MATH 150
Series
Cheung Man Wai
Tutorial 9
The purpose of this note is to find two independent solutions around
x
0
of the
following
P
(
x
)
y
+
Q
(
x
)
y
+
R
(
x
)
y
= 0
,
(1)
where
P
(
x
),
Q
(
x
) and
R
(
x
) are polynomials with no common factors. We shall
consider two cases respectively, say
x
0
is an
ordinary point
(if
P
(
x
0
) = 0) or a
singular point
(otherwise) of (1).
1. Ordinary points
We determine two power series solutions for
y
=
y
(
x
) around
x
0
.
Let us see
an example.
Example 1.
y
+
k
2
x
2
y
= 0
,
x
0
= 0
,
k
a constant
.
With
y
(
x
) =
∞
n
=0
a
n
x
n
,
we find that
y
(
x
) =
∞
n
=2
n
(
n

1)
a
n
x
n

2
and the equation becomes
∞
n
=2
n
(
n

1)
a
n
x
n

2
+
k
2
∞
n
=0
a
n
x
n
+2
= 0
.
By shift of index one has
∞
n
=0
(
n
+ 2)(
n
+ 1)
a
n
+2
x
n
+
k
2
∞
n
=2
a
n

2
x
n
= 0
.
1
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The coefficient of each power of
x
vanishes, thus
a
2
=
a
3
= 0
,
(
n
+ 2)(
n
+ 1)
a
n
+2
+
k
2
a
n

2
= 0
,
n
≥
2
.
The recurrence relation can be equivalently written as
a
n
+4
=

k
2
(
n
+ 4)(
n
+ 3)
a
n
,
n
≥
0
.
Four sequences of coefficients, starting with
a
0
, a
1
, a
2
and
a
3
decouple. In par
ticular the sequences are
a
0
, a
4
, a
8
, a
12
, . . .
;
a
1
, a
5
, a
9
, a
13
, . . .
;
a
2
, a
6
, a
10
, a
14
, . . .
;
a
3
, a
7
, a
11
, a
15
,
. . . .
Since
a
2
=
a
3
= 0
,
by recurrence relation the last two sequences vanish identi
cally. Starting with
a
0
, we have
a
0
,
a
4
=

k
2
3
·
4
a
0
,
a
8
=
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 Spring '09
 T.Qian
 Math, Polynomials, Factors, Characteristic polynomial, Complex number, Zagreb, Recurrence relation, Regular singular point

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