HW4solutions - Differential Equations Assignment #...

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Unformatted text preview: Differential Equations Assignment # 4(Solutions) Variation of Parameters and Mechanical Vibrations 1. Use Variation of Parameters to solve the following non-homogeneous initial value problems: (a) y 00- 3 y + 2 y = 2 e t , y (0) = 0, y (0) = 0 Two homogeneous solutions to this problem are given by: y 1 = e t , y 2 = e 2 t The Wronskian of these solutions is given by: W ( y 1 ,y 2 ) = e t e 2 t e t 2 e 2 t = e 3 t The variation of parameters formula gives the solution to this problem as: y =- y 1 Z t y 2 ( s ) F ( s ) W ( y 1 ,y 2 ) ds + y 2 Z t y 1 ( s ) F ( s ) W ( y 1 ,y 2 ) ds y =- e t Z t e 2 s (2 e s ) e 3 s ds + e 2 t Z t e s ( s + 1) e 3 s ds y =- 2 e t Z t 1 ds + 2 e 2 t Z t e- s ds y =- 2 te t + 2 e 2 t- e- t + 1 y =- 2 te t- 2 e t + 2 e 2 t (b) y 00- 2 y + y = t + 1, y (0) = 0, y (0) = 0 Two homogeneous solutions to this problem are given by: y 1 = e t , y 2 = te t The Wronskian of these solutions is given by: W ( y 1 ,y 2 ) = e t te t e t e t + te t = e 2 t 1 The variation of parameters formula gives the solution to this problem as: y =- y 1 Z t y 2 ( s ) F ( s ) W ( y 1 ,y 2 ) ds + y 2 Z t y 1 ( s ) F ( s ) W ( y 1 ,y 2 ) ds y =- e t Z t se s ( s + 1) e 2 s ds + te t Z t e s ( s + 1) e 2 s ds y =- e t Z t ( s 2 + s ) e- s ds + te t Z t ( s + 1) e- s ds Integration by parts on each of these integrands yields the solution as: y =- e t- t 2 e- t- 3 te- t- 3 e- t + 3 + te t- te- t- 2 e- t + 2 y = t 2 + 3 t + 3- 3 e t +- t 2- 2 t + 2 te t y = t + 3- 3 e t + 2 te t (c) y 00- y = e- t , y (0) = 0, y (0) = 0 Two homogeneous solutions to this problem are given by: y 1 = e t , y 2 = e- t The Wronskian of these solutions is given by: W ( y 1 ,y 2 ) = e t e- t e t- e- t = 2 The variation of parameters formula gives the solution to this problem as: y =- y 1 Z t y 2 ( s ) F ( s ) W ( y 1 ,y 2 ) ds + y 2 Z t y 1 ( s ) F ( s ) W ( y 1 ,y 2 ) ds y =- e t Z t e- s ( e- t ) 2 ds + e- t Z t e s ( e- s ) 2 ds y =- e t 2 Z t e- 2 s ds + e- t 2 Z t 1 ds Integration on each of these integrands yields the solution as: y =- e t 2- e- 2 t 2 + 1 2 + e- t 2 [ t- 0] y = e- t 4- e t 4 + te- t 2 2 (d) y 00 + y = sin( t ), y ( π ) = 0, y ( π ) = 0 Two homogeneous solutions to this problem are given by: y 1 = cos( t ) , y 2 = sin( t ) The Wronskian of these solutions is given by: W ( y 1 ,y 2 ) = cos( t ) sin( t )- sin( t ) cos( t ) = 1 The variation of parameters formula gives the solution to this problem as: y =- y 1 Z t y 2 ( s ) F ( s ) W ( y 1 ,y 2 ) ds + y 2 Z t y 1 ( s ) F (...
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This homework help was uploaded on 04/08/2008 for the course MATH 527 taught by Professor Boucher during the Fall '07 term at New Hampshire.

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HW4solutions - Differential Equations Assignment #...

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