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Chapter 6
yields the roots
r
=1,
−
2, and 2. Thus a general solution has the form
y
(
x
)=
c
1
e
x
+
c
2
e
−
2
x
+
c
3
e
2
x
.
Next, we Fnd constants
c
1
,
c
2
,and
c
3
such that the solution satisFes the initial conditions.
Di±erentiating
y
(
x
) and substituting the initial conditions, we obtain the system
y
(0) =
(
c
1
e
x
+
c
2
e
−
2
x
+
c
3
e
2
x
)±
±
x
=0
=
c
1
+
c
2
+
c
3
=
−
4
,
y
0
(0) =
(
c
1
e
x
−
2
c
2
e
−
2
x
+2
c
3
e
2
x
)±
±
x
=0
=
c
1
−
2
c
2
c
3
=
−
1
,
y
0
(0) =
(
c
1
e
x
+4
c
2
e
−
2
x
c
3
e
2
x
)±
±
x
=0
=
c
1
c
2
c
3
=
−
19
.
Solving yields
c
1
=1
,c
2
=
−
2
3
=
−
3
.
With these coeﬃcients, the solution to the given initial problem is
y
(
x
e
x
−
2
e
−
2
x
−
3
e
2
x
.
21.
By inspection,
r
= 2 is a root of the characteristic equation,
r
3
−
4
r
2
+7
r
−
6 = 0. ²actoring
yields
r
3
−
4
r
2
r
−
6=(
r
−
2)(
r
2
−
2
r
+3)=0
.
Therefore, the other two roots are the roots of
r
2
−
2
r
+3=0,which are
r
±
√
2
i
so 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 University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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