1 Solutions
41
6
4
2
0
2
4
6
4
3
2
1
0
1
2
3
4
t
Graph of
y
(
t
) =
t
+
ct
3
for various
c
2. Every solution satisfies
y
(0) = 0
. There is no contradiction to Theorem
5 since, in standard form, the equation is
y
0
=
2
t
y
1 =
F
(
t; y
)
and
F
(
t; y
)
is not continuous for
t
= 0
.
20.
1. If
F
(
t; y
) =
y
2
then
F
y
(
t; y
) = 2
y
. Both are continuous on any rectan
gle that contains
(
t
0
; y
0
)
. Hence Theorem 5 applies and implies there
is a unique solution on an interval that contains
t
0
.
2.
y
(
t
) = 0
is a solution defined for all
t
;
y
(
t
) =
1
1
t
is a solution defined
on
(
1
;
1)
.
21.
No. Both
y
1
(
t
)
and
y
2
(
t
)
would be solutions to the initial value problem
y
0
=
F
(
t; y
)
,
y
(0) = 0
. If
F
(
t; y
)
and
F
y
(
t; y
)
are both continuous near
(0
;
0)
, then the initial value problem would have a unique solution by
Theorem 5.
22.
No, Both
y
1
(
t
)
and
y
2
(
t
)
would be solutions to the initial value problem
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 Fall '08
 BELL,D
 Topology, Boundary value problem, Lefthandedness, Lipschitz continuity

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