Question 1.
[4 points] Consider the differential equation
dx
dt
=
x

a
2
x
3
.
(a) Find all equilibria.
(b) Determine the stability of each equilibrium point.
(c) Draw the phaseline diagram for the differential equation.
Solution.
a) Equilibria occur when
f
(
x
) =
x

a
2
x
3
= 0. Factoring, we have
x
(1

a
2
x
2
) =
0 so
x
= 0
,
±
1
a
.
•
Version 1
:
a
= 2, so the answer is
x
= 0
,
±
1
2
.
•
Version 2
:
a
= 3, so the answer is
x
= 0
,
±
1
3
.
•
Version 3
:
a
= 6, so the answer is
x
= 0
,
±
1
6
.
b)
Differentiating, we have
f
0
= 1

3
a
2
x
2
f
0
(0) = 1
>
0
f
0
(
±
1
) = 1

3
a
2
1
<
0
0
1
a

1
a
Figure 1: 1(c):
x
= 0 is unstable; the other two points are stable.
Question 2.
[6 points] Consider the equations
•
Version 1
: 2
x
3

8
x
2
+ 11
x

6 = 0
.
•
Version 2
: 2
x
3

10
x
2
+ 15
x

9 = 0
.
•
Version 3
: 2
x
3

12
x
2
+ 19
x

12 = 0
.
1
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(a) Show that
x
1
= 2 (
x
1
= 3) [
x
1
= 4] is a solution of the equation.
(b) Use long division to find
x
2
and
x
3
, the other two solutions.
(c) Calculate ¯
x
2
/x
3
, where ¯
w
is the complex conjugate of
w
.
(d) Find the four roots of
x
4
=
c
2
in the form
a
+
ib
, where
a
and
b
are real.
(e) Express each root in part (d) in the form
re
iθ
with
r >
0.
Solution.
a)
•
Version 1
: When
x
= 2, we have 2(2
3
)

8(2
2
) + 11(2)

6 = 0.
•
Version 2
: When
x
= 3, we have 2(3
3
)

10(3
2
) + 15(3)

9 = 0.
•
Version 3
: When
x
= 4, we have 2(4
3
)

12(4
2
) + 19(4)

12 = 0.
b)
Long division factors the equation as
(
x

a
)(2
x
2

4
x
+ 3) = 0
.
Thus, using the quadratic formula, the other two solutions satisfy
x
=
1
±
p
16

4(2)(3)
4
=
4
±
√
8
i
4
= 1
±
i
√
2
(You might notice some similarity between this and an assignment question!)
c)
It actually doesn’t matter which solution is
x
2
and which is
x
3
, as the answer will be
the same. If
x
2
= 1

i
√
2
, then ¯
x
2
= 1 +
i
√
2
and thus
¯
x
2
x
3
=
1 +
i
√
2
1 +
i
√
2
= 1
d)
If
x
4
=
c
4
, then we have
x
2
=
c
2
or
x
2
=

c
2
. From
x
2
=
c
2
, we have
x
=
±
c
. From
x
2
=

c
2
, we have
x
=
±
ci
.
•
Version 1
:
c
= 5, so
x
=
±
5,
±
5
i
.
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 Spring '07
 MUNTEANU
 SEPTA Regional Rail, 11:11, Eigenvalue, eigenvector and eigenspace, Tier One, Scaled Composites, Scaled Composites White Knight

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