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Exercises 1.2
–10
10
–0.5
0.5
C
=2
C
=1
C
=0
.
5
C
=0
C
=
−
0
.
5
C
=
−
1
C
=
−
2
Figure 1–A
: Graphs of the functions
y
=
Ce
3
x
+1for
C
=0,
±
0
.
5,
±
1, and
±
2.
17.
DiFerentiating
φ
(
x
), we ±nd that
φ
0
(
x
)=
±
2
1
−
ce
x
²
0
=
³
2(1
−
ce
x
)
−
1
´
0
=2
(
−
1) (1
−
ce
x
)
−
2
(1
−
ce
x
)
0
ce
x
(1
−
ce
x
)
−
2
.
(1.2)
On the other hand, substitution of
φ
(
x
)for
y
into the righthand side of the given equation
yields
φ
(
x
)(
φ
(
x
)
−
2)
2
=
1
2
2
1
−
ce
x
±
2
1
−
ce
x
−
2
²
=
2
1
−
ce
x
±
1
1
−
ce
x
−
1
²
=
2
1
−
ce
x
1
−
(1
−
ce
x
)
1
−
ce
x
=
2
ce
x
(1
−
ce
x
)
2
,
which is identical to
φ
0
(
x
) found in (1.2).
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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 University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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