We now need to start looking into determining a particular
solution for nth order
differential equations. The two methods
that we’ll be looking at are the same as those that we looked at in the 2nd
order chapter.
In this section we’ll look at the method of Undetermined
Coefficients and this will be a fairly short section. With one small extension, which we’ll see in
the lone example in this section, the method is identical to what we saw back when we were looking at
undetermined coefficients in the 2nd order differential equations
chapter.
Given the differential equation,
if 
is an exponential function, polynomial, sine,
cosine, sum/difference of one of these and/or a product of one of these then we
guess the form of a particular solution using the same guidelines that we used
in the 2nd order material. We
then plug the guess into the differential equation, simplify and set the
coefficients equal to solve for the constants.
The one thing that we need to recall is that we first need
the complementary solution prior to making our guess for a particular
solution. If any term in our guess is in
the complementary solution then we need to multiply the portion of our guess
that contains that term by a t. This is where the one extension to the method
comes into play. With a 2nd
order differential equation the most we’d ever need to multiply by is . With higher order differential equations this
may need to be more than .
The work involved here is almost identical to the work we’ve
already done and in fact it isn’t even that much more difficult unless the
guess is particularly messy and that makes for more mess when we take the
derivatives and solve for the coefficients.
Because there isn’t much difference in the work here we’re only going to
do a single example in this section illustrating the extension. So, let’s take a look at the lone example
we’re going to do here.
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Example 1 Solve
the following differential equation.
Solution
We first need the complementary solution so the
characteristic equation is,
We’ve got a single root of multiplicity 3 so the
complementary solution is,
Now, our first guess for a particular solution is,
Notice that the
last term in our guess is in the complementary solution so we’ll need to add
one at least one t to the third
term in our guess. Also notice that
multiplying the third term by either t
or will result in a new term that is still in
the complementary solution and so we’ll need to multiply the third term by in order to get a term that is not contained
in the complementary solution.
Our final guess
is then,
Now all we need
to do is take three derivatives of this, plug this into the differential
equation and simplify to get (we’ll leave it to you to verify the work here),
Setting
coefficients equal and solving gives,
A particular
solution is then,
The general
solution to this differential equation is then,
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Okay, we’ve only worked one example here, but remember that
we mentioned earlier that with the exception of the extension to the method
that we used in this example the work here is identical to work we did the 2nd
order material.