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Consider the recurrence relation6" for 2 2_Find the general solution. Find the solution given aoand 01-201-! 301-4...

Question

Consider the recurrence relation6" for 2 2_Find the general solution. Find the solution given aoand 01-201-! 301-4

Consider the recurrence relation 6" for 2 2_ Find the general solution. Find the solution given ao and 01 -201-! 301-4



Answers

Find all solutions of the recurrence relation $a_{n}=$ $5 a_{n-1}-6 a_{n-2}+2^{n}+3 n .[\text { Hint: Look for a particular solution}$
of the form $q n 2^{n}+p_{1} n+p_{2},$ where $q, p_{1},$ and $p_{2}$ are constants. $]$

You're given a recurrence relation from your ass Defying to the solution of this relation their occurrence relation is a N equals 2 a.m. minus one plus three times two to the end. This is a linear, non homogeneous recurrence Relation the associate ID When your ma Jenness recurrence relation is a N equals 2 a.m. minus one And here we have the characteristic equation is ar minus two equals zero said the characteristic route. It's simply r equals two we have that the general solution to this homogeneous equation has the form Alfa Times two to the end where Alfa is some constant, we have the FN. The function of end is three times two to the end three's a polynomial of degree zero and we have it too, is a characteristic group of multiplicity one and so it follows. That's the general form. Very particular solution. It's going to be end times p zero times two to the end and to find the value the coefficients we put particular solution back into the non homogeneous equation. So we have p zero end times two to the end is equal to two times p zero times n minus one times two to the N minus one plus three times two to the end. If we divide those sides by to to the n minus one, we get two he not and is equal to to p zero and minus two p zero plus six in writing this on one side we have zero is equal to two p zero ends, cancel out and we have negative two p zero plus six. And so it follows that p zero is going to be equal to three so that the particular solution is three n times two to the end and therefore we have the general solution for this non homogeneous equation is the general solutions. The modernise equation must particular solution and this is equal to Alfa Times two to the end, plus three end times two to the end.

So here we have the t r Function T is gonna be equal to end at times t squared if I did it on over to and initial condition T one is equal to six. So first put him in a substitute and equals two to the power of K. And so this becomes t of to the power of K is gonna be equal to two to the power of K Times T is squared to the power of K over too. Now, which, of course, is gonna be equal to two out of K times T square Heard it too. K minus one. Um and they're gonna make another substitution here. Where were you? Uh, a Okay, equal blob thio t two to the power of K. And when we do that, we get that this is equal to Bob, too. Two to the power of K Times T squared to K minus one. Yeah, we can use some of our log rules. We get love two times two to the K plus Ah, two times allowed. Two of key to the Caymans. One square. Okay. And this is gonna be equal to okay. Plus two, uh, two tee times two to the cameras, one power. Um and then this is equal to K plus two. Ah said K minus one. Yes, that's gonna be a lot easier to work with. What we've made The substitution, sze and for simplicity, we can really write this as, uh, a n a k would be a n is equal to end plus 2 a.m. I s water. I shouldn't be in the same form that we're used to you. I wanted to translate our initial conditions as well, so we would get so gonna rewrite. To get a N is equal to to a and minus one plus on. And then we would have a knot is equal to log to t to the two to the power of zero, right, which is gonna be just one two to the power becomes t one, which we know. Um, from what we're given is that that is really just steaks. That's what we want to find the roots characteristic equation here on when we do that, we get that our homogeneous portion is gonna be equal to. And so I would say our square is equal to two. Are is equal to two when we set an equal to our own and my 21 s of the Alfa Times two to the power of n And then we want to move on to the particular solution. Uh, in this case, our function is just equal to end. Uh, which is the same as one? Well, to find that by one of the power of an and one is not a route. So we'll take that form of our particular solution but should be equal to P one times and plus p not times one of the part of end, uh, which is just p one times and plus p not. And then now we need Thio Saul for this And make sure it satisfies ari current relation here. And so we will rewrite it. Um, we have a n equals two times a and minus one plus n, but put in our particular solution p one times n plus p not and for a m. We'll put it in for a M s. One is well, so we get to times p one times and minus one plus peanut plus n. We can extend that. So it's gonna be two p one times and the minus two p one plus two p not plus n. I think in some tracks from the left hands high and the rice we would get zero is equal to p one times end minus two p one plus p not plus. And so I get up the ends and the not ends we get. Zero is equal to P one plus one times n plus negative two p one plus peanut, and these coefficients must be equal to zero. So we'll just set them equals zero and solve. And when we do that, we have that p one plus one is equal zero and negative. Two p one plus peanut is equal zero. So the first one's pretty easy, right? So here we get that p one is equal to negative one. We put negative one in here and we end up with two. Plus, peanut is equal to zero. Assis means that peanut is equal to negative too. So we'll put that been to our particular solution, right? Sue? Our particular solution was looking like p one times on the plus p not and so we know this will be negative and minus two. The entire solution then, is going to be our home on genius portioned plus our particular portion. So that is going to be Alfa Times two to the power and minus minus and minus two. So we'll use our initial conditions to evaluate it at an equal zero and fine Alfa. And so at n equals zero we get Alfa minus two, which was equal to a knot was going to be lot two of six. Okay, so that means Alfa is equal to God. Thio six plus two eso es and is going to be We'll have to six plus two times two to the power and minus end minus two. Okay, but now we need to put this back in two terms of tea. Pray and so a N is really equal to log thio t to to the end. But we had called it okay previously, so that is finding their way. Eso we'll say log two t uh, two to the power of K. It's gonna be equal to love to six times to power and plus two to the n plus one. So I just before I had both those multiplied by that to the part of N. And now I have just multiply it out those components a minus and minus two. OK, we could take the exponential of both sides. So we get T T k is equal to two to the power of Glod to six times to the end, plus two plus one minus and minus two. I suppose we shouldn't use thes ks. Okay. And when we do that, this is going to be equal to six to the power of two out of K times two to the power of twos of Howard K plus one times two to the power of negative K times two to the power of negative too. And over the algebra this is equal to, um, can put thes below rights would be six the tower of two to the power of K times two part of two to power K plus one over two to the power of K a times two squared, and that is going to be equal to six to the powder toot out of K times four to the power of two to the power of K over four times two to the power of canceling up here for about a two to the power of two. Uh, so definitely messy. For the moment, we have t two to the power of K being equal to six to the power of two to the part of K times four to the power of two to the power of K over four times two to the power of care. That's enough. We let end be equal to two to the power of K. So that was our first substitution we made. We end up with T power tea. Then it is equal. Thio, what we have up here. Ah, we get six and times for the power of n over four times n which will simplify as six end times for the power N minus one over and

