differentialequations-4

5:03-6:02Let me give you a very simple example which will explain most of these things. It’s the equation, it’s a version of the Cooling Law, which applies at very high temperature and it rusn. It’s cooling law except it’s internal and external temperatures are very, what is important is not its first power …. But the fourth power.. so it’s a constant, the difference is now the external temperature, so there weren’t be so many captital t’s M the fourth power minus the T to the fourth power. So T is the internal temperature, the thing we are interested in. and M is the external constant which I will assume now it’s an external constant temperature.

6:03-7:03 so this is validate for big temperature differences. …. Law breaks down. Now the free to solve that equation is to just stand and there are difficulties ….. because it deals with fourth power. But before you do that, one should scale, I will show you how to scale. While you scale by…. Relating T to M ,  so the new variable I am going to use is T1 equals T divided by M. this is now dimensionless because M to the fourth is the units of temperature ………….celces, degrees, whatever it is 

7:04- 8:17 as does T therefore, by taking the ratio, there it become there is no units attach to it. So this is dimensionless. So how actually do I change the variables in the equation? Watch this, it’s utterly truly idea and utterly important. Don’t slog around doing it this way to stuff it in and divide first. Instead, do the inverse, in other word, write T equal MT1. the reason being T is the one facing you in the equation. …………..T you want to substitute for. So if you do it, the new equation will be what, the T …. Since this is the constant, the left hand side becomes dT1 over dT times M equals K times M to the fourth minus M to the fourth T1 to the fourth so I am gonna factor out the M to the fourth and make it one minus T1 to the fourth, OK?.

8:18-9:08now I can divide through by M and get rid of one of those and so the new equation is dT1 dT is equal to k I am going to give a new name k1. essentially, it’s the same equation, it’s no harder to solve and no easier to solve than the original one. But it has been simplified as least one thing is better, it looks better so to compare those two, I will put this one green and this one green too. Just ….. indicate these are the same equations.

9:09-10:02 notice now T1 is dimensionless so I do not have to even ask when I saw this equation please tell me what the units of temperatures are, what kinds of, how you are measuring temperature makes no difference to this equation. K1 still has units, what units does it have……  it’s been simplified now it’s has the units of .. since this is dimensionless and this is dimensionless, so it has the units inverse time. So K1, was that units involve both degrees and seconds before. Now it’s inverse time, as it’s units. And moreover, there is one less constant, so one less constant in the equation. 

10:03-11:08 it just looks better. Sure, This business, I think you know, k1,the passes, forming K1 out of k M cubed is called lumping constant. I think they use that standard terminology in physics and engineering, constant. Try to get all the constant together, like this, and you make them they lumped, they are lumped for you and you just give the lump a new name. So that’s an example of scaling. Watch out for when you can use that. For example, it would have probably be a good thing to use on in the first problem. When you are handling this problem of drug elimination, holmon elimination production, sort of the thing, you can lump constant and as wish to exend on the solution… a neater …without so many constants in it. .

11:09-12:23 ok, now, let’s go to the serious stuff where we are actually gonna make changes of variables which we hope rande unsolvable equations become solvable. Now I am going to make do that by making substitutions. But I think it’s quite important to watch out that there are two different kinds of substitutions. There is direct substations, that’s where you introduce a new variable, I donot know how to write this on the board so I will just write it schematically. The new, it’s the one which it says the new variable, is equal to the some combination of the old variables. The other kind of substitution is inverse, it’s just the reverse, here you say the old variables, are the some combination of the new, often you have to stick into some old variables, too. But the basic is what it appears on the left hand side. Is it the new var that appear on LHS by itself or is it the old variable that appears on the LHS.

