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Consider the DE:2kx -2y = 0, (x > 0) where k is a constant, usC the substitution y=X lo prove In both cases the DE has oscillatory or non - oscillatory solutions...

Question

Consider the DE:2kx -2y = 0, (x > 0) where k is a constant, usC the substitution y=X lo prove In both cases the DE has oscillatory or non - oscillatory solutions according ask>} or ks;I7]

Consider the DE: 2kx -2y = 0, (x > 0) where k is a constant, usC the substitution y=X lo prove In both cases the DE has oscillatory or non - oscillatory solutions according as k>} or ks; I7]



Answers

Let $A x=0$ be a homogeneous system of $n$ linear equations in $n$ unknowns that has only the trivial solution. Prove that if $k$ is any positive integer, then the system $A^{k} \mathbf{x}=\mathbf{0}$ also has only the trivial solution.

Spoke about the question at least seven. We need to prove that this differential equation is the situation of the solution to this particular differential equation. So we have to find the first elevated over here. So let's differentiate this. Uh we're here and this is a this is a rational uh functions. We're going to use caution rules and the question tool. We have denominator square, we have denominator times differentiation of human. In fact, before moving ahead like this. Why don't we take the constant? Because the constant is a square K. Because it's given that A which is positive and they are constant. So let's take a a square out to ease our things. So we just have to differentiate. And by the way, the differentiation won't be with respect to exit will be with respect to T. Because in this case the variables are Y. And so we have diva over DT is equal to a squared is taken on since it is a constant. So we have the validity of T over one plus a t uh, a square K is taken out. Then we have the denominator square which is one plus eight. KT whole square. We have denominator times one was 80 80 times differentiation on numerator. So the differentiation of tea with respect is just one minus. We're using quotient rule here minus numerator which is city. And differentiation of denominator which is one plus A The so this is equal to a square K. And we have one plus a Katie over here minus 30 times differentiation of one is just zero. A K is a conference that comes outside in the differentiation of tea with respect to T. Is just one over one plus a key. The whole square. So this becomes equal to a square K times. Let's open a strike to simplify this. So this is one plus a KT minus a Katie over one plus a K A D old square. Fortunately these two terms cancel out. And what are we left with? That's right it over here. This means that de vie over DT is equal to a square K times square three times over one plus E old square. All right. So although these two doesn't look similar, but we need to use some relations and we need to prove it. All right. So, we first off noticed that this is just in terms of wire. So we have to get rid of this expression post and which we can do actually, how can we do that? Let's uh, let's try to write the value of A Square K over in five. We can replace Let's try to replace this expression in terms of why? So we're gonna make some effort here. Let's write, write. The question was why? The initial expression once, what is going to is quick 80/1 plus a Katie. This means that 1/8, 1 or one plus Phil divide both sides by a square Katy. So we have y over a square. Katie is equal to 1/1 plus eight Katie. And we square both sides. We have wire square over a square Katie. Whole square is equal to one over a one for one plus eight E whole square. And so we got the value of 1/1 plus 8 80 Whole Square. So let's substitute that over here and continue the black one. So if we continue that, we have levi over DT is equal to a square K. Is as it is. And 1/1 plus 80 80 square is replaced by Y square over a square K. T. Well, where, So this can be written as if you can provide us. We have a square K. Y square over eighties to four K square D square. So some of the expressions will be cancer. So we left with via square in the numerator and a square Katie square in the denominator. So we need to replace uh this a square Katie square or perhaps just to T the expression but T square only. So that also we can do, we'll try to find the value of T. From here. So let's write the green one once again. So this this expression we are writing and we are actually crossword deploying this. So we have wire times one plus 8 80 is equal to a squared Katie. So if you open up the parentheses, we have Y plus a K A T Y. As you go to a square Katie, let's move this one word of the right. So we have wide support to and let's take 8 80 out as a common factor. So we are left with a minus y. This means that the value of a square gaity is why over a minus y. And from here we can also say that a minus y is equal to y over a square Katie. We have just done a cross multiplication. And uh if we uh if we move this time over to the again by the cross multiplication, so we have a square K times a minus Y is equal to y over the Now this would be very helpful sense. Over here we have one over a square case outside and we have y over the whole square, which can be written as this expression. So we have a square three times a minus y. And this complete expressions under square over a square cake. So, if you open up the parentheses with square, we have areas to four K square a minus y whole square over a square cake. So finally we are left with is where K times a minus y whole square, which is exactly what we need. Not Exactly. There is an additional expression of a square. Uh let's reach ago book, ideally it should be just K times a minus y whole square. So uh if we do a wide square, then we have areas support. Let me just quickly recheck the walk. So, over here, we made a mistake, we made a typo that from here. When we are taking this the storm down. Uh when we are moving this over, dividing both sides by a minus Y. This expression is a Katie, not a square Katie. So there will be no square over here and hence there will be no square, and there will be no square over here as well. So this was a Katie. If there is no square over here, then there is no square over here. It means that this will just be a square, so just be a square and the squareness squares cancers. Finally, we won't have any expression of a square over here as well. So finally, we get the value of devour D tsk times a minus five holes. Where, which is exactly what we need. Thank you.

