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OK. Welcome to introductory calculus. I will start with some practical information,
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and then I'll tell you a little bit about the syllabus and what we will cover in this
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course and then give you some examples of differential equations from physical sciences.
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And then a little bit of integration towards the end, so for for many of you, this might be the easiest course here that you take in Oxford.
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But I think things will get progressively harder.
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So maybe in a couple of weeks, it will be interesting to everybody if today's lecture might be a bit too easy for some of you.
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OK, so let me tell you some practical information. So we have 16 lectures.
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The lecture notes are online. Online, these are the lecture notes, these were written by Kath Wilkins.
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She taught this course for a few years until last year, so we'll just follow them.
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I guess I should have introduced myself at some points of the lecture in.
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So you can call me down, my name is Daniel thorugh and will meet on.
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So we'll meet twice a week. Today's special just because he's the first week we'll meet on Mondays and Wednesdays at 10 a.m., So not too early.
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And you'll have eight problem sheets. So the first two problem sheets are online.
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The problem sheets you'll cover in for tutorials in your college.
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OK, so for our our tutorial.
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What do I? So I said the lecture notes on a line, the reading list is also line.
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So see online. The book that I like is Mary Boluses Mathematical Methods in physical sciences.
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You know this book? And most of your colleagues should have a copy of of this, if not, the university does as well.
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So this this book is quite concise, and it has various examples from physics and engineering and science,
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and it also has the added advantage that if if unlike the other books on on the reading list,
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if you drop this one on your foot, you might be able to to walk without seeing an orthopaedic surgeon.
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All right, so that's any any questions about this?
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OK, now, syllabus. So the first half of the course, about about seven or eight lectures will be devoted to differential equations.
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So this is about seven eight lectures. So if two kinds ordinary differential equations, all these and partial differential equations.
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Our is. So I'll give you I'll give you some examples very soon, we'll look at fairly easy examples of differential equations.
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We'll learn some techniques. It's it's a combination solving them.
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It's a combination of science and art. You have to do some educated guesses at some point, but it's it's quite an interesting and very useful subject.
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And then after that, we'll talk about lying and double integrals, line integrals and double intervals.
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And the reason these are useful is because we will be able to compute arc length.
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Of of cars and areas.
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Various regions in the plane or surfaces.
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So this is maybe three lectures. And then finally, we'll do calculus of functions.
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In two variables. So this should be viewed as a gentle introduction into multivariable calculus.
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So amongst amongst the things that will do, we'll look at various surfaces, gradients, normal vectors.
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We'll look at Taylor's theorem into variables, critical points and a little bit of Lagrange multipliers.
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Which are useful for optimisation problems. OK.
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Now, there is a lot of interaction between this course and other brilliant courses that you will take.
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So intro calculus. Well, be directly useful.
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Well, obviously multivariable calculus, as I said.
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In a way, it's a little bit unfair. We we set set up the work we do some examples for in introduction calculus.
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But then the really cool results and theorems you prove in multivariable calculus.
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So we just do a little bit of the groundwork towards that.
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You also do. These are also useful in dynamics and in B, the ease of you will do the next term for the series and please.
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Now there is a lot of interaction between internal calculus and analysis, particularly analysis tool, which is what you do next term.
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So there will be quite a few results. From analysis that will just fade and not prove, maybe prove some particular examples and so on.
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But real rigorous proof seal doing analysis next term.
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But then it all comes together when you revise or for your exams in Trinity.
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OK. So that's that, of course, in part A.
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There will be lots of applied mathematics options that will continue this differential equations is a big option, fluid and waves, et cetera.
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So this is a very useful course. It's also mandatory. So you have you have to be.
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Yes. OK. So now let me give you some examples of where all these might appear.
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OK, so all these. So what what is a different ordinary differential equation?
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So this is an equation. Involving an independent variable, let's call it X and a function of X.
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Which we call it. Why? So why this would be the dependent variable and the derivatives of why?
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With respect to X. So for example, DUI, the X D squared, y squared,
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etc. So the order of the highest derivative that occurs when you call that the order of the differential equation.
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So, for example, the simplest so the simplest kind of, Audie, would be something of the foreign DUI.
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The X equals some function in X.
