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Chapter 6
The roots of the auxiliary equation,
r
2
+4
r
+ 5 = 0, for this second order equation are
r
=
−
2
±
i
. Therefore,
{
w
1
(
x
)
,w
2
(
x
)
}
=
±
e
−
2
x
cos
x, e
−
2
x
sin
x
²
form a fundamental solution set. Integrating, we get
v
1
(
x
)=
Z
w
1
(
x
Z
e
−
2
x
cos
xdx
=
e
−
2
x
(sin
x
−
2cos
x
)
5
,
v
2
(
x
Z
w
2
(
x
Z
e
−
2
x
sin
=
−
e
−
2
x
(2 sin
x
+cos
x
)
5
,
where we have chosen integration constants to be zero. Thus, functions
f
(
x
e
2
x
,
y
1
(
x
v
1
(
x
)
f
(
x
e
−
2
x
(sin
x
−
x
)
5
e
2
x
=
sin
x
−
x
5
,
y
2
(
x
v
2
(
x
)
f
(
x
e
−
2
x
(2 sin
x
x
)
5
e
2
x
=
2sin
x
x
5
are three linearly independent solutions to the given diFerential equation.
35.
±irst, let us evaluate the Wronskian of the system
{
x,
sin
x,
cos
x
}
to make sure that the result
of Problem 34 can be applied.
W
[
x,
sin
x,
cos
x
]=
³
³
³
³
³
³
³
³
x
sin
x
cos
x
1c
o
s
x
−
sin
x
0
−
sin
x
−
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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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