I am really not an expert in mathematics. I'd like to solve the problem
$$ 2y_n = 11 y_{n-1} -20y_{n-2}+12y_{n-3}+2 $$
with $y_0 = \frac{9}{8}$, $y_1 = \frac{-5}{4}$ and $y_2 = \frac{-1}{2}$
Could you explain to me how to solve such problem?
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keep plugging in the values and get the next one ... – Jul 31 '17 at 21:02
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Let's use symbolic differentiation to get a homogeneous recurrence relation. Note that we have both $$2y_n-11y_{n-1}+20y_{n-2}-12y_{n-3}-2 = 0$$ and $$2y_{n-1}-11y_{n-2}+20y_{n-3}-12y_{n-4}-2 = 0$$ Subtracting the left-hand side of the second equation from the first gives us $$2y_n-13y_{n-1}+31y_{n-2}-32y_{n-3}+12y_{n-4} = 0$$ This has the characteristic equation $$2r^4-13r^3+31r^2-32r+12 = (r-2)^2(2r-3)(r-1) = 0$$ so our solution will have the form $$y_n = (c_1+c_2n)2^n+c_3(3/2)^n+c_4$$ At this point, you can plug in values for $y_n$ to get a linear system in the coefficients, which you can solve to find $$y_n = \left(-\frac{327}{8}+\frac{33n}{4}\right)2^n+44\cdot(3/2)^n-2$$
Michael L.
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I used the rational root theorem to check for possible rational roots (if none or only one of the roots had been rational, I would probably just have introduced constants $r_1$, $r_2$, etc. to represent the others). Fortunately, they were all rational. Of course, it's probably easier to do this. – Michael L. Jul 31 '17 at 21:26