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Unformatted text preview: Series solutions near Ordinary points – PS March 22, 2011 Problem 1 Find the recurrence relation in the differential equation be near the specified point. Also find the first 3 terms in each of the two solutions (unless the series terminates earlier) (1 + x ) y 00 + y = 0 , x = 0 . (1) Solution : In this problem, P ( x ) = 1 + x , and P ( x = 0) = 1 6 = 0 and therefore x = 0 is an ordinary point, and hence we have series solution which appears in the form y = ∞ X n =0 a n ( x x ) n = ∞ X n =0 a n x n . Differentiating solution form (with respect to x ), we have y = ∞ X n =1 a n nx n 1 ,y 00 = ∞ X n =2 a n n ( n 1) x n 2 . Next, plug these back into the differential equation (1 + x ) ∞ X n =2 a n n ( n 1) x n 2 + ∞ X n =0 a n x n = 0 . Following the steps we talked about, we shouldn’t have any terms multiplied by the sum mations. Therefore, distribute (1 + x ) into the first summation (which will result in 2 summations): ∞ X n =2 a n n ( n 1) x n 2 + ∞ X n =2 a n n ( n 1) x n 1 + ∞ X n =0 a n x n = 0 . Next, we inspect to see what the first term of each summation is: ∞ X n =2 a n n ( n 1) x n 2 = 2 a 2 x + ··· , ∞ X n =2 a n n ( n 1) x n +1 = 2 a 2 x 1 + ··· , (shortest summation) ∞ X n =0 6 a n x n +1 = 6 a x + ··· . We need to make all the summations of the same “length”. Remember that our options are 1 c HF, 2011 Series solutions near Ordinary points – PS March 22, 2011 1. Add zeros to the shorter ∑ ’s. 2. Take out terms from the larger ∑ ’s or a combination of above....
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 Spring '10
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 Summation, Recurrence relation, n=0, Summations

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