This preview shows page 1. Sign up to view the full content.
Section 5.5
Legendre’s Diﬀerential Equation
89
Another faster way to see the answer is to simply note that
P
3
is an odd function, so
its integral over any symmetric interval is 0. There is yet another more important
reason for this integral to equal 0. In fact,
±
1

1
P
n
(
x
)
dx
= 0
for all
n
±
=0
.
This is a consequence of orthogonality that you will study in Section 5.6.
9.
This is Legendre’s equation with
n
(
n
+ 1) = 30 so
n
= 5. Its general solution
is of the form
y
=
c
1
P
5
(
x
)+
c
2
Q
5
(
x
)
=
c
1
1
8
(63
x
5

70
x
3
+15
x
c
2
(
1

15
x
2
+30
x
4
+
···
)
=
c
1
(63
x
5

70
x
3
x
c
2
(
1

15
x
2
x
4
+
)
In Fnding
P
5
(
x
), we used the given formulas in the text. In Fnding the Frst few
terms of
Q
5
(
x
), we used (3) with
n
= 5. (If you are comparing with the answers in
your textbook, just remember that
c
1
and
c
2
are arbitrary constants.)
13.
This is Legendre’s equation with
n
(
n
+ 1) = 6 or
n
= 2. Its general solution
is
y
=
c
1
P
2
(
x
c
2
Q
2
(
x
). The solution will be bounded on [

1
,
1] if and only
if
c
2
= 0; that’s because
P
2
is bounded in [

1
,
1] but
Q
2
is not. Now, using
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details