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Last time I spent solving a system of equations dealing with the chilling of this hard-boiled egg being put in a nice bath.
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We call T1 the temperature of the yolk and T2 the temperature of the white.
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What I'm going to do is revisit that same system of equations.
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Basically today, the topic for today is to learn to solve that system of equations by a completely different method.
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It's the method that's normally used in practice.
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Elimination is used mostly by people who have forgotten how to do it any other way.
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Now, in order to make it a little more general, I'm not going to use the variable, dependent variables, T1 and T2, because they suggest temperature a little too closely.
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Let's change them to neutral variables. I'll use x equals T1, and for T2, I'll just use y.
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I'm not going to re-derive anything. I'm not going to re-solve anything. I'm not going to repeat anything of what I did last time.
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Except to write down, remind you of what the system was.
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In terms of these variables, the system that we derive are using that particular conductivity constants 2 and 3 respectively.
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The system was this one, minus 2x plus 2y, and the y prime was 2x minus 5y.
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We solve this by elimination.
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We've got a single second-order equation, which we solve with constant coefficients, which solve the usual way.
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From that, I derive what the x was. From that, we derive what the y was, and then I put them all together.
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I'll just remind you what the final solution was when written out in terms of arbitrary constants.
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C1 times e to the negative T plus, I forget, C2, e to the negative 60, and y was c1 over 2, e to the negative T plus minus 2c2, e to the negative 60.
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That was the solution we got.
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Then I went on to put in initial conditions, but we're not going to do that. Explore that. I expect of it today. We will in a week or so.
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This was the solution. The general solution, because it allowed it to have two arbitrary constants in it.
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What I want to do now is revisit this and do it by a different method, which makes heavy use of matrices.
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That's a prerequisite for this course. I'm assuming that you reviewed a little bit about matrices.
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It's in your book. Or your book very puts in a nice little review section.
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2 by 2 and 3 by 3 will be good enough for 18.03, mostly because I don't want you to calculate all night on bigger matrices, bigger systems.
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Nothing serious. Matrix multiplication, solving systems of linear equations, and by N systems. I'll remind you at the appropriate places today of what it is you need to remember.
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I haven't figured out the color coding for this lecture yet, but let's make the system in green and the solution can be in purple.
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Invisible purple. But I have a lot of it.
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Let's abbreviate the system using matrices. I'm going to let xy, I'm going to make a column vector out of xy.
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Then you differentiate a column vector by differentiating each component.
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I can write the left hand side of the system as xy prime.
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About the right hand side. Well, I say I can write the matrix of coefficients to negative 2 to negative 5 times xy.
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I say that this matrix equation says exactly the same thing as that green equation.
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Therefore, it's legitimate to put it up in green too.
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Let's see. The top here is x prime. What's the top here? After I multiply these two, I get a column vector.
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What's its top entry? It's negative 2x plus 2y. There it is.
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The bottom entry, the same way, is 2x minus 5y, just as it is down there.
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Now, what I want to do is, maybe I should translate the solution. What does the solution look like?
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We got that too. xy equals, how am I going to write this as a matrix equation?
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Actually, this is, if I told you to use this, use matrices, use vectors, the point at which you might be most hesitant is this one right here.
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The next very next step. Because how you should write it is extremely well concealed in this notation.
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What the point is, this is a column vector and the sum of the, I'm adding together two column vectors and what's in each one of the column vectors.
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Both, think of these two things as a column vector, pull out all the scalars from them that you can.
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Well, you see that c1 is a common factor of both entries. And so is e to the negative t, that function.
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Now, if I pull both of those out of the vector, what's left of the vector? Well, you can't even see it.
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What's left is a 1 up here and a 1 half there.
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So, I'm going to write that in the following form. I'll pull out the c1 that's the common factor in both and put that out front.
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Then I'll put in the guts of the vector, even though you can't see it, the column vector 1, 1 half.
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And then I'll put the other scalar factor, scalar function, in back.
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The only reason for putting one of these in front and one in back is visual, so to make it easy to read. There is no other reason.
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You can put the c1 here, you can put it here, you can put the e-negative t in front.
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If you want to, but people will fire you. Don't do that. Write it the standard way because that's the way that it's easiest to read.
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The constants out front, the functions behind and the column vector of numbers in the middle.
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The other one will be written how? Well, here, that one's a little more transparent. C2, 1, 2, and the other thing is e to the negative 16.
