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Chapter 4
By substituting
π
+ arctan
(
√
3
/
2
)
for
φ
in the last equation above and by requiring that
t
be
greater than zero, we obtain
t
n
=
(
π/
3) + (
n
−
1)
π
−
arctan
(
√
3
/
2
)
2
√
3
,n
=1
,
2
,
3
,... .
We see that the solution curve given by equation (4.13) above will touch the exponential
curves
y
(
t
)=
±
±
p
7
/
12
²
e
−
2
t
when we have
r
7
12
e
−
2
t
sin
±
2
√
3
t
+
φ
²
=
±
r
7
12
e
−
2
t
,
where
φ
=
π
+ arctan
(
√
3
/
2
)
. This will occur when sin
(
2
√
3
t
+
φ
)
=
±
1. Since sin
θ
=
±
1
when
θ
=(
2) +
mπ
for any integer
m
, we see that the times
T
m
, when the solution touches
the exponential curves, satisfy
2
√
3
T
m
+
φ
=
π
2
+
mπ
⇒
T
m
=
(
2) +
mπ
−
φ
2
√
3
,
where
φ
=
π
+ arctan
(
√
3
/
2
)
and
m
is an integer. Again requiring that
t
be positive we see
that
y
(
t
) touches the exponential curve when
T
m
=
(
2) + (
m
−
1)
π
−
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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