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Unformatted text preview: 5.2.1. Consider y 00 y = 0, x = 0. We seek a series solution about x = 0. i.e. We are looking for a solution of the form y ( x ) = X n =0 a n x n . (1) It follows then that y 00 ( x ) = X n =0 n ( n 1) a n x n 2 = X n =2 n ( n 1) a n x n 2 . (The last equality follows from the fact that the first two terms of the series were 0.) Shifting indices to start from n = 0, we have y 00 ( x ) = X n =0 ( n + 2)( n + 1) a n +2 x n . (2) Plugging (1) and (2) into our differential equation, we see X n =0 (( n + 2)( n + 1) a n +2 a n ) x n = 0 . Thus as the equality above must hold for all x , ( n + 2)( n + 1) a n +2 a n = 0 for all n . That is, we have the recurrence relation a n +2 = a n ( n + 1)( n + 2) . Note for n even, a n = a n 2 ( n 1) n = a n 4 ( n 3)( n 2)( n 1) n = = a 1 2 n = a n ! . Similarly for n odd, a n = a n 2 ( n 1) n = a n 4 ( n 3)( n 2)( n 1) n = = a 1 2 3 n = a 1 n ! . We use theorem 5.3.1 to find a lower bound for the radius of convergence. In the language of the theorem, p = 0 = x n and q = 1 = 1 + x n . These trivial series clearly have an infinite radius of convergence, and so our series has a radius of convergence of at least the minimum of and . i.e. We have an infinite radius of convergence. We are free to vary a and a 1 as we please. A reasonable way to get two linearly independent solutions is to first take a = 1 and a 1 = 0 and then to take a = 0 and a 1 = 1....
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 Fall '07
 COSTIN
 Differential Equations, Equations

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