Were given a recurrence relation with initial conditions and were asked to solve this recurrence relation with these initial conditions, they're currents. Relation is a N equals 6 a.m. minus one minus 12 a.m. minus to plus 8 a.m. minus three and the initial conditions are a zero equals negative. Five Anyone calls for a two equals 88. This is a linear, homogeneous, recursive relation. And so the characteristic equation is our cubed minus six r squared plus 12 or minus eight equals zero. And either you confected this by hand or use a calculator, but find that if you plug in r equals two, you obtain a solution to this equation. So we have that we get ar minus two times This is going to be r squared minus four are plus four equals zero and every factor further he gets ar minus two cubed equals zero. So the characteristic roots is three is to with a multiplicity of three. And since the roots are all too, it follows that the general solution is a N equals alfa one times two to the end, plus out the end end times two to the end, plus Alfa three times and squared times two to the end. This is by a serum from the book and with our initial conditions we obtain the system of equations. Negative five equals out. The one plus outfit too well nigh Alfa to and we have that four, which is a one is equal to to Alfa one plus to Alfa to plus two out of three and we have that a two which is 88. This is equal to four Alfa one plus two times for his eight Alfa two plus two Squared is four times four is 16 out of three and so we see that we obtain who want to use matrix form. We have that 100 222 for 8 16 and then on the right side we have negative five for 88. Is there augmented matrix and then we subtract two of row one from road to So we get 100 and then this will be to minus two is zero who might a zero a zero to minus 00 and then we get four minus two times negative five is four plus 10 which is 14 and will subtract four of row one from rose three. So we get four minus four is zero eight minus zero is 8 16 minus zero a 16 and 88 minus four times negative. Five. This is 88 plus 20 which is 108 which this reduces to the matrix. 100 zero is about him By two we get 11 seven in dividing by eight, we get one to and then this is 108 invited by eight. And so we have 100 negative five zero and the Muslim tracked Wrote to from row three, Sweat 011 seven and 01 minus 10 to minus one is one and we have 108 over. Eight minus seven. That's 108 This is the same as 54/4, which is the same as 27/2. So 27 over. Two minus seven. Which is the same as 27 minus 14. Over to just 13. Over to and finally subtract. There are three from road to we get ones years. Your negative 501 one minus one is zero seven minus 13 halves is 14 halves minus 13 halves, which is 1/2 and then 001 13 half's. And so solution to our system is a one after one equals negative. Five outfit two equals 1/2 and Alfa three equals 13 halfs. And so the particular solution is a M equals negative five times two to the end, plus 1/2 n times two to the end, plus 13 halves times end squared times two to the end.

