Chapter 5
17.
Expressed in operator notation, this system becomes
(
D
2
+ 5) [
x
]
−
4[
y
]
=
0
,
−
[
x
] + (
D
2
+ 2) [
y
]
=
0
.
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
]
=
0
,
−
(
D
2
+ 5
)
[
x
] +
(
D
2
+ 5
) (
D
2
+ 2
)
[
y
]
=
0
.
Adding these two equations together yields the differential equation involving the single func
tion
y
(
t
) given by
(
D
2
+ 5)(
D
2
+ 2)
−
4
[
y
] = 0
⇒
(
D
4
+ 7
D
2
+ 6
)
[
y
] = 0
.
The auxiliary equation for this homogeneous equation,
r
4
+ 7
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 find a function
x
(
t
) that satisfies the system of differential equations given in
the problem. To do this we solve the second equation of the system of differential equations
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Cos, Expression, C3 sin

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