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Chapter 5
17.
Expressed in operator notation, this system becomes
(
D
2
+5)[
x
]
−
4[
y
]=0
,
−
[
x
]+(
D
2
+2)[
y
.
In order to eliminate the function
x
(
t
), we apply the operator (
D
2
+5) to the second equation.
Thus, we have
(
D
2
+5
)
[
x
]
−
4[
y
,
−
(
D
2
)
[
x
]+
(
D
2
)(
D
2
+2
)
[
y
.
Adding these two equations together yields the diFerential equation involving the single func
tion
y
(
t
)g
ivenby
±
(
D
2
+5)(
D
2
+2)
−
4
²
[
y
⇒
(
D
4
+7
D
2
+6
)
[
y
.
The auxiliary equation for this homogeneous equation,
r
4
r
2
+6=(
r
2
+1)(
r
2
+6)=0,
has roots
r
=
±
i
,
±
i
√
6. Thus, a general solution is
y
(
t
)=
C
1
sin
t
+
C
2
cos
t
+
C
3
sin
√
6
t
+
C
4
cos
√
6
t.
We must now ±nd a function
x
(
t
) that satis±es the system of diFerential equations given in
the problem. To do this we solve the second equation of the system of diFerential equations
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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