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So, there's our solution. That's going to need a lot of purple, but I have it.
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And now I want to talk about how the new method of solving the equation. It's based just on the same idea that the way we solved second-order equation, yes, question.
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Is what?
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Oh, here, sorry, this should be negative, too. Thanks very much.
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Well, I'm going to use as a trial solution. Remember when we had a second-order equation with constant coefficients?
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The very first thing I did, I said, okay, we're going to try a solution of the form, e to the, lambda, e to the RT.
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Why that? Well, because oil is thought of it and it's been known for two or three hundred years that that's the thing you should do.
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Well, this hasn't been known nearly as long because matrices were only invented around 1880 or so.
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And people didn't really use them to solve differential equations, systems of differential equations, until the middle of this, the last century, 1950, 1960.
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If you look at books written in 1950, they won't even talk about systems of differential equations.
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Or they'll talk very little anyway, and they won't solve them using matrices.
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So, this is only 50 years old, I mean, my god, that's...
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I'm not a mathematics, that's very up to date. Okay. Particularly elementary mathematics.
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Anyway, I'm going to... The method of solving is going to use as a trial solution.
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Now, the form of the... If you will left your own devices, you might say, well, let's try.
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So, this equals some constant times e to the lambda 1t and y equals some other constant times e to the lambda 2t.
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Okay, we'll go in. Now, if you try that, it's a sensible thing to try, but it will turn out not to work.
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And that's the reason I've written out this particular solution.
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You can see what solutions look like. The essential point is, here's my solution, the basic solution I'm trying to find.
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Here's another one. The form is a vector of constants, a column vector of constants, but the exponential factor, they both use the same exponential factor.
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That's the point. In other words, I should not use here in my trial solution to different lambdas.
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I should use the same lambda.
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And so, the way to write the trial solution is, xy equals two unknown numbers, that, or that, or whatever, times e to an unsingled unknown exponential exponent factor.
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Let's call it lambda t. It's called lambda, it's called r, it's called m. I've never seen it called anything, but that was one of those three things.
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But I'm using lambda, your book uses lambda. It's a common choice. Let's stick with it.
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Okay, now, what's the next step? Well, we plug in to the system. So, substitute into the system.
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What are we going to get? Well, let's do it. First of all, I have to differentiate. The left hand side asks me to differentiate this. How do I differentiate this?
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A vector times a function. Well, the column vector acts as a constant, and I differentiate that. That's lambda e to the lambda t.
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It's the xy prime is a1, a2 times e to the lambda t times lambda. Now, it's ugly to put the lambda afterwards, because it's a number, but I said you should put it in front.
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Again, to make things easier to read. But this lambda comes from differentiating e to the lambda t and using the chain rule. So, this much is the left hand side. That's the derivative xy prime.
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I differentiate the x and I differentiate the y. How about the right hand side? Well, the right hand side is negative 2, 2, 2, negative 5 times what? Well, times xy, which is a1, a2, e to the lambda t.
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Now, the same thing that happened a month or a month and a half ago happens now. The whole point of making that substitution is that the e to the lambda t, the function part of it, drops out completely.
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And one is left with what? An algebraic equation to be solved for lambda a1 and a2.
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In other words, by means of that substitution, and it basically uses the fact that the coefficients are constant, what you've done is reduce the problem of calculus of differential equations of solving differential equations to solving algebraic equations.
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And so, this is the only method there is, unless you do numerical stuff. You reduce the calculus to algebra. The Laplace transform is exactly the same thing. All the work is algebra.
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You turn the original differential equation into an algebraic equation for y of s. You solve it and then you use more algebra to find out what the original y of t, little y of t was.
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It's no different here. Okay, so let's solve this system. Okay, now the whole problem with solving this system. First of all, what is the system? Let's write it out explicitly.
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Well, it's really two equations, isn't it? The first one says lambda a1 is equal to negative 2a1 plus 2a2. That's the first one.
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The other one says lambda a2 is equal to 2a1 minus 5a2.
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If you only want to classify that, that's two equations in three variables, three unknowns. The a1, a2 and lambda are all unknown. And unfortunately, if you want to classify them correctly, they are nonlinear equations because they made nonlinear by the fact that you've multiplied two of the variables.
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Well, if you sit down and try to hack away at solving those without a plan, you're not going to get anywhere. It's going to be a mess. Also, two equations and three unknowns is indeterminate.
