98
1 Solutions
30.
(
1
;
1
)
31.
(3
;
1
)
32.
(
1
;
2)
,
(
2
;
0)
,
(0 2)
,
(2
;
1
)
33.
The initial condition occurs at
t
= 0
which is precisely where
a
2
(
t
) =
t
2
has a zero. Theorem 6 does not apply.
34.
In this case
y
(
t
0
) = 0
and
y
0
(
t
0
) = 0
. The function
y
(
t
) = 0
,
t
2
I
is a
solution to the initial value problem. By the uniqueness part of Theorem
6
y
= 0
is the only solution.
35.
The assumptions say that
y
1
(
t
0
) =
y
2
(
t
0
)
and
y
0
1
(
t
0
) =
y
0
2
(
t
0
)
. Both
y
1
and
y
2
therefore satisfies the same initial conditions. By the uniqueness
part of Theorem 6
y
1
=
y
2
.
Section 4.2
1.
dependent;
2
t
and
5
t
are multiples of each other.
2.
independent
3.
independent
4.
dependent;
e
2
t
+ 1 =
e
1
e
2
t
and
e
2
t
3
=
e
3
e
2
t
, they are multiples of each
other.
5.
independent
6.
dependent;
ln
t
2
= 2 ln
t
and
ln
t
5
= 5 ln
t
, they are multiples of each other.
7.
dependent;
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 Fall '08
 BELL,D
 Boundary value problem

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