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All right. Hello. My name is Peter,

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and I'm going to be lecturing differential equations this term. So the lecture

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notes and problem sheets and everything for this course in the usual place. And they're also quite a few

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good textbooks for this material. I mean, some of those I mentioned on the website as well.

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So I'd recommend you look out for those. Now, just one

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quick. So general point. And quite often it might be that

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I write down. You see, my handwriting isn't that great. Maybe I'll write down something that you can't read or

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that doesn't seem to make sense. Or I might say something that just, you know, doesn't make sense. So that happens.

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It be great if you just yell out and asked me a question rather than starting a chat amongst all your neighbours.

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So I don't you. I find it very distracting of other people. So chatting all around you. So. So, yeah. If you have any questions

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ready queries, any points, then just a shout out and ask me and rather than starting a conversation going.

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Right. All right. Good. So most this course is about

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second order linear. We value problems. And I'm going to

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use BTP as a shorthand for that. And I'm just do a few examples

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just to sort of illustrate what some of the issues are. And here's the first one, by the way.

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So here in sort of henceforward and he's a prime as shorthand for differentiation.

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Hope that's okay. So this is a very simple bound value problem.

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Of second idea. And I've got two the conditions. And we can easily solve this using elementary methods

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for the first year. So here we go.

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So how do we solve an equation like this? I guess we have to look for a particular solution.

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And then the complementary function. This case of security is just one, right?

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Easy.

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That said, should be standard staff from last year. So the general solution is a sign

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of a particularly in school and a complementary function in this complementary function. We have two arbitrary

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constants

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that straight for men to work out, see one and see two. Obviously plug in band conditions. Okay.

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Oh, okay. Also, I'm going to use BSE for Švanda conditions and hope that's okay.

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Well, here's one I worked out earlier.

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Okay.

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Easy, right? I'm happy so far. Yeah, well,

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it is gonna get harder. I promise. Here's an example to you.

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So I'm going to place up one by 10x ten, I guess isn't defined.

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Valla XP too big. That's just how.

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So he's going to do this on Xscape and zero to Piver four. So

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in principle is exact, the same kind of Boundy Valley problem. I've got second equation, too bad conditions,

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but good luck spotting the particular integral to this. All right. So in principle,

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the same method should work for how are we supposed to find a particular interval for this?

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Is somebody trying to spot it now? Everyone does it while they enter the lecture, I'll be impressed.

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I sample three.

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It's the same Odey that we started with. And I've just tweaks the bad conditions slightly.

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So instead of why Prime applies to and I've got white pies, two principles, same

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approach will work, right? No, this this problem has no solutions.

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And you can easily cheque that yourself as we try to follow the particular range go plus complementary function

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approach, you just end up with a contradiction. It's impossible to satisfy both those editions conditions.

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So finally, I'm going to do the same. I'm just gonna change this two to a one, so exactly

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the same.

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Okay. So everything the same apart, that final band coalition and this problem

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has an infinite number of solutions.

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So I've got non-unique solution and actually I'll just write this down.

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So just one plus a sign X works for any A.

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OK, so this is just supposed to illustrate what the

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issues are with Boundy Valley problems. And so I'm just gonna write this

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down. What are the main questions that we're going to be trying to answer in this bit of the course?

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And I say a couple of issues that we encountered here. The first is

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that, um, unless the right hand side is something very simple, um,

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how are we going to support a particular integral? And isn't there a way to sort of just construct

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the particular integral? And the answer is, yeah, there is.

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Car P.I. for particular integral second OP suddenly one.

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So like without star,

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without somehow trying to sort of guess it or spot it, how can we actually construct it in a systematic way?

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And then the second question, which perhaps is more important is, um, when does a family value problem

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like this have a solution and when is this mission unique?

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Okay. Everything clear so far, okay.

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Right. And so first things first. I'm just gonna introduce some

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notations of jargon that I'm gonna be using throughout the course. OK.

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And so I'm mostly gonna be considering second order and

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differential equations, maybe some exceptions, but mostly.

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Oh, yes. Again, Odia is only the the equation.

