3f-fall2010-exam_3_solutions

3f-fall2010-exam_3_solutions - Math 3F Exam #3 Solutions...

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Unformatted text preview: Math 3F Exam #3 Solutions Laney College, Fall 2010 Fred Bourgoin 1. Use power series to solve the initial-value problem: y - xy - y = 0 , y (0) = 1 , y (0) = 1 . Since the initial conditions are given at x = 0, it makes sense to center your series at x = 0. Write your answer in sigma notation. (22 points) Solution. Let y = n =0 a n x n . Then y = n =1 na n x n 1 , y = n =2 n ( n- 1) a n x n 2 , and y - xy - y = n =2 n ( n- 1) a n x n 2- x n =1 na n x n 1- n =0 a n x n = n =0 ( n + 2)( n + 1) a n +2 x n- n =1 na n x n- n =0 a n x n = n =0 ( n + 2)( n + 1) a n +2 x n- n =0 na n x n- n =0 a n x n = n =0 [ ( n + 2)( n + 1) a n +2- na n- a n ] x n = 0 , so ( n + 2)( n + 1) a n +2- na n- a n = 0 for all n , and we have the recurrence relation a n +2 = a n n + 2 ....
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3f-fall2010-exam_3_solutions - Math 3F Exam #3 Solutions...

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