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Ordinary Diff Eq Exam Review Solutions 39

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

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1 Solutions 41 -6 -4 -2 0 2 4 6 -4 -3 -2 -1 0 1 2 3 4 t Graph of y ( t ) = t + ct 3 for various c 2. Every solution satisfies y (0) = 0 . There is no contradiction to Theorem 5 since, in standard form, the equation is y 0 = 2 t y 1 = F ( t; y ) and F ( t; y ) is not continuous for t = 0 . 20. 1. If F ( t; y ) = y 2 then F y ( t; y ) = 2 y . Both are continuous on any rectan- gle that contains ( t 0 ; y 0 ) . Hence Theorem 5 applies and implies there is a unique solution on an interval that contains t 0 . 2. y ( t ) = 0 is a solution defined for all t ; y ( t ) = 1 1 t is a solution defined on ( 1 ; 1) . 21. No. Both y 1 ( t ) and y 2 ( t ) would be solutions to the initial value problem y 0 = F ( t; y ) , y (0) = 0 . If F ( t; y ) and F y ( t; y ) are both continuous near (0 ; 0) , then the initial value problem would have a unique solution by Theorem 5. 22. No, Both y 1 ( t ) and y 2 ( t ) would be solutions to the initial value problem
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