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Ex: FijJ 4L Sol-hxLL Dv;clbl Pn blov o<r<a 2<0 <T i- L sehi_ cerclr rai= 17 | 4(e,e), 4te) 8 -2 ulno) > ? Wrm) L5 Sol; ulr;0) _ 2 ar"si- (0) 0 &...

Question

Ex: FijJ 4L Sol-hxLL Dv;clbl Pn blov o<r<a 2<0 <T i- L sehi_ cerclr rai= 17 | 4(e,e), 4te) 8 -2 ulno) > ? Wrm) L5 Sol; ulr;0) _ 2 ar"si- (0) 0 'swn 0ftc) si-(ne) d & T

Ex: FijJ 4L Sol-hx LL Dv;clbl Pn blov o<r<a 2<0 <T i- L sehi_ cerclr rai= 17 | 4(e,e), 4te) 8 -2 ulno) > ? Wrm) L5 Sol; ulr;0) _ 2 ar"si- (0) 0 ' swn 0 ftc) si-(ne) d & T



Answers

$$t y^{\prime \prime}-2 y^{\prime}+t y=0 ; \quad y(0)=1, \quad y^{\prime}(0)=0$$
[Hint: $$\mathscr{L}^{-1}\left\{1 /\left(s^{2}+1\right)^{2}\right\}(t)=(\sin t-t \cos t) / 2$$.]

In this video, we're gonna go through the solution to question 60 from chapter 4.5. So after determine whether a combination of the method of undetermined coefficients with the principle of superposition can be used to solve this differential equation. First we look at the left outside was clear to see that each one of these terms is a video coefficient term. So all that's okay so far at the right Inside, we have these three terms that we need to deserving whether we can use the principal or determined coefficients to find a particular integral. That starts this particular solution for just looking at these terms. No, exactly clear yet. So it's try on DDE. I write them in a different way. So this first term we can weaken right sign of tea as 1/2 times by e t i t minus eats the minus I t Where I hear is this square of minus one. So then you'll see that if we multiply it was by this by t squared times Easy, my t. We're gonna get some at Terminus of the form t squared times. He's a part of something. It's the power of some constant times T because it will be a complex constant about Wi Fi way Know that we can deal with that. So this term is okay. What about this too? Where we can write this in a similar way. We see that cause T is equal to 1/2 times eat the i t close e to the minus. I see. And then if we most by this by 80 then we're gonna get 80 times by Ah, yeah, it's the power of constant times T and we know how to deal with this using the method of when it's over coefficients. So this is fine. What about this turn? Well, again, we need to write this in a different way, so tend to the teeth. This is looking kind of something similar to a E to the t. But rather than e to the T is a tent city. So if we take the exponential off the walk with both sides so that inverse operations so we could do that And I'm sorry, not both sides but off this tend to the tea. So we take E to the natural log 10 to the T and then using the algebra of logarithms. We can take this to you. It's a factor at the front. Then we're left with, uh, e to the t times natural 10. So this is just e to the some constant times by T. Onda We know how to deal with those ah terms using the method of from determined coefficients. So therefore we can use the method of undetermined coefficients together with principal superposition. Just all this question.

Hello. We have probably number 129. Okay and this question is X equal to cause in verse one dynasty is square by one plus T squared X. Equal to cause universe one minus T. Is required by one. Plastic squired. And why it calls to. Okay and why Cause to second verse one plus T squared by one. Minister square taken worse one plus T squared by one minus. Do you square Where T is between 0-1? So do you have any x? He is between 0-1. And so we need to find divide by dx. Okay. No this is course in verse that has signed universe. Okay. This is a formula that we have to remember that cause universe X equal to. Uh huh. Second verse one by X. No this whole thing we can get is why equal to so I called to because in verse one of -D. Square by one place P square. So this is X. If you differentiate will be getting do you have any X equal to one? So They should be equal to one. And and cities option number eight. Thank you.

Hello. We have probably number 129. Okay and this question is X equal to cause in verse one dynasty is square by one plus T squared X. Equal to cause universe one minus T. Is required by one. Plastic squired. And why it calls to. Okay and why Cause to second verse one plus T squared by one. Minister square taken worse one plus T squared by one minus. Do you square Where T is between 0-1? So do you have any x? He is between 0-1. And so we need to find divide by dx. Okay. No this is course in verse that has signed universe. Okay. This is a formula that we have to remember that cause universe X equal to. Uh huh. Second verse one by X. No this whole thing we can get is why equal to so I called to because in verse one of -D. Square by one place P square. So this is X. If you differentiate will be getting do you have any X equal to one? So They should be equal to one. And and cities option number eight. Thank you.

Okay. Good day. Ladies and gentlemen, today we're looking at problem number 11 here. And the question is whether or not we can apply the domesticated of undetermined coefficients Thio this, uh, ordinary different show equation here. And so I'm not really gonna, um, go through all the different parts of the undetermined coefficients cause I think there's probably 5 to 6 distinct cases. But it is important, I think, for you two know each of those cases because they tell you how to go about solving, um, the, uh, ordinary differential equations. They'll tell you how to solve a bunch of cases of, or a bunch of, um, ordinary difference, your equations on particular ones with constant coefficients here. And so the first thing is to realize that, um oops, sorry about that. So if I take three different, distinct functions, I think one of the year of two of two here at three of tea Now I'm going to really look at this case by case in each case. So in the first case, this one here is, um and it is in fact, a proper form. It is one of the cases, and I'm not sure which But if you look, if you flip through, you'll see the thing cases. And this is in fact, one of the cases, um, in the 2nd 1 here again is also a case again. I don't know exactly what's important, but it is, in fact, one of the cases covered by the, um, undetermined coefficients. But the 3rd 1 is not and in particular one over tea is not a polynomial. It is Tito the negative first, and that is not, um, covered by any of the cases. So in particular than, um, since you have one I mean, really, you can't get rid of this one over t s o. The end of the final answer is that it's not applicable, and it's not applicable because there's no case that covers us. And the older way you could solve this. Using the undetermined coefficients is to break this into threes. In cases here, I'm solve each one and then applies the the superposition principle. And in this case, you can't because one of those, uh, you know, is not solvable using that method. Uh huh. But it doesn't mean that there's not other methods to solve it. It's just not solvable using this method really, all the collections after. So there's no need to go any further with. So, um, again, I would just mention that, um, it's a good idea to have in the back your mind. What thes, um the what the undetermined coefficients method involves and sort of go through each of the steps because it's it's actually fairly involved. There's quite a few different steps, and there's a bunch of different cases. So there kind of TVs, but still probably could know. Uh, okay, so that's it for this problem. Thank you very much. I haven't.


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