We now need to address nonhomogeneous systems briefly. Both of the methods that we looked at back in
the second order differential equations chapter can also be used here. As we will see Undetermined Coefficients is almost
identical when used on systems while Variation
of Parameters will need to have a new formula derived, but will actually be
slightly easier when applied to systems.
Undetermined
Coefficients
The method of Undetermined Coefficients for systems is
pretty much identical to the second order differential equation case. The only difference is that the coefficients
will need to be vectors now.
Let’s take a quick look at an example.
Example 1 Find
the general solution to the following system.
Solution
We already have the complementary solution as we solved
that part back in the real eigenvalue section. It is,
Guessing the form of the particular solution will work in
exactly the same way it did back when we first looked at this method. We have a linear polynomial and so our
guess will need to be a linear polynomial.
The only difference is that the “coefficients” will need to be vectors
instead of constants. The particular
solution will have the form,
So, we need to differentiate the guess
Before plugging into the system let’s simplify the
notation a little to help with our work.
We’ll write the system as,
This will make the following work a little easier. Now, let’s plug things into the system.
Now we need to set the coefficients equal. Doing this gives,
Now only is unknown in the first equation so we can
use Gaussian elimination to solve the system.
We’ll leave this work to you to check.
Now that we know we can solve the second equation for .
So, since we were able to solve both equations, the
particular solution is then,
The general solution is then,

So, as you can see undetermined coefficients is nearly the
same as the first time we saw it. The
work in solving for the “constants” is a little messier however.
Variation of
Parameters
In this case we will need to derive a new formula for
variation of parameters for systems. The
derivation this time will be much simpler than the when we first saw variation of parameters.
First let X(t) be
a matrix whose i^{th} column
is the i^{th} linearly
independent solution to the system,
Now it can be shown that X(t)
will be a solution to the following differential equation.
This is nothing more than the original system with the
matrix in place of the original vector.
We are going to try and find a particular solution to
We will assume that we can find a solution of the form,
where we will need to determine the vector . To do this we will need to plug this into the
nonhomogeneous system. Don’t forget to
product rule the particular solution when plugging the guess into the system.
Note that we dropped the “(t)” part of things to simplify the notation a little. Now using (1) we
can rewrite this a little.
Because we formed X
using linearly independent solutions we know that det(X) must be nonzero and this in turn means that we can find the inverse of X. So, multiply both sides by the inverse of X.
Now all that we need to do is integrate both sides to get .
As with the second order differential equation case we can
ignore any constants of integration. The
particular solution is then,


(2)

Let’s work a quick example using this.
Example 2 Find
the general solution to the following system.
Solution
We found the complementary solution to this system in the
real eigenvalue section. It is,
Now the matrix X
is,
Now, we need to find the inverse of this matrix. We saw how to find inverses of matrices
back in the second linear algebra review section
and the process is the same here even though we don’t have constant
entries. We’ll leave the detail to you
to check.
Now do the multiplication in the integral.
Now do the integral.
Remember that to integrate a matrix or vector you just
integrate the individual entries.
We can now get the particular solution.
The general solution is then,

So, some of the work can be a little messy, but overall not
too bad.
We looked at two methods of solving nonhomogeneous
differential equations here and while the work can be a little messy they
aren’t too bad. Of course we also kept
the nonhomogeneous part fairly simple here.
More complicated problems will have significant amounts of work
involved.