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So the wide X equals four works, you can solve that by direct integration, so this can be solved.
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So y equals so y you can think of Y as being the entity derivative of F of X and then we can use integration.
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That's the simplest kind of differential equation that we can have,
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and this is the reason why we'll start the course by reviewing a little bit of integration techniques.
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But more interesting, there could be more interesting differential equations.
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So let me give you some examples from. From physical sciences.
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So, for example, from mechanics, this is something that you have all seen.
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You can have Newton's second law.
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Which says that the force is the mass times, the acceleration, so a is the acceleration.
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But then the acceleration is a derivative, is the derivative of the velocity with respect to time.
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So that's already a differential equation, but it could be a second or that differential equation,
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if you think that V is B R B T, where R is the displacement?
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Then you get, for example, a is the R v squared r d squared, which is a second order differential equation so that.
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That's an easy example of how differential equations appear in mechanics.
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Well, you could also have differential equations in.
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Of engineering or if you have an electrical circuit.
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So if I take a simple one, so a simple series circuit, which, for example, an R L C circuit,
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which means that it has the following components it has R stands for resistor, so it has a registered R.
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It has an inductor L with Inductance L, and it has a capacitor with the party capacitance C.
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And it has a source of. Voltage, something like a battery v, so here I am, so I have a capacitor with C capacitance.
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And the register. With our assistance and an inductor with L and inductance.
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So here are our L and C our constants, they're independent of time.
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But then I'm interested, for example, in the current across the circuit, so this is the current.
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Across I of T is the current across the circuit, which is a function of the time.
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So in terms of differential equations, t the time would be the independent variable.
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And this i of T, for example, is a dependent variable.
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I can also have Q of T, which is the charge across.
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Capacitor. Sorry.
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On the capacitor and the relation between the two of them is that I is the Q, the T.
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So kick-offs law says that.
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The total voltage is zero around the circuit, which in another way, the voltage V from the battery, which is a function of T.
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Equals the voltage across the resistor, plus the voltage across the inductive plus the voltage on the capacitor,
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and now you're right, each one of them. The voltage across the resistor by ohms law is R times-I of T.
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The one on the capacitor is just one over three times the charge.
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And for the inductor is El the constant times DADT, which is faraday's.
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Hello. So now I can express, for example, so I have an equation involving V I.
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And Q, but I is the Q DP so I can rewrite everything in terms of Q, for example.
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So I can get a differential equation in Q, which will be simply.
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So this would be the leading term DDT, so l times the IDP becomes L times d squared, humidity squared, plus our eye is our times.
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Cue the 80 plus one over C times Q equals V, so that is a second order differential equation.
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That appears in electrical circuits, you.
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So it's second order, because the highest derivative is of second order.
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It has constant coefficients because the constants are L, R and C are constant.
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And it's what we'll call in homogeneous because this doesn't have to be zero.
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So those are the type of differential equations that we can we can study.
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And there are many other examples, so I'll leave one as an exercise for you.
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And is the exercise? So I'll tell you, the problem is the rate at which a radioactive.
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Substance, because.
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Is proportional. To the remaining.
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A number of. Number of attempts.
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So I want you to as an exercise to write the differential equation that describes this situation.
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OK, so we'll come back to things like this later.
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So what? So the question is, what's the what's the differential equation?
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OK. So as as you progress along in this course, in the mathematics course here you will encounter very,
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very interesting and sophisticated differential equations in applied mathematics.
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So we're just scratching the surface a little. All right, now, going back to what I what I said before,
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the simplest kind of ODA is the idea X equals f of X, which you can solve by direct integration.
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So let me review a couple of facts about integration.
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So one of the most useful techniques, which I'm sure most of you are quite familiar with is integration by parts.
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OK, so where does integration by parts come from? Well, it comes from the product rules, product or life needs rule.
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If you want to sound fancy. For derivatives.
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So if I have two functions, FMG and I multiply them and then I differentiate them, so I have four times.
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Prime is f prime g plus f g prime.
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Which means that. F g f times, g prime equals f times g prime minus f prime times G.
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And if I integrate both sides. Then I end up with the integration by parts, which is.
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F times, G Prime B X, if they're functions of X, equals F.
