Separable Differential Equations and Exponential Series

Uh, I think so, maybe. Oh, good. You know, I love that. I'm gonna actually. And don't worry about Lucas here. If he if you wants to keep the recording on, he might take a peek. What's happening in calculus? And I love directly talking to you. And I begin with usual. Are you turning on your recording? Uh, yeah, beautiful. And any questions from the previous sessions? Yeah, I couldn't really solve.Which one are you referring to?We're using the idea of an integrating factor, and try to see how we bundle up the left-hand side of the equation into a complete derivative. And we know if we want to do that, we hypothesize. One second, this is too big. We hypothesize, I want to make it appear out of the product rule. So imagine that we're going to multiply the y by f to begin with, and when you do this, when you take the derivative, what you're getting is a y f prime, and then plus the f prime y. Oh, sorry.RepeatRepeat it.To fit into this native x squared and a prime equal to a one, they are inconsistent with each other. But we could scale them though. I can try to find the f prime over f. The ratio would fit into the correct ratio, which is one over the native x squared. Correct. Then if I can do so, what did we accomplish? Then at least you could say this is.Equal to this y times the f derivative, and then divided by whatever the actual scaling factor, which will be a function of x. It's a it's arbitrary function including x here. We may have brought in more stuff; they're only proportional to each other, and they're not exactly equal to what we want to find. Do you see my idea? Is the f like a substitution? Oh no, f is actually a function we're looking for.In order to force this into the y f, the complete derivative, so that the product rule could really accommodate that the appearance of both y and y prime. Then Eddie, and could we find what is the solution to this?We take the integral of both sides. Indeed, we take integral because these are functions of x now. And what do we have coming out of the right side? f prime. Oh, at right side. Ah, yes. One over x. That's very good. That's antiderivative. And what is the antiderivative on the left? Natural log of g, uh, natural log of absolute value of g. Where is the g? Oh, uh, f. Sorry, yeah, very good, very good. Eddie, you are still sick, aren't you? Yeah. And does it hurt if I make you talk? Uh, it's not too bad. It's it's alright. Okay, and.I'm so sorry, and I'll try to be light on you. Elaine likes, she'll do the talking. So Elaine, do you agree? These are the two sides of the equation. Yeah. Alright, well then, why don't we isolate it by exponentiating both sides now? That's our choice of f, and what we have is absolute value of the f, really equal to the e to the one over x power. And if we do so, okay, then we do know this is going to lead to different.Alternative f's, however, right now we're doing by construction. This is where you don't need to worry about losing one solution. We don't need to be thorough. We just need to choose such f. So remember what is the purpose of introducing f that legitimizes my directly leading up to f just equal to. This is my choice. I can choose this one. I don't need to choose both, because all we care about is actually start with that one here, and once we found our f and.We begin to do this now. Try to begin with this. When you take the derivative, see what we're getting. How does that help us solve the original differential equation? Well, it does if we found the correct f now. So if we take the derivative, we're using both the product rule and the chain rule. First of all, we're going to actually fix the y, take the derivative of the other side, chain rule, and exponential yields exponential, and multiply by the derivative of the inner function, which is the negative one over the x squared.That's power rule. So now we're done with the step. We're fixing the y. Now we add on to. Now we fix the other component. We take the derivative y prime. Voila. Does it look like what we have up there? Sort of, not quite. So we have to really change that into. If I begin to work with the original equation now, I'm coming back over here, and then this is indeed the sort of combination we're looking at, but.It actually give us the e of the y of the x y prime, but we need to scale it, divide by a bunch of stuff. Why don't you guys work out what do I put onto the denominator? TensorFlow.Where did the e to the one over x y prime come from? It comes from here. That's how we solve the R F. What did you put it on top here? Ah, isn't that how we designed how we're gonna use that F? Oh, oh yeah. This is F. We're starting by really conjuring up, really wishfully conjuring up that product rule. Oh, okay.``` ```Penso work. We've done all the work now. You just need to translate it onto the paper.中文嗯。``Èddie, are you working out? Hands work.中文嗯。