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Now. So this is a bit abstract at the moment, but think of the France based

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be Kelly Ale. It looks like a two. It's supposed to be a curly L.

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All right. And let me just define what else is.

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OK, so this is of the most general second order linear differential operator

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that we could have set at some linear combination of the second first derivatives

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of Y and Y itself. And these P two P, one P zero kind of any

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Allwood functions we like. I guess I'm going to assume they're continuous, at least

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in practise, you know, in all the examples will do. They will be

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almost always they'll be analytic in fact.

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And I'm gonna to assume that why is differentiable enough times for this to make sense.

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Okay. And again, in practise, we're nearly always gonna be looking for solutions that are

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analytic.

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And then the final thing, which might be a bit mysterious to start with, is gonna seem that P

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two is not zero. So that's.

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And by that, I mean it's not able to zero anywhere.

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So not just that it's not the zero function, but it's everywhere. Non-zero, I think.

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And this may be even thought about this before, but maybe you can see if P t was zero anyway. Then

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this second Autodefensas equation better kind of turn into a first order, one that actually called correspond

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to a singular point of this idea. Again, I'm gonna come back that late in the course.

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Just cheque. I've got to.

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Okay, now there's, um, two sort of cases of this that we're gonna

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it's gonna be hard to separate those out. So first day of F is zero. Then this equation is called

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homogeneous. Right. That is not zero. It's called in homogeneous or non homogeneous.

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So, like, am I gonna label those separately? Because you're gonna be referring to those two different cases a lot. Okay.

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So Iraq is the zero function that's called in as called homogeneous. I better get that right.

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So

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so that is going to be referred to throughout as H for the homogeneous problem.

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Mm hmm. Right. And the case where F is not

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zero is called in homogeneous.

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And for some reason, this conventionally enabled n fence or non homogeneous.

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OK. Exactly. Yeah,

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and I'm gonna be people firing quite a lot to these two different cases, h and then right

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now a differential equations, one last term, you would mostly be looking

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at what are called initial value problems. All right.

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I've been surprised to find there's an acronym for that as well.

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And IVP is I'm so what I mean by that is if I've got a second order

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idea. What I would normally do is give you what Y and Y primed are at some point.

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So typically do

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so at some point. X equals A I will tell you what Y and why prime da. I think you've got to work

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out what Y is everywhere else right now for that problem.

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You can use Picards theorem from different equations, one that guarantees that

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a unique solution exists, at least locally. OK. Oh at least likely provided P

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two is not zero and maybe you can see why. Now if you try to apply Picard same appeal to a zero, you'd

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have a problem. Okay.

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Ray.

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Now pick out same guarantees, local existence

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and uniqueness. And in fact, but linear equations like this and you might remember this from, you

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know, a time ago. Actually, it satisfies a global lipshitz condition. And you can

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show this region exist everywhere, can provide a Peter is not zero.

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And so, in fact, provide pictures, not zero. The existence, uniqueness of global.

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But for boundy value problems. That's not true.

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And I've already shown you two examples where that's not true, where you instead of specifying why

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and why print at one point we had bad indications at two different points. And you can either get non-existence or non

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uniqueness.

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Mike. OK,

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so I'm just gonna quote some basic properties, again, from Elementary from last

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year, basically, or even from a level, I think.

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And the first is if I've got a second order in homogeneous

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Odey, well, you could do this trick that I did in the very first example. You could break down the general solution

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into a particular integral plus complementary function. Okay.

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It's going to break it up like there, so I come write the general solution as a particular integral

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like Y P.I. plus the complimentary function which I call y S.F.

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And what do I mean by that? So why P.I. is any function at all

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that satisfies the inherent genius problem?

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OK.

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So a particular age group is any solution that you can find by any means necessary of the inherited

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genius problem and why S.F. is the general solution a homogeneous problem? You've got rid of the Yaffe on the

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right hand side. Hopefully that Will

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has some sort of familiarity to it.

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And so the second thing I thought about, that's why S.F. thing complementary function

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well for a second order, linear ODP and the space of solutions

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of the homogeneous problem is a vector space of dimension, too. Okay.

