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In Exercises $13-16,$ find a basis for the solution space of the difference equation. Prove that the solutions you find span the solution set. $$y_{k+2}-25 y_{k}=0$$

the set $\left[ \left\{ 5 ^ { x } , ( - 5 ) ^ { k } \right\} \right]$ forms basis for the solution space of the difference equation.

Calculus 3

Chapter 4

Vector Spaces

Section 8

Applications to Difference Equations

Vectors

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in this example, we have a difference equation that's provide here. It's homogeneous due to the equal zero portion that we see, and we're going to let h denote the subset that which contains all solutions to this equation. Now what we want to do is find a basis for age. And in order to do that, we need to convert our difference equation seen here into an auxiliary equation. That auxiliary equation will be R squared, coming from Y to the power of K Plus two minus 25 we set this equal to zero. But this immediately implies that solutions to the AUC's early equation R r equals five and negative five. So from this work, we can now go back to the original difference equation and assert that the set, which is going to be five to the power of K negative fighter the power of K contains solutions to the equation. And when we say the equation, we're referring to the original difference equation here. So now that we have a set of solutions, were one step closer to determining if we have a basis to do that, let's take a look at the form of these solutions notice with five to the power K and a negative on the inside. These are not multiples of each other. If the negative here was on the outside the parentheses, then they would be in multiples of each other. So it's a state next that these solutions are not multiples off each other. And so we know whenever we have two vectors coming from a vector space H in this case and we have that they're not multiples of each other. We can conclude that they are linearly independent so we can write. And so the set is Lee nearly independent? Well, next before we determining fully determined whether this set is a basis, let's consider this difference equation here. The difference equation is of order to where to came from this to that is right here, which gave us our squared in the auxiliary equation. So now that the difference equation is of order of to this implies that the dimension of age is also to But then if we have a set, as we do here, which is already known to be linearly independent and are set, contains two vectors where the dimension of the target spaces to that immediately implies that the SAT must span H But then, with linear independence and spanning, we can conclude that the set fight of the power of K negative fighter the power of K is a basis for the set H and that completes our solution to this problem.

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