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Unformatted text preview: 1 Solutions 7 22. 23. 24. We ﬁrst calculate Observe that
deﬁned are so is not deﬁned at
and
. so the two intervals where is 25. The denominator of
is deﬁned are is when
and
. . Thus the two intervals where 26. This is a diﬀerential equation we can solve by simple integration: We get
.
. 27. Integration gives
28. Integration (by parts) gives
29. Observe that . . Integration gives 30. We integrate two times. First,
. .
. Second, 31. We integrate two times. First,
.
32. From Problem 19 the general solution is
. It follows that . Second,
. At
and we get
. ...
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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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