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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 2 FALL 2007 Some of these solutions were written by Professor Dan Ocone. 4.3.6 (a) The equation is 2 x 2 y + 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 summationdisplay n =0 [ ( r + n ) a n + a n 4 ] x r + n = 0 , (13.1) where as usual, a k = 0 for integers k > 0, and where ( r ) = 2 r ( r 1) + r = r (2 r 1). (If you dont like the trick of introducing a 1 , a 2 , etc., you can write this instead as ( r ) a = 0; ( r + 1) a 1 = 0; ( r + 2) a 2 = 0; ( r + 3) a 3 = 0; summationdisplay n =4 [ ( r + n ) a n + a n 4 ] x r + n = 0 . ) The coefficient corresponding to n = 0 in (13.1) is set to 0 by choosing r to satisfy the indicial equation ( 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 (13.1), and set the coefficients for n 1 equal to zero, to get the recursion equation a n = 1 ( n + (1 / 2)) a n 4 = 1 n (2 n + 1) a n 4 = 0 , n 1 . (13.2) Since a n 4 = 0 if n < 4, this recursion relation implies a 1 = a 2 = a 3 = 0. Next, taking n = 4 in (13.2), 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 parenleftbigg 1 x 4 36 + x 8 4896 + a 12 x 12 + parenrightbigg . The recursion relations can be solved explicitly using the Gamma function to get y 1 ( x ) = x 1 / 2 summationdisplay k =0 ( 1) k (9 / 8) (32) k k !( k + (9 / 8)) x 4 k . To find a solution corresponding to r 2 set r = 0 in (13.1), and set the coefficients for n 1 equal to zero, to get the recursion equation a n = 1 ( n ) a n 4 = 1 n (2 n 1) a n 1 = 0 , n 1 . (13.3) As before, since a n 4 = 0 if n < 4, this recursion relation implies a 1 = a 2 = a 3 = 0. Next, taking n = 4 in (13.3), a 4 = a / 28. Again, a 5 = a 6 = a 7 = 0 because by the recursion formula they are multiples of a 1 , a 2 , and a 3 respectively, and, continuing, all 1 642:527 SOLUTIONS: ASSIGNMENT 2 FALL 2007 coefficients are zero except those corresponding to indices n = 4 k for integers k . For n = 8, a 8 = a 4 / 120 = a / (28 120) = a / 3360. Setting a = 1, y 2 ( x ) = 1 x 4 28 + x 8 3360 + . The recursion relations can be solved explicitly using the Gamma function to get y 2 ( x ) = summationdisplay k =0 ( 1) k (7 / 8) (32) k k !( k + (7 / 8)) x 4 k ....
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This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.
 Fall '07
 SPEER

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