2360_2010Fall_hw2

2360_2010Fall_hw2 - Homework#2 Section 1.5 3 y = y 4 3 y(0...

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Unformatted text preview: Homework #2 Section 1.5 3. y = y 4 / 3 , y (0) = 0 (a) In this case, f ( t,y ) = y 4 / 3 f y ( t,y ) = 4 3 y 1 / 3 . Both f and f y are continous for all t and y , so by Picard’s theorem we conclude that the differential equation has a unique solution for any initial condition y ( t ) = y . In particular, there will be a unique solution passing through y (0) = 0, which we know to be y ( t ) = 0. (b) Picard’s conditions hold in the entire ty plane. 6. y = tan y , y (0) = π 2 (a) In this case, f ( t,y ) = tan y f y ( t,y ) = sec 2 y Both f and f y are continous for all t and for all y 6 = ± π 2 , ± 3 π 2 ,... , so by Picard’s theorem we conclude that the differential equation has a unique solution for any initial condition y ( t ) = y such that y 6 = ± π 2 , ± 3 π 2 ,... . In particular, there is no solution to the IVP with y (0) = π 2 . (b) Picard’s conditions hold over any rectangle with y-values between two successive odd multiples of π 2 . 8. y = y y- t , y (1) = 1 (a) In this case, f ( t,y ) = y y- t f y ( t,y ) =- t ( y- t ) 2 . Both f and f y are continuous for all t and y except when y = t , where neither function exists. Hence, there is a unique solution passing through y ( t ) = y except when t = y . The initial value problem with y (1) = 1 is not defined. 1 (b) Picard’s conditions hold for the entire ty plane except along the line y = t . Thus, it holds for any rectangle that does not contain any part of y = t ....
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This note was uploaded on 10/24/2010 for the course APPM 2360 at Colorado.

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2360_2010Fall_hw2 - Homework#2 Section 1.5 3 y = y 4 3 y(0...

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