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So what I mean by that is that the complementary function I can write has.

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I can write just in terms of two basic functions, basically.

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Exactly. I guess what I did for that very simple first example. And so I

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see why see to end arbitrary constants. And then why when and why to. I guess. Ready set of basis

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functions for the vector space. A basic ready to solutions that are linearly independent.

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Okay.

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Okay, so more or less familiar.

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Point any questions so far?

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Good point. I'm so you just a bit more about Libya independence.

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Again, this is gonna be kind of important going forward.

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And Occy, my rights is getting smaller after Naghmeh. If I keep doing that

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and and let be is very closely related that I think called the ront skin.

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And I guess a bit about that as well.

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Okay, so we remember what it means for two functions, or in general two

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elements of a vector space to be linear, linearly independent or linearly

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dependent. I guess maybe it's easy to write down the situation for them

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to be linearly dependent.

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So these two functions would be literally dependent. If you could take some linear combination

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of them, that adds up to zero sum, non-trivial linear combination of them.

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OK, so if access some constancy, want to see

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to not both zero such that C1 while unclassy T.Y. two adds up to the zero

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function. And I guess Elenita

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independent if and no such see one to see two exist.

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Okay.

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That's just the definition of what we mean by linearly dependent. Now, let's suppose this was true.

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So I suppose they are daylily dependent. And then remember, I'm assuming

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why one and why two are twice defensible, actually. So I can certainly differentiate. They suspect two

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X

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and say, I'm sorry, I see one. Why on policy to Y two is zero. Then also

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they must say, must be true for the derivatives or buy one of Y two as well.

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So actually now I could think of this as like a linear system of equations for C one and C two.

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And I guess we know then that we can get non-trivial solutions. So

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you want to see two even only if the determinant of that matrix is zero. OK.

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And of. That determines what's called the wrong skin.

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Okay, so so

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that's what I mean by the ront skin, the determinants of that matrix.

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And then here, I guess I'm going to slip between. It's a slight Abuzer notation, I guess. But and

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here I've kind of expressed W as being a of by linear functional of y one of my

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two. So if you give it to functions, it spits out another function. Okay.

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But it's also help with sometimes just to think of W as being a function of X. Okay.

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Right.

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So I mean by that is I actually now evaluate why one and why two. Then I get W is a function of X.

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Okay. So it's just I have two different ways of thinking. The same quantity.

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And so I guess what I've argued then is that on

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the eve, why, when and why to linearly dependent, then the Ront

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scheme must be zero. That's what I've shown. Yeah.

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Again, I hope you don't mind this. All right. L.D. literally dependent.

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That's clear to say myself. Some

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ink. So be shown that if that's true, that implies that the

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wrong skin is zero. And

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I guess we can also use this as a contrapositive of that.

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So,

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OK. And this isn't going to upset you when I say it is not zero.

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Then why? What about why two must be linearly independent. OK. And I can use L

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I for that. Is that clear? I said, turning the previous identity

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on its head now,

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it would be really nice if we could turn that implication into an even only if. But it's not quite

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true in the other direction. And actually, you can quite easily come up with

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counterexamples and.

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So here's a counterexample.

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So here's two functions that sort of defined in a piecewise way and actually

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the kind of examples that you can easily dream up for this Jamele, about this sort of form.

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So these are functions that are Leavesley cheque, their linearly independent.

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So the first thing is these two functions, either to find this funny piecewise way. There are actually

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three times differentiable, but something funny happens at the Origin, but it's only the fourth derivative.

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That's discontinuous.

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OK. And then secondly,

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you can easily cheque sort of an exercise you don't believe me. Is it these two functions, ELENITA Independent.

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So it's impossible to find a non-trivial combination of them that adds up to zero function.

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But finally, the real killer is the wrong skin. These two functions a zero.

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So although the one scale is zero. These functions are linearly independent. So it

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means we can't put this implication the other way.

