Systems of Differential Equations (Part 8. Real, Distinct Eigenvalues)
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Real, Distinct Eigenvalues
It’s now time to start solving systems of differential
equations. We’ve seen that solutions to the system,



will be of the form
where λ and are eigenvalues and eigenvectors of the
matrix A. We will be working with 2 x 2 systems so this
means that we are going to be looking for two solutions, and ,
where the determinant of the matrix,


is nonzero.
We are going to start by looking at the case where our two
eigenvalues, and are real and distinct. In other words they will be real, simple
eigenvalues. Recall as well that the eigenvectors for simple
eigenvalues are linearly independent.
This means that the solutions we get from these will also be linearly
independent. If the solutions are
linearly independent the matrix X
must be nonsingular and hence these two solutions will be a fundamental set of
solutions. The general solution in this
case will then be,



Note that each of our examples will actually be broken into
two examples. The first example will be
solving the system and the second example will be sketching the phase portrait
for the system. Phase portraits are not
always taught in a differential equations course and so we’ll strip those out
of the solution process so that if you haven’t covered them in your class you
can ignore the phase portrait example for the system.
Example 1 Solve
the following IVP.
Solution
So, the first thing that we need to do is find the
eigenvalues for the matrix.
Now let’s find the eigenvectors for each of these.
We’ll need to solve,
The eigenvector in this case is,
We’ll need to solve,
The eigenvector in this case is,
Then general solution is then,
Now, we need to find the constants. To do this we simply need to apply the
initial conditions.
All we need to do now is multiply the constants through
and we then get two equations (one for each row) that we can solve for the
constants. This gives,
The solution is then,

Now, let’s take a look at the phase portrait for the system.
Example 2 Sketch
the phase portrait for the following system.
Solution
From the last example we know that the eigenvalues and
eigenvectors for this system are,
It turns out that this is all the information that we will
need to sketch the direction field. We
will relate things back to our solution however so that we can see that
things are going correctly.
We’ll start by sketching lines that follow the direction of
the two eigenvectors. This gives,
Now, from the first example our general solution is
If we have c_{2}
= 0 then the solution is an exponential times a vector and all that the
exponential does is affect the magnitude of the vector and the constant c_{1} will affect both the
sign and the magnitude of the vector.
In other words, the trajectory in this case will be a straight line
that is parallel to the vector,
So the line in the graph above marked with
If we now turn things around and look at the solution
corresponding to having c_{1}
= 0 we will have a trajectory that is parallel to
Notice that we could have gotten this information with
actually going to the solution. All we
really need to do is look at the eigenvalues. Eigenvalues that are negative will
correspond to solutions that will move towards the origin as t increases in a direction that is
parallel to its eigenvector. Likewise,
eigenvalues that are positive move away from the origin as t increases in a direction that will
be parallel to its eigenvector.
If both constants are in the solution we will have a
combination of these behaviors. For
large negative t’s the solution
will be dominated by the portion that has the negative eigenvalue since in
these cases the exponent will be large and positive. Trajectories for large negative t’s will be parallel to
Solutions for large positive t’s will be dominated by the portion with the positive
eigenvalue. Trajectories in this case
will be parallel to
In general, it looks like trajectories will start “near”
In this case the equilibrium solution (0,0) is called a saddle point and is unstable. In this case unstable means that solutions move away from it as t increases.

So, we’ve solved a system in matrix form, but remember that
we started out without the systems in matrix form. Now let’s take a quick look at an example of
a system that isn’t in matrix form initially.
Example 3 Find
the solution to the following system.
Solution
We first need to convert this into matrix form. This is easy enough. Here is the matrix form of the system.
This is just the system from the first example and so
we’ve already got the solution to this system. Here it is.
Now, since we want the solution to the system not in matrix
form let’s go one step farther here.
Let’s multiply the constants and exponentials into the vectors and
then add up the two vectors.
Now, recall,
So, the solution to the system is then,

Let’s work another example.
Example 4 Solve
the following IVP.
Solution
So, the first thing that we need to do is find the
eigenvalues for the matrix.
Now let’s find the eigenvectors for each of these.
We’ll need to solve,
The eigenvector in this case is,
We’ll need to solve,
The eigenvector in this case is,
Then general solution is then,
Now, we need to find the constants. To do this we simply need to apply the
initial conditions.
Now solve the system for the constants.
The solution is then,

Now let’s find the phase portrait for this system.
Example 5 Sketch
the phase portrait for the following system.
Solution
From the last example we know that the eigenvalues and
eigenvectors for this system are,
This one is a little different from the first one. However it starts in the same way. We’ll first sketch the trajectories
corresponding to the eigenvectors.
Notice as well that both of the eigenvalues are negative and so
trajectories for these will move in towards the origin as t increases. When we sketch the trajectories we’ll add
in arrows to denote the direction they take as t increases. Here is the
sketch of these trajectories.
Now, here is where the slight difference from the first
phase portrait comes up. All of the
trajectories will move in towards the origin as t increases since both of the eigenvalues are negative. The issue that we need to decide upon is
just how they do this. This is
actually easier than it might appear to be at first.
The second eigenvalue is larger than the first. For large and positive t’s this means that the solution for this eigenvalue will be
smaller than the solution for the first eigenvalue. Therefore, as t increases the trajectory will move in towards the origin and do
so parallel to
Adding in some trajectories gives the following sketch.
In these cases we call the equilibrium solution (0,0) a node and it is asymptotically
stable. Equilibrium solutions are
asymptotically stable if all the trajectories move in towards it as t increases.

Note that nodes can also be unstable. In the last example if both of the
eigenvalues had been positive all the trajectories would have moved away from
the origin and in this case the equilibrium solution would have been unstable.
Before moving on to the next section we need to do one more
example. When we first started talking
about systems it was mentioned that we can convert a higher order differential
equation into a system. We need to do an
example like this so we can see how to solve higher order differential
equations using systems.
Example 6 Convert
the following differential equation into a system, solve the system and use
this solution to get the solution to the original differential equation.
Solution
So, we first need to convert this into a system. Here’s the change of variables,
The system is then,
where,
Now we need to find the eigenvalues for the matrix.
Now let’s find the eigenvectors.
We’ll need to solve,
The eigenvector in this case is,
We’ll need to solve,
The eigenvector in this case is,
The general solution is then,
Apply the initial condition.
This gives the system of equations that we can solve for
the constants.
The actual solution to the system is then,
Now recalling that,
we can see that the solution to the original differential
equation is just the top row of the solution to the matrix system. The solution to the original differential
equation is then,
Notice that as a check, in this case, the bottom row
should be the derivative of the top row.

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