12:24-13:45Now right here we have an example. If I did it as a direct substitution, I would written T1 equals T over M, that’s the way I defined the new variable, which of course you have to do if you introducing it. But when I actually did the substitution, I did the inverse substitution. Namely, I used T equals T1, M times T1, and the reason for doing that because the capital T is facing me in the equation and I had to have sth to replace it with. Now, you have all seen this in calculus, this distinguish, but that might have been a year and half ago. Just let me remind you, typically in calculus, for example, when you want to do this type of integral, x time the square root of one minus x square dx, the substitution you would use for that is u equals one minus x square, right? And you calculate and you observe that this x dx more likes mix up with du apart from the constant factor they are. So this would be an example of direct substitution. You put it in and convert the integral into an integral of u.

13:46-14:45What would be an example of inverse substitution? I take away the x, and I was skilled to set to do this integral, then you know the right thing to do is not to start with u but start with x, and write x with sin or cos u. so this is a direct substitution in that integral, but this integral calls for an inverse substitution in order to be able to do it. And notice it looks practically the same but of course you know from your experience they are not. They are very different. Ok, so I am going to watch for that distinguishing as I do these examples. The first one I want to do is an example with direct substitution.

14:46-17:12So applies to the equation of the form y prime equals to, there are two term, two kinds of terms on the right side. A times, er… let’s use p of x, p of x time y plus q of x time any power of what so ever of y. notice for example n is zero, what kind of equation will this be. Y to the end will be one, and this will be a linear equation which you already know how to solve. So n is equal to zero which we already known how to solve, so let’s assume that n is not equal to zero, so we have a new territory. If we put n=1, then you can separate variables so that was not too exciting. But nontherless it will be included in what I am saying now. but if it is 2 or 3, n also can be 1/2, so n is anything even with zero, which is quite silly. Any number, could be negative num, n=-5 will be fine also. This kind of equation to give the name is the Bermouilli equation, named after which bermouilli I have not a slightly idea. There are three or four of them and they fought each other and they are all smart. The key trick, if you like, to solve any Bermouilli equation, let me talk one of thing I should talk, the most important thing is what’s missing, it must not have any pure x terms in it, and that goes for constant term either, in other words, it must look exactly like this. Everything is multiplied by y or power of y, two terms. So for example if I add a one to this, the equation will become unduable. It’s very easy to contemnated it into an equation that’s unsolvable. It’s gotta look just like that. Now you gotta one in your homework, actually you got several, part one and part two has equations on it… this is practical in some sense.

17:13----18:03The idea is to divide by y to the n, ignore all formulas that you are given, just remember when you see sth that looks like this, or sth that you can turn it into sth that looks like this, divide by y to the n’s power, no matter what n is. So y prime over y to the power of n is equal to p of x time one over y to the power of n-1 plus q of x. well, that certainly doesn’t look even better than where i started with, in your terms you probably even looks worse because it has those ys in the denominator and who wants to see them there.  

18:04---19:11 but look at it in this very slightly transformed Bermouilli equation is a linear equation struggling to be free. Where is it? Why is it trying to be a linear equation? Make a new variable, call this hunk of it the new variable, let’s call it V, so v is equal to 1 over y to the n-1, or if you like it, you can think of it as y to the 1-n. What’s V prime? So this is the direct substitution that I am going to use but of course the problem is where I am going to use on it. Well, there is a little miracle happens, what’s the derivative of this is 1-n times y to the negative n times y prime. In other words, up to this constant factor 1-n, it’s exactly LHS of the equation.

19:12--- 20:00well, let’s make n not equal to 1, as I said, you can separate variables with n equals to one. What’s the equation then turned into? A Berrnouilli equation divided by this way is turned into equation 1-n, sorry, V prime divided by 1-n is equal to p of x times v plus q of x, it’s linear.

20:01—21:06and now solve it as linear equation, notice now, it’s not in standard form, not in standard linear form. To do that, you gonna have to put the p on the other side, that’s ok, that term on the other side, solve it. And atthe end, don’t forget you put in the V, it wasn’t in the original problem, so you have to convert the answer back in terms of y. It will come out in terms of V but you have to convert it in terms of y. let’s do a really simple example just to illustrate the method, to illustrate the factor that I don’t want you to memorize the formulas. Learn methods not final formulas. So suppose the equation is y prime equals y over x minus y square. That’s a Berrnouilli equation.   

