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Then, with
A
1,0
A
3,0
, and substituting from (3.207) for
A
6,0
, the second and
fourth equations are found to be identical. Solving the first three equations in terms
of
A
4,1
yields
=
1
μ
A
1,2
=−
A
3,2
+
A
4,1
,
(3.210)
A
2,1
=−
q
1
(1)
A
3,2
+
q
2
(1)
A
4,1
,
(3.211)
p
1
(1)
ρ
0
2γ
+
ω
A
1,0
+
ρ
0
A
6,0
. (3.212)
1
2
2
A
3,2
=
+
2
Ω
−
2
m
ω
Ω
+
p
2
(1)
A
4,1
Now
A
1,0
,
A
6,0
,
A
4,1
may be regarded as the three free constants determining the
three regular solutions at the geocentre for
n
1. No further eigenvalues are
encountered in the recursions for higher terms in the power series expansions.
For the regular solution generated by the free constant
A
1,0
, from equations
(3.210) through (3.212), we have
⎝
=
⎠
⎝
⎠
A
1,2
A
2,1
A
3,2
A
4,1
−
1
p
1
(1)
2γ
+
ω
=
ρ
0
−
q
1
(1)
1
0
2
2
+
2
Ω
−
2
m
ω
Ω
A
1,0
.
(3.213)
From (3.207),
A
5,1
0. For numerical integration, the y-
variables are again replaced by a system of
z
-variables for
n
=
4π
G
ρ
0
A
1,0
for
A
6,0
=
1. For the regular
solutions generated by the free constants
A
1,0
and
A
6,0
the new
z
-variables are just
those defined in (3.187) for α
=
=
1. For the regular solution generated by
the free constant
A
4,1
they are those defined in (3.187) for α
=
n
−
2
=−
1. The power
series expansions for the regular solution generated by the free constant
A
1,0
then
have the forms
n
=
A
1,2
r
2
A
1,4
r
4
z
1
(
r
)
=
y
1
(
r
)
=
A
1,0
+
+
+···
,
(3.214)
A
2,1
r
2
A
2,3
r
4
z
2
(
r
)
=
r
y
2
(
r
)
=
+
+···
,
(3.215)
A
3,2
r
2
A
3,4
r
4
z
3
(
r
)
=
y
3
(
r
)
=
A
1,0
+
+
+···
,
(3.216)
A
4,3
r
4
z
4
(
r
)
=
r
y
4
(
r
)
=
+···
,
(3.217)
A
5,3
r
2
A
5,5
r
4
z
5
(
r
)
=
y
5
(
r
)/
r
=
4π
G
ρ
0
A
1,0
+
+
+···
,
(3.218)
A
6,2
r
2
A
6,4
r
4
z
6
(
r
)
=
y
6
(
r
)
=
+
+···
.
(3.219)
For the regular solution generated by the free constant
A
6,0
, from equations
(3.210) through (3.212), we have
⎝
⎠
⎝
⎠
A
1,2
A
2,1
A
3,2
A
4,1
−
ρ
0
/
p
1
(1)
−
ρ
0
q
1
(1)/
p
1
(1)
ρ
0
/
p
1
(1)
0
=
A
6,0
.
(3.220)
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