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P(t) Use E{p(t)} E{p(t)} L{p(t)} 3 =1 - the 49 definition of the 83 %l - Laplace transform L{plt)} 83Question2 dt, to find L{p(t)}:...

Question

P(t) Use E{p(t)} E{p(t)} L{p(t)} 3 =1 - the 49 definition of the 83 %l - Laplace transform L{plt)} 83Question2 dt, to find L{p(t)}:

P(t) Use E{p(t)} E{p(t)} L{p(t)} 3 =1 - the 49 definition of the 83 %l - Laplace transform L{plt)} 83 Question 2 dt, to find L{p(t)}:



Answers

In Exercises 91-98, find the Laplace Transform of the function.
$$f(t)=e^{a t}$$

This question asks us to find the blessed transform. That's what l c. Off function effort D. It's constant. One in a blessed transform is defined, I think just above this question, uh, give him a capital F s. So it's a new function. The end across from zero to infinity e to the minus esty ft. Bt. So in this gaze into powerful zero affinity, you need to t minus esty CT because F it's just constant one that this is an improper, integral, but the only reason it's inappropriate because be infinitely bound. So we write it. This, um, this is a straightforward, integral. We have done many times during the exercises for this section. This is one over my guess. Even the miners Esty Green zero game. So it is the limits be going to infinity won over mine. Is this to the mine? Says B. Maya's one over Mines s zero. So that's just one. And evaluating its limit, we find actual answer becomes one over s class transform because one function is one of us. It may seem that LaPlace transform was going on a random construction That's not very interesting, but it turns out to be quite useful in solving certain differential equations, which are things you find in physics. Quite love

This question asks us to find the blessed transform off a function f which equals D squared. So the last transform justified, defined Somewhere near where this question is asked is the integral from zero to infinity each of the minus esty times fft Richards D squared, Petey. Now, this is an improper integral so he needs to brighten assess the limits. In this case, the only improper points is the infinite prevent. So you can just write this past this integral D team. So we're gonna do this using integration by Marx. Um, I'm gonna do that fairly quickly. I will write down most of the steps in between, but I'm not gonna write down u equals C equals D. V equals et cetera. So this is C squared. Won over mines that's in the mountains is T zero and B ah plus zero to be eso. This plus is using the minus coming from the anti derivative um to d you too divine SST DT the limits. That's because to infinity ups, let's put to be in here. So we get beats weird. One of our minds s need to be minds as being, um they were gonna do it again. Only this got this time for they interpret with just the t. So we're gonna get, uh, to over S t times one over minus S E to the minus s t between zero and B and again. Plus, because we were already absorbing the mine site. From there, you get to over s squared. Uh, no t anymore taking derivative of tea was gone. U t minus esty DT. So we're almost ready to start taking the limits so it gets, uh, b squared, defined by minus s. You mind, Mrs B Plus to be divided by mines squared entreaty. Miles says being Plus, uh, finally we gets two divided by minds s pubes me to the minus s t between zero b. So this becomes the limits as being goes to infinity. Well, be squares divided by miners s plus to be divided by minus squared plus two divided by minus is huge. All times each of minds is being. And then finally we get plus two divided by this cute, which is the term coming from the zero there. Now, um, let's quickly do this limit on next page just to see how it works, we get the limits of B to infinity off B squared, divided by mines s plus to be divided by, uh, minus squared. Plus two divided by mines is cute if I remember correctly. Yes. Divided by the Jedi has B. So this is a classic local tall limits. Sam, we're gonna take the differential on the sides. So this is the limit as B goes to infinity. Oh, to be divided by minus s plus two divided by minus squared zero divided by s DSB, which is the limit as biggest to infinity again by look it all off two divided by minus s divided by s squared each adsb, which is zero. Because now I thought this constant on the bottom bends to infinity. Element is zero, which means that plus transform is too divided by S Cube. So it's a It's a bit of work. Um, I know that it's really hard. It's just you have to keep track of everything and do the final limit. Property

All right. So if you want to find a little lost transformation of the form of F of T is equal to t plugging it into the equation. We're gonna end up with f s is equal to one over s square, this people to us to the power of negative.