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You can solve three equations and three unknowns and get definite answer, but two equations and three unknowns usually have an infinity of solutions.
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Well, at this point, is the only idea that's required? Well, this was a little idea, but I assume one would think of that.
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The next one is, the idea that's required here is, I think, not so unnatural. It is not to view these a1, a2 and lambda as equal.
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Not all variables are created equal. Some are more equal than others. A1 and a2 are definitely equal to each other. And let's relegate lambda to the background.
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In other words, I'm going to think of lambda as just a parameter. I'm going to demote it from the status of variable to a parameter. If I demoted it further, it would just be an unknown constant. That's as bad as you can be.
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So I'm going to focus my attention on the a1, a2 and sort of view the lambda as a nuisance. Now, as soon as I do that, I see that these equations are linear if I just look at them as equations in a1 and a2.
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And moreover, they're not just linear, they are homogenous. Because if I think of lambda just as a parameter, I should write the equations. I should rewrite them this way.
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I'm going to subtract this and move the left hand side to the right side. It's going to look like minus 2 minus lambda times a1 plus 2a2 is equal to 0.
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And the second one is going to be 2a1 plus what's the coefficient? Minus 5 minus lambda a2. That's a pair of simultaneous linear equations for determining a1 and a2. And the coefficients involve a parameter lambda.
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Now, what's the point of doing that? Well, now the point is whatever you learned about linear equations, you should have learned the most fundamental theorem of linear equations.
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The main theorem is that you have a square system of homogenous equations. This is a 2 by 2 systems that would square. It always has a trivial solution of course, a1, a2 equals 0.
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Now, we don't want that trivial solution because if a1 and a2 are 0, then so are x and y0. Now, that's a solution unfortunately said no interest.
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If the solution were x and y are 0 corresponds to the fact that this is an ice bath, the yolk is at 0, the white is at 0, and it stays that way for all time until the ice melts.
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So that's the solution we don't want. We don't want the trivial solution. Well, when does it have a non-trivial solution? Non-trivial. That means non-zero in other words.
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The solution, if and only if this determinant is 0. In other words, what we find is there's a condition on lambda.
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By using that theorem on linear equations, what we end up with is a condition that lambda must satisfy. In equation and lambda, in order that we'd be able to find a non-zero values for a1 and a2.
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So let's write it out. I'll recopy it over here. What was a negative 2 minus lambda? 2 here was 2 and minus 5 minus lambda.
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So let's, this is an equation. You have to expand the determinant. In other words, we're trying to find out for what values of lambda is this determinant 0. Those will be the good values which lead to non-trivial solutions for the a's.
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So this is the equation 2. So this is lambda plus 2. See if I multiply this is minus that and minus that. The product of the 2 minus 1 is plus 1. So it's lambda plus 2 times lambda plus 5. That's the product of the 2 diagonal elements.
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Minus the product of the 2 anti diagonal elements that's 4 is equal to 0. And if I write that out, so what is that? That's the equation.
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Lambda squared plus 7 lambda, right? 5 lambda plus 2 lambda. And then the constant term is 10 minus 4 which is 6.
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Ah, how many of you have long memories, long enough memories, two day memories that remember that equation. When we did the elimination, when I did the method of elimination, it led to exactly the same equation except it had r's instead of lambda.
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And this is equation therefore is given the same name and another color. Let's make it salmon.
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So this is our and it's given the same name of this is called the characteristic equation. For this method. All right, we now I'm going to use the work from last time. You know, you factor this from the factorization.
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We get its roots easily enough. The roots are lambda equals negative 1 and lambda equals negative 6 by factory equation. Now what am I supposed to do?
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It's, you have to keep the different parts of the method together. Okay, now I found the only values of lambda for which I will be able to find non zero values for the a 1 and a 2.
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For each of those values of lambda, I now have to find the corresponding a 1 and a 2. So let's do the one at a time.
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So let's take first lambda equals negative 1. My problem is now to find a 1 and a 2.
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Now where am I going to find them from? Well, from that system of equations over there. I'll recopy it over here. What is the system? Okay, if lambda is negative 1 minus 2, the hardest part of this is dealing with the multiple minus signs, which, but you've had experience with that determinants, you know all about that.
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Okay, so in other words, there's the system of equations over there. The question is, let's recopy them here. So minus 2 minus minus 1 makes minus minus, well that makes minus 1.
