Chapter 4
3.
The auxiliary equation,
r
2
−
6
r
+10=0,hasroots
r
=
(
6
±
√
6
2
−
40
)
/
2=3
±
i
.So
α
=3,
β
=1
,and
z
(
t
)=
c
1
e
3
t
cos
t
+
c
2
e
3
t
sin
t
is a general solution.
5.
This diferential equation has the auxiliary equation
r
2
+4
r
+6 = 0. The roots oF this auxiliary
equation are
r
=
(
−
4
±
√
16
−
24
)
/
2=
−
2
±
√
2
i
.W
es
e
etha
t
α
=
−
2and
β
=
√
2. Thus,
a general solution to the diferential equation is given by
w
(
t
)=
c
1
e
−
2
t
cos
√
2
t
+
c
2
e
−
2
t
sin
√
2
t.
7.
The auxiliary equation For this problem is given by
4
r
2
−
4
r
+26=0
⇒
2
r
2
−
2
r
+13=0
⇒
r
=
2
±
√
4
−
104
4
=
1
2
±
5
2
i.
ThereFore,
α
=1
/
2and
β
=5
/
2. Thus, a general solution is given by
y
(
t
)=
c
1
e
t/
2
cos
±
5
t
2
²
+
c
2
e
t/
2
sin
±
5
t
2
²
.
9.
The associated auxiliary equation,
r
2
−
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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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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