We now need to take a look at the second method of
determining a particular solution to a differential equation. As we did when we first saw Variation of Parameters we’ll go
through the whole process and derive up a set of formulas that can be used to
generate a particular solution.
However, as we saw previously when looking at 2nd
order differential equations this method can lead to integrals that are not
easy to evaluate. So, while this method
can always be used, unlike Undetermined Coefficients, to at least write down a
formula for a particular solution it is not always going to be possible to
actually get a solution.
So let’s get started on the process. We’ll start with the differential equation,
and assume that we’ve found a fundamental set of solutions, ,
for the associated homogeneous differential equation.
Because we have a fundamental set of solutions to the
homogeneous differential equation we now know that the complementary solution
The method of variation of parameters involves trying to
find a set of new functions, so that,
will be a solution to the nonhomogeneous differential equation. In order to determine if this is possible,
and to find the if it is possible, we’ll need a total of n equations involving the unknown
functions that we can (hopefully) solve.
One of the equations is easy. The guess, (2),
will need to satisfy the original differential equation, (1). So, let’s start taking some derivatives and
as we did when we first looked at variation of parameters we’ll make some
assumptions along the way that will simplify our work and in the process
generate the remaining equations we’ll need.
The first derivative of (2) is,
Note that we rearranged the results of the differentiation
process a little here and we dropped the part on the u and y to make this a
little easier to read. Now, if we keep
differentiating this it will quickly become unwieldy and so let’s make as
assumption to simplify things here.
Because we are after the we should probably try to avoid letting the
derivatives on these become too large.
So, let’s make the assumption that,
The natural question at this point is does this even make
sense to do? The answer is, if we end up
with a system of n equations that we
can solve for the then yes it does make sense to do. Of course, the other answer is, we wouldn’t
be making this assumption if we didn’t know that it was going to work. However to accept this answer requires that
you trust us to make the correct assumptions so maybe the first answer is the
best at this point.
At this point the first derivative of (2)
and so we can now take the second derivative to get,
This looks an awful lot like the original first derivative
prior to us simplifying it so let’s again make a simplification. We’ll again want to keep the derivatives on the
to a minimum so this time let’s assume that,
and with this assumption the second derivative becomes,
Hopefully you’re starting to see a pattern develop
here. If we continue this process for
the first derivatives we will arrive at the following
formula for these derivatives.
To get to each of these formulas we also had to assume that,
and recall that the 0th derivative of a function
is just the function itself. So, for
Notice as well that the set of assumptions in (4)
actually give us equations in terms of the derivatives of the
unknown functions : .
All we need to do then is finish generating the first
equation we started this process to find (i.e.
into (1)). To do this we’ll need one more derivative of
the guess. Differentiating the derivative, which we can get from (3),
to get the nth derivative
This time we’ll also not be making any assumptions to
simplify this but instead just plug this along with the derivatives given in (3)
into the differential equation, (1)
Next, rearrange this a little to get,
Recall that are all solutions to the homogeneous
differential equation and so all the quantities in the [ ] are zero and this
reduces down to,
So this equation, along with those given in (4),
give us the n equations that we
needed. Let’s list them all out here for
the sake of completeness.
So, we’ve got n
equations, but notice that just like we got when we did this for 2nd
order differential equations the unknowns in the system are not but instead they are the derivatives, . This isn’t a major problem however. Provided we can solve this system we can then
just integrate the solutions to get the functions that we’re after.
Also, recall that the are assumed to be known functions and so they
along with their derivatives (which appear in the system) are all known
quantities in the system.
Now, we need to think about how to solve this system. If there aren’t too many equations we can
just solve it directly if we want to.
However, for large n (and it
won’t take much to get large here) that could be quite tedious and prone to
error and it won’t work at all for general n
as we have here.
The best solution method to use at this point is then Cramer’s Rule. We’ve used Cramer’s Rule several times in
this course, but the best reference for our purposes here is when we used it
when we first defined Fundamental Sets
of Solutions back in the 2nd order material.
Upon using Cramer’s Rule to solve the system the resulting
solution for each will be a quotient of two determinants of n x n
matrices. The denominator of each
solution will be the determinant of the matrix of the known coefficients,
This however, is just the Wronskian of as noted above and because we have assumed
that these form a fundamental set of solutions we also know that the Wronskian
will not be zero. This in turn tells us
that the system above is in fact solvable and all of the assumptions we
apparently made out of the blue above did in fact work.
The numerators of the solution for will be the determinant of the matrix of
coefficients with the ith
column replaced with the column . For example, the numerator for the first one,
Now, by a nice property of
determinants if we factor something out of one of the columns of a matrix then
the determinant of the resulting matrix will be the factor times the
determinant of new matrix. In other
words, if we factor out of this matrix we arrive at,
We did this only for the first one, but we could just as
easily done this with any of the n
solutions. So, let represent the determinant we get by replacing
the ith column of the
Wronskian with the column (0,0,0,…,0,1) and the solution to the system can then
be written as,
Wow! That was a lot
of effort to generate and solve the system but we’re almost there. With the solution to the system in hand we
can now integrate each of these terms to determine just what the unknown
functions, we’ve after all along are.
Finally, a particular solution to (1) is
then given by,
We should also note that in the derivation process here we
assumed that the coefficient of the term was a one and that has been factored into
the formula above. If the coefficient of
this term is not one then we’ll need to make sure and divide it out before
trying to use this formula.
Before we work an example here we really should note that
while we can write this formula down actually computing these integrals may be
all but impossible to do.
Okay let’s take a look at a quick example.
Example 1 Solve
the following differential equation.
The characteristic equation is,
So we have three real distinct roots here and so the
complimentary solution is,
Okay, we’ve now got several determinants to compute. We’ll leave it to you to verify the
following determinant computations.
Now, given that
we can compute each of the . Here are those integrals.
Note that we didn’t include the constants of integration
in each of these because including them would just have introduced a term
that would get absorbed into the complementary solution just as we saw when we were dealing with 2nd
order differential equations.
Finally, a particular solution for this differential
equation is then,
The general solution is then,
We’re only going to do a single example in this section to
illustrate the process more than anything so with that we’ll close out this