Chapter 5
From the second equation we get
x
0
−
y
0
=0
⇒
(
x
−
y
)
0
⇒
x
(
t
)
−
y
(
t
)=
c
1
⇒
x
(
t
c
1
+
c
2
e
−
2
t
,
and a general solution is given by
x
(
t
c
1
+
c
2
e
−
2
t
,y
(
t
c
2
e
−
2
t
.
5.
Writing this system in operator notation yields the system
(
D
−
1)[
x
]+
D
[
y
]=5
,
D
[
x
]+(
D
+1)[
y
]=1
.
(5.1)
We will frst eliminate the ±unction
x
(
t
), although we could proceed just as easily by eliminat
ing the ±unction
y
(
t
). Thus, we apply the operator
D
to the frst equation and the operator
−
(
D
−
1) to the second equation to obtain
D
(
D
−
1)[
x
D
2
[
y
]=
D
[5] = 0
,
−
(
D
−
1)
D
[
x
]
−
(
D
−
1)(
D
y
−
(
D
−
1)[1] = 1
.
Adding these two equations yields
{
D
(
D
−
1)
−
(
D
−
1)
D
}
[
x
±
D
2
−
(
D
2
−
1)
²
[
y
⇒
0
·
x
+1
·
y
=1
⇒
y
(
t
)=1
.
To fnd the ±unction
x
(
t
), we will eliminate
y
±rom the system given in (5.1). There±ore, we
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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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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