This is the final case that we need to take a look at. In this section we are going to look at
solutions to the system,
where the eigenvalues are repeated eigenvalues. Since we are going to be working with systems
in which A is a 2 x 2 matrix we will
make that assumption from the start. So
the system will have a double eigenvalue, λ.
This presents us with a problem. We want two linearly independent solutions so
that we can form a general solution.
However, with a double eigenvalue we will have only one,
So, we need to come up with a second solution. Recall that when we looked at the double root case with the second order
differential equations we ran into a similar problem. In that section we simply added a t to the solution and were able to get a
second solution. Let’s see if the same
thing will work in this case as well.
We’ll see if
will also be a solution.
To check all we need to do is plug into the system. Don’t forget to product rule the proposed
solution when you differentiate!
Now, we got two functions here on the left side, an
exponential by itself and an exponential times a t. So, in order for our
guess to be a solution we will need to require,
The first requirement isn’t a problem since this just
says that λ is an eigenvalue and it’s eigenvector is . We already knew this however so there’s
nothing new there. The second however is
a problem. Since is an eigenvector we know that it can’t
be zero, yet in order to satisfy the second condition it would have to be.
So, our guess was incorrect.
The problem seems to be that there is a lone term with just an
exponential in it so let’s see if we can’t fix up our guess to correct
that. Let’s try the following guess.
where is an unknown vector that we’ll need to
determine.
As with the first guess let’s plug this into the system and
see what we get.
Now set coefficients equal again,
As with our first guess the first equation tells us nothing
that we didn’t already know. This time
the second equation is not a problem.
All the second equation tells us is that must be a solution to this equation.
It looks like our second guess worked. Therefore,
will be a solution to the system provided is a solution to
Also this solution and the first solution are linearly
independent and so they form a fundamental set of solutions and so the general
solution in the double eigenvalue case is,
Let’s work an example.
Example 1 Solve
the following IVP.
Solution
First find the eigenvalues for the system.
So, we got a double eigenvalue. Of course that shouldn’t be too surprising
given the section that we’re in. Let’s
find the eigenvector for this eigenvalue.
The eigenvector is then,
The next step is find . To do this we’ll need to solve,
Note that this is almost identical to the system that we
solve to find the eigenvalue. The only
difference is the right hand side. The
most general possible is
In this case, unlike the eigenvector system we can choose
the constant to be anything we want, so we might as well pick it to make our
life easier. This usually means
picking it to be zero.
We can now write down the general solution to the system.
Applying the initial condition to find the constants gives
us,
The actual solution is then,
Note that we did a little combining here to simplify the
solution up a little.

So, the next example will be to sketch the phase portrait
for this system.
Example 2 Sketch
the phase portrait for the system.
Solution
These will start in the same way that real, distinct
eigenvalue phase portraits start.
We’ll first sketch in a trajectory that is parallel to the eigenvector
and note that since the eigenvalue is positive the trajectory will be moving
away from the origin.
Now, it will be easier to explain the remainder of the
phase portrait if we actually have one in front of us. So here is the full phase portrait with
some more trajectories sketched in.
Trajectories in these cases always emerge from (or move
into) the origin in a direction that is parallel to the eigenvector. Likewise they will start in one direction
before turning around and moving off into the other direction. The directions in which they move are
opposite depending on which side of the trajectory corresponding to the
eigenvector we are on. Also, as the
trajectories move away from the origin it should start becoming parallel to
the trajectory corresponding to the eigenvector.
So, how do we determine the direction? We can do the same thing that we did in the
complex case. We’ll plug in (1,0) into
the system and see which direction the trajectories are moving at that
point. Since this point is directly to
the right of the origin the trajectory at that point must have already turned
around and so this will give the direction that it will traveling after
turning around.
Doing that for this problem to check our phase portrait
gives,
This vector will point down into the fourth quadrant and
so the trajectory must be moving into the fourth quadrant as well. This does match up with our phase portrait.
In these cases the equilibrium is called a node and is unstable in this
case. Note that sometimes you will
hear nodes for the repeated eigenvalue case called degenerate nodes or improper
nodes.

Let’s work one more example.
Example 3 Solve
the following IVP.
Solution
First the eigenvalue for the system.
Now let’s get the eigenvector.
Now find ,
The general solution for the system is then,
Applying the initial condition gives,
Note that we didn’t use t=0 this time! We now need
to solve the following system,
The actual solution is then,

And just to be consistent with all the other problems that
we’ve done let’s sketch the phase portrait.
Example 4 Sketch
the phase portrait for the system.
Solution
Let’s first notice that since the eigenvalue is negative
in this case the trajectories should all move in towards the origin. Let’s check the direction of the
trajectories at (1,0)
So it looks like the trajectories should be pointing into
the third quadrant at (1,0). This
gives the following phase portrait.
