative the solution will die out as t increases.

So, if the real part is positive the trajectories will spiral out from the origin and if the real part is negative they will spiral into the origin.  We determine the direction of rotation (clockwise vs. counterclockwise) in the same way that we did for the center.

In our case the trajectories will spiral out from the origin since the real part is positive and

will rotate in the counterclockwise direction as the last example did.

Here is a sketch of some of the trajectories for this system.

Here we call the equilibrium solution a spiral (oddly enough…) and in this case it’s unstable since the trajectories move away from the origin.