Negative e to the one over x over x squared y plus y prime e to the one x over e to the one over x y. My goodness, what are we talking about?Exactly what you're gonna put down the denominator. You know the numerator comes out as this now, and that is really close to what we want, which is this. They're already proportional to each other. Otherwise, why did we find such a f to begin with? We just need to see what scaling factors we need. And surely, Elaine, you're not gonna have any y on the denominator. Because if we do, we're in trouble. Because then it is a the whole problem has not become finding the derivative of the new function now. We need the denominator to be consistent with only functions of x, so that you can push them to the other side. It becomes separable. That's the whole spirit. I think I misunderstood your question. Oh, okay. And what do we have on the denominator?e to the one over x. Let's see. Eddie, what do you have? Do you even have a pencil at all? Are you writing it down? Yeah. Oh, beautiful.The other one is negative one.x squared, right? Very good. This is really negative, and one over the x squared. You can write it like this. That's sufficient to cancel out this unwanted additional coefficient. We only want a single y here, and once we pull that out, what the other component i should becomes minus x squared. So in a way, you can rewrite it here. This is a negative e to the one over the x over the x squared, and then inside.You got the y minus y prime x squared, and this component here is what we need. But the whole spirit we have we know for sure we accomplished something because that extra it's only a function of x. If it's only a function of x now, you can divide it up to the other side. We have sufficiently cleaned up this part here. Is everything clear so far? Okay, how does that help now to solve.The entire equation, and at this moment, we could translate it to the other side. What we're having is e to the one one over the x and the y. This whole thing derivative actually equals to, no matter how nasty it looks, but it's only a function of x, and that's a significant improvement. What we're getting is actually the three and the e to the one over the x and the x squared, and then, ah, minus.The e to the minus x, e to the one over x. Double check and see whether this is what you are getting. If it is, give me a signal.``` I got the same thing. Edit you. Yeah, excellent. Well, it's our life better, yes, significantly. However, nightmarish.It is. We have officially solved the equation now, and we just need to find the antiderivative. At this moment, we're ready to take off the derivative and say e to the one over the x y now really just equal to the entire antiderivative. It's a whole bundle, which is going to be the three of the e one over the x x squared. I'm going to write them separately. Okay, you you'll see why, and the x e of the one over the x dx, and we're finding the two antiderivatives, because.In that indefinite integral, each one would consist of a object constant c. It's ready there. Now, after you finish the integral and the derivative, we can divide that over. And in fact, one of these is unintegrable. What does that actually mean? In fact, for mathematicians here, this is a legit answer. There are a category of the functions where, for example, the Bell curve, the so-called normal distribution in statistics. It looks like a e of the ninety x squared dx.We're finding the anti-derivative. It's not because we're not smart enough. You simply cannot write it, and well, in any compact form. There's not a analytic way so that you can write it out as a f of x. Now, the best we can do is to keep the anti-derivative. This is fine; it just called the integral expression, because it's no longer integrable. You can do a power expansion, and write the answer in itself as an expansion. How do we do that? How do we do that in a.Second I shall show you, okay. But coming back over here, and similarly, this is also unintegrable, meaning by the time we're getting here, we're done. And no matter what, this is a legit function of x. It's just the antiderivative here. We have isolated our y. Once we're going to multiply both sides by that e to the negative one over the x. Now, but this one though is integrable. In fact, we have just seen how come this additional one over the x squared can come out of the chain.例如，艾迪做了类似的事情。所以，我们可以知道，什么是什么的反导数。``Ànd keep me posted of your mental process. Well, it has to have a. An exponential function in it. I solely agree. Otherwise, you wouldn't be able to produce that. A very good start, Elaine. I'm gonna go with what you say, and let's put it in there. The power is e to the one x because chain rule.You mean e to the one or x? Yeah. Hmm, I agree. I've already written down. Because you already indicated this much, I agree. And how do we clean up the miscellaneous? Negative. Uh, and then we we can pull out the three. Yeah, brilliant. Elaine is recognizing this actual term over here. It's really the natural.To result out of the chain rule, so we wouldn't do without it. That's it. This is the antiderivative of the first component, and in fact, when you do so, you can see. Oh, and eventually when we divide that over, and the part of the final answer just minus three. By the way, it is correct. That is actually part of our antiderivative. Our solution to the y, you can see it because when you actually have a three here and the second term, when we take the derivative of the y.The three would vanish here, and that negative three—sorry, negative three—which is exactly rendering the other side of the equation. We're good. Okay, now let's see. And I said the second component can't be dealt with further, meaning there's not a compact way that you can just write out what this antiderivative is. But it is a basically infinite power. And I want to justify it for you: why in certain circumstances.In fact, if you really look at all the functions, the majority of the functions, the antiderivatives cannot be written as a single function. Majority. I just said that when we are students, we encounter only an island of order, which is we're only given the good functions where we could integrate as a toy. So I'm giving you two prime examples today, and the bell curve we shall reencounter again and again. This is a function