Example 1 Find
the series solution around for the following differential equation.
Solution
Recall that we can only find a series solution about if this point is an ordinary point, or in
other words, if the coefficient of the highest derivative term is not zero at
. That is clearly the case here so let’s
start with the form of the solutions as well as the derivatives that we’ll
need for this solution.
Plugging into the differential equation gives,
Now, move all the coefficients into the series and do
appropriate shifts so that all the
series are in terms of
.
Next, let’s notice that we can start the second series at since that term will be zero. So let’s do that and then we can combine
the second and third terms to get,
So, we got a nice simplification in the new series that
will help with some further simplification.
The new second series can now be started at and then combined with the first series to
get,
We can now set the coefficients equal to get a fairly
simply recurrence relation.
Solving the recurrence relation gives,
Now we need to start plugging in values of n and this will be one of the main
areas where we can see a somewhat significant increase in the amount of work
required when moving into a higher order differential equation.
Okay, we can now break the coefficients down into 4 sub
cases given by ,
,
and for We’ll give a semidetailed derivation for and then leave the rest to you with only
couple of comments as they are nearly identical derivations.
First notice that all the terms have in them and they will alternate in
sign. Next notice that we can turn the
denominator into a factorial, to be exact, if we multiply top and bottom
by the numbers that are already in the numerator and so this will turn these
numbers into squares. Next notice that
the product in the top will start at 1 and increase by 4 until we reach . So, taking all of this into account we get,
and notice that
this will only work starting with as we won’t get out of this formula as we should by plugging
in .
Now, for the derivation is almost identical and so
the formula is,
and again
notice that this won’t work for
The formula for
is again nearly identical except for this
one note that we also need to multiply top and bottom by a 2 in order to get
the factorial to appear in the denominator and so the formula here is,
noticing yet
one more time that this won’t work for .
Finally, we
have for
Now that we have all the coefficients let’s get the solution,
Collecting up the terms that contain the same coefficient
(except for the first one in each case since the formula won’t work for
those) and writing everything as a set of series gives us our solution,
