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MATH150T9

# MATH150T9 - MATH 150 Series Cheung Man Wai Tutorial 9 The...

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MATH 150 Series Cheung Man Wai Tutorial 9 The purpose of this note is to find two independent solutions around x 0 of the following P ( x ) y + Q ( x ) y + R ( x ) y = 0 , (1) where P ( x ), Q ( x ) and R ( x ) are polynomials with no common factors. We shall consider two cases respectively, say x 0 is an ordinary point (if P ( x 0 ) = 0) or a singular point (otherwise) of (1). 1. Ordinary points We determine two power series solutions for y = y ( x ) around x 0 . Let us see an example. Example 1. y + k 2 x 2 y = 0 , x 0 = 0 , k a constant . With y ( x ) = n =0 a n x n , we find that y ( x ) = n =2 n ( n - 1) a n x n - 2 and the equation becomes n =2 n ( n - 1) a n x n - 2 + k 2 n =0 a n x n +2 = 0 . By shift of index one has n =0 ( n + 2)( n + 1) a n +2 x n + k 2 n =2 a n - 2 x n = 0 . 1

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The coefficient of each power of x vanishes, thus a 2 = a 3 = 0 , ( n + 2)( n + 1) a n +2 + k 2 a n - 2 = 0 , n 2 . The recurrence relation can be equivalently written as a n +4 = - k 2 ( n + 4)( n + 3) a n , n 0 . Four sequences of coefficients, starting with a 0 , a 1 , a 2 and a 3 decouple. In par- ticular the sequences are a 0 , a 4 , a 8 , a 12 , . . . ; a 1 , a 5 , a 9 , a 13 , . . . ; a 2 , a 6 , a 10 , a 14 , . . . ; a 3 , a 7 , a 11 , a 15 , . . . . Since a 2 = a 3 = 0 , by recurrence relation the last two sequences vanish identi- cally. Starting with a 0 , we have a 0 , a 4 = - k 2 3 · 4 a 0 , a 8 =
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MATH150T9 - MATH 150 Series Cheung Man Wai Tutorial 9 The...

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