✔ 最佳答案
y{t} - y{t-1} = 6x{t-1)
x{t} = y{t-1} + 2
So, y{t} - y{t-1} = 6 [ y{t-2} + 2 ]
y{t} = y{t-1} + 6 y{t-2} + 12
which is not homogeneous.
Subtracting it by y{t-1} = y{t-2} + 6 y{t-3} + 12
we have,
y{t} = 2 y{t-1} + 5 y{t-2} - 6 y{t-3} which is homogeneous
The characteristic equation is,
r^3 - 2r^2 - 5r + 6 = 0.
Note that since r = 1 is one of the solution.
Divide it by (r-1), we have,
(r-1)(r^2 - r - 6) = 0
(r-1)(r+2)(r-3) = 0
r = 1, -2, 3
Therefore, the general solution is,
y{t} = A + B(-2)^t + C 3^t
Given, y{0} = 1 , x{0} = 1/6, we have y{1} = 2, y{2} = 20
As a result,
y{0} = 1 = A + B + C
y{1} = 2 = A - 2B + 3C
y{2} = 20 = A + 4B + 9C
which we have 3 equations with 3 unknowns.
Upon solving, we have A = -2, B = 1, C = 2.
Therefore, the solution is,
y{t} = -2 + (-2)^t + 2 (3^t)
Since x{t} = y{t-1} + 2
x{t} = -2 + (-2)^(t-1) + 2 [ 3^(t-1) ] + 2
= (-2)^(t-1) + 2 [ 3^(t-1) ]
2010-10-05 16:58:03 補充：
Solving
y{0} = 1 = A + B + C
y{1} = 2 = A - 2B + 3C
y{2} = 20 = A + 4B + 9C
( 1 1 1 | 1 ) ( 1 0 0 | -2 )
( 1 -2 3 | 2 ) ~ ( 0 1 0 | 1 )
( 1 4 9 | 20) ( 0 0 1 | 2 )
Which gives A = -2, B = 1, C = 2
