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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. $$y_{k+2}-y_{k+1}+\frac{2}{9} y_{k}=0$$

the set $\left[ \left( \frac { 2 } { 3 } \right) ^ { k } , \left( \frac { 1 } { 3 } \right) ^ { k } \right\}$ 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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for this example, we have a second order difference equation that were considering, and what we're going to do here is for the set H, which is the set of all solutions. We want to find a basis. So to do that, we're going to convert our difference equation into theocracy Ilary equation, which is R squared minus are plus 2/9 equal zero. To solve this equation, I'll take the step of multiplying both sides by nine. So we now have nine R squared minus nine R plus two is zero, and this is becomes an equation which weaken factor pretty quickly. Nine r squared becomes three R and three are in these groups. And if we place a minus one and a minus to the factory, ization is complete. So we now have solutions to the auxiliary equation. They are r equals 1/3 coming from the first factor and 2/3 from the second factor. What this tells us then, is that this set, which is going to be 1/3 to the power K and 2/3 to the power of K, is a set of solutions. Two are provided difference equation. Now, what's special about this set of solutions is that it is linearly independent. Weaken say this set is linearly independent in a church, since the signal's are not multiples of each other. This method for determining independence works with having to signals. But for three signals we would have to look at a Casarotto matrix and then go from there. So now let's pause for a moment. We have a difference equation, which is of order to do to that set our that subscript there. And H is a subspace that contains all the solutions. This tells us that the dimension of H is equal to equal to two, and this implies because this set is linearly independent, that 1/3 to the power of K and 2/3 to the power of K is a basis four h because if the dimension of H is too and we already have a linearly independent set that immediately implies this set spans to form a basis

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