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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 shouldnt 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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This note was uploaded on 05/02/2011 for the course GE 207K taught by Professor None during the Spring '10 term at University of Texas at Austin.
 Spring '10
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