Ordinary Diff Eq Exam Review Solutions 39

Ordinary Diff Eq Exam Review Solutions 39 - 1 Solutions 41...

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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.

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