In the previous section we introduced the
Wronskian to help us determine whether two solutions were a fundamental set of
solutions. In this section we will look
at another application of the Wronskian as well as an alternate method of
computing the Wronskian.
Let’s start with the application. We need to introduce a couple of new concepts
first.
Given two nonzero functions f(x) and g(x) write down
the following equation.
Notice that c = 0
and k = 0 will make (1)
true for all x regardless of the
functions that we use.
Now, if we can find nonzero constants c and k for which (1)
will also be true for all x then we
call the two functions linearly
dependent. On the other hand if the
only two constants for which (1) is true are c = 0 and k = 0 then we call the functions linearly independent.
Example 1 Determine
if the following sets of functions are linearly dependent or linearly
independent.
(a) [Solution]
(b) [Solution]
Solution
(a)
We’ll start by writing down (1)
for these two functions.
We need to determine if we can find nonzero constants c and k that will make this true for all x or if c = 0 and k = 0 are the only constants that will
make this true for all x. This is often a fairly difficult
process. The process can be simplified
with a good intuition for this kind of thing, but that’s hard to come by,
especially if you haven’t done many of these kinds of problems.
In this case the problem can be simplified by recalling
Using this fact our equation becomes.
With this simplification we can see that this will be zero
for any pair of constants c and k that satisfy
Among the possible pairs on constants that we could use
are the following pairs.
As I’m sure you can see there are literally thousands of
possible pairs and they can be made as “simple” or as “complicated” as you
want them to be.
So, we’ve managed to find a pair of nonzero constants
that will make the equation true for all x
and so the two functions are linearly dependent.
[Return to Problems]
(b)
As with the last part, we’ll start by writing down (1)
for these functions.
In this case there isn’t any quick and simple formula to
write one of the functions in terms of the other as we did in the first
part. So, we’re just going to have to
see if we can find constants. We’ll
start by noticing that if the original equation is true, then if we
differentiate everything we get a new equation that must also be true. In other words, we’ve got the following
system of two equations in two unknowns.
We can solve this system for c and k and see what we
get. We’ll start by solving the second
equation for c.
Now, plug this into the first equation.
Recall that we are after constants that will make this
true for all t. The only way that this will ever be zero
for all t is if k = 0!
So, if k = 0 we must also
have c = 0.
Therefore, we’ve shown that the only way that
will be true for all t
is to require that c = 0 and k = 0.
The two functions therefore, are linearly independent.
[Return to Problems]

As we saw in the previous examples determining whether two
functions are linearly independent or linearly dependent can be a fairly involved
process. This is where the Wronskian can
help.
Fact
Given two functions f(x)
and g(x) that are differentiable on
some interval I.
(1) If for some x_{0}
in I, then f(x) and g(x) are linearly independent on the
interval I.
(2) If f(x) and g(x) are linearly dependent on I then W(f,g)(x) = 0 for all x in the interval I.

Be very careful with this fact. It DOES NOT say that if W(f,g)(x) = 0 then f(x)
and g(x) are linearly dependent! In fact it is possible for two linearly
independent functions to have a zero Wronskian!
This fact is used to quickly identify linearly independent
functions and functions that are liable to be linearly dependent.
Example 2 Verify
the fact using the functions from the previous example.
Solution
(a)
In this case if we compute the Wronskian of the two
functions we should get zero since we have already determined that these
functions are linearly dependent.
So, we get zero as we should have. Notice the heavy use of trig formulas to
simplify the work!
(b)
Here we know that the two functions are linearly
independent and so we should get a nonzero Wronskian.
The
Wronskian is nonzero as we expected provided . This is not a problem. As long as the Wronskian is not identically
zero for all t we are okay.

Example 3 Determine
if the following functions are linearly dependent or linearly independent.
(a) [Solution]
(b) [Solution]
Solution
(a)
Now that we have the Wronskian to use here let’s first check that. If its nonzero then we will know that the
two functions are linearly independent and if its zero then we can be pretty
sure that they are linearly dependent.
So, by the fact these two functions are linearly
independent. Much easier this time
around!
[Return to Problems]
(b) We’ll do the same thing here as we did in the first
part. Recall that
Now compute the Wronskian.
Now, this does not say that the two functions are linearly
dependent! However, we can guess that
they probably are linearly dependent.
To prove that they are in fact linearly dependent we’ll need to write
down (1)
and see if we can find nonzero c
and k that will make it true for
all x.
So, it looks like we could use any constants that satisfy
to make this zero for all x. In particular we could
use
We have nonzero constants that will make the equation
true for all x. Therefore, the
functions are linearly dependent.
[Return to Problems]

Before
proceeding to the next topic in this section let’s talk a little more about
linearly independent and linearly dependent functions. Let’s start off by assuming that f(x) and g(x) are linearly dependent.
So, that means there are nonzero constants c and k so that
is true for all x.
Now, we can solve this in either of the following two ways.
Note that this can be done because we know that c and k are nonzero and hence the divisions can be done without worrying
about division by zero.
So, this means that two linearly dependent functions can be
written in such a way that one is nothing more than a constant times the
other. Go back and look at both of the
sets of linearly dependent functions that we wrote down and you will see that
this is true for both of them.
Two functions that are linearly independent can’t be written
in this manner and so we can’t get from one to the other simply by multiplying
by a constant.
Next, we don’t want to leave you with the impression that
linear independence and linear dependence is only for two functions. We can easily extend the idea to as many
functions as we’d like.
Let’s suppose that we have n nonzero functions, f_{1}(x),
f_{2}(x),…, f_{n}(x).
Write down the following equation.
If we can find constants c_{1},
c_{2}, …, c_{n} with at least two nonzero so that (2)
is true for all x then we call the
functions linearly dependent. If, on the
other hand, the only constants that make (2)
true for x are c_{1} = 0, c_{2 }=
0, …, c_{n} = 0 then we call the functions linearly independent.
Note that unlike the two function case we can have some of
the constants be zero and still have the functions be linearly dependent.
In this case just what does it mean for the functions to be
linearly dependent? Well, let’s suppose
that they are. So, this means that we
can find constants, with at least two nonzero so that (2) is
true for all x. For the sake of argument let’s suppose that c_{1} is one of the nonzero
constants. This means that we can do the
following.
In other words, if the functions are linearly dependent then
we can write at least one of them in terms of the other functions.
Okay, let’s move on to the other topic of this section. There is an alternate method of computing the
Wronskian. The following theorem gives
this alternate method.
Abel’s Theorem
If y_{1}(t)
and y_{2}(t) are two
solutions to
then the Wronskian of the two solutions is
for some t_{0}.

Because we don’t know the Wronskian and we don’t know t_{0} this won’t do us a lot of
good apparently. However, we can rewrite
this as
where the original Wronskian sitting in front of the
exponential is absorbed into the c
and the evaluation of the integral at t_{0}
will put a constant in the exponential that can also be brought out and
absorbed into the constant c. If you don’t recall how to do this go back
and take a look at the linear, first order differential equation section as we did something similar there.
With this rewrite we can compute the Wronskian up to a
multiplicative constant, which isn’t too bad.
Notice as well that we don’t actually need the two solutions to do
this. All we need is the coefficient of
the first derivative from the differential equation (provided the coefficient
of the second derivative is one of course…).
Let’s take a look at a quick example of this.
Example 4 Without
solving, determine the Wronskian of two solutions to the following
differential equation.
Solution
The first thing that we need to do is divide the
differential equation by the coefficient of the second derivative as that
needs to be a one. This gives us
Now, using (3)
the Wronskian is