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Okay, what's the other coefficient? It is just played all 2. Good. So this is my first equation. And when I substitute lambda equals negative 1 for the second equation, what do you get?
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2 a 1 plus negative 5 minus negative 1 makes negative 4. There's my system for finding that will find me a 1 and a. Now what's the first thing you notice about it? Well, you immediately notice this system is fake.
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Because this second equation is twice the first one. So something's wrong. No, something's right. If that did not happen, if the second equation were not a constant multiple of the first one, then the only solution of the system would be a 1 equals 0, a 2 equals 0.
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Because the determinant of the coefficients would not be 0. The whole function of this exercise was to find the value of lambda might negative 1 for which the system would be redundant and therefore would have a non-trivial solution.
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So in other words, calculate the system out. Just as I've done here, you have an automatic check on the method. If one equation is not a constant multiple of the other, you made a mistake.
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You don't have the right value of lambda or you substitute it into the system wrong, which is that's frankly a more common error. You substitute the system wrong. Go back, recheck first the substitution. And if you convince that's right, then recheck where you got lambda from.
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Okay, but here everything's going fine. So we can now find out what the value of a 1 and a 2 are. Now this is usually since you don't have to go through a big song and dance for this.
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Most of the time you'll have 2 by 2 equations and now and then 3 by 3. But 2 by 2, all you do is since we really have the same equation twice, I can assign to get a solution, I can assign one of the variables any value and then simply solve for the other.
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The natural thing to do is to make a 2 equal 1, then I won't need fractions, then a 1 will be 2. So a solution is.
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I'm only trying to find one solution. Any constant multiple of this would also be a solution, as long as it wasn't 0, 0, which is the trivial one.
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And therefore, so this is a solution to this system of algebraic equations and the solution to the whole system of differential equations, then, is this is only the a 1, a 2 part.
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I have to add to it. As a factor, the e to the will lambda is negative 1 and therefore e to the minus t. There's our purple thing.
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See how I got it? Starting with the trial solution, I first found out through this procedure what the lambda has to be, then I took the lambda, found what the corresponding a 1 and a 2 that went with it and then made up my solution out of that.
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Now, quickly, I'll do the same thing for lambda equals negative 6. Each one of these must be treated separately. There are separate problems. You're looking for separate solutions.
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Okay, for lambda equals negative 6, what do I do? Out of my equations, look now. Well, the first one is minus 2 minus negative 6 makes plus 4.
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So it's 4 a 1 plus 2 a 2 equals 0. And then I hold my breath while I calculate the second one to see if it comes out to be a constant multiple.
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So I get 2 a 1 plus negative 5 minus negative 6. That makes plus 1. Negative 5 minus negative 6 is plus 1. And indeed, 1 is a multiple, a constant multiple of the other.
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We really only have one equation there. And therefore, the a 1, a 2 for the a 1, I'll just write down immediately now what the solution is to the system.
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So, the a 1, a 2 will be 1. Now, it's more natural to make a 1 equal 1 and then solve to get an integer for a 2. If a 1 is 1, then a 2 is negative 2. And I should multiply that by e to the negative 6 t because this is negative 6 is the corresponding value.
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So, this is my other one. And now, there's a superposition principle which if I get a chance, I'll prove for you at the end of the hour. If not, you'll have to do it for yourself for homework.
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So, once you have, since this is a linear system of equations, once you have two separate solutions, neither multiple or constant multiple of the other, you can multiply each one of these by a constant and it will still be a solution.
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And that gives the general solution. So, the general solution is the sum of these two. An arbitrary constant, I'm going to change the name since I don't want to confuse it with the c1 I used before.
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Times, the first solution which is 2 1 e to the negative t plus c2 and other arbitrary constant times 1 negative 2 e to the minus 6 t.
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Now, you notice that's exactly the same solution I got before. The only difference is that I've renamed the arbitrary constants.
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The relation between them is that what I'm calling 1 1 1 half, what I c1 over 2 was, I'm now calling c1 tilde and c2, I'm calling c2 tilde.
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If you have an arbitrary constant, it doesn't matter whether you divide it by 2, it still just as arbitrary a constant. It covers all values in other words.
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Well, I think you'll agree that's a different procedure. It's a different procedure. Yet it has only one coincidence. Namely, it's like elimination goes this way and it comes to the answer.
