Quiz #5
Math 20D
March 6
Two problems (
§
5.4, #5 and
§
5.6, #3).
SOLUTIONS.
1. (
§
5.4, #5. 10 points) Find all singular points of the given equation and determine whether each one is
regular or irregular:
(
1

x
2
)
2
y
00
+
x
(1

x
)
y
0
+ (1 +
x
)
y
= 0
.
Solution
: We rewrite this as (since
(
1

x
2
)
2
= (1

x
)
2
(1 +
x
)
2
)
0 =
y
00
+
x
(1

x
)
(1

x
2
)
2
y
0
+
1 +
x
(1

x
2
)
2
y
=
y
00
+
x
(1

x
) (1 +
x
)
2
y
0
+
1
(1

x
)
2
(1 +
x
)
y.
The singular points are
x
=

1 and
x
= 1
.
(You may write
p
(
x
) =
x
(1

x
)(1+
x
)
2
and
q
(
x
) =
1
(1

x
)
2
(1+
x
)
.
)
(a) For
x
=

1 we have
(
x
+ 1)
p
(
x
) = (
x
+ 1)
x
(1

x
) (1 +
x
)
2
=
x
(1

x
) (1 +
x
)
is not analytic near
x
=

1
.
(We do not need to compute (
x
+ 1)
2
q
(
x
)
.
) So
x
=

1 is an irregular
singular point.
(b) For
x
= 1 we have
(
x

1)
p
(
x
) = (
x

1)
x
(1

x
) (1 +
x
)
2
=

x
(1 +
x
)
2
is analytic near
x
= 1
.
And we have
(
x

1)
2
q
(
x
) = (
x

1)
2
1
(1

x
)
2
(1 +
x
)
=
1
1 +
x
which is also analytic near
x
= 1
.
So
x
= 1 is a regular singular point.
2.
§
5.6, #3.
y
=
∞
X
n
=0
a
n
x
r
+
n
y
0
=
∞
X
n
=0
(
r
+
n
)
a
n
x
r
+
n

1
y
00
=
∞
X
n
=0
(
r
+
n

1) (
r
+
n
)
a
n
x
r
+
n

2
.
Then
xy
00
=
∞
X
n
=0
(
r
+
n

1) (
r
+
n
)
a
n
x
r
+
n

1
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Relabel
xy
00
=
∞
X
n
=

1
(
r
+
n
) (
r
+
n
+ 1)
a
n
+1
x
r
+
n
.
Then
0 =
xy
00
+
y
=
∞
X
n
=

1
(
r
+
n
) (
r
+
n
+ 1)
a
n
+1
x
r
+
n
+
∞
X
n
=0
a
n
x
r
+
n
= (
r

1)
r
·
a
0
x
r

1
+
∞
X
n
=0
((
r
+
n
) (
r
+
n
+ 1)
a
n
+1
+
a
n
)
x
r
+
n
.
Indicial equation (from
n
=

1 case):
(
r

1)
r
= 0
.
Roots:
r
1
= 0
r
2
= 1
.
The recurrence relation
(
r
+
n
) (
r
+
n
+ 1)
a
n
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Mohanty
 Math, xy, Frobenius method, Regular singular point, n=0

Click to edit the document details