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(2ccs: Z sing 4icct > 0" Find the z7c Iength furcazi #) Find anorhis descriplion ef ihe bclt tat uss ac Ieuth # te petnttehnr Encatl Enalk- Cho %atottze &qu...

Question

(2ccs: Z sing 4icct > 0" Find the z7c Iength furcazi #) Find anorhis descriplion ef ihe bclt tat uss ac Ieuth # te petnttehnr Encatl Enalk- Cho %atottze "{etatktes Enltoa pubted Vele?kFdm MatIr Tood Cnti €0e # "4] 0,D. n [e 003

(2ccs: Z sing 4icct > 0" Find the z7c Iength furcazi #) Find anorhis descriplion ef ihe bclt tat uss ac Ieuth # te petntt ehnr Encatl Enalk- Cho %atottze "{etatktes Enltoa pubted Vele?kFdm MatIr Tood Cnti € 0e # "4] 0,D. n [e 003



Answers

$$z^{\prime \prime}-2 z^{\prime}-2 z=0 ; \quad z(0)=0, \quad z^{\prime}(0)=3$$

Good day, ladies and gentlemen. Ah, today we want to discuss the application of fear, Um, five to the given initial value problem. And now what You're what you're, um, five involves is, um, essentially, I mean, it's looking at, um, differential equations off the given form. So in this now we're allowing the coefficients, um, of Z shit coefficients three functions of time now. So instead of just fixed numbers, um, they're coefficients of time and particularly their non home of genius. And we have an initial value point. Uh, in this case is zero and left. So what you first have to remember is that you want to get these functions P of T. Q of T and G F T wish a theory mart serum is discussing. You have to divide through by the coefficient of the or is he coefficient of Z double prime here in particular? Sometimes you'll see it referred to his A of t. So you divide the entire expression by a of t And what we end up with this pft being the coefficient of ze prime. And in this case, it's one over tea. Um, two of keeping the coefficient of Z. And of course, that's one over T squared. And G f T is the non homogeneity here, which is co signed t over T Square. And now you'll notice, uh, is that the largest intervals of continuity off all three of these functions are concerned here with all three of these guys is negative infinity to zero and zero to infinity. Now, of course in particular, that the T zero value here is in. If he is not in those intervals, it's as, ah, it's at the dis continuity. And so what that tells us, then, is that, dear? I'm five. Sorry if you can't see that. But what? That's what that bottom statement says is that here on five does not apply. So, um, in the end, our our results are results. Here Is that because the initial point is a dis continuity of the function of one of the functions in this case, all of the functions P of T. Q of T and G. A. T. And I do want to point out that it just takes one of these guys to be discontinuous at that point. Um, there. Then we can't We can't fair and five's not applicability. Now, I also mentioned that just because they're in five isn't applicability all here? Yeah, says nothing. I mean, that doesn't tell you that. Ah, you can't find, uh, unique. You can't find unique solutions at this point. It's on Lee that, um this'll ce this. The room doesn't guarantee such solutions, and that's all it says. So it says nothing about what is actually possible. It just tells you that the steering is not applicant pool. Okay, um, hopefully that's clear enough. So in particular, I might use the example of if I change the initial value here to one. Okay, So if I change my initial value to one, um then, in that case, one is in this interval, so the serum would then guarantee the existence of unique solutions on this interval here. So, I mean, in that case, you can draw a conclusion in the case. That zero all you can conclude is that you can't conclude really anything. You just could include that There. The serum says nothing. So, um so that's that's all you can really conclude so again, you can't. You you can't really say that solutions don't exist because you don't know. Maybe they do. Maybe they don't. But because it's just continuous there, you can't draw any conclusions. So, um, that is and that's all you can really do. And I think it's worth I'm pointing this out because I know it is kind of a common. It is kind of a common mistake, and I made it myself in the past that you see something like that. You know, like the this continuity here at zero, and your intuition might be to say, Well, then the firm says, You know that solutions don't exist there whenever, but it's important to remember that it doesn't tell you that it just doesn't tell you anything. So it's kind of a small little point to observe, but I still think it's worth pointing out anyhow. That's it for this problem. Thank you very much.

