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Chapter 4
Therefore, the auxiliary equation has a double root
−
2 and a root 2. The functions
e
−
2
t
,
te
−
2
t
,and
e
2
t
form a linearly independent solution set. Therefore, a general solution in this
problem is
z
(
t
)=
c
1
e
−
2
t
+
c
2
te
−
2
t
+
c
3
e
2
t
.
45.
By inspection, we see that
r
= 2 is a root of the auxiliary equation,
r
3
+3
r
2
−
4
r
−
12 = 0.
Dividing the polynomial
r
3
r
2
−
4
r
−
12 by
r
−
2 yields
r
3
r
2
−
4
r
−
12 = (
r
−
2)
(
r
2
+5
r
+6
)
=(
r
−
2)(
r
+2)(
r
+3)
.
Hence, two other roots of the auxiliary equation are
r
=
−
2and
r
=
−
3. The functions
e
−
3
t
,
e
−
2
t
e
2
t
are three linearly independent solutions to the given equation, and a general
solution is given by
y
(
t
c
1
e
−
3
t
+
c
2
e
−
2
t
+
c
3
e
2
t
.
47.
First we ±nd a general solution to the equation
y
0
−
y
0
= 0. Its characteristic equation,
r
3
−
r
= 0, has roots
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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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