21:07—22:11I could have concealed it by writing x y prime minus y equals to negative y square, then it wouldn’t instantly look like a Berrnouille equation. But if you have to stare it for a while, you will say, hey it’s a Berrnouille equation. So I am handing it to you on the silver plate as it were. So what we do? Divided through by y square, so it’s y prime over y square equals one over x time one over y minus 1, so now, the substitution I am going to make is for this thing. V equals 1 over y. so direct substitution, V prime is going to be -1 over y square time y prime, don’t forget to use the Chain rule when you differentiate with respect because the differentiation with respect of x, of course not with respect of y.

22:12—24:05so what’s this thing, on the LHS, it got a negative sign. So it’s minus V prime equals, 1/x stays 1/x and 1/y, so V/x-1, so let’s put this into standard form. In standard form, this thing will look like, first imagining the whole thing times -1 and put this term on the other side. And it will turn into V prime plus V/x is equal to one. So that’s a linear equation in the linear standard form we asked to solve and the solution isn’t very hard. The integrating factor is well, I integrate 1/v that is logX, and e to the log x, which is just x itself, so I should multiply it through by x to be able to integrate it. Some of you, I would hope, could see that right away that if you multiply through by x it will look good. So after we do multiply through by x, we will get xv prime, maybe I should not skip the step. You are still learning this, I should not skip the steps. So it becomes xv prime plus v equal to x after multiply through by the integrating factor. So this is (xv) prime, I quickly checked it should be it equal to, equals to x, and therefore xv is equal to 1/2 x square plus a constant.

24:06—25:53and therefore V is equal to 1/2 x plus c over x. you could leave it in that form or you can combine terms, that doesn’t matter much. Am I done? The answer is no I am not done because nobody reading this answer would know what V was. V wasn’t in the original problem, it was y in the original problem, therefore, I have to, the ratio is one is the reciprocal of the other, therefore, I have to turn this expression upside down. Well, if you are going to turn it upside down, you probably should make it look better. Let’s rewrite it as x square plus 2c, combing fractions, I think they call it in high school or elementary school, plus 2C, how’s that. X square plus 2C divided by 2 x. Now, 2c, you don’t call a constant 2c, it’s just a arbitrary, call it C1, my answer will be y equals 2x divided by x square plus an arbitrary, to indicate it different from that one, I will call it c1, c1 is two times the old one but that doesn’t really matter. So there is the solution. It has an arbitrary constant in it but you know it’s not an additive arbitrary constant. The arbitrary constant is tucked into the solution. If you have to satisfy an initial condition, you will take this form and you will start with this form, figure out what c1 was in order to satisfy the initial condition.

25:54—26:55 thus the Bernoulli equation is solved, the first Bernoulli, isn’t that exciting? There is the equation and there is its solution. Now the one I ask you to solve in the problem set in part two is no hard than this except I ask you some hard questions about it, not very hard but a little hard. But I hope you find them interesting questions, you have already had the experiment evidence from the first problem set, the problem that is to explain experiment evidence by actually solving the equation. I did find it interesting, but maybe that is just my highest hope.

26:56—28.11 I’d like to turn to the second method where the second class of equation which requires inverse substitution. And also the equation which is called Homogenous, highly overworked word in differential equations and mathematic in general, but unfortunately it’s just right word to describe them. So these are homogenous first order—ODE’s now I already used word one context in talking about the linear equations, zero is the right hand side, this is different but nonetheless, the two uses of the words have the common source. Homogenous differential equation, homogenous you speak, is y prime equal, it’s the question what the right hand side looks like, now the simplest way to say it is you should be able to write the right hand side as a function of combined variable of y/x. In other words, after fooling around with the right hand side a little bit, you should be able to write it so that every time a variable appears, it’s always in the combination of y/x.

