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exam2_sol

# exam2_sol - Exam 2 Solutions 1(a If a solution is in the...

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Exam 2 Solutions 1] (a) If a solution is in the form y x = n = 0 A n x n , then substituting into the equation gives us 1 x 2 n = 2 n n 1 A n x n 2 2x n = 0 n A n x n 1 2 n = 0 A n x n = 0 , or n = 2 n n 1 A n x n 2 n = 0 2 2n n n 1  A n x n = 0 , or n = 0 n 2  n 1 A n 2 x n  2 2n n n 1  A n x n = 0 . So the condition on the coefficients A n ( n from 0 to ) is n 2  n 1 A n 2  2 2n n n 1  A n = 0 , or n 2  n 1 A n 2 − n 2  n 1 A n = 0 . This, A 0 and A 1 are arbitrary, all even A n are determined by A 0 , and all odd A n are determined by A 1 . The first independent solution is: A 0 1 x 2 x 4 / 3 x 6 / 5 x 8 / 7 ... . The second independent solution is just A 1 x (all the higher odd coefficients are zero).

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exam2_sol - Exam 2 Solutions 1(a If a solution is in the...

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