Chapter 6
5.
We can factor the auxiliary equation,
r
3
+3
r
2
+28
r
+ 26 = 0, as follows:
r
3
r
2
r
+26 = (
r
3
+
r
2
)+(2
r
2
+2
r
)+(26
r
+ 26)
=
r
2
(
r
+1)+2
r
(
r
+ 1) + 26(
r
+1)=(
r
+1)(
r
2
r
+ 26) = 0
.
Thus either
r
+1=0
⇒
r
=
−
1or
r
2
r
+26=0
⇒
r
=
−
1
±
5
i
. Therefore, a
general solution is given by
y
(
x
)=
c
1
e
−
x
+
c
2
e
−
x
cos 5
x
+
c
3
e
−
x
sin 5
x.
7.
Factoring the characteristic polynomial yields
2
r
3
−
r
2
−
10
r
−
7=(
2
r
3
r
2
)+(
−
3
r
2
−
3
r
−
7
r
−
7)
=2
r
2
(
r
+1)
−
3
r
(
r
−
7(
r
r
+ 1)(2
r
2
−
3
r
−
7)
.
Thus the roots of the characteristic equation, 2
r
3
−
r
2
−
10
r
−
7 = 0, are
r
⇒
r
=
−
1
,
2
r
2
−
3
r
−
7=0
⇒
r
=
3
±
p
3
2
−
4(2)(
−
7)
4
=
3
±
√
65
4
,
and a general solution is
y
(
x
c
1
e
−
x
+
c
2
e
(3+
√
65)
x/
4
+
c
3
e
(3
−
√
65)
x/
4
.
9.
In the characteristic equation,
r
3
−
9
r
2
+27
r
−
27 = 0, we recognize a complete cube, namely,
(
r
−
3)
3
= 0. Thus, it has just one root,
r
= 3, of multiplicity three. Therefore, a general
solution to the given di±erential equation is given by
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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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