Okay. To find the general solution to this differential equation here. First, I'm going to rewrite it into Ah, differential. Operator form. That's the squared. Minus two M uh, D plus M squared minus k squared. Then we have time Are applied to why is equal to zero. So then now the auxiliary creation p of our it's gonna be equal to, um r squared minus two m r plus m squared minus k squared. We said that equal to zero. Now, we're trying to find the solutions with respect to our So this is ah, quadratic form quadratic equation. Sweeney is the quadratic formula here. Are it's gonna be equal to and then ah, negative. Be so that's gonna be to em. Plus or minus plus or minus square toe B squared. So this square, it is gonna be four m squared, minus four. Uh, A is one, and then c is m squared minus case. Good. Okay, all over too. So then we have two m divided by two. This is going to be able to em and then we have a plus or minus plus or minus. So this four m squared minus four m squared. We're gonna cancel out this negative makes this become positive. So we all we're left with is a four k squared squared of that. It's just going to be equal to, uh to K. Um, since we have a two k ah, and then are divided by two, that's just gonna be plus or minus k oops, plus or minus K like that. So our general solution is gonna look like why of X is equal to C one e to the M plus K eggs plus plus C two e two the and then M minus K x. Okay, so this is what our general solution is going to look like. Uh, here. Now we also need to show that the solution can be written in the form with coach and sinche. So remember the definition of Coche X is equal to e to the X plus e to the negative X divided by two, and cinch X is equal to e to the X minus e to the negative X all over, too. So we're going to rewrite this in terms of this here. So first we noticed that we can factor out on E or a C one e to be, uh this becomes m X plus K x. Right? So since this is an addition here, we can rewrite it as either the m x times e to the k X. So we have eat of the M X e to the K X then plus C two, e two the m x e to the negative Negative K X. All right, so we'll have, um uh well, we can back around each of the MX and then we'll have C one. Um and I'm sure you're gonna replace our constant here with be one first and be too for now. Um, just so that we can later we will rewrite Ah, it in terms of C one and C two once we have Kocian cinch. So I'm gonna have ah c one e to the k X plus C two e to the negative k X or Sorry B one plus B two e to the negative K X right. So I want to rewrite this in terms of cinch and coach Remember synching coach. Let's write that on the next page here. So if we just have a C one, uh, coach K X plus C to cinch que x here. Um, if we want to set that equal to our, uh, b one e to the k x plus bead to e to the negative K X So we'll replace the synching coche with our formula. So we have C one. Then we have eat of the X plus e to than Sorry, each of the B to the K X plus e to the negative k X all over too, plus C two and then eat the k X minus e to the negative K X all over all over Ah, to like So I move this down a little. Um, so let's rewrite some of this. So we have C one divided by two e to the k x plus. See, um, so I see one divided by two e to the negative k x plus C c two divided by two eats of an A K X and then minus C two divided by two e to the native K X Let's group the K X and the negative K x together. So first, this, um, the K X. So then we get C one, uh, or we just get 1/2 C one plus C two b to the K X plus and then group these two terms Together we'll get plus 1/2 and then see one minus C two e to the negative K x and we want this evil to be our equal to R B one e to the K X plus B two e to the negative K x here, So will magical, efficient swill have 1/2 C one plus C two equals B 1 1/2 see one minus C two is able to be to so Ah, we have 1/2 C one plus See to is equal to be one and then 1/2 see one minus C two is equal to be to We can add these two equations here. If we do that, let's let me let me factor out the 1/2 actually see one plus 1/2 C two is able to be one see one minus 1/2 C two is able to be to So now we're going to add the two. So that means these cancel out and then here, When we add this, we just get C one. It's going to be to be one plus B two like that? Um uh, it only to put down pregnancies. Then if we try to solve for C two. Okay, so 1/2 B one plus 1/2 be too. I'm going to just plug it into the 1st 1 Plus 1/2 see too, equals one I z Sorry is able to be one. So I have subtract thes two over here. So then I get 1/2 see to is equal to So be one B one, um, minus be one minus 1/2 B one is 1/2 V one and then we have minus 1/2 B two. So then multiply by two on both sides. We get that C two. It's one of equal to be one minus B two. So that means if we, um right, uh, see, one is going to be one plus B two and C t equals B one minus b two. That means okay, B one e to the k x que x plus beat to e to the negative. K X is equal to, And then we'll have B one plus b two e to the k X plus e to the negative k x all over, too. And then plus be one minus B two times e to the k X minus e to the native K X all over, too. Let's verify this. Okay, so we'll have ah b one be one divided by two e to the K X. Right. Um, so we also need a, uh, plus. And then we have also plus 1/2 B one Ah B one B two e to the K X right here. Uh, yeah, Let's just let's let's write this out just to be sure. And we put this on another page over here. Okay, so I want to write out this side here. This is equal to. So I'll first multiply this. Be one here. So be one be one divided by to eat the k x plus B one divided by two e to the k x or sorry, negative K x And then this be to multiply out now. So plus B two divided by two e to the k X plus B two divided by two e to the negative K x And then now the right hand side. We have a plus that will distribute this be one. So we get be one divided by two e to the K X and then ah minus B one divided by two e to the negative K x then distributing this negative B two, we get minus B two divided by two e to the K X and then plus B two divided by two e to the negative K X So now this 1/2 or B 1/2 e to the negative K X, um, this cancels out with this term here. So this cancels out with this and this cancels out with this here. So now this term and this term here combine into just B one each of the K X and then this term combines with this term to become plus B two e to the negative K x So we verify this. This does work. So going back to the original problem, why of X is equal to e to the M X and then we have B one e to the K X plus B two e to the negative K X. This is equal to e to the M X times. Then we have B one B one plus B two and then ah e to the k X plus e to the negative k x all over too. Plus B one minus B two times e to the k X minus e to the native K X all over, too. Member, this is Coach K X kohsh. Um, So what we have is e to the M X and then we'll call this see one so we just get C one Coach que x and then ah, this. Ah, see one. This is Coach K X, and then this is going to be C two here, So plus C two and this is cinch que x So see to, uh, pussy to sign or a cinch Que x like that. Okay, so now we've shown that it can take the form of this solution. Ah, here.

