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Unformatted text preview: Homework 3, part 2: Solutions and remarks on selected problems Greenberg, 4.3, 6(a). The equation is 2 x 2 y 00 + xy + x 4 y = 0. (We have multiplied through by an additional factor of x to simplify bookkeeping.) Substitution of ∑ ∞ a n x r + n in the equation leads to ∞ X n =0 [ F ( r + n ) a n + a n 4 ] x r + n = 0 , (1) where as usual, a k = 0 for integers k > 0, and where F ( r ) = 2 r ( r 1) + r = r (2 r 1). The coefficient corresponding to n = 0 in (1) is set to 0 by choosing r to satisfy the indicial equation F ( r ) = r (2 r 1) = 0. The two roots are r 1 = 1 / 2 and r 2 = 0. To find a solution corresponding to r 1 set r = 1 / 2 in (1), and set the coefficients for n ≥ 1 equal to zero, to get the recursion equation a n = 1 F ( n +(1 / 2)) a n 4 = 1 2 n ( n + 1 / 2) a n 1 = 0 , n ≥ 1 . (2) Since a n 4 = 0 if n < 4, this recursion relation implies a 1 = a 2 = a 3 = 0. Next, taking n = 4 in (2), a 4 = a / 36. Then a 5 = a 6 = a 7 = 0 because by the recursion formula they are multiples of a 1 , a 2 , and a 3 respectively. Next, for n = 8, a 8 = a 4 / (136) = a / (36 · 136) = a / 4896. It is clear that only every fourth coefficient will be nonzero. Setting a = 1, y 1 ( x ) = x 1 / 2 1 x 4 36 + x 8 4896 + a 12 x 12 + ··· ! . The recursion relations can be solved explicitly using the Gamma function to get y 1 ( x ) = x 1 / 2 ∞ X k =0 ( 1) k Γ(9 / 8) (32) k k !Γ( k +(9 / 8)) x 4 k . To find a solution corresponding to r 2 set r = 0 in (1), and set the coefficients for n ≥ 1 equal to zero, to get the recursion equation a n = 1 F ( n ) a n 4 = 1 2 n ( n 1 / 2) a n 1 = 0 , n ≥ 1 . (3) As before, since a n 4 = 0 if n < 4, this recursion relation implies a 1 = a 2 = a 3 = 0....
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This note was uploaded on 09/29/2009 for the course 650 527 taught by Professor Danocone during the Fall '06 term at Rutgers.
 Fall '06
 DanOcone

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