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Economics Dynamics Problems 133

# Economics Dynamics Problems 133 - 2 n 1 − 2 n y ∗ = 12...

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Discrete dynamic systems 117 In this example, 1 + a + b = 0 and so it is not possible to use y as a solution. In this instance we try a moving fixed point, ny . Thus ( n + 2) y 5( n + 1) y + 4 ny = 4 3 y = 4 y = − 4 3 . . . ny = 4 n 3 For the complementary component we need to solve the homogeneous equation y n + 2 5 y n + 1 + 4 y n = 0 whose characteristic equation is x 2 5 x + 4 = 0 with solutions r , s = 5 ± 25 16 2 = 5 ± 3 2 i.e. r = 4 and s = 1. Hence y n = c 1 4 n + c 2 1 n (4 n / 3) = c 1 4 n + c 2 (4 n / 3) Example 3.15 y n + 2 + y n + 1 2 y n = 12 y 0 = 4 and y 1 = 5 The particular solution cannot be solved for y (since 1 + a + b = 0) and so we employ ny ( n + 2) y + ( n + 1) y 2 ny = 12 ( n +
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Unformatted text preview: 2 + n + 1 − 2 n ) y ∗ = 12 y ∗ = 12 3 = 4 Hence, ny ∗ = 4 n = y p . The complementary component is derived by solving the characteristic equation x 2 + x − 2 = ( x + 2)( x − 1) = giving r = 1 and s = − 2. Giving the complementary component o± y c = c 1 r n + c 2 s n = c 1 (1) n + c 2 ( − 2) n = c 1 + c 2 ( − 2) n Hence, the general solution is y n = y c + y p = c 1 + c 2 ( − 2) n + 4 n...
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