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Exercises 5.2
This homogeneous linear equation with constant coeﬃcients has the characteristic equation
r
2
−
2
r
+ 5 = 0 with roots
r
=1
±
2
i
. Thus a general solution is
y
=
c
1
e
t
cos 2
t
+
c
2
e
t
sin 2
t.
Therefore,
x
=
(
c
1
e
t
cos 2
t
+
c
2
e
t
sin 2
t
)
0
−
(
c
1
e
t
cos 2
t
+
c
2
e
t
sin 2
t
)
=
(
c
1
e
t
cos 2
t
−
2
c
1
e
t
sin 2
t
+
c
2
e
t
sin 2
t
+2
c
2
e
t
cos 2
t
)
−
(
c
1
e
t
cos 2
t
+
c
2
e
t
sin 2
t
)
=2
c
2
e
t
cos 2
t
−
2
c
1
e
t
sin 2
t.
15.
In operator form, the system becomes
−
2
z
+(
D
−
5)[
w
]=5
t,
(
D
−
4)[
z
]
−
3
w
7
t.
We multiply the Frst equation by 3, the second equation by (
D
−
5), and add the resulting
equations.
{−
6+(
D
−
5)(
D
−
4)
}
[
z
]=3(5
t
)+(
D
−
5)[17
t
]=
−
70
t
+17
⇒
(
D
2
−
9
D
+14
)
[
z
−
70
t
.
Solving the characteristic equation,
r
2
−
9
r
+ 14 = 0, we obtain
r
,
7. Hence, a general
solution to the corresponding homogeneous equation is
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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