28:12—29:03Let me give some examples. For example, suppose y prime is x square y divided by x cube plus y cube, that doesn’t look like the right form but it is. Imaging dividing the top and bottom by x cube, what would you get? The top will be y/x if you divided by x cube; and if I divide the bottom by x cube also, which of course that doesn’t change the value of the facture, as they say in the pre-school, 1+y/x, so this is the way how it start out looking, but you said ha, that is a homogenous equation because you can see looking that way.

29:04—30:04 How about another homogenous equation? How about xy’= root(square(x)+square(y))? Is that a homogenous equation? Of course it is, otherwise why would I be talking about it? You divide through by x, you can tuck inside the radical square root if you remember to square when you do that, and become of square root of x square over x square, which is one, plus y square over x square, it’s homogenous. Now, you might say: hey this looks like that you have to be rather clever to figure out the equation to be homogenous, is there some other way? Yeah, there is some other way and it’s the some other way which explains why it’s called homogenous.

30:05—31:08 You can think of this way. It’s an equation which is, in modern speak, invariant under the operation zoom. What is zoom? Zoom is you increase the scale equally on both axis. So the zoom operation is the one which says x into a times x, and y into a times y, in other words, you changed the scale on both axis by the same factor a. Now what I can say is, Gee, maybe I shouldn’t write like this. Why don’t we say we introduce, how about this, which I think is a change of variables, write like that, and you can put here a equal sign if you don’t now this meaningless arrow means.

31:09—32:29so I am making this change of variables and I am describing it as an inverse substitution, but of course it wouldn’t make any difference if it’s exactly the same as the direct substitution I started out with. And the scaling, the only difference is I am not using different scales on both axis, I am expanding them both equally and that’s what I mean by zoom. Now what happens to the equation? What happens to dy/dx? Well, dx is adx1, dy is ady1, and therefore, the ration, dy/dx, is the same as the dy1/dx1. So the left hand side becomes dy1/dx1, and right hand side becomes F of, well, y/x is the same as, since I scale them equally, this is the same is y1/x1, so it’s y1/x1. And the matter of fact is, I simply everywhere I have a x, I change it into x1, and wherever I have a y, I change it into a y1, which what’s in name, it’s a identical equation.

32:30— 33:58 So I haven’t changed equation at all by zoom transformation, and that’s what makes a homogenous. But that’s not an uncommon use of the word homogenous. It’s when you say space is homogenous, that means, I don’t know, means you know, I am giving the trouble in. I have to let it go since I haven’t prepared any better explanation. This is a pretty good one. Ok, suppose we got a homogenous equation, how do we solve it? So here is our equation F of y/x. Well, what substitution you would like to make? Obviously, we should make a direct substitution z=y/x, so why did you say this would be an example of inverse substitution? Because I want to confuse you, but look, that’s fine. If you write it in that form, you would know exactly what to do with the right hand side and this is why everybody loves to do that.

33:59—35:10 Unfortunately you have to substitute into the left hand side as well. And I can testify from many years of looking, thinking hard of the examination papers, what happens if you try to make a direct substitution like this is o, I need z’. z’ equals, I ‘d better use the Quotion Rule and you know, it comes this long and then either a long pause what I do now? Because it’s not at all obvious what to do at that point; or much worse, two pages of frank calculations and giving it in total despair. Now the reason for that is because you try to make a direct substitution. All you have to do instead is to treat it as an inverse substitution, make y=zx. What’s the motivation in doing that? It’s clear from the equation, this goes through all the mathematics, whenever you have to change a variable, look at what you have to substitute for, and focus your attention on that.

