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Unformatted text preview: ACM 95b/100b Problem Set 4 Solutions Prepared by: Lei Zhang February 7, 2007 Total: 134 points Include grading section: 2 points Problem 1 (32 points) Consider the following initial value problem with variable coefficients y 00 + xy + 2 y = 0 , y (0) = 4 , y (0) = 1 . (1) a) (10 pts) Find the recurrence relation for the coefficients in a power series solution expanded around x = 0. b) (12 pts) Find the first three nonzero terms of each of the linearly independent solutions. c) (10 pts) Set the two constants to satisfy the initial conditions and plot several approximations to y ( x ) for different truncations of the series solution. Solution 1 a) Since this equation admits an analytic solution on ( , ), we know that we can write y as a power series about x = 0: y ( x ) = X n =0 a n x n . This implies that y ( x ) = X n =1 na n x n 1 y 00 ( x ) = X n =2 n ( n 1) a n x n 2 . Substituting into (1), we have X n =2 n ( n 1) a n x n 2 + X n =1 na n x n + 2 X n =0 a n x n = 0 . Adjusting so that all powers of x are equal and we can combine the sums into one term, we find X n =0 [( n + 1)( n + 2) a n +2 + ( n + 2) a n ] x n = 0 , from which we can conclude ( n + 1)( n + 2) a n +2 + ( n + 2) a n = 0 , or a n +2 = a n n + 1 . 1 0.5 1 1.5 2105 5 10 Figure 1: Series solutions to the IVP of Problem 4 of increasing accuracy are plotted in color, with a more exact numerical solution shown in black. b) Since the recurrence relation couples every other coefficient, a and a 1 are arbitrary. Setting a = 1, a 1 = 0, we obtain a solution (with y (0) = 1 and y (0) = 0) y 1 ( x ) = 1 x 2 + 1 3 x 4 + O ( x 6 ) . Setting a = 0 and a 1 = 1, we obtain a linearly independent solution (with y (0) = 0 and y (0) = 1) y 2 ( x ) = x 1 2 x 3 + 1 8 x 5 + O ( x 7 ) . c) The solution y ( x ) = 4 y 1 y 2 = 4 x 4 x 2 + 1 2 x 3 + 4 3 x 4 1 8 x 5 + O ( x 6 ) , with y 1 and y 2 given above fits the initial data. Solutions beginning with y ( x ) = 4 and adding one term successively until this solution are plotted along with a more exact numerical solution in Figure 1.successively until this solution are plotted along with a more exact numerical solution in Figure 1....
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 Winter '09
 NilesA.Pierce

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