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Section A.6 The Method of Frobenius
A197
Solutions to Exercises A.6
1.
For the equation
y
±±
+(1

x
2
)
y
±
+
xy
=0,
p
(
x
)=1

x
2
and
q
(
x
)=
x
are both
analytic at
a
= 0. So
a
= 0 is a ordinary point.
5.
For the equation
x
2
y
±±

e
x
)
y
±
+
xy
=0
,
p
(
x
1

e
x
x
2
,x
p
(
x
1

e
x
x
;
q
(
x
1
x
2
q
(
x
x.
p
(
x
) and
q
(
x
) are not analytic at 0. So
a
= 0 is a singular point. Since
xp
(
x
) and
x
2
q
(
x
) are analytic at
a
= 0, the point
a
= 0 is a regular singular point. To see
that
xp
(
x
) is analytic at 0, derive its Taylor series as follows: for all
x
,
e
x
=1
+
x
+
x
2
2!
+
x
3
3!
+
···
1

e
x
=

x

x
2
2!

x
3
3!
+
=
x
±

1

x
2!

x
2
3!
+
²
1

e
x
x
=

1

x
2!

x
2
3!
+
.
Since
1

e
x
x
has a Taylor series expansion about 0 (valid for all
x
), it is analytic at
0.
9.
For the equation 4
x
2
y
±±

14
xy
±
+ (20

x
)
y
,
p
(
x

7
2
x
p
(
x

7
2
,p
0
=

7
2
;
q
(
x
20

x
4
x
2
2
q
(
x
)=5

x
4
,q
0
=5
.
p
(
x
) and
q
(
x
) are not analytic at 0. So
a
= 0 is a singular point. Since
xp
(
x
) and
x
2
q
(
x
) are analytic at
a
= 0, the point
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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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