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This question is taken directly from "Difference Equations: An Introduction with Applications",second edition, by Walter G. Kelley and Allan C. Peterson:

Solve $$8y(t+2)-6y(t+1)+y(t)=2^t$$ via the annihilator method.

Could someone show this in great detail, especially finding the homogenous equation solution for $2^t$?

Matt Groff
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3 Answers3

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With constant coefficients these are similar to differential equations.

The roots of $ 8 \lambda^2 - 6 \lambda + 1 = 0$ are $\lambda = 1/2, 1/4.$ Instead of $e^{\lambda t},$ you just take $\lambda^t.$ So our general solution with a $0$ on the right hand side is $$ A \left( \frac{1}{2} \right)^t + B \left( \frac{1}{4} \right)^t. $$ Furthermore, any two solutions to the problem with $2^t$ on the right hand side differ precisely by such a sum.

Next, we get some good luck with $2^t.$ As we can check (and have already shown in general) the number 2 is not one of the $\lambda.$ As a result, if we add in $C \cdot 2^t$ to $y(t),$ we get something nonzero, in particular $$ C \cdot 8 \cdot 4 \cdot 2^t - C \cdot 6 \cdot 2 \cdot 2^t + C \cdot 2^t = C \cdot ( 32 - 12 + 1 ) \cdot 2^t = 21 \cdot C \cdot 2^t. $$ We then need $21 C = 1.$ So, with completely general constants $A,B,$ we get $$ y(t) = A \cdot \left( \frac{1}{2} \right)^t + B \cdot \left( \frac{1}{4} \right)^t + \left( \frac{1}{21} \right) \cdot 2^t $$

Will Jagy
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    Very clean and nice, + 1! – Amzoti Mar 03 '13 at 03:55
  • @WillJagy: According to the book, the annihilator method first puts the equation in operator form; so I get $(8E^2-6E+1)y(t)=2^t$. Then we are supposed to solve $2^t$ in operator form again, which is where I'm confused. For example, another problem is $y(t+2)-7y(t+1)+6y(t)=t$. The book has this restated as $(E^2-7E+6)y(t)=t$. Then it says $t$ satisfies the homogeneous equation $(E-1)^2t=0$. I'm having troubles now finding what $2^t$ satisfies in this form. Could you walk me through this with explanation? – Matt Groff Mar 03 '13 at 14:52
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Three standard operators. Identity $$ If(t) = f(t). $$ Translation $$ Ef(t) = f(t+1) $$ Difference $$ \Delta f(t) = f(t+1) - f(t). $$ Relation $$ \Delta = E - I. $$ Your book is using $1$ instead of $I,$ no real problem. $$ E 2^t = 2^{t+1} = 2 \cdot 2^t $$ In general, $$ E \lambda^t = \lambda^{t+1} = \lambda \cdot \lambda^t. $$ $$ \Delta \lambda^t = ( \lambda - 1) \cdot \lambda^t. $$ For a constant funtion of $t,$ $$ E 1 = 1,$$ $$ I 1 = 1.$$ $$ \Delta 1 = E1 - I1 = 0.$$ $$ Et = t+1, $$ $$ It = t,$$ $$ \Delta t = t+1 - t = 1.$$ So $$ \Delta^2 t = 0.$$

Will Jagy
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  • Excellent! I found it on my own, but this is what I was looking for. So not to take away from your answer, +1. Their "annihilator method" uses your translation operator to transform the rhs into an equation. They then combine two equations with shift operators. Anyways, thank you very much! – Matt Groff Mar 03 '13 at 20:53
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Just in case anyone's wondering, here's what I get via Kelley and Peterson's "Annihilator Method":

The equation gives us an equation in shift operators (i.e. Jagy's translation operator): $$(8E^2-6E+I)y(t)=0$$. $$8(E-1/2)(E-1/4)=0$$ which is similar to Dr. Jagy's answer. Then $2^t$ satisfies the equation: $$(E-2)(2^t)=0$$

So now we have, combining the two results: $$8(E-1/2)(E-1/4)(E-2)y(t)=0$$

...which leads to... $$y(t)=c_1(1/2)^t+c_2(1/4)^t+c_3(2)^t$$

We then see that $c_1(1/2)^t+c_2(1/4)^t$ satisfies the homogeneous portion of the equation. So we substitute $y(t)=c_3(2)^t$ into the original equation/question: $$8c_3(2)^{t+2}-6c_3(2)^{t+1}+c_3(2)^t=2^t$$ $$c_3=1/21$$ Then we plug this in: $$y(t)=c_1(1/2)^t+c_2(1/4)^t+\frac{1}{21}(2)^t$$

Matt Groff
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