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Chapter 5
Thus, a general solution to the given system is
u
(
t
)=
c
1
−
1
2
c
2
e
−
t
+
1
2
e
t
+
5
3
t,
v
(
t
c
1
+
c
2
e
−
t
+
5
3
t.
9.
Expressed in operator notation, this system becomes
(
D
+2)[
x
]+
D
[
y
]=0
,
(
D
−
1)[
x
]+(
D
−
1)[
y
]=s
i
n
t.
In order to eliminate the function
y
(
t
), we will apply the operator (
D
−
1) to the Frst equation
above and the operator
−
D
to the second one. Thus, we have
(
D
−
1)(
D
x
D
−
1)
D
[
y
]=(
D
−
1)[0] = 0
,
−
D
(
D
−
1)[
x
]
−
D
(
D
−
1)[
y
]=
−
D
[sin
t
−
cos
t.
Adding these two equations yields the di±erential equation involving the single function
x
(
t
)
given by
±
(
D
2
+
D
−
2)
−
(
D
2
−
D
)
²
[
x
−
cos
t
⇒
2(
D
−
1)[
x
−
cos
t.
(5.3)
This is a linear Frst order di±erential equation with constant coeﬃcients and so can be solved
by the methods of Chapter 2.
(See Section 2.3.)
However, we will use the methods of
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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