Fall 2003 Math 308/501–502
6 Theory of HigherOrder Linear ODEs
6.1 Basic Theory of Linear ODEs
Wed, 24/Sep
c
±
2003, Art Belmonte
Summary
We give an overview of the theory of
n
th order linear differential
equations; here
n
≥
1. Of course, the case where
n
=
1 was dealt
with in Section 2.3, whereas
n
=
2 is dealt with in Chapter 4.
Using summation and matrix notation along with a smattering of
linear algebra concepts, we treat the general case once and for all.
In the following,
I
=
(
a
,
b
)
is an open real interval.
TERMINOLOGY
A
linear
ODE of order
n
has the form
n
X
k
=
0
a
k
(
x
)
y
(
k
)
(
x
)
=
b
(
x
)
.
Here the
a
k
depend only on
x
(the independent variable), not on
y
(the dependent variable). The equation has
constant coefﬁcients
if the
a
k
are constants; otherwise, it has
variable coefﬁcients
.If
b
(
x
)
or
g
(
x
)
is zero on
I
, then the equation is
homogeneous
;
otherwise it is
nonhomogeneous
.
With the
a
k
and
b
continuous on
I
and
a
n
(
x
)
6=
0on
I
,divideto
obtain the
standard form
y
(
n
)
(
x
)
+
n
X
j
=
1
p
j
(
x
)
y
(
n

j
)
(
x
)
=
g
(
x
)
or
L
[
y
]
(
x
)
=
g
(
x
)
,where
L
=
D
n
+
∑
n
j
=
1
p
j
D
n

j
is called a
linear differential operator
; i.e.,
L
"
m
X
i
=
1
y
i
#
=
m
X
i
=
1
L
[
y
i
]and
L
[
cy
]
=
cL
[
y
]. (This follows from properties of differentiation.)
DEFINITIONS
Let
f
1
,...,
f
n
be
n
functions that are differentiable
(
n

1
)
times.
The
Wronskian matrix
is
n
×
n
square array of derivatives
M
=
f
1
f
2
···
f
n
f
0
1
f
0
2
f
0
n
.
.
.
.
.
.
.
.
.
.
.
.
f
(
n

1
)
1
f
(
n

1
)
2
f
(
n

1
)
n
,
Its determinant
W
is called the
Wronskian
[determinant].
The
m
functions
f
1
f
m
are
linearly dependent
on
I
if there
exist constants
c
1
,...
c
m
, not all zero, such that
m
X
k
=
1
c
k
f
k
(
x
)
=
0
for all
x
∈
I
. Otherwise, the functions are
linearly independent
.
(Note that in the linearly dependent case, one function is a linear
combination of the other
m

1 functions.)
THEOREMS
Existence and Uniqueness
Let
g
(
x
)
and
p
k
(
x
)
be continuous
on
I
, an interval containing
x
0
. For any choice of constants
γ
k
,
there exists a
unique
solution
y
(
x
)
on the
entire
interval
I
to the
initial value problem
y
(
n
)
(
x
)
+
n
X
j
=
1
p
j
(
x
)
y
(
n

j
)
(
x
)
=
g
(
x
),
y
(
k
)
(
x
0
)
=
γ
k
,
k
=
0
n

1
.
Representation of Solutions (Homogeneous Case)
Suppose
that
y
1
y
n
are
n
solutions on
I
of
y
(
n
)
(
x
)
+
n
X
j
=
1
p
j
(
x
)
y
(
n

j
)
(
x
)
=
0
,(
*
)
where the
p
k
are continuous on
I
. If the Wronskian of the
y
k
is
nonzero at
some
point
x
0
∈
I
,then
every
solution of (*) on
I
may
be expressed as
y
(
x
)
=
n
X
k
=
1
c
k
y
k
(
x
)
,wherethe
c
k
are constants.