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1
Lecture 18
1.1
Nonhomogeneous linear ODE of order
n
Let
L
=
D
n
+
p
1
(
x
)
D
n
1
+
p
2
(
x
)
D
n
2
+
:::
+
p
n
1
(
x
)
D
+
p
n
(
x
)
D
0
:
Consider nonhomogeneous ODE
y
(
n
)
+
p
1
(
x
)
y
(
n
1)
+
p
2
(
x
)
y
(
n
2)
+
:::
+
p
n
(
x
)
y
(
x
) =
f
(
x
)
, or
L
[
y
] =
f:
The complementary homogeneous equation is
y
(
n
)
+
p
1
(
x
)
y
(
n
1)
+
p
2
(
x
)
y
(
n
2)
+
:::
+
p
n
(
x
)
y
(
x
) = 0
, or
L
[
y
] = 0
:
Solutions of complementary equation are called complementary solutions
y
c
(
x
)
. So,
L
[
y
c
] = 0
. Let
y
p
(
x
)
homogeneous equation
L
[
y
] =
f
. Hence,
L
[
y
p
] =
f
. Then
y
(
x
) =
y
c
(
x
) +
y
p
(
x
)
is a solution of the nonhomogeneous equation. Indeed,
L
[
y
c
+
y
p
] =
L
[
y
c
] +
L
[
y
p
] = 0 +
f
(
x
) =
f
(
x
)
:
Notice, that if
y
c
is a general solution of the complimentary equation, then,
there are
n
constants in the formula for
y
c
. Thus,
y
(
x
) =
y
c
(
x
) +
y
p
(
x
)
contains
n
constants
C
1
; C
2
; :::; C
n
. It can be showed that by selecting the
constants
C
1
; C
2
; :::; C
n
, we can solve a given initial value problem.
Example 1
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This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Staff
 Differential Equations, Equations

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