This preview shows page 1. Sign up to view the full content.
Chapter 5
We use the second equation to express
z
in terms of
x
and
y
.
z
=
−
x
+
D
[
y
]
.
(5.10)
Substituting this expression into the other two equations, we obtain
(
D
−
1)[
x
]
−
2
y
+(
−
x
+
D
[
y
]) = 0
,
−
4
x
+4
y
D
−
5)[
−
x
+
D
[
y
]] = 0
⇒
(
D
−
2)[
x
]+(
D
−
2)[
y
]) = 0
,
−
(
D
−
1)[
x
D
2
−
5
D
+4)[
y
]=0
.
(5.11)
Now we eliminate
x
by multiplying the Frst equation by (
D
−
1), the second equation – by
(
D
−
2), and adding the results. This yields
±
(
D
−
1)(
D
−
2) + (
D
−
2)(
D
2
−
5
D
+4)
²
[
y
⇒
±
(
D
−
2)(
D
2
−
4
D
+3)
²
[
y
⇒{
(
D
−
2)(
D
−
1)(
D
−
3)
}
[
y
.
The roots of the characteristic equation, (
r
−
2)(
r
−
1)(
r
−
3) = 0, are
r
=1
,
2, and 3. Thus,
a general solution for
y
is
y
=
c
1
e
t
+
c
2
e
2
t
+
c
3
e
3
t
.
With
h
:=
x
+
y
, the Frst equation in (5.11) can be written in the form
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details