35:11—36:11I need to know where y’ is? Ok, then I’d better know where y is. If I know what y is , do I know what y’ is? Of course, y’ = z’x plus z times the derivative of this factor, which is one. And how I’ve turned, without one scope, the equation is now become z’x+z=F(x). I don’t know. Can I solve that? Sure, that can be solved because this is x times dz/dx and put the z on the other side, it’s F(z)-z, and now this side is just a function of z, separate variables, and the only thing to watch out for is that at the end, the z was your business, you got to put the answers back in terms of x and y.

36:12— 37:56 ok, so let’s see a working example of this, since I haven’t done any modeling yet in this period, so let’s make a model, differential equation model, it’s a physical situation, which will be solved by a equation, and guess what, it will turn out to be homogenous. Ok, so the situation just follows, we are in the Caribbean somewhere; at a little isolated island with a light house on it. That’s the origin, and the bean of light shine from the light house. The bean of light could rotate the way with the light house spins, but this particular bean is being controlled by a guy in the light house aiming where he wants. And the reason he is interested in aiming where he wants is because there is a drug boat, here, which is just being caught in the bean of light. So drug boat, which is just being caught by the bean of light and feels it’s better to escape. Now the light house keeper wants to keep the drug boat light shining absolutely so that the US coast guarder, AD cops can zoom over it in and do whatever they do to the drug boat, I don’t know. So the drug boat immediately has to follow an escape strategy, and the only one that occurs in the situation: must go further away, of course, from the light house. On the other side, it doesn’t seem sensitive to do it on the straight line because the bean will keep shining on it.

37:57—39:11so he fixes his boat on some angle, let’s say, so it goes off so the angle stays 45 degrees. So the angle between the bean and, have to draw the bean a little less so it looks like a 45-degree angle, so the angle between the bean and the boat path is always 45 degrees. The boat goes at a constant 45-degree angle to the bean, hoping thereby to escape from the light house. On the other hand, of course, the light house guy keeps the bean on the boat, so it’s not clear, it’s a good strategy but this is a differential equation’s class. The question is what’s the path of the boat? What’s the boat’s path? And the obvious question is why this a problem of differential question at all? In other words, looking at this, you might scratch your head and think of different ways to solve it but why, what adjust it going to be a differential equation. 

39:12—40:10the answer is you are looking for a path, I am looking for the answer is going to be a curve. A curve means a function; we are looking for a unknown function, in other words. And what type of information do you have about the function? The only information we have about the function is something about its slope, that slope is making a constant 45-degree angle with the light house bean. Its slope makes a known angle. Well if you are trying to make it into a function, and what you know is something about its slope, that is a problem of differential equations.

40:11—41:22 ok, let’s try to solve it. Let’s see. Well, let me draw just a little bit so here is the horizontal, let’s introduce the coordinates, so there is the horizontal and here is the boat to indicate where I am with respect to the picture. So here is the boat, here is the beam and the path of the boat is going to make a 45 degree angle with it so this is the path that we are talking about. And now let’s label what I know. Well this angle is 45 degrees. This angle I don’t know but of course I can calculate it easily enough because it has to do with, if I know the coordinates of this point (x,y), and of course that horizontal angle, I know the slope of this line and that angle will be related to the slope.

41:23—42:33 so let’s call this alpha. And now what I want to know is what the slope of the whole path is. So y’, so let’s call y=y(x) the unknown function whose graph is going to be the boat’s path, unknown graph. What’s its slope? Well, its slope is the tangent of the sum of those two angles, (alpha+45 degrees). Now what I know? Well, I know the tangent of the alpha is how much, that’s y/x, in other words, if this is the point of x/y, this is the angle, it makes with the horizontal, if you think it over here, this is parallel so this angle is the same as that one. And the slope is y/x, so the tangent of the angle is y/x.

