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Unformatted text preview: k 2 × (2) = 0 k 1 + k 2 = 0 Since c 1 = 3, k 1 × 3 1 + k 2 × (2) 1 = 3 3 k 12 k 2 = 3 Solving the system of equations in k 1 and k 2 , we get k 1 = 3 5 k 2 =3 5 Solution is c n = 3 5 · 3 n3 5 (2) n 2 • 13 See text. • 14 Characteristic equation is r 2 =4 r4 r 2 + 4 r + 4 = 0 ( r + 2) 2 = 0 r =2 is a double root Solution has the form s n = c 1 (2) n + c 2 n (2) n Since s = 0, c 1 (2) + c 2 (0)(2) = 0 c 1 = 0 Since s 1 =1, c 1 (2) 1 + c 2 (1)(2) 1 =12 c 12 c 2 =1 Solving the system of equations in c 1 and c 2 , we get c 1 = 0 c 2 = 1 2 Solution is s n = 1 2 n (2) n 3...
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 Fall '06
 Carter
 Equations, Elementary algebra, characteristic equation, FK

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