In this chapter we will be looking exclusively at linear
second order differential equations. The
most general linear second order differential equation is in the form.
In fact, we will rarely look at nonconstant coefficient
linear second order differential equations.
In the case where we assume constant coefficients we will use the
following differential equation.
Where possible we will use (1)
just to make the point that certain facts, theorems, properties, and/or
techniques can be used with the nonconstant form. However, most of the time we will be using (2)
as it can be fairly difficult to solve second order nonconstant coefficient
differential equations.
Initially we will make our life easier by looking at
differential equations with g(t) =
0. When g(t) = 0 we call the differential equation homogeneous and when we call the differential equation nonhomogeneous.
So, let’s start thinking about how to go about solving a
constant coefficient, homogeneous, linear, second order differential
equation. Here is the general constant
coefficient, homogeneous, linear, second order differential equation.
It’s probably best to start off with an example. This example will lead us to a very important
fact that we will use in every problem from this point on. The example will also give us clues into how
to go about solving these in general.
Example 1 Determine
some solutions to
Solution
We can get some solutions here simply by inspection. We need functions whose second derivative
is 9 times the original function. One
of the first functions that I can think of that comes back to itself after
two derivatives is an exponential function and with proper exponents the 9
will get taken care of as well.
So, it looks like the following two functions are
solutions.
We’ll leave it to you to verify that these are in fact
solutions.
These two functions are not the only solutions to the
differential equation however. Any of
the following are also solutions to the differential equation.
In fact if you think about it any function that is in the
form
will be a solution to the differential equation.

This example leads us to a very important fact that we will
use in practically every problem in this chapter.
Principle of
Superposition
Note that we didn’t include the restriction of constant
coefficient or second order in this.
This will work for any linear homogeneous differential equation.
If we further assume second order and one other condition
(which we’ll give in a second) we can go a step further.
If and are two solutions to a linear, second order
homogeneous differential equation and they are “nice enough” then the general
solution to the linear, second order differential equation is given by (3).
So, just what do we mean by “nice enough”? We’ll hold off on that until a later section. At this point you’ll hopefully believe it
when we say that specific functions are “nice enough”.
So, if we now make the assumption that we are dealing with a
linear, second order differential equation, we now know that (3)
will be its general solution. The next
question that we can ask is how to find the constants c_{1} and c_{2}. Since we have two constants it makes sense,
hopefully, that we will need two equations, or conditions, to find them.
One way to do this is to specify the value of the solution
at two distinct points, or,
These are typically called boundary values and are not
really the focus of this course so we won’t be working with them.
Another way to find the constants would be to specify the
value of the solution and its derivative at a particular point. Or,
These are the two conditions that we’ll be using here. As with the first order differential
equations these will be called initial conditions.
Example 2 Solve
the following IVP.
Solution
First, the two functions
are “nice enough” for us to form the general solution to
the differential equation. At this
point, please just believe this. You
will be able to verify this for yourself in a couple of sections.
The general solution to our differential equation is then
Now all we need to do is apply the initial
conditions. This means that we need
the derivative of the solution.
Plug in the initial conditions
This gives us a system of two equations and two unknowns
that can be solved. Doing this yields
The solution to the IVP is then,

Up to this point we’ve only looked at a single differential
equation and we got its solution by inspection.
For a rare few differential equations we can do this. However, for the vast majority of the second
order differential equations out there we will be unable to do this.
So, we would like a method for arriving at the two solutions
we will need in order to form a general solution that will work for any linear,
constant coefficient, second order differential equation. This is easier than it might initially
look.
We will use the solutions we found in the first example as a
guide. All of the solutions in this example
were in the form
Note, that we didn’t include a constant in front of it since
we can literally include any constant that we want and still get a
solution. The important idea here is to
get the exponential function. Once we
have that we can add on constants to our hearts content.
So, let’s assume that all solutions to
will be of the form
To see if we are correct all we need to do is plug this into
the differential equation and see what happens.
So, let’s get some derivatives and then plug in.
So, if (5) is to be a solution to (4)
then the following must be true
This can be reduced
further by noting that exponentials are never zero. Therefore, (5)
will be a solution to (4) provided r is a solution to
This equation is typically called the characteristic equation for (4).
Okay, so how do we use this to find solutions to a linear,
constant coefficient, second order differential equation? First write down the characteristic equation,
(6),
for the differential equation, (4). This will be a quadratic equation and so we
should expect two roots, r_{1}
and r_{2}. Once we have these two roots we have two
solutions to the differential equation.
Let’s take a look at a quick example.
Example 3 Find
two solutions to
Solution
This is the same differential equation that we looked at
in the first example. This time
however, let’s not just guess. Let’s
go through the process as outlined above to see the functions that we guess
above are the same as the functions the process gives us.
First write down the characteristic equation for this
differential equation and solve it.
The two roots are 3 and 3. Therefore, two solutions are
These match up with the first guesses that we made in the
first example.

You’ll notice that we neglected to mention whether or not
the two solutions listed in (7) are in fact “nice enough”
to form the general solution to (4). This was intentional. We have three cases that we need to look at
and this will be addressed differently in each of these cases.
So, what are the cases?
As we previously noted the characteristic equation is quadratic and so
will have two roots, r_{1}
and r_{2}. The roots will have three possible
forms. These are
 Real, distinct roots, .
 Complex root, .
 Double roots, .
The next three sections will look at each of these in some
more depth, including giving forms for the solution that will be “nice enough”
to get a general solution.