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F times, g minus f prime times g, d x.
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OK, so this is the version, the indefinite integrals version.
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You can have a definite intervallo version where you put the limits of integration.
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So let me spell it out. So this is the definite into integrals version.
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All right. So let's do a couple of examples.
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So the first example. So suppose I want to integrate X squared sign X.
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The. Now. So this would solve so this would give this gives the solution.
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To DUI, the X equals X squared sine X.
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OK, so integration by parts, you have to decide which one is off and which ones?
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G. Now clearly I would like to decrease. The power here, I know I can never get rid of the sign by differentiation.
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So then maybe this, then I have to do this Earth and this is G Prime, which means that G is minus cos x.
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So if I call this interval, I I is.
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X squared times minus six and then minus the derivative of F, which is two x times minus cos x.
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The ex. So this is minus x squared cos x plus two times x x x.
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And now again, this should be F and this should be G Prime.
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Score across X plus two times.
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X Sign X. Minus two times.
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Sign X the X, so please try to follow through what I'm doing and let me know if I make a mistake,
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this is this is kind of my nightmare to do integrate into clubs like this while I'm being filmed.
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This is not exactly what I like to do. So two x sign x and then minus korsak, then plus C.
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Is this. So, so. Plus plus.
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Thank you. Good. As I said.
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So see here, denotes. A constant.
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Because we're doing indefinite integral. All right, let's do another example.
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So again, an indefinite integral to X minus one times Alan X squared plus one.
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The X. OK. What do you think?
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How which one should be f and which ones should be G or G Prime?
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Say that again. Right.
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So this I want to differentiate to get rid of the logger, so I should call this, which means that this is going to be different.
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Thank you. And that makes G X squared minus six.
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So this becomes X squared minus six times l n x squared plus one minus the integral.
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Of X squared minus six times the derivative of the natural log of X squared plus one, which is two x over x squared plus one d x.
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So I'm finally this term. What do I do here?
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Good. So we do long division. So let's rewrite it first.
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This is x squared minus x l n x squared plus one and then minus two x cubed minus x squared over x squared plus one d x.
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So I have to remember how to do long division.
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So how x cubed minus x. Now.
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Depending how you learn this, you will draw the long division in different ways.
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So you just do it your way, and I'll I'll do it my way.
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So that's X and minus six cubed. Minus X then.
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Well, minus x squared minus six, and that's minus one.
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OK, so this means that x cubed minus x squared over x squared plus one equals x minus one.
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Plus, minus six plus one over x squared plus one.
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Did you get the same thing? Good. OK, so let's call this integral, Jay.
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And now we come to you, Jay. The integral of X minus one plus or minus X plus one over x squared plus one.
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The X, which equals X squared minus half x squared minus X.
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And then. How do I integrate this term?
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The fraction? So I should split x over x squared plus one.
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The ex. Yeah.
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And let me write the last term plus. The X over x squared plus one.
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So this one, the last term, we should recognise that what is it?
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Are 10 or 10 years, depending how you want to.
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The notices are 10 of X, just 10 inverse of X and what do we do with this?
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We can substitute, yeah, let let, let's do that so that we remember how to do substitutions.
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You might just look at it and know what it is, right, but just to review substitution if I said you equals x squared plus one.
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Then deal equals to X the X deal, the X equals to X, which means that this is one half.
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But the you over you. Which is one half line of you.
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She is one half a line of x squared plus one that you might have guessed.
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Just because you have enough practise, some of you.
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OK, so now let's put them all together, so Jay is one half x squared minus X.
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Minus one half, Alan X squared plus one.
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Plus. Ten inverse of X and some constant.
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Which means that the original integral, the.
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The integral in the beginning. Which I should have called I so that I don't have to roll down the balls but equals.
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X squared minus x l and x squared.
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Plus one. Minus twice this, so minus x squared plus two x plus Ellen.
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X squared plus one. Minus 10 in verse six and then.
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Plus to see. Thank you.
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Any other mistakes? All right.
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OK. So that's a that's an integral.
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There are cases when integration by parts will not simplify either of the two functions FMG.
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But what happens is if you do it twice, then you sort of come back to.
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What you started with, so the typical example. And.
