Unformatted text preview: 40 1 Solutions Comparing
to order 3. to the exact solution, we see that the series agree up 17. Let
. Then
. Let
and
be arbitrary real numbers. Then by the mean value theorem there
is a number
in between
and
such that
It follows that
is Lipschitz on
any strip. Theorem 10 implies there is a unique solution on all of .
18. Let
. Since
and
are continuous on
it follows that
is continuous on the strip
.
Further more
is bounded: i.e. there is a number such that
,
for all
. Let
and
and
be arbitrary real numbers.
Then It follows that
10,
interval
. is Lipschitz with Lipschitz constant . By Theorem
has a unique solution on the entire 19. 1. First assume that
standard form becomes . Then is linear and in
. An integrating factor is
and multiplying both sides by
gives
. This simpliﬁes to
. Now integrate to get
or
. We observe that this
solution is also valid for
. Graphs are given below for various
values of . ...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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