Hello. Today we're going to find the LaPlace transform of are given function the hyperbolic sign of 80. So to do this, we're going to plug in our function into the Austrians form uh, the integral from zero to infinity. Either Negative s t of the function, Petey. So we're gonna be plugging this function into here now, Before you do that, it'll be easier to work with. Hyperbolic sign. Do you put it into another form? Either 80 minus e to the negative. 80 over too. This would be easy to work with because we already have e inner integral. And we could just combine them from there. So we have the integral from zero to infinity. Ethan Negative s a t e a and T minus e the negative a of t over too DT. Now, the first thing I want to do is I want to get this over to I'm gonna bring that half out to in front of the integral begin. Do we can deal with this later and I'm going to combine or this story I can distribute this e the negative s t that either 80 the negative e to the negative 18 right. We could combine their powers to obtain e thinking. Negative s plus a T minus e negative s minus a t dt. No, we're We're dealing with a with an end popper, Integral. The reason I know this is because we have an infinity in her upper bound of are integral. And the way we deal with improper Negroes is by making it into a limit. So I'm going to make it a limit of B and maybe goes towards infinity and replace the infinity. But being and everything else will be the same. Great. No, we can, um we can integrate, So we'll have the same half the limits, and then the integral we're gonna take the integral of this. Do that we find, Let's find Let's first find the anti derivative of E the negative s plus 80 which it says, e that a negative s plus a tea with divided by the coefficient of tea. Great. That now find the integrity of of the next. What should be hee thinking? It s minus 80 over negative s nato s minus. Hey, great. You know, since we're integrating, I'm going to be evaluating this from from zero to be. And these are Arlette, uh, balance of integration. So let's plug in, um, this book and what we have. So we still have the have we still have the limit? Now, when I plug in, first can plug in being, so I'm going to get e beginner s plus a e over negative s plus a minus. You some parentheses here. E two negative s minus a e o negative s minus a and then we subtract What happens when you plug in zero? So either negative s plus a I'm zero overnegative s plus a my e negative s minus a time. Zero. Well, we're negative s minus a, um so we can see right away that this top part of the fraction is going to be one, because E Because zero times negative s plus a will still be zero. And either zero is one. And by the same reasoning, this be one as well. So let's, uh let's write that out. Well, quick. So the first half will still be the same. Well, look at that. Next. No, Why? I'm writing this when you're looking for the loss, transform you really interested? And when it's going to converge, Right? But they were just then. It's not much of a hope. So when we want this to converge, we have to look at this part in this part. So hee to some power. Um, so be goes towards infinity, right? If you plug in being to infinity, all sorts of things gonna happen, right, we can have pto infinity e to the er You depends. So let me go back quick. So you have e the negative s plus a times a plug in B for book and infinity for being okay. So there's three things that can happen one, um, s plus a It's positive. And that's positive. This means that I'll be e to deposit affinity and this diverges, which is a no go. The second case, it's when then is when um s equals a the best equals a. Then we'll have e to the zero times Affinity. Uh, this is Indeterminant and determine it and is also no go. We don't look at this either, but we do want to look at the case Were negative s plus a It's negative, right? Because this will mean that will have eaten and negative affinity and this becomes zero. So it converges Great. Um, so we want negative s plus a to be less than zero. So that means that we want a dog less than s now for the same reasoning. We also want this to be less than zero. You want negative s minus eight to be less than zero. So we'll have negative a be left in s right. So under these conditions, well, our ah function converge, right? And if s is great and a and s is greater than negative, eh? That's the same thing as saying s is greater than you have to value bay. So that's the assumption going into the next part of the question. A Sioux s is greater in the absolute value of a great So what that means is we have 1/2 the limit. SP goes towards infinity. Uh, now the first part, like we went over before, it will be either naked affinity minus one over negative s minus day or e negative affinity. And the other part was won over native s plus a minus want over and negative s my saying so then this becomes zero. This become zero and this whole thing will become zero Great. No. Oh, and since we already plug and B, you don't really need this limit anymore. Now, since this whole thing become zero, there's no more bees to like to mess with. All we have left is this half and this part's we have minus half off, one over negative s plus a minus one over negative s minus a. And I want to combine this into one fraction. So I multiply both sides. Oh, the first fraction by the denominator of the other faction. So my mother, by this I negative s minus a negative s mine is a and I multiply this part by negative s plus a multiply this part negative s plus a and I'll keep these imprint disease. So here we're going to distribute this out. Um, and what we'll get is so on top will have negative s minus day when the bottom will have s squared minus a squared and here will have negative s plus a and the bottom You have s squared minus a square. And so since they have the same the non leader, we can combine them. I'm the one denominator and we'll combine the new readers Now. Here. If you notice you have negative s minus negative s So these cancel. I love negative, eh? Minus positive A. So this will give you so give you minus to a on top. And then we could do some canceling the Negus These negative cancel thes twos will cancel. So what we're left with is a over s squared, minus a squared, and this is our solution.


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