Four sequences of coefficients, starting with
a
0
, a
1
, a
2
and
a
3
decouple. In
particular 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
identically. Starting with
a
0
, we have
a
0
,
a
4
=

k
2
3
·
4
a
0
,
a
8
=
k
4
3
·
4
·
7
·
8
a
0
,
a
12
=

k
6
3
·
4
·
7
·
8
·
11
·
12
a
0
;
and starting with
a
1
,
a
1
,
a
5
=

k
2
4
·
5
a
1
,
a
9
=
k
4
4
·
5
·
8
·
9
a
1
,
a
13
=

k
6
4
·
5
·
8
·
9
·
12
·
13
a
1
.
Finally the general solution can therefore written as
y
(
x
) =
a
0
y
0
(
x
) +
a
1
y
1
(
x
)
,
where
y
0
(
x
)
=
1

k
2
x
4
3
·
4
+
k
4
x
8
3
·
4
·
7
·
8

k
6
x
12
3
·
4
·
7
·
8
·
11
·
12
+
· · ·
=
1 +
∞
X
m
=0
(

1)
m
+1
(
k
2
x
4
)
m
+1
3
·
4
·
7
·
8
· · ·
(4
m
+ 3)(4
m
+ 4)
,
y
1
(
x
)
=
x

k
2
x
5
4
·
5
+
k
4
x
9
4
·
5
·
8
·
9

k
6
x
13
4
·
5
·
8
·
9
·
12
·
13
+
· · ·
=
x
"
1 +
∞
X
m
=0
(

1)
m
+1
(
k
2
x
4
)
m
+1
4
·
5
·
8
·
9
· · ·
(4
m
+ 4)(4
m
+ 5)
#
.
2
Remark.
By the relations
y
(0) =
a
0
, y
0
(0) =
a
1
, the solution of Example 1
2