Exercises 5.5
By solving these equations simultaneously, we find
c
1
=
−
1
,
c
2
=
−
1
,
c
3
=
−
1
,
and
c
4
= 0
.
Therefore, the solution to this springmassdashpot system is
x
(
t
) =
−
e
−
t
−
te
−
t
−
cos
t,
y
(
t
) =
e
−
t
+
te
−
t
−
cos
t.
7.
In operator notations,
(
D
2
+ 5
)
[
x
]
−
2
y
= 0
,
−
2
x
+
(
D
2
+ 2
)
[
y
] = 3 sin 2
t.
Multiplying the first equation by (
D
2
+ 2) and the second equation by 2, and adding the
results, we obtain
(
D
2
+ 2
) (
D
2
+ 5
)
−
4
[
x
] = 6 sin 2
t
⇒
(
D
4
+ 7
D
2
+ 6
)
[
x
] = 6 sin 2
t
⇒
(
D
2
+ 1
) (
D
2
+ 6
)
[
x
] = 6 sin 2
t .
(5.35)
Since the characteristic equation, (
r
2
+ 1)(
r
2
+ 6) = 0, has the roots
r
=
±
i
and
r
=
±
i
√
6, a
general solution to the corresponding homogeneous equation is given by
x
h
(
t
) =
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Sin, The Roots, D2

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