Unformatted text preview: 1 Solutions 23. yes, 97 . 24. Parts (1) and (2) are done by computing
where
,
, or
. Then by Theorem 3, every function of the
form
is a solution to
, where
and
are constants. If we want a solution to
with
and
, then we need to solve for
and : These equations give
the answers for Part (3). , . Particular choices of and give (3)a.
(3)b.
(3)c.
(3)d.
25.3)a.
(
(3)b.
(3)c.
(3)d.
26.3)a.
(
(3)b.
(3)c.
(3)d.
27. Write the equation in the standard form: Then , , and . These three functions are all continuous on the intervals
and
. Thus, Theorem 6
shows that if
then the unique solution is also deﬁned on the
interval
, and if
, then the unique solution is deﬁned
on
.
28. Maximal intervals are
29. where , , ...
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 Fall '08
 BELL,D
 Expression, unique solution, Maximal intervals

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