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DIFFERENTIAL EQUATIONS,
Math 21b, O. Knill
Homework: section 9.3: 6,28,34,42,44,16*,20*
LINEAR
DIFFERENTIAL
EQUATIONS
WITH
CONSTANT
COEFFICIENTS.
Df
=
T f
=
f
0
is
a
linear map on the space of smooth functions
C
∞
.
If
p
(
x
) =
a
0
+
a
1
x
+
...
+
a
n
x
n
is a polynomial, then
p
(
D
) =
a
0
+
a
1
D
+
...
+
a
n
D
n
is a linear map on
C
∞
(
R
) too. We will see here how to ±nd the general solution
of
p
(
D
)
f
=
g
.
EXAMPLE. For
p
(
x
) =
x
2

x
+ 6 and
g
(
x
) = cos(
x
) the problem
p
(
D
)
f
=
g
is the di²erential equation
f
00
(
x
)

f
0
(
x
)

6
f
(
x
) = cos(
x
).
It has the solution
c
1
e

2
x
+
c
2
e
3
x

(sin(
x
) + 7 cos(
x
))
/
50, where
c
1
, c
2
are
arbitrary constants. How do we ±nd these solutions?
THE IDEA. In general, a di²erential equation
p
(
D
)
f
=
g
has many solution. For example, for
p
(
D
) =
D
3
, the
equation
D
3
f
= 0 has solutions (
c
0
+
c
1
x
+
c
2
x
2
). The constants come from integrating three times. Integrating
means applying
D

1
but since
D
has as a kernel the constant functions, integration gives a one dimensional
space of antiderivatives (we can add a constant to the result and still have an antiderivative). In order to solve
D
3
f
=
g
, we integrate
g
three times. We will generalize this idea by writing
T
=
p
(
D
) as a product of simpler
transformations which we can invert. These simpler transformations have the form (
D

λ
)
f
=
g
.
FINDING THE KERNEL OF A POLYNOMIAL IN D. How do we ±nd a basis for the kernel of
T f
=
f
00
+2
f
0
+
f
?
The linear map
T
can be written as a polynomial in
D
which means
T
=
D
2

D

2 = (
D
+ 1)(
D

2).
The kernel of
T
contains the kernel of
D

2 which is onedimensional and spanned by
f
1
=
e
2
x
.
The kernel
of
T
= (
D

2)(
D
+ 1) also contains the kernel of
D
+ 1 which is spanned by
f
2
=
e

x
.
The kernel of
T
is
therefore two dimensional and spanned by
e
2
x
and
e

x
.
THEOREM: If
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