Ah, good day, ladies and gentlemen. Today, we're going to be looking at problem number 39 here, which is to Seoul, Um, equation number 39. And again, this is ah, ordinary differential equation as whole genius linear and has constant, constant coefficients. It's 1/3 order. So, um, that means we should have three distinct linearly independent solutions. Um, this problem actually involves a few steps, and it's important to handle all of them, I think. First off, um, Azi lt's as you did in the in the second order. You look at the auxiliary function of our and what you'll find is pretty easily you'll end up with this. Now, again, it's 1/3 order polynomial. And when you look at the roots, you should get, um, Well, you're going to get two distinct roots. The second negative too, is actually a double route. And, um, are one Here is a single root. Um, and so what does that mean? Well, um, I'm going to show Is that, um, these the functions? Um Why one? Why to and why? Three. Here are the three, um, literally independent solutions. 2 39 But to do that, I first have to actually show that these guys are being to the independent. And, um, now to do this, I have to admit, um ah, I think to do this I had to think about it a bit. And I'm not I mean, I don't know. I kind of used, um, the method they used, um, in in problem number 36. Ah, but it's a little different in this time. Notice that these three functions are not straightforward exponential. So you can't just use, um 36 measures as your proof You have to actually have to go through. And, um, first off, what I did is I just just I assume for the moment that I had three. Constance C one C two and see free satisfying this expression. Okay, um, and now what would that mean? Well, okay. So immediately, it doesn't mean much to me. So what I did is I just multiplying through by each of the negative to t here to get this expression here. Um, now I want thio. Okay, s so Yeah, so I multiplied through by either negative to t to get this. Um and then ah, you can differentiate this expression here. Um, because it has to hold for all t. I just differentiated that here, and that's what I got here. So, really, this is just the derivative of this function. Um, 39 c is really derivative of 39. Be okay. So I wanted Thio make a comment here quickly before I go further that it's important when you're doing this kind of thing. Just a hint. I noticed that I have three constants. Okay, so I have C one. I have C two and C three. And to answer this question, I need to show that each of these have to be zero. Okay, so in particular than I need three distinct equations. Because I have three variables of you will on the variables being the constant. See, once he doing c three, um, so eso I have to somehow or another find these expression, and that's kind of what I need to do. So, um well, one way to do it is, of course, directly from here. I get this expression here. Um, and since I know that this expression has told for all tea while I just differentiated to get the 2nd 1 um, noting importantly. Ah, of course. The derivative of this guy here. Um, I made a mistake the first time, and when I did it, I didn't do the derivative. Correct. So I made about st. So it's important that you do the derivative correctly. Okay, so now with this guy, um, since it holds for all tea while I can use teak zero, and, um, I get in these 1st 2 equations, OK, so I just want to try and emphasize that t equals zero. Gives me these 1st 2 expressions. Um, and this 3rd 1 39 d here is separate. And it's really just, um the, uh 39 c multiplying through by eating negative four or eat of fourty. So, really, this this last one is just, um, found by manipulating the 39. See there. Okay, so I know I don't wantto imply that these 1st 2 I'm just plugging take zero into, um, these two expressions, which in particular eliminates, uh, this one and this one here because the tea is there and you'll notice from here. Then, um, I can substitute, um c three Ah, to be four c two. Okay. And I'm gonna plug thatch into this expression. Okay? And what does that do? Well, when you do that, um, you'll see that since the C three is four, see to that these two terms cancel, um, and then you end up with just for all T you end up with this guy, the 16. Ah, see, to t to be zero, which, of course, means then the C two has to be zero and a C to a zero, then, well, the other numbers Air zero. So, um, then that tells me that, in fact, I get linear independence on. Admittedly, this is a case where, um ah, I had to play around with it. So, um, you know, it wasn't straightforward how to do it. So took me a little bit of manipulation and sink maggots. Okay, so, um, of course now we know that we have three linearly independence independent functions, but it's not enough to get the general solution, because we have to know that they are in fact solutions. But I'm not I just inviting you. Check that. Their solution. That's pretty straightforward. Do, um, and finally, um, uh, the conclusion here is really six, which is since we have step six years since we have three linearly independent solutions through 1/3 order OD. Ah, that gives us the general solution right here. So it's, um, a bit of Ah, you know, ah, process. But in truth, um, if you want to answer the question completely, you really have to show each card. So, um Okay, so that's, uh that's that's the solution for this problem. Um, the trick really is the linear independence and that that that took me a bit to figure out. So, uh, okay.


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