where the antiderivative simply cannot be found.I mean, it can't be written out as something without either an infinite summation or this this infinite integral, a sign. So how do we do such a thing that knowing that we cannot really find the antiderivative? You do your power expansion. Elaine, why don't you expand the integral for me into its Maclaurin expansion? By definition, it's e to something power. It falls into the e.To a bundle. And how do we? This is really coming from definition now. How do we resolve that e to that star power? You take the function at zero. Yes. And then add it to the first derivative at zero, and then so on. Yes. And however, we also know that expansion by definition. And give me the result, please. It's one of the expansions that you really want to keep in mind.You just know it by memory. Addy, do you know what is expansion of e to something? It's x plus x squared over oh wait x plus x squared over two factorial plus. You mean the one? Oh yeah, and then what is the next term? x squared over two factorial.哎，问题是。no， it's x, it's x, it's x. Remember, what this exponent is an even x. The exponent as a whole is already negative x squared. You're doing expansion not according to x, but according to this independent variable. You're expanding e to some power. Oh, Elaine, if you want to take the derivative, then it's a lot of a redundant.And announces her work because of the chain rule. That's why I don't want you to think like that. Negative x squared. Yeah, you just look at this negative x squared itself as a whole, and you go by. We're doing an expansion according to the star term. It is so much easier because even if you want to go by taking higher derivatives, they don't involve a chain rule. So we're just looking at the one plus whatever that exponent, which is a minus x squared.An um plus that exponent over the two factorial, right? And Eddie, could you give me the next two terms, please.```Please talk to us. Show the mental process, or in case you're stuck somewhere.I want to know where.中文嗯。Minus x to the six over three factorial. Hmm, minus i to the six over the third factorial. That's it. And the next one be adding x to the eighth power over the fourth factorial. Da da da, adding infinitum. A on the test, you you will want to be able to write the general formula for it. Generically, what This is each term, so this is really if you do it as a sigma, it would actually equal to one. Oh, actually, there's no one. It is simply going to be the uh negative x squared k's power over the k factorial, fitting into the very definition of the micro expansion of the exponential function e to something going from zero to infinity. Correct. We're finding what is anti derivative of this. Oh, that's no problem. It's just a polynomial. We all know.How to do that? Go ahead and find what is the integral.```````````````Very good, and Eddie, can you give a confirmation? Is that also what you're getting? I'm also getting that. Beautiful. So the issue is: this is a perfect answer already. We just don't know how to call that a function. Sorry, remember we defined functions before by really calling this power expansion into a exponential function. Maybe we can do something like here. Well, maybe you can, except for it's not a standard notation. This is simply the antiderivative of the bell curve. We call that.Cumulative, the so-called PDF. It's a statistical parlance here, referring to it's a probability density function, but it's cumulative in the sense it has been integrated. So be familiar with this function. This is where we have to stop. That's what it is. And if you want to write it out as really something like a summation sign to make it look more compact here, fine, we can do that. And I'm going to try you because you do need.This scale here to do it on your AP and also in college. So can I actually bundle this up into a general term? Meaning, I want to write it as summation. Meaning, you have to be careful by exactly identifying the pattern.And they keep me posted.In case you encounter any small barriers along the way.```Good work. When did you get so good at manipulating the summation patterns? I don't know. But okay, good. That shows mastery. Very good.Okay, perfect. Ah, be careful though. Don't let a minor detail ruin your. Your your your solution, and would it be correct if you do plug in n equal to zero? Oh yes, it is actually sorry, it is it is correct. I thought n starts with one, but this is perfect. It is truly perfect. Yeah, we're good. Ah. Except for no, the way I wrote it, I do the two and minus one. But yeah, yeah, yours is fine. Yours is fine because start from the zero now and mine starts with a one now. The both are fine except for I do the two and minus one. You do the two and plus one, and it matches with the correct sign. Perfect, Elaine Bravo. This is very good, and that's as much as you can do. This is a perfect answer. It's a legit one for the integral for the anti derivative of the bell curve. Now, Eddie, are you getting the same?Yes, beautiful. Then we're going to actually try likewise. And how do we go about this one now? Our task is to integrate find antiderivatives of the x and e one over the x. The x now. We also do a micro expansion, except for it's not a micro expansion of x. We can't do it simply because at x equal zero, the function value. If you're going to regard this as a function of x now and try to do the micro expansion thereof, it's.Because even the f of a zero, it's undefined. It's infinity. So we actually do all we need is a power expansion. So we just do the negative power expansions. And again, regard this total power here as a whole, and we're going to look at its McLaurin expansion. If that's the case, that's sufficient first to change this whole integral, this antiderivative into an antiderivative of the summation of a lot of power.Majority of those are actually negative, and I'm gonna do it on the board. You do your pencil work, and match with mine. See whether you fully agree.We're expanding e to the one over x, right? Yes. And also bring in that actual x power on the outside. We're expanding the entire integrand.