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=
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(k+s)z
k+s-1
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=
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(k+s)(k+s-1)z
k+s-2Подставляя (2.17) в уравнение (2.16) находим
z(1-z)
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(
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z
k+s 
+[
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-(
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+
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+1)z]
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(
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z
k+s 
-
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z
k+s=0,
z(1-z)
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(
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z
k+s-1(k+s)(k+s-1))+[
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-(
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+
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+1)z]
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(
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z
k+s-1(k+s))-
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zk+s=
=
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(
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z
k+s-1(k+s)(k+s-1))-
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(
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z
k+s(k+s)(k+s-1))+
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(
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z
k+s-1 
(k+s))-
-
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z
k+s(
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+
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+1)(k+s))-
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z
k+s
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=
=
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z
k+s-1(k+s)(k+s-1+
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)-
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z
k+s(s+k+
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)(s+k+
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)=0,
откуда для определения показателя s и
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получается система уравнений
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s(s-1-)=0,
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(s+k)(s+k-1+
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) -
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(s+k-1+
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)(s+k-1+
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)=0,
k=1,2,…,
первое из которых дает s=0 или s=1-
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1) Предположим, что
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0,-1,-2,… и выберем s=0
Тогда для вычисления коэффициентов
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получим реккурентное соотношение
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=
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k=1,2,…,
откуда, если принять
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=1, следует
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=
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k=0,1,2,…,
где для сокращения записи введено обозначение
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=
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(
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+1)…(
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+k-1),
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=1, k=1,2,…,
Таким образом первое частное решение уравнения (2.16) при
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0,-1,-2,… будет
u=
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= F(
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,
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,
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,z)=
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z
k,
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<1 (2.18)
2) Аналогично, выбирая s=1-
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получаем в предположении, что
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2,3,4,…
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=
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k=1,2,…,
откуда, если взять
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=1 находим
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=
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k=0,1,2,…,
Таким образом, при
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2,3,4,… уравнение (2.16) имеет второе частное решение
u=
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=
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=
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F(1-
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+
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,1-
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+
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,2-
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,z), (2.19)