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Verify that the signals in Exercises 1 and 2 are solutions of the accompanying difference equation..$$2^{k},(-4)^{k} ; y_{k+2}+2 y_{k+1}-8 y_{k}=0$$

the signal $y _ { k } = ( - 4 ) ^ { k }$ is a solution of the difference equation $y _ { k + 2 } + 2 y _ { k + 1 } - 8 y _ { k } = 0$

Calculus 3

Chapter 4

Vector Spaces

Section 8

Applications to Difference Equations

Vectors

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In this example, we have a second order linear difference equation that's provided, which is also homogeneous. What we're going to do in this example is take first y que equals two to the power of K and show that this is a solution. To do that, we're going to be substituting this sequence y que into the difference equation displayed here will obtain. That's why K plus two will be two to the Power K plus two plus two times why K plus one is two to the power of K plus one minus eight times y que, which is two to the power of K. Now, one thing we can do here is factor out to to the power of K. If we do that, we're left with two squared plus two times to the power of one minus eight. Then this becomes altogether ah four plus four, my estate resulting in zero so altogether this equals zero and that tells us that why K equals to the power K is indeed a solution. Next, let's try a different potential solution. Let's consider why K equals this time negative four to the power of K in this case If we substitute into the difference equation, we get first. Why K Plus two, which is negative for to the power of K plus two. Then the next part of the difference equation is plus two times y to the power of K plus one. But that's negative. Four to the Power K plus one the next we have minus eight times y que But why K is altogether negative four to the power of K. Now, if we can show this one equals zero, we would have verified a second solution. So again, a decent strategy here is to factor out negative four Power K. This leaves us behind negative four to the power of to then from the second term will have plus two times negative four to the power of one and then just a negative eight for the second term. Now, if we simplify on the inside, we have negative four square to 16 plus two times negative. Four is negative. Eight minus eight. And so inside the brackets we have a zero. It's all together. We have zero again, and that tells us that the seconds sequence is also a solution to this given difference equation.

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