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InaUtnut Uac[lerl(4) Uea graphlng culculate fomt neaha numnellicul Yg ttt t7- decleMearenannuneanthettonmacuitnt Au adndnurel Cal YallesLceImm...

Question

InaUtnut Uac[lerl(4) Uea graphlng culculate fomt neaha numnellicul Yg ttt t7- decleMearenannuneanthettonmacuitnt Au adndnurel Cal YallesLceImm

Ina Utnut Uac[lerl (4) Uea graphlng culculate fomt neaha numnellicul Yg ttt t7- decleMe arenannunean thetton macuitnt Au adnd nurel Cal Yalles Lce Imm



Answers

$$\mathscr{L}\left\{t^{3}-t e^{t}+e^{4 t} \cos t\right\}$$

Hello and welcome to this video solution of numerous. Here we are given a link comprehension. So here in statement one were given that and iron or a on roasting with sodium carbonate and lying in the presence of air gives to compound PNC. And you're from option the second part the solution be in concentrated. It's alan reaction with production furrows and it gives a blue color on precipitated of compound their cost solution of C. On treatment with considerate statistical gives a yellow colored compound E. And the compound even treat it with a seal gives an orange red compound F. Which is used as an ox raise anything. So you have to make certain predictions to come into the solution. So let me show you how it's made. So first of all we take for if he oh she had to poetry thus eight and a two freedom global neck And with air that is given seven or 2 good line gives you to a free two or three that's eight the name so Seattle four. That's it. It's a frustration. Next we have if we two or three reacting with concentrate sales that is given gives you who official three Last three H 2. Right now this official tree reacts with who before if he 10 6 that is protection fellow sign a great this gives you the blue solution pushing blue solution of if the four if he CN six whole trade plus 12 cases. Mm Next From three we have two any he wanted to see our well for That's 8/10 of food gives you we need to the odd 47 is the yellow color solution. Greatness in A. Two so four. That's it too. No you have in A to see our two or seven. Okay, last case here is the orange color substance of people pr two or seven that's 2010. So this is the so let us state what is A what is B. The first we have me is If you see a 23, this is a Next we have every two or 3 SCB or Yes. So this suits also be ready And in in 204 this is C. Right BNC. That is mentioned. Yes. Now this pushing blue books. Mhm. Yeah. Yeah. Whoa it is. Mhm. Okay. To see again this election with the this one your local er this is E and we have orange. What are some common? It does if Yeah. Right. I hope this is clear to you and have a very good rest of the day. 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.

Okay, so here we are doing Let's see the limit as t goes to zero. Okay, this kind of long one. But we got signed t over tea. I another factor. And we got 10 square T over. Signed to T J Linus t cubed minus eight over t plus two. Okay, we're just gonna Francis. Okay, so we got her eye term. I got r J term. We've got our K term. The easiest thing here, I think, is just going to break these up. So we'll look at these as, uh, I and then Jay and then Okay, so the 1st 1 will start with his eye. And here we're gonna be looking at the limit as T goes zero of Sign T over Costa over just tea. Sorry about that. And if we plug in zero right into these, we're going to see that we get zero over zero. No problem, because we can do Lobito halls. So we're gonna take the derivative of the top. We should come out crew sign, T, and we're gonna take the derivative of the bottom which should come out to be one now only plug in co sign of zero actually give us one over one, which is okay, We'll come back to that in a second. But for now, that's all we really need there. So now we're gonna do the limit as t goes zero of this nasty one in the middle. And this is Tan's where t over signed of to take now for this one again, If we pulled in zero Well, we're gonna get kind of a zero over zero on the top and then we're going to get a zero on the bottom. If you can imagine plugging zero into the sign that's on the top of tangent that's gonna give us a zero. The coastline is gonna be one. So it's zero, and then the sign of two zeros on the bottom is still there. So we get another zero over zero. That's an indeterminate form. So again, we're gonna pilo petals. So take the derivative of the top, which is going to be too tan t seeking squared t and then on bottom, we're gonna have to co sign of two t as the dirt, that one. And that was a really bad too. So I'm just gonna erase it and rewrite it except me. Okay, so there's a two for some reason written in black. Awesome. So we've got that to tangent. T c can score t over to co sign duty. Now, we're gonna clean this up a bit, and we're just gonna substitute in a sign over co sign for Tanja, and we're gonna substitute one over co sign for each of these sequins. So also, we're gonna get rid of these two tubes. So what we're gonna wind up with is one sign I'm on top over co sign of actually, it's too T Times co sign cubed of tea. Now, all this stuff on the bottom, these are all one's co sign of two time zeros. One coast on cubed of zero is also one. So we're going to get a one on the bottom and on the top. Sign of zero is gonna be zero. We got zero over one is zero cool. The next one is is kind of the easiest. So this is the land of, uh, t 00 of Alice E T cubes minus eight over t plus two. Okay. Um oops. Sorry about that. And we are just going to try to plug zero straight into this. I should mention because there's a minus sign here. We're actually going through a negative on the front there just to make this easy for ourselves. So if we plug zero straight into this, what? We're going to get his negative negative eight over to notice that the T cubed begins, a zero in the tea becomes a zero. So we're actually gonna wind up with negative negative eight to buy them too, Which is before. So our final answer here putting everything together is gonna be one I zero J's was for Jerry. And in fact, just to clean this up a little bit, we can actually get rid of the eye and our side won. And in giant foam, for some reason, we can just highlight that answer.

As we know that, as we know that under root burk is equal to and is equal to reflective index is equal to reflective index, which is demands. And lace. Richard diamonds and diamonds and Elise went pretty so diamonds and so diamonds and will be equal to am not L not he not option D is the correct answer for this problem. I hope you understand the solution.


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