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Unformatted text preview: a n = 6 a n19 a n2 + 4 n ; a = 4 ,a 1 = 4 Well start by solving r 26 r + 9 = 0 and get r = 3 , 3. Thus, the homogeneous part has solution c 1 3 n + c 2 n 3 n . We now guess p n = An + B and plug in: An + B6( A ( n1) + B ) + 9( A ( n2) + B ) = 4 n ( A6 A + 9 A ) n + ( B + 6 A6 B18 A + 9 B ) = 4 n 4 An + (4 B12 A ) = 4 n So A = 1 which makes B = 3. Thus, the general solution is a n = c 1 3 n + c 2 n 3 n + n + 3. Plugging in initial conditions will give 4 = c 1 +3 so c 1 = 1 and 4 = 3 c 1 +3 c 2 +4, so c 2 =1. Thus, the solution is a n = 3 nn 3 n + n + 3...
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This note was uploaded on 08/31/2009 for the course MATH 55 taught by Professor Strain during the Summer '08 term at University of California, Berkeley.
 Summer '08
 STRAIN
 Math

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