18.03 PS10 Solutions—Spring 09
1a.
Char. eq. (
λ
+ 1)
2
= 0 =
⇒
λ
=

1;
A

λI
=
A
+
I
=

λ
1

1

2

λ
=
1
1

1

1
.
We see by inspection that (
A
+
I
)
1

1
=
0
, so
x
1
=
e

t
1

1
is a solution. We guess that
the other independent solution will have the form
x
2
= (
a
t
+
b
)
e

t
for column vectors
a
=
a
1
a
2
,
b
=
b
1
b
2
. Plugging in
x
2
0
=
A
x
2
, we obtain (
A
+
I
)
a
= 0 and (
A
+
I
)
b
=
a
.
Take
a
=
1

1
. Given that choice of
a
, the condition on
b
is
b
1
+
b
2
= 1, so
b
=
1
0
works.
x
1
=
e

t
1

1
and
x
2
=
1

1
t
+
1
0
e

t
are two independent solutions. Note that the answer is far from unique.
1b.
The 2ndorder equation is
x
00
+ 2
x
0
+
x
= 0
.
Two solutions are
x
=
e

t
, y
=
te

t
.
x
x
0
=
e

t
1

1
=
x
1
;
y
y
0
=
te

t
1

1
+
e

t
0
1
=
x
2

x
1
.
2ac.
See Figure 1 for the diagram. Table follows.
Region
Geom. type
Stability
Eigenvalues
(
T, D
) definition
1
saddle
N/A (unstable)
λ
1
<
0
< λ
2
D <
0
2
node
stable = sink
λ
1
< λ
2
<
0
T <
0
,
0
< D < T
2
/
4
3
spiral
stable = sink
r
±
si
,
r <
0
T <
0
, D > T
2
/
4
4
spiral
unstable = source
r
±
si
,
r >
0
T >
0
, D > T
2
/
4
5
node
unstable = source
0
< λ
1
< λ
2
T >
0
,
0
< D < T
2
/
4
2b.
The parabola is given by (i)
λ
1
=
λ
2
, (ii)
s
= 0, and (iii)
D
=
T
2
/
4.
2d.
The equation for the boundary is
T
= 0
, D >
0. In terms of the eigenvalues, one can
say that they must be purely imaginary or that they are complex with
r
= 0. As we move
across the boundary from
T <
0 to
T >
0, we begin with spirals pointing inward.
The
spirals get tighter and tighter until, at the boundary point
T
= 0, they stop pointing inward
at all and become ellipses. Then as
T
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 Spring '09
 unknown
 Classless InterDomain Routing, bt eat cos, eat cos bt, et 1

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