```中文嗯。``` ```中文```中文嗯。I got the same thing, and at you. Me too. Alright, now same drills. And however, these are negative powers, but we could easily find the antiderivatives term by term. Let's finish that last bit. And if we do so, what would be the antiderivative that comes up?好的。``````中文嗯。 ``Ì got the same thing, and Eddie, you. Me too. We notice here it's not only a power. The anti-derivative this time is not made up of just power functions anymore. There's that quirky natural log, which obviously come out of this. And we do know if you're integrating the rational function family, sometimes we leave the family and get into either the inverse tricks or.These are natural logs. That's because of the reciprocal rule. Meaning, when you take the derivative inverse function, you end up with the reciprocal of the derivative of the original function evaluated at y. That reciprocal would give you a derivative which is a rational function with something on the denominator, and yet the antiderivative is not within the same family. This quirky appearance here, it's actually marking this antiderivative, and even the rest here. There's no way to bundle them up into.An analytic function, meaning you can't replace it with a function name that you just call them. And again, this is actually the best we can do. And again, you can rewrite it as a summation sign. It looks like that's a trivial exercise now. So Hans, this is the entire solution to this differential equation that I gave you. And what today I want to really characterize it and give it a categorization. In the future, when you see something you recognize, oh, it forces.Into which kind and what would be the method that's being called for? And Elaine, I wouldn't blame you. How not why for not being able to solve it? You're not supposed to as a high schooler, and on AP, you surely will not encounter such a differential equation. And the the most difficult ones you will encounter. We have already covered yesterday. Those are separable. Meaning you could always actually separate out two sides of the equation, and you could always be sure to know that they are integrable.Meaning you would you will you won't encounter the case where the antiderivative itself doesn't really have a analytic expression for it. They always do. Are we comfortable now? Yeah. Okay. Now at this moment, I want to give you a bit of a theoretical background because I like talking with a systematic parlance here. And again, as a high schooler, you don't need to know this. However, to be a mature mathematician, I would.Say, putting these into categories. It's a it's a good move. It's a mature move. So, what kind of a differential equation this is? We actually call that linear, first order, first order, nonhomogeneous, nonhomogeneous. What does that mean here? I'm going to explain them term by term. Linear simply means you got actually linear combination meaning highest up to the high the first power now.It's linear regarding the y, the y derivative, the y is a function we're solving. Okay, we're solving for y as a function of x now. It's a linear combination of the y, y derivative, y second derivative, all the way to as high as you care, maybe even up to the case derivative. They're linear regarding these. Now we are permitted to include nonlinear functions of x anywhere as coefficients. So, for example, it is linear. If I go. For the y third derivative, and divided by the e to the x, and plus, and this is going to be a two x minus the natural log of the x times the y second derivative, and minus a x cubed y equals to, let's say six e to the x. This is a linear differential equation. It is linear because we're free to look at everything else if it doesn't have a y in it.It's just a bunch of coefficients. So whatever these, they just become coefficients now. If you focus on the appearance of the y, triple derivative, second derivative of y now, and they all appear in linear superposition or sum of each other. So this is linear. It checks out. That's quadratic. But it is linear. I I'm looking at y the root times y a minus x cubed, uh y squared, and.It's not linear. Very good because I wrote a y, right? Excellent. But what if I just write that that's equal to the y to x? Is this linear? Yeah. No, it's not linear because we're multiplying the y with the y derivative. The total power in that term is second, so it's like if I have a linear function regarding both x and y, and then you can't have the term x y in the binomial expansion.Because we actually call that second power term. It's matter of terminology. A next session remind me to finish the definition of the linear first order non homogeneous differential equation. Would you please? Thank you, Elaine. You're doing really well. I did you too, and I would appreciate if you talk a bit more. But maybe not today. Get get better first. Take care. Bye. Keep up your good work. That was really good. Thank you. Thank you.

Separable Differential Equations and Exponential Series

Separable Differential Equations and Exponential Series

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