You're given a recurrence relation in the initial conditions. When you're asked to find all solutions of this relation with these conditions, the relation is a N equals 7 a.m. minus one minus 16 a. M minus two plus 12 a.m. minus three plus n times four to the end. And the initial conditions are a zero equals negative, too. A one equals zero day two equals five. This is a when you're not homogeneous. Recurrence Relation The associated Linear homogeneous recurrence Relation is a unequal 7 a.m. minus one minus 16 and minus two plus 12 a n minus three into the characters to equation is aren't to the fourth minus seven are cute plus 16 Are this sorry they should be are to the third minus seven r squared plus 16 r minus 12 equals zero. And we have that The roots of this characteristic equation. There are few families to find out from plugging r equals two. We get eight minus 28 negative. 20 plus 32 is 12 minus 12 0 So we have it. Two is a root and using synthetic division, we obtained AR minus two times R squared minus five r plus six equals zero. This could be further factored as R minus two squared times ar minus three equals zero. And so we have that characteristic roots are to with a multiplicity of to and three. Therefore, we have that the general solution to the linear Ma Jenness currents relation is a one times two to the end, plus 82 times end times two to the end, plus a three times three to the end and you have a particular we have f of m is end of the fourth times four to the end for just end times four to the end. We have that and is a polynomial of degree. One and four is not the characteristic route, so it follows that general form for a particular solution. It's going to be Do you want n plus p zero times four to the end and to find the values of the coefficients, pluck the particular solution back into the anonymous homogeneous equation. From the left hand side, we have p one n plus p zero times four to the end. On the right hand side, we have 7 a.m. minus one, which is seven Times p, one times n minus one plus p zero times four to the end minus one minus 16 A n minus two Just minus 16 times he one times n minus two plus p zero times four to the end, minus two and plus 12 and minus three. This is plus 12 times p one times and minus three close P zero times 40 and minus three and lastly plus end times four to the end. Dividing both sides by afford to the n minus three, you get four cubed, which is 64 times p one n plus 64 ki zero is equal to here. We have seven times four is 28. He won and minus 28 p one plus 28 p zero minus. This is actually mistake. That should be, but by Ford GM minus three, that leaves Foursquare 16 so it's actually seven times 16 sets, four times 28 virtues. 80 plus 3200 and 12. So we have 112 p. One n minus 112 p. One plus 112 p zero and then we have minus 16 times. This is just four, which is minus 64 p one n. This is an N minus. Choose this is gonna be plus two times 64 which is 128 p one and minus another 64 p zero plus. And here this simply 12 p one n minus. Three times 12 is 36 p one and plus 12 b zero plus and 40 and over four to the N minus three is for acute. This is going to be 64 times and and so the west side, we have zero in on the right side. We have 112 minus 64 times two, 20 1 12 minus 64 minus 6400 and 20 128. That's a negative. So instead, if I move all the terms to the left hand side, we have 64 plus 64 is 1 20 Hate, Then subtract from that 1 24 you get four p one times in. We also have minus 64 times n and then for the constant term from living everything to the left you have. Plus, this is going to be first with the P one terms we have 112 minus 1 28 which is negative. Negative. 16 p one. Then adds that 36 he's plus 20 p one men. Four p zero You have 64 p zero. We add to that another 64 p zero to give 128 p. Zero and subtract from that 124 p zero. So we have plus four p zero and so it follows the coefficients of this polynomial or zero. So we have that p one is equal to 64 before which 16 we have that p zero is equal to negative. 20/4, which is negative. Five he won. This is negative five times 16 inches negative 50 minus 30 which is negative hating. And so we have that The particular solution is 16 n minus a times four to the end. And so it follows that the general form for a solution to the non homogeneous equation is going to be the general form for the solution to the modernise equation. Plus the particular solution. This gives us Alfa one times two to the end plus outfit two times end times two to the end plus Alfa three times three to the end. And then add to that the particular solution, which is 16 n minus. A. It's four to the end. And now, using our initial conditions, we can find the values of Alfa one out to an Alfa three. We have it. A zero is negative. Two. This tells us that negative, too. Sequel to out for one plus out to time zero, which is Euro plus Outfit three plus then 16 00 says his negatives. 80 Time is 4 to 0 is just one. We also have the equations. A one is equal to zero, so we have it. Zero is equal to to a one plus to a to plus three a three and then 16 times one minus 80 Being negative 64 times forward to the first, just for and lastly we have a two is equal to five. It's that five is equal to four Alfa one plus two times two squared, which is eight Alfa to plus three Squared, which is nine Alfa three and then plus 16 times two is 32 minus 80. That's minus 48 times four squared. So you can solve this by hand or use a computer algebra system. It's a moment pause here after solving you'll teen that Alfa one is equal to 17. Alfa two is equal to 39 over to in doubt for three is equal to 61 and therefore follows that the only solution to this recurrence relation with these initial conditions is 17 times to dbn plus 39 times two to the N minus one times end, plus 61 times three to the end and then add to that particular solution which we have is plus 16 n minus 80 times, four to the end. This is our answer.


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