Fundamental Sets of
Solutions
The time has finally come to define “nice enough”. We’ve been using this term throughout the
last few sections to describe those solutions that could be used to form a
general solution and it is now time to officially define it.
First, because everything that we’re going to be doing here
only requires linear and homogeneous we won’t require constant coefficients in
our differential equation. So, let’s
start with the following IVP.
Let’s also suppose that we have already found two solutions
to this differential equation, y_{1}(t)
and y_{2}(t). We know from the Principle of Superposition
that
will also be a solution to the differential equation. What we want to know is whether or not it
will be a general solution. In order for
(2)
to be considered a general solution it must satisfy the general initial
conditions in (1).
This will also imply that any solution to the differential
equation can be written in this form.
So, let’s see if we can find constants that will satisfy
these conditions. First differentiate (2)
and plug in the initial conditions.
Since we are assuming that we’ve already got the two
solutions everything in this system is technically known and so this is a system
that can be solved for c_{1}
and c_{2}. This can be done in general using Cramer’s
Rule. Using Cramer’s Rule gives the
following solution.
where,
is the determinant of a 2x2 matrix. If you don’t know about determinants that is
okay, just use the formula that we’ve provided above.
Now, (4) will give the solution to
the system (3). Note that in practice we generally don’t use
Cramer’s Rule to solve systems, we just proceed in a straightforward manner and
solve the system using basic algebra techniques. So, why did we use Cramer’s Rule here then?
We used Cramer’s Rule because we can use (4)
to develop a condition that will allow us to determine when we can solve for
the constants. All three (yes three, the
denominators are the same!) of the quantities in (4) are
just numbers and the only thing that will prevent us from actually getting a
solution will be when the denominator is zero.
The quantity in the denominator is called the Wronskian and is denoted as
When it is clear what the functions and/or t are we often just denote the Wronskian
by W.
Let’s recall what we were after here. We wanted to determine when two solutions to (1)
would be nice enough to form a general solution. The two solutions will form a general
solution to (1)
if they satisfy the general initial conditions given in (1) and
we can see from Cramer’s Rule that they will satisfy the initial
conditions provided the Wronskian isn’t zero.
Or,
So,
suppose that y_{1}(t) and y_{2}(t) are two solutions to (1)
and that . Then the two solutions are called a fundamental set of solutions and the
general solution to (1)
is
We know now what “nice enough” means. Two solutions are “nice enough” if they are a
fundamental set of solutions.
So, let’s check one of the claims that we made in a previous
section. We’ll leave the other two to
you to check if you’d like to.
Example 1 Back
in the complex root section we made the claim
that
were a fundamental set of solutions. Prove that they in fact are.
Solution
So, to prove this we will need to take the Wronskian
for these two solutions and show that it isn’t zero.
Now, the exponential will never be zero and (if it were we wouldn’t have complex roots!)
and so . Therefore, these two solutions are in fact
a fundamental set of solutions and so the general solution in this case is.

Example 2 In
the first example that we worked in the Reduction of Order
section we found a second solution to
Show that this second solution, along with the given
solution, form a fundamental set of solutions for the differential equation.
Solution
The two solutions from that example are
Let’s compute the Wronskian of these two solutions.
So, the Wronskian will never be zero. Note that we can’t plug t = 0 into the Wronskian. This would be a problem in finding the
constants in the general solution, except that we also can’t plug t = 0 into the solution either and so
this isn’t the problem that it might appear to be.
So, since the Wronskian isn’t zero for any t the two solutions form a fundamental
set of solutions and the general solution is
as we claimed in that example.

To this point we’ve found a set of solutions then we’ve
claimed that they are in fact a fundamental set of solutions. Of course, you can now verify all those
claims that we’ve made, however this does bring up a question. How do we know that for a given differential
equation a set of fundamental solutions will exist? The following theorem answers this question.
Theorem
Consider the differential equation
where p(t) and q(t) are continuous functions on some
interval I. Choose t_{0} to be any point in the
interval I. Let y_{1}(t) be a solution to the differential equation that
satisfies the initial conditions.
Let y_{2}(t)
be a solution to the differential equation that satisfies the initial
conditions.
Then y_{1}(t)
and y_{2}(t) form a
fundamental set of solutions for the differential equation.

It is easy enough to show that these two solutions form a
fundamental set of solutions. Just
compute the Wronskian.
So, fundamental sets of solutions will exist provided we can
solve the two IVP’s given in the theorem.
Example 3 Use
the theorem to find a fundamental set of solutions for
using t_{0}
= 0.
Solution
Using the techniques from the first part of this chapter
we can find the two solutions that we’ve been using to this point.
These do form a fundamental set of solutions as we can
easily verify. However, they are NOT
the set that will be given by the theorem.
Neither of these solutions will satisfy either of the two sets of
initial conditions given in the theorem.
We will have to use these to find the fundamental set of solutions
that is given by the theorem.
We know that the following is also a solution to the
differential equation.
So, let’s apply the first set of initial conditions and
see if we can find constants that will work.
We’ll leave it to you to verify that we get the following
solution upon doing this.
Likewise, if we apply the second set of initial
conditions,
we will get
According to the theorem these should form a fundament set
of solutions. This is easy enough to check.

So, we got a completely different set of fundamental
solutions from the theorem than what we’ve been using up to this point. This is not a problem. There are an infinite number of pairs of
functions that we could use as a fundamental set of solutions for this problem.
So, which set of fundamental solutions should we use? Well, if we use the ones that we originally
found, the general solution would be,
Whereas, if we used the set from the theorem the general
solution would be,
This would not be very fun to work with when it came to
determining the coefficients to satisfy a general set of initial conditions.
So, which set of fundamental solutions should we use? We should always try to use the set that is
the most convenient to use for a given problem.