1
Lecture 19
For a number
a
, consider the following operator:
(
D
&
a
) [
f
]
j
x
=
f
0
(
x
)
&
af
(
x
)
:
First, we prove the following property of some operators. We claim that,
for every pair of real numbers,
(
D
&
a
) (
D
&
b
) = (
D
&
b
) (
D
&
a
)
:
Indeed,
(
D
&
a
) (
D
&
b
) [
f
] = (
D
&
a
) [(
D
&
b
) [
f
]] = (
D
&
a
) [
f
0
&
bf
]
= (
f
0
&
bf
)
0
&
a
(
f
0
&
bf
) =
f
00
&
bf
0
&
af
0
&
abf:
Likewise,
(
D
&
b
) (
D
&
a
) [
f
] = (
D
&
b
) [
f
0
&
af
] = (
f
0
&
af
)
0
&
b
(
f
0
&
af
)
=
f
00
&
af
0
&
bf
0
&
abf:
Hence,
(
D
&
a
) (
D
&
b
) = (
D
&
b
) (
D
&
a
)
:
Now
L
[
y
] = 0
()
(
D
&
r
0
)
m
0
(
D
&
r
1
)
m
1
:::
(
D
&
r
l
)
m
l
[
y
]
= (
D
&
r
1
)
m
1
:::
(
D
&
r
l
)
m
l
(
D
&
r
0
)
m
0
[
y
] = 0
:
Therefore, every solution of the equation
(
D
&
r
0
)
m
0
[
y
] = 0
will be a solution
of the equation
L
[
y
] = 0
. As a trial solution of the ODE
(
D
&
r
0
)
m
0
[
y
] = 0
,
take
y
(
x
) =
u
(
x
)
e
r
0
x
;
where the function
u
(
x
)
is to be found
:
We have
(
D
&
r
0
) [
u
(
x
)
e
r
0
x
] = (
u
(
x
)
e
r
0
x
)
0
&
r
0
(
u
(
x
)
e
r
0
x
)
=
u
0
e
r
0
x
+
r
0
ue
r
0
x
&
r
0
(
u
(
x
)
e
r
0
x
) =
u
0
e
r
0
x
:
Likewise,
(
D
&
r
0
)
2
[
u
(
x
)
e
r
0
x
] = (
D
&
r
0
) [
u
0
(
x
)
e
r
0
x
] =
u
00
(
x
)
e
r
0
x
and
:::
(
D
&
r
0
)
m
0
[
u
(
x
)
e
r
0
x
] = (
D
&
r
0
)
&
u
(
m
0
&
1)
e
r
0
x
±
=
u
m
0
e
r
0
x
;
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documentwhence
u
(
m
0
)
e
r
0
x
= 0
,
whence
u
(
m
0
)
(
x
)
&
0
)
u
(
x
) =
C
0
+
C
1
x
+
:::
+
C
m
0
&
1
x
m
0
&
1
:
So, if
r
=
r
0
is a root of multiplicity
m
0
, then
y
(
x
) =
&
C
0
+
C
1
x
+
:::
+
C
m
0
&
1
x
m
0
&
1
±
e
r
0
x
is a solution of the ODE
L
[
y
] = 0
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Staff
 Differential Equations, Real Numbers, Equations, Complex number, general solution, iy, JZ

Click to edit the document details