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Winter 2006 - Reynold's Class - Exam 2 (Version A)

# Winter 2006 - Reynold's Class - Exam 2 (Version A) - Exam 2...

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Unformatted text preview: Exam 2, Version A Math 20D, Lecture C, Winter 2006 6 March 2006 Name: ID #: Section Time: Read all of the following information before starting the exam: • Show all work, clearly and in order, if you want to get full credit. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct). • Justify your answers algebraically whenever possible to en- sure full credit. When you do use your calculator, sketch all relevant graphs and explain all relevant mathematics. • Circle or otherwise indicate your final answers. • Please keep your written answers brief; be clear and to the point. I will take points off for rambling and for incorrect or irrelevant statements. • This test has 4 problems and is worth 100 points. It is your responsibility to make sure that you have all of the pages! • Good luck! # Score 1 2a 2b 2c 2d 3a 3b 4a 4b Total 1. [ 25 points ] (Variation of Parameters) Given the two linearly independent solutions y 1 ( t ) = t, y 2 ( t ) = te t , to the linear, second-order homogeneous equation, t 2 y 00- t ( t + 2) y + ( t + 2) y = 0 , t > , determine the general solution to the non-homogeneous ordinary differential equation, t 2 y 00- t ( t + 2) y + ( t + 2) y = t 3 e 2 t , t > . Solution: In order to use the variation of parameters method, the differential equation must be in the form y 00 + p ( t ) y + q ( t ) y = g ( t ) . Since t > , we may divide every term by t 2 to convert the equation to the proper form, y 00- t + 2 t y + t + 2 t 2 y = te 2 t , Thus our non-homogeneous term is g ( t ) = te 2 t (5 pts for using the correct g ( t ) ). Given the two solutions y 1 ( t ) = t , y 2 ( t ) = te t , the Wronskian is given by (3 pts for correct Wronskian) W ( y 1 , y 2 )( t ) = y 1 y 2- y 1 y 2 = te t + t 2 e t- te t = t 2 e t . We now have all of the necessary components to use the variation of parameters formula for the particular solution Y ( t ) (7 pts for correct setup, 5 points for correct Y ( t ) ): Y ( t ) =- y 1 ( t ) Z y 2 ( t ) g ( t ) W ( y 1 , y 2 )( t ) dt + y 2 ( t ) Z y 1 ( t ) g ( t ) W ( y 1 , y 2 )( t ) dt (1) =- t Z te t ( te 2 t )...
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Winter 2006 - Reynold's Class - Exam 2 (Version A) - Exam 2...

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