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p(x) =
(3/2−x)
x
q(x) =
(−1/2−x/2)
x
2
x
0
= 0
p(x) q(x)
p
0
= lim
x→0
x
(3/2 − x)
x
= 3/2, q
0
= lim
x→0
x
2
(−1/2 − x/2)
x
2
= −1/2.
r(r − 1) + 3/2r − 1/2 = 0,
r
1
= 1/2 r
2
= −1 r
1
r
2
3/2,
y
1
y
2
y
1
=
∞
X
k=0
c
(1)
k
x
k+1/2
, y
2
=
∞
X
k=0
c
(2)
k
x
k−1
.
y
1
2x
2
∞
X
k=0
c
(1)
k
(k + 1/2)(k − 1/2)x
k−3/2
+ 3x
∞
X
k=0
c
(1)
k
(k + 1/2)x
k−1/2
−
−2x
2
∞
X
k=0
c
(1)
k
(k + 1/2)x
k−1/2
− x
∞
X
k=0
c
(1)
k
x
k+1/2
−
∞
X
k=0
c
(1)
k
x
k+1/2
= 0.
c
(1)
k
x
1/2
0 ≡ 0, c
(1)
0
x
3/2
2c
(1)
1
3
2
·
1
2
+ 3c
(1)
1
3
2
− 2c
(1)
0
1
2
− c
(1)
0
− c
(1)
1
= 0;
x
5/2
2c
(1)
2
5
2
·
3
2
+ 3c
(1)
2
5
2
− 2c
(1)
1
3
2
− c
(1)
1
− c
(1)
2
= 0;
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
x
k+1/2
c
(1)
k
(k + 1/2)(k − 1/2) + 3c
(1)
k
(k + 1/2) − 2c
(1)
k−1
(k − 1/2)−
− c
(1)
k−1
− c
(1)
k
= 0.
c
(1)
1
=
2
5
c
(1)
0
, c
(1)
2
=
2
7
c
(1)
1
=
2
2
5 · 7
c
(1)
0
, . . . , c
(1)
k
=
2c
(1)
k−1
2k + 3
=
2
k
c
(1)
0
5 · 7 · . . . · (2k + 3)
.
50
(3/2−x) (−1/2−x/2)
Êîýôôèöèåíòû p(x) = x
è q(x) = x2
èìåþò îñîáåííîñòü â òî÷êå x0 = 0,
ñëåäîâàòåëüíî, p(x) è q(x) ðàçëàãàþòñÿ â ðÿä (4.14). Ïî òåîðåìå (2) èìååòñÿ õîòÿ áû îäíî
ðåøåíèå â âèäå îáîáùåííîãî ñòåïåííîãî ðÿäà.
Ïî ôîðìóëå (4.17)
(3/2 − x) (−1/2 − x/2)
p0 = lim x = 3/2, q0 = lim x2 = −1/2.
x→0 x x→0 x2
Ñîñòàâèì îïðåäåëÿþùåå óðàâíåíèå (4.16):
r(r − 1) + 3/2r − 1/2 = 0,
îòêóäà r1 = 1/2 è r2 = −1. Ðàçíîñòü r1 è r2 ðàâíà 3/2, íå ÿâëÿåòñÿ öåëûì ÷èñëîì,
ñëåäîâàòåëüíî, èç óòâåðæäåíèÿ (1) ïîëó÷àåòñÿ, ÷òî èìåþòñÿ äâà ëèíåéíî íåçàâèñèìûõ
ðåøåíèÿ y1 è y2 â âèäå îáîáùåííûõ ñòåïåííûõ ðÿäîâ:
∞
X ∞
X
(1) (2)
y1 = ck xk+1/2 , y2 = ck xk−1 . (4.23)
k=0 k=0
Ïîäñòàâëÿåì y1 â óðàâíåíèå (4.21):
∞
X ∞
X
(1) (1)
2x2 ck (k + 1/2)(k − 1/2)xk−3/2 + 3x ck (k + 1/2)xk−1/2 −
k=0 k=0
∞
X ∞
X ∞
X
2 (1) k−1/2 (1) (1)
−2x ck (k + 1/2)x −x ck xk+1/2 − ck xk+1/2 = 0.
k=0 k=0 k=0
(1)
Ïîëó÷àåì ñèñòåìó óðàâíåíèé íà ck :
(1)
ïðè x1/2 : 0 ≡ 0, ò.å c0 - ïðîèçâîëüíîå;
(1) 3 1 (1) 3 (1) 1 (1) (1)
ïðè x3/2 : 2c1 · + 3c1 − 2c0 − c0 − c1 = 0;
2 2 2 2
(1) 5 3 (1) 5 (1) 3 (1) (1)
ïðè x5/2 : 2c2 · + 3c2 − 2c1 − c1 − c2 = 0;
2 2 2 2
.....................................................................
(1) (1) (1)
ïðè xk+1/2 : ck (k + 1/2)(k − 1/2) + 3ck (k + 1/2) − 2ck−1 (k − 1/2)−
(1) (1)
− ck−1 − ck = 0.
Òîãäà êîýôôèöèåíòû èìåþò âèä
(1) (1)
(1) 2 (1) (1) 2 (1) 22 (1) (1) 2ck−1 2k c0
c1 = c0 , c2 = c1 = c ,..., ck = = .
5 7 5·7 0 2k + 3 5 · 7 · . . . · (2k + 3)
