Chapter 5
Thus, a general solution to the given system is
u
(
t
) =
c
1
−
1
2
c
2
e
−
t
+
1
2
e
t
+
5
3
t,
v
(
t
) =
c
1
+
c
2
e
−
t
+
5
3
t.
9.
Expressed in operator notation, this system becomes
(
D
+ 2)[
x
] +
D
[
y
]
=
0
,
(
D
−
1)[
x
] + (
D
−
1)[
y
]
=
sin
t.
In order to eliminate the function
y
(
t
), we will apply the operator (
D
−
1) to the first equation
above and the operator
−
D
to the second one. Thus, we have
(
D
−
1)(
D
+ 2)[
x
] + (
D
−
1)
D
[
y
] = (
D
−
1)[0] = 0
,
−
D
(
D
−
1)[
x
]
−
D
(
D
−
1)[
y
] =
−
D
[sin
t
] =
−
cos
t.
Adding these two equations yields the differential equation involving the single function
x
(
t
)
given by
(
D
2
+
D
−
2)
−
(
D
2
−
D
)
[
x
] =
−
cos
t
⇒
2(
D
−
1)[
x
] =
−
cos
t.
(5.3)
This is a linear first order differential equation with constant coeﬃcients and so can be solved
by the methods of Chapter 2.
(See Section 2.3.)
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Derivative, Elementary algebra

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