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Is. I equals the to e to the X, Senex, the X.
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All right. So maybe we don't need to go through the entire calculation.
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This is in the lecture notes as well. But how would you solve it?
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Right. So you do you do it? So for example, I can say that this is gee prime and this is off.
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And then I integrate, I get calls and then I do it again, and I will end up with some expression minus.
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They seem to grow and then I sold for it.
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So you, you do this and you get the answer to be something like one half e to the X sign X minus six, then plus.
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Constant. OK.
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So another type of examples, which is which are more difficult,
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are the ones which you cannot solve in just one go, but you have to find a recursive formula.
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So I'll just do an example like that. You've you've seen other examples before.
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So this is when we get a reduction or if you want to call it a recursive formula.
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So I start I suppose, I'm looking at this interval.
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Co-sign to the end, ex, the ex.
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Now I want to label this integral I n because I'm going to get a formula of I n in terms of i n minus one or I n minus two, et cetera.
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Now, there is not much choice here, what you should call and.
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Mother, you should call. So I'm going to just do it, so this is coarse and minus one x times call six the X.
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So this is Earth and this is g prime. Then we get costs and minus one x sine x minus.
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The integral now I need to differentiate, if so and minus one costs and minus two x and then minus sine X.
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And then another sign X the X. It equals costs and minus one x sine X.
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Minus and minus one times, or maybe I'll make it a plus.
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Costs. And minus two x sine squared x the X.
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So if I write it like that. What do you do now?
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He right sine squared as one minus cos squared X.
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Which then gives you costs and minus one x sine X.
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Plus and minus one, the integral, of course, and minus two x the X.
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Minus and minus swan, the interval, of course, and Square X the X.
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So now I recognise that this is the integral, of course, and minus two is I sob and minus two.
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And the integral, of course, and this is in. So I have an equal that.
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So if I sold for, I end. We got I and equals.
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So I get an I and equals course and minus one x sine X plus and minus one i n minus two.
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Which gives me the recursive formula. I n equals one over n course and minus one x sine x plus and minus one over n i minus.
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So this is true for all and greater than or equal to two.
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OK. Now, if I want to know all of these integrals, I mean, using this formula.
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What else do I need to know? I zero and I one because it drops down by two.
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So let's compute zero and one. So we also need I zero, and I won, so I zero would be just the integral the X, which is X plus a c.
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And I Wan is the integral, of course, x the X, which is signing X plus C.
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And now with this, you can you can get any integral you want.
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For example, if you want to get, I don't know, ice.
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I six. You just follow that and you get that it's one sixth course.
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To the fifth. Sine X plus five over six times I for.
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Which is one of the six course. Sine X plus five over six times.
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I four is one fourth. Of course.
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Cubed X Sine X Plus.
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Three, four, two. Then what's eye to eye to his one half cos x sign x plus one half I zero i zero six.
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So you put. You substitute this in there and they get an I six is one sixth course to the fifth sine x plus.
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Five over 24 cores, cubed sine X.
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Plus. Five times, three times, one over six times, four times to cause sine X plus.
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Five times three over. I want over six times, four times to X Plus see.
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The. So it has I think I think you can you can probably cook up a general formula using this example, you see how it goes.
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So if I asked you to write. A Psalm involving all the terms, I think you can you can get the coefficients of each time inductively.
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Good. OK, so this is a quick review of integration by parts if if you're not.
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Fully comfortable with these examples or similar examples, then get up,
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get together an integration textbook and do a few more examples with integration by parts, substitutions and so on.
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Because differential equity, what we will do in solving differential equations will learn a lot of techniques.
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But ultimately you will have to integrate some, some some functions, so you should be able to do that.
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So what we learnt is how do you how to reduce the problem to integrating various functions?
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But you'll have to be able to do that. OK, so we discussed about the simplest kind of odds.
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Which can be solved just by direct integration. The next simplest.
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All these. Are the so-called separable.
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Oh, these. So we had the case did X equals f of X, which you can just integrate, the next case would be the I.D. X equals A of X times B of Y.
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So what I mean by that is that this is a function only index.
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Only, and similarly, B of Y is a function of.
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Why only? If you have a situation like that, then you can reduce it to the direct integration with one simple trick.