42:34—43-57 How about the tangent of 45 degrees, that’s one. And there is a formula, this is the hard part, you have to know the formula exist and you look it up if you forgot it, relating the tangent giving you the tangent of the sum of two angles. And you could if you like you clever derive it from the formula for the sine and cosine of the sum of the two angles, but one peak is worth a thousand synthesizers. So it is, the tangent of alpha plus the tangent of 45 degrees. Let me write it all out in great details so at least u will learn the formula, 1 minus the tangent of alpha times the tangent of 45 degrees this will work for the tangent of the sum of another two angles, that’s the formula. So how’s that? What do I get then? y’ is equal to the tangent of alpha, which is y/x, I like that combination, plus 1, divided by (1-y/x*1). There is no reason doing anything to it but let’s make it a little prettier and thereby, make it less obvious it’s a homogenous equation. It’s, by multiply both top and bottom by x, it looks prettier. (y+x)/(y-x) equals y’, that’s our differential equation.

43:58—45:36 But notice, the last step to make it pretty has undone the good work. It’s fine if you immediately recognize this being a homogenous equation because you could divide top and bottom by x. But here it’s a lot clear that it’s a homogenous equation because it has already been written in the right form. Ok, let’s solve it now since we know what to do. We are going to use a new variable z=y/x. and as I wrote up there, for y’, we will substitute z’x+z. And with that, let’s solve. All the equation becomes. z’x+z is equal (z+1)/(1-z), put z on the other side, we want to separate variables so has to put all the z on one side. so this is going to be: xdz/dx equals this thing minus z, which is (z+1)/(1-z)-z. And now as you realize putting it on the other side, I am going to turn it up-side-down. Just as before, if you have to turn sth up-side-down, it’s better to combine the terms and make it one tiny little fraction. Otherwise, you are in for a quite a lot mess if you don’t do this nicely.

45:37—47:12 (z+1)-z that’s all about z, the numerator is (1+z square)/(1-z), and so the equation is dz, what does it put the other side and turn it up-side-down, so that will be (1-z)/(1+z square) on the left hand side, and on the right hand side is dx/x. Well, that’s ready to be integrated just as it stands. The right hand side is going to be logx, the left hand side is the sum of two terms. The integral of (1+z square) is arctan(z), the derivative of this is 1/(1+z square). How about the term z/(1+z square), that integral is a logarithm, it’s more or less of the log(1+z square), if I differentiate it, I will get 1/(1+z square) times 2z, but I wish to have negative here instead. So I should put a minus sing, and multiply by 1/2 to make it come out right, and this is logx on the right hand side and pluc c, put it as an arbitrary constant.

47:13—48:39 And now what? Let’s now fooling around with it a little bit. The arctan, I am going to simultaneously, no, two steps, I have to remember your innocents. Though probably a lot of you are better calculators than I am. I am going to change this, use as many of the laws of logarithm as possible, I am going to put this to the exponent, and put this on the other side, that’s gonna turn it into the ln(square root(1+z square))+ln(x)+c. and now I am gonna go back and remember z=y/x. So this becomes the arctan(y/x) equals, now I combine the logarithm, this is the lnx time the ln square root, make one logarithm out of it, and then put z=y/x, and you will see it if you do that, ln(x*square root (1+square(y/x))), what is that? Well, if I put this over x square and take it out and cancelled it and what you left is ln(square root (x square + y square)) plus a constant.

48:40—49:30 now technically you solve the equation but not morally, because I mean, my god, what a mess, incredibly bad, you know tells me absolutely nothing. Well, what is it screaming? Change me too~~~(professor being sarcastic). What a coincidence, what is the arctan(y/x)? theta (⊙)in polar coordinates, it’s theta. What this is are so the curve is theta equals ln of r + a constant and I could make that even look a little better if I exponentiate everything, exponentiate both side, combine this as the usual way and what you get is r is equal to some other constant times e to the theta.  

49:31—that’s the curves. It’s called expenetial spiral, and that’s what, that’s where our boat goes in and notice probably if I set up the problem in polar coordinate from the beginning, nobody would be able to solve but if anybody who did it will get that answer immediately.  Thanks.