For the ordinary differential equation
find the first 3 terms of a non-zero series solution about satisfying
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.
This is a standard question in this course. If you have forgotten how to do this type of question, check your class notes or homework for a similar example.
Is an ordinary point or a singular point?
Compare the ODE with
We see that . Since when , is a singular point.
Is a regular singular point or an irregular singular point?
Again compare the ODE with
We see that , and . Now,
Therefore, is a regular singular point.
Should we consider series solutions of the form
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
Following the comments in the hints, we see that is a regular singular point. Therefore, we should look for series solutions of the form
We have that
(Please note that the both of the sums for and start at .)
Substitute these into the ODE
we get that
We should shift the index of so the order of matches with the other terms. To do so, we should replace every with . Then
Then we have
which, after separating the terms, can be rewritten as
Look at the above equation, because the right hand side is 0, the coefficient of the term must be 0 as well. Hence,
Now, if , then all the subsequent will be zero as well and we only get the trivial solution . (Why? You can verify that yourself after we finish the question. Set and you will see the other terms will be forced to be 0 by the recursion relation.) In other words, we must have
giving us two roots and . (Note: In the above, is known as the indicial equation.)
By the same reason as , we have that
If we Isolate , we get the recurrence relation
After simplifying and factoring the expression in the denominator, the above expression can be rewritten as
which may be easier to do calculations with.
Now recall that from the indicial equation, we found and .
In the case , .
Note that the above equation does not satisfy due to the term, so the expression corresponding to is not what the question is asking for.
In the case , .
Note that the above equation satisfies . In other words, this is the one the question is asking for.
It remains to find and in terms of . To find , substitue and into the recurrence relation. We get that
To find , substitue and into the recurrence relation. We get that
Therefore, we have