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If B y is not zero, then you divide by it and you get one over B of y d y the X equals a x.
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And now you can integrate just as we did before.
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So you'll get then the integral. Yes.
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So the left hand side is the integral DUI over B of Y and the right hand side is a D X two.
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And now you have to direct integration switch.
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Hopefully we can we can solve the type of integrals that we have in this course will be the
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time for which you can apply integration by parts or some other techniques and solve them.
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If I if I were to write an arbitrary function there and ask you how to integrate it, then we can't do that in in a closed formula.
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OK, so here here's an example. Find the general solution.
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To the separable. Differential equation.
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So the hint is already in the problem that this is a separable differential equation
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x times y squared minus one plus y x squared minus one d y d x equals two zero.
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And. X is between zero and one to avoid.
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Some some issues. About continuity or what not?
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OK. How do you separate this differential equation?
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I'll do separate variables. Correct.
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So I think you're. About two steps ahead of me, so I will first.
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But it's correct. But let me do it step by step.
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So what I will do is first isolate that.
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So I have y x squared minus one d y x equals.
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Minus x y squared minus one.
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And then separate the variables, as the name suggests, you have y over.
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Y squared minus one DUI, d x equals minus x over x squared minus Y.
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OK. What do we do now? Correct.
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So if we look at this, then so we integrate, let's integrate.
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Well, let me write one one more, so we integrate this and we get why over y squared minus one d y equals minus x over x squared minus one d x.
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So now we could do a substitution, as we did before, but I think we know how to do it.
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This looks like the derivative of a logarithm. So if I differentiate a land of X squared minus one.
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Then I get to X over X squared minus one.
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So except X is between zero and one, so maybe it's better to write this as one x over one minus x squared and get get rid of the minus.
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Sign. So then I'll do one minus x squared minus two x over one minus x squared.
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So then this is minus 11 of one minus x squared.
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And the half. And then plus, see?
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Whereas here I'll have to put absolute values because I don't know, it's one half.
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It's a long way squared. Minus one.
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In absolute value. Right. Now, the easiest way to write this is to get rid of the logarithm by moving this to the other side,
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using the properties of the logarithm an exponential rate.
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So let's do that. So we have one half. If I move the logarithm in X to the left hand side, then I use the property.
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Well, it doesn't matter much. Equals C.
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Which means that. The equation will be y squared minus one times one minus six squared.
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Absolute value equals it would be easy to see squared.
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Or E to to see which I can just call.
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Another kind of sea. And this would be a positive no.
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So the equation then that we get is. So the answer then is.
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Is that? Where a sea.
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It's positive, but I can't relax that this is always positive,
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because one minus x squared is always positive because I'm assuming X is between zero and one.
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But I can rewrite the answer in a nicer form by dropping.
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The absolute value and requiring and dropping the assumption on C.
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So another way of doing this is.
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Well, for you uniformity, I'll write it as one minus y squared equals C, so one minus y squared times one minus x squared equals C.
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No assumption on C except. So she could be both positive or negative.
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Except in this for me, it looks like it can't be zero. Right here, I got an exponential which is never zero.
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So this is positive. I drove the absolute value, and the cost is that now see.
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Can also be negative. But somehow zero is missing.
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How is that possible? That doesn't look like solid mathematics.
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Yes. That's right.
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OK, so where did I lose that case? Right here.
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So I divide it by white squared minus one. I have I did that.
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Ignoring the case when y squared minus one is zero.
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So note, so let's call this story here, so in star.
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Why minus one is why squared minus one is not zero.
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But if we need to allow that because it is possible for white it to be plus or minus one, for example.
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If why is the cost and function one, then this is zero.
326
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The idea is zero, so that's OK. So if we allow, if y is plus plus minus one is included.
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In the solution. If we allow.
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See, to be zero. You know, in the answer.
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So then the bottom line is that the answer is. This implicit equation in Y and X, where a C can be any constant.
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Good. So be careful the when when you divide by the functioning, why, as I said, here you can do that if you know that's not zero.
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But sometimes you get solutions from it being zero, so you have to be careful there.
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All right. That's the end of the first lecture, I'll see you tomorrow for for the second lecture and we'll do more differential equations.