There has to show that the set of all solutions to the non homogeneous differential equation shown here is not a subspace of the vector space of functions that are continuous. Have at least two continuous derivatives on the interval up, so we can see that this effort eggs is non zero. It means that our zero victory, which is the zero function y zero, does not exist. Zero zero function is not a solution, which means that this is not a subspace.

Ah, good day, ladies and gentlemen. Today we're looking at a problem number 29 from section 4.7. And it is asking us to consider too linearly independent solutions to the to the differential equation like this and are and we want to show that s o It's a linear differential differential equation of this form. On some interval you need to be. And we want to prove that, um, both Why wanted White? You cannot both be zero at some value of tea. Okay, so this is a question that is asking us to prove something. And the proof method we use again, eyes a proof by contradiction, which means that we assume Okay, so we assume both that. Why want it? Why, too your are, in fact, linearly independent. So we're assuming this is true. And we're also assuming that at some value of tea, there they are, both zero. Okay. And now what happens then? Well, if they're both zero, let's let's just calculate the wrong skin and evaluate at that poin t. Well, um, this is zero, and this is zero and zero times. Any number, of course, is equal to zero. So we have zero minus zero and, of course, us equal to zero. Now the key here is that the Ron skiing one of the properties of the Ron skiing is if it is zero. If w why wonder why, too? Is zero at any value of tea, Just any value, just one value. It means that the two functions here are linearly dependent. Okay, so if you don't know that again, that's something you need to know about the round skin. That it's it's one of the important. It's is one of the most important properties of the round ski in here. And this pertains on Lee to, um, solutions off differential equations. Amazingly, um, that that works. So what we've done now is we've shown that there exists a fixed value of tea such flat. The ron skin is zero. And of course what? Like I said, that means that they are linearly depended or in particularly not linearly independent. So you can't you can't assume something is literally independent. And that, um, is that you can't assume that they're both linearly independent and this together because we get a contradiction. We we've shown that they're they're not linearly independent So you can't. I mean, you can't be literally independent and not not linearly independent at the same time. So something has to be wrong. And, of course, what it is is it's this for a year because this is our second assumption of people. Well, um, that this cannot be true. So therefore, they cannot both be zero at the same point. Um, yeah. So So this is really just a nice I mean, Proust by contradiction are really nice. Can be really nice little proofs, but it just turns out, uh, that therefore, since they cannot, this cannot hold. That's the idea. So, uh, that is it for here? Um, sometimes proust, like contradiction and stuff like that take ahh. Bit of getting your head around. So I might be worthwhile to think about it in your own time. And I just want to remind you again that the basic idea is we made two assumptions on the functions in our proof. We said, Well, we're gonna assume, and this is this is the what I might call, like the first major assumption. So the one that we have to I mean, we have to have something. Has to be absolutely true, which is that they are. We're assuming that they are linearly independent solutions to this differential equation. Okay. And then in the course of the proof I'm saying, Well, I'm gonna assume that I give this statement here, all right? And then what it leads me to is a contradiction. And since this leads me to a contradiction, it cannot be true with this assumption. And therefore, these two statements cannot beach both through at the same time. So that that's basically the idea here. And, um, that's it for today. Thank you very much.


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