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Exercises 5.5
which is a quadratic in
r
2
.So
r
2
=
−
23
±
√
529
−
240
6
=
−
23
±
17
6
.
Since
−
20
/
3and
−
1 are negative, the roots of the characteristic equation are
±
iβ
1
and
±
iβ
2
,
where
β
1
=
r
20
3
,β
2
=1
.
Hence
x
(
t
)=
c
1
cos
β
1
t
+
c
2
sin
β
1
t
+
c
3
cos
β
2
t
+
c
4
sin
β
2
t.
Solving the Frst equation of the system given in (5.28) for
y
,weget
y
(
t
1
4
(
D
2
+4
)
[
x
]=
1
4
±(
−
β
2
1
)
c
1
cos
β
1
t
+
(
−
β
2
1
)
c
2
sin
β
1
t
+
(
−
β
2
2
)
c
3
cos
β
2
t
+
(
−
β
2
2
)
c
4
sin
β
2
t
²
.
Next we substitute into the initial conditions. Setting
x
(0) =
−
1,
x
0
(0) = 0 yields
−
1=
c
1
+
c
3
,
0=
c
2
β
1
+
c
4
β
2
.
±rom the initial conditions
y
(0) = 0,
y
0
(0) = 0, we get
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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