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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
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