Now that we’ve got some of the basics out of the way for systems
of differential equations it’s time to start thinking about how to solve a
system of differential equations. We
will start with the homogeneous system written in matrix form,
where, A is an n x n
matrix and is a vector whose components are the unknown
functions in the system.
Now, if we start with n
= 1 then the system reduces to a fairly simple linear
(or separable) first order differential equation.
and this has the following solution,
So, let’s use this as a guide and for a general n let’s see if
will be a solution.
Note that the only real difference here is that we let the constant in
front of the exponential be a vector.
All we need to do then is plug this into the differential equation and
see what we get. First notice that the
So upon plugging the guess into the differential equation we
Now, since we know that exponentials are not zero we can
drop that portion and we then see that in order for (2) to
be a solution to (1)
then we must have
Or, in order for (2) to
be a solution to (1),
r and must be an eigenvalue and eigenvector for the
Therefore, in order to solve (1) we
first find the eigenvalues and eigenvectors of the matrix A and then we can form solutions using (2). There are going to be three cases that we’ll
need to look at. The cases are real,
distinct eigenvalues, complex eigenvalues and repeated eigenvalues.
None of this tells us how to completely solve a system of
differential equations. We’ll need the
following couple of facts to do this.
- If and are two solutions to a homogeneous
is also a
solution to the system.
that A is an n x n matrix and suppose that ,
…, are solutions to a homogeneous system, (1). Define,
words, X is a matrix whose ith column is the ith solution. Now define,
We call W the Wronskian. If then the solutions form a fundamental set of solutions and the
general solution to the system is,
Note that if we have a fundamental set of solutions then the
solutions are also going to be linearly independent. Likewise, if we have a set of linearly
independent solutions then they will also be a fundamental set of solutions
since the Wronskian will not be zero.