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Unformatted text preview: 1 Solutions 41 4
3
2
1
0
−1
−2
−3
−4
−6 −4 Graph of
2. Every solution satisﬁes −2 0
t 2 4 6 for various
. There is no contradiction to Theorem 5 since, in standard form, the equation is
is not continuous for and . 20. 1. If
then
. Both are continuous on any rectangle that contains
. Hence Theorem 5 applies and implies there
is a unique solution on an interval that contains .
2. is a solution deﬁned for all ;
on is a solution deﬁned . 21. No. Both and
would be solutions to the initial value problem
,
. If
and
are both continuous near
, then the initial value problem would have a unique solution by
Theorem 5. 22. No, Both and
would be solutions to the initial value problem
,
. If
and
are both continuous near
, then the initial value problem would have a unique solution by
Theorem 5. 23. For we have and for we have we calculate . For
. To compute this limit we show the left hand and right hand limits agree. We get It follows that for
for and so ...
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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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