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All right. But there is so in general, this invitation does go in both

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directions. But in the case, luckily for us, I guess in the case is we're really interested in

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when, why, when and why not just any two old functions that you pull out of the air like these two.

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But they are actually two solutions. I have a second order idea. And in that case, this indication

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does go in both directions.

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Why I want to buy you a street. The second already. Then a patient does go both

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ways.

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And. So I'm just I'm going to prove this. I'm talking a state of proposition.

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A proposition of two halves.

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So let's take why, when and why to be any two solutions of our second order

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homogeneous Odey. And then there's two things

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that we can then say. And the

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first thing is, so the wrong skin, basically, either

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it's Zerai, beware or it's zero. No way.

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Why does that matter? Because I guess it wouldn't make sense for two functions to be kind of linearly

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independent some places, but not in others.

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And the second thing is that.

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I'm going to stay this way.

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So w zero. Even only four, I went away to an elderly dependent.

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OK. And Sandy just proved these one at a time. Now, I'll probably wrap up after that.

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And so for the first bit, we know that why, when and why to

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both solutions of our homogeneous idea. So let me just write that out explicitly.

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OK.

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So we know that why, when and why to both satisfy the same ODAC.

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And so the way I think about this is I now I want to eliminate this P0

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term. OK, so that's sort of my little way of remembering what to do here.

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All right. So what I mean is I'm going to do y one time this four minus Y two times this

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one. Okay.

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You've got Claire. So it's all done.

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Okay. So why won't times the first one minus Y two times the second one.

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And now this. Right. This is the wrong skin right here.

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All right. Actually, this is the derivative

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of their own skin. So if you imagine differentiating

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this with spec two X, you see the first of this cancel Y one prior March primaries, Y to

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prime Y one prime. You set up with the second remitted. And so actually W

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satisfies this first order idea.

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And actually, you can easily solve our Audi.

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All right. So you can to think about this as

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being like an integrating factor or by taking this time over the other side

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and separating the variables. I'm that you again using elementary methods.

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Again, note we need PE to not be zero.

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Now we know that exponential and can't ever be zero. All right. So

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this constant zero, it isn't the constant zero, then WCF everywhere. If the constant

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isn't zero w zero nowhere. Okay.

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So why the WAC, right? Know it's not zera everywhere.

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OK.

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Mm hmm.

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All right. For the second part, and we've done

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that one, okay. And so I think we need to do is the other way.

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We need to do that way. Okay. That's what remains to be done.

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And

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so we've done the.

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So what I need to do is the use of vindication going in the right direction. So let's suppose W is zero

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and I'm trying to show that, then why, when and why do you have to be linearly dependent?

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All

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right. So first things first, I'm going to assume that why one is not

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the zero function.

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Because if it was, then why, when and why two would trivially be liberty dependent. Okay.

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Okay. Because you could just take C to be zero and see one to be anything. Yeah.

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Okay. So if one wants at the zero function, then they must exist. Somebody of X, Y, Y one is not zero.

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Now, here's the trick.

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So I'm going to define why I have to be this particular linear combination of why one and why two.

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So it's why one of a way to maximise Y to a Y, one of X. Okay.

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Then what we can see is that if I put X equals item, this Y is zero.

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That's clear. Right.

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And also actually I differentiate this and input X equals I then I get the ront

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skin evaluates that X equals I. Okay.

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And by assumption the wrong skin is zero. Right.

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And so you've got this function and both why and it's first derivative.

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A zero X equals a. And also because it's a linear accommodation of Y one and Y two.

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Y satisfies a homogeneous equation.

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It satisfies this second order differential equation with Ziya on the right hand side.

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So then by Picards Theorem, the solution of this initial value

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problem is existence is unique and it's just zero. Okay.

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So I've got this homogeneous ody with zero initial conditions. The sweep of that problem is

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just Weich zero and by Picards there in that switch is unique.

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I'm at Ben tells us that there does exist a non-trivial combination of y one of Y

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two that adds up to zero. Okay.

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All right. All right. So I think that's all for today,

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I say tomorrow. Going to get into more actual solving problems

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and more techniques to solving things.