As with 2^{nd} order differential equations we can’t
solve a nonhomogeneous differential equation unless we can first solve the
homogeneous differential equation. We’ll
also need to restrict ourselves down to constant coefficient differential
equations as solving nonconstant coefficient differential equations is quite
difficult and so we won’t be discussing them here. Likewise, we’ll only be looking at linear differential
equations.
So, let’s start off with the following differential
equation,
Now, assume that solutions to this differential equation
will be in the form and plug this into the differential equation
and with a little simplification we get,
and so in order for this to be zero we’ll need to require
that
This is called the characteristic
polynomial/equation and its roots/solutions will give us the solutions to
the differential equation. We know that, including
repeated roots, an n^{th}
degree polynomial (which we have here) will have n roots. So, we need to go
through all the possibilities that we’ve got for roots here.
This is where we start to see differences in how we deal
with n^{th} order
differential equations versus 2^{nd} order differential equations. There are still the three main cases: real
distinct roots, repeated roots and complex roots (although these can now also
be repeated as we’ll see). In 2^{nd} order differential equations each
differential equation could only involve one of these cases. Now, however, that will not necessarily be
the case. We could very easily have
differential equations that contain each of these cases.
For instance suppose that we have an 9^{th} order
differential equation. The complete list
of roots could have 3 roots which only occur once in the list (i.e. real distinct roots), a root with
multiplicity 4 (i.e. occurs 4 times
in the list) and a set of complex conjugate roots (recall that because the
coefficients are all real complex roots will always occur in conjugate pairs).
So, for each n^{th}
order differential equation we’ll need to form a set of n linearly independent functions (i.e. a fundamental set of solutions) in order to get a general
solution. In the work that follows we’ll
discuss the solutions that we get from each case but we will leave it to you to
verify that when we put everything together to form a general solution that we
do indeed get a fundamental set of solutions.
Recall that in order to this we need to verify that the Wronskian is not zero.
So, let’s get started with the work here. Let’s start off by assuming that in the list
of roots of the characteristic equation we have and they only occur once in the list. The solution from each of these will then be,
There’s nothing really new here for real distinct roots.
Now let’s take a look at repeated roots. The result here is a natural extension of the
work we saw in the 2^{nd} order case. Let’s suppose that r is a root of multiplicity k
(i.e. r occurs k times in the
list of roots). We will then get the
following k solutions to the
differential equation,
So, for repeated roots we just add in a t for each of the solutions past the first one until we have a
total of k solutions. Again, we will leave it to you to compute the
Wronskian to verify that these are in fact a set of linearly independent
solutions.
Finally we need to deal with complex roots. The biggest issue here is that we can now
have repeated complex roots for 4^{th} order or higher differential
equations. We’ll start off by assuming
that occurs only once in the list of roots. In this case we’ll get the standard two
solutions,
Now let’s suppose that has a multiplicity of k (i.e. they occur k times in the list of roots). In this case we can use the work from the
repeated roots above to get the following set of 2k complexvalued solutions,
The problem here of course is that we really want realvalued
solutions. So, recall that in the case where
they occurred once all we had to do was use Euler’s formula on the first one and
then take the real and imaginary part to get two real valued solutions. We’ll do the same thing here and use Euler’s
formula on the first set of complexvalued solutions above, split each one into
its real and imaginary parts to arrive at the following set of 2k realvalued solutions.
Once again we’ll leave it to you to verify that these do in
fact form a fundamental set of solutions.
Before we work a couple of quick examples here we should
point out that the characteristic polynomial is now going to be at least a 3^{rd}
degree polynomial and finding the roots of these by hand is often a very
difficult and time consuming process and in many cases if the roots are not
rational (i.e. in the form ) it can be almost impossible to find
them all by hand. To see a process for
determining all the rational roots of a polynomial check out the Finding Zeroes of Polynomials
page in my Algebra notes. In practice
however, we usually use some form of computation aid such as Maple or
Mathematica to find all the roots.
So, let’s work a couple of example here to illustrate at
least some of the ideas discussed here.
Example 1 Solve
the following IVP.
Solution
The characteristic equation is,
So we have three real distinct roots here and so the
general solution is,
Differentiating a couple of times and applying the initial
conditions gives the following system of equations that we’ll need to solve
in order to find the coefficients.
The actual solution is then,

So, outside of needing to solve a cubic polynomial (which we
left the details to you to verify) and needing to solve a system of 3 equations
to find the coefficients (which we also left to you to fill in the details) the
work here is pretty much identical to the work we did in solving a 2^{nd}
order IVP.
Because the initial condition work is identical to work that
we should be very familiar with to this point with the exception that it
involved solving larger systems we’re going to not bother with solving IVP’s
for the rest of the examples. The main
point of this section is the new ideas involved in finding the general solution
to the differential equation anyway and so we’ll concentrate on that for the
remaining examples.
Also note that we’ll not be showing very much work in
solving the characteristic polynomial.
We are using computational aids here and would encourage you to do the
same here. Solving these higher degree
polynomials is just too much work and would obscure the point of these
examples.
So, let’s move into a couple of examples where we have more
than one case involved in the solution.
Example 2 Solve
the following differential equation.
Solution
The characteristic equation is,
So, we have two roots here, and which is multiplicity of 3. Remember that we’ll get three solutions for
the second root and after the first one we add t’s only the solution until we reach three solutions.
The general solution is then,

Example 3 Solve
the following differential equation.
Solution
The characteristic equation is,
So, we have one real root and a pair of complex roots each with multiplicity 2. So, the solution for the real root is easy
and for the complex roots we’ll get a total of 4 solutions, 2 will be the normal solutions and two will be the
normal solution each multiplied by t.
The general solution is,

Let’s now work an example that contains all three of the
basic cases just to say that we that we’ve got one work here.
Example 4 Solve
the following differential equation.
Solution
The characteristic equation is
In this case we’ve got one real distinct root, ,
and double root, ,
and a pair of complex roots, that only occur once.
The general solution is then,

We’ve got one final example to work here that on the surface
at least seems almost too easy. The
problem here will be finding the roots as well see.
Example 5 Solve
the following differential equation.
Solution
The characteristic equation is
So, a really simple characteristic equation. However, in order to find the roots we need
to compute the fourth root of 16 and that is something that most people
haven’t done at this point in their mathematical career. We’ll just give the
formula here for finding them, but if you’re interested in seeing a little
more about this you might want to check out the Powers and Roots section of
my Complex Numbers
Primer.
The 4 (and yes there are 4!) 4^{th} roots of 16
can be found by evaluating the following,
Note that each
value of k will give a distinct 4^{th}
root of 16. Also, note that for the 4^{th}
root (and ONLY the 4^{th} root) of any negative number all we need to
do is replace the 16 in the above formula with the absolute value of the
number in question and this formula will work for those as well.
Here are the 4^{th}
roots of 16.
So, we have two sets of complex roots : and . The general solution is,

So, we’ve worked a handful of examples here of higher order
differential equations that should give you a feel for how these work in most
cases.
There are of course a great many different kinds of
combinations of the basic cases than what we did here and of course we didn’t
work any case involving 6^{th} order or higher, but once you’ve got an
idea on how these work it’s pretty easy to see that they all work pretty in
pretty much the same manner. The biggest
problem with the higher order differential equations is that the work in
solving the characteristic polynomial and the system for the coefficients on
the solution can be quite involved.