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Section A.5 The Power Series Method, Part II
A191
Solutions to Exercises A.5
1.
For the diﬀerential equation
y
±
+2
xy
=0
,
p
(
x
) = 1 is its own power series
expansion about
a
= 0. So
a
= 0 is an ordinary point. To solve, let Let
y
=
∞
±
m
=0
a
m
x
m
y
±
=
∞
±
m
=1
ma
m
x
m

1
.
Then
y
±
+2
xy
=
∞
±
m
=1
ma
m
x
m

1
+2
x
∞
±
m
=0
a
m
x
m
=
∞
±
m
=1
ma
m
x
m

1
+
∞
±
m
=0
2
a
m
x
m
+1
=
∞
±
m
=0
(
m
+1)
a
m
+1
x
m
+
∞
±
m
=1
2
a
m

1
x
m
=
a
1
+
∞
±
m
=1
[(
m
+1)
a
m
+1
+2
a
m

1
]
x
m
.
So
y
±
+2
xy
= 0 implies that
a
1
+
∞
±
m
=1
[(
m
+1)
a
m
+1
+2
a
m

1
]
x
m
=0
;
a
1
=0
(
m
+1)
a
m
+1
+2
a
m

1
=0
a
m
+1
=

2
m
+1
a
m

1
.
From the recurrence relation,
a
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Unformatted text preview: a 1 = a 3 = a 5 = ··· = a 2 k +1 = ··· = 0; a is arbitrary; a 2 =2 2 a =a , a 4 =2 4 a 2 = 1 2! a , a 6 =2 6 a 4 =1 3! a , a 8 =2 8 a 6 = 1 4! a , . . . a 2 k = (1) k k ! a . So y = a ∞ ± k =0 (1) k k ! x 2 k = a ∞ ± k =0 (x 2 ) k k ! = a ex 2 ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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