9.2*Special cases
For the two special cases I will just give
the solution. It requires a substantial amount of algebra to
study these two cases.
9.2.1 Two equal roots
If the indicial equation has two equal
roots,
, we have one solution of the form
|
(9.17) |
The other solution takes the form
|
(9.18) |
Notice that this last solution is always
singular at
, whatever the value of
!
9.2.2 Two roots differing by an
integer
If the indicial equation that differ
by an integer,
, we have one solution of the form
|
(9.19) |
The other solution takes the form
|
(9.20) |
The constant
is determined by substitution, and in a few relevant
cases is even
, so that the solutions
can be of the generalised series
form.
9.2.3 Example 1
Find two independent solutions of
|
(9.21) |
near
. The indicial equation is
, so we get one solution of the series form
|
(9.22) |
We find
We add terms of equal power in
,
|
(9.24) |
Both of these ways give
|
(9.25) |
and lead to the recurrence relation
|
(9.26) |
which has the solution
|
(9.27) |
and thus
|
(9.28) |
Let us look at the second solution
|
(9.29) |
Her I replace the power series with a symbol,
for convenience. We find
Taking all this together, we have,
If we now substitute the series expansions
for
and
we get
|
(9.32) |
which can be manipulated to the form
Here there is some missing
material
9.2.4 Example 2
Find two independent solutions of
|
(9.33) |
near
.
The indicial equation is
, so that we have two roots differing by an
integer. The solution for
is
, as can be checked by substitution. The other solution
should be found in the form
|
(9.34) |
We find
We thus find
We find
|
(9.37) |
On fixing
we find
|
(9.38) |