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This method goes a completely different route and comes to the answer except it's not quite like that. They walk like this and then they come within viewing distance of each other to check that both they're using the same characteristic equation and then again they go this separate ways and end up with the same answer.
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So, they're something special about the values. You cannot get away from those two values of lambda. Somehow they're really intrinsically connected. They occur as the exponential coefficients and they're intrinsically connected with the problem of the egg that we started with.
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Now, what I'd like to do is, very quickly, sketch how this method looks when I remove all the numbers from it. In some sense, it becomes a little clearer what's going on. That will give me a chance to introduce the terminology that you need when you talk about it.
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Let's see. What do you got notes?
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Let's try to write it down in general.
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I'll first write it out by 2 by 2. So, I'm just going to sketch. So, the system looks like x, y equals, let's put it still up in colors.
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Except now, instead of using twos and fives, I'll use a, b, c, d. The trial solution will look how? The trial is going to be a1, a2. That I don't have to change the name of.
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I'm going to substitute in. And what's the result of substitution is, is going to be lambda, a1, a2. I'm going to skip a step and pretend that the e to the lambda t's have already been canceled out, is equal to,
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a, b, c, d times a1, a2. What does that correspond to? That corresponds to the system as I wrote it here. And then we wrote it out in terms of two equations.
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And what was the resulting thing that we ended up with? Well, you write it out, you move the lambda to the other side, and then the homogeneous system is, we'll look in general how? Well, we could write it out. It's going to look like a minus lambda, b, c, d minus lambda. That's just how it looks there.
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And the general calculation is the same, times a1, a2 is equal to zero. Now, that is solvable. This is solvable.
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Non-trivial, in other words, has a non-trivial solution. If and only if the determinant, this determinant is zero. The determinant of coefficients is zero.
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Okay, let's now write that out, calculate out once and for all what that determinant is. Okay, it's going to be, or let's, I'll write it out here. A minus lambda, times d minus lambda, that's the product of the diagonal elements.
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Minus the anti diagonal, minus b, c is equal to zero. And let's calculate that out. It's lambda squared minus a lambda, minus d lambda, plus a, d is the constant term from here, negative b, c from there.
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So, plus a, d minus b, c, where have I seen that before? So this equation is the general, sorry.
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That equation is the general form using letters of what we calculated using the specific numbers before. So, again, I'll code it the same way with, what's that color, is that salmon?
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Now, most of the calculations will be for two by two systems. I advise you with the strongest possible terms to remember this equation. You can write down this equation immediately from the matrix. You don't have to go through all this stuff.
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God's sakes don't say, let the trial solution be blah, blah, blah, blah, blah, blah, blah, blah, blah. You don't want to do that. I don't want you to repeat the derivation of this every time you go through a particular problem.
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It's just like in solving second-order equations. You immediately write down the second-order equation, you immediately write down its characteristic equation, then you factor it, you find its roots, and you construct the solution. It takes a minute. The same thing, this takes a minute too.
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Now, what is this for the, remember, what's the constant term? AD minus BC, what's that? Matrix as A, B, C, D, AD minus BC is it's D, E, you're terminating. This is the determinant of that matrix A. I didn't give the matrix a name, did I? Okay, I will now give the matrix a name A.
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What's this? Well, you're not supposed to know that until now. I'll tell you. This is called the trace of A. Put that down in your little books.
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The abbreviation is trace A, and the word is trace. The trace of a square matrix is the sum of it, the elements down its main diagonal.
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There are three by three, there would be three terms in that A, I don't know what your A, whatever you're up to. But here it's A plus D. The sum of the diagonal elements.
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So you can immediately write down this characteristic equation. Let's give it what's more its name. So this is the characteristic equation of what? Of the matrix now, not of the system, of the matrix.
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You have a 2 by 2 matrix. You can immediately write down its characteristic equation. Watch out for this sign. Minus, that's a very common error to leave out the minus sign. You must put in the minus sign because that's the way the formula comes out.
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It's roots. It's a quadratic equation. It will have roots. Lambda 1, Lambda 2, which, for the moment, let's assume a real and distinct.
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Okay, for the enrichment of your vocabulary, those are called the eigenvalues. There's something which belongs to the matrix A. There are two secret numbers that don't...
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You can calculate from the coefficients A, B, and C, but they're not in the coefficients. You can't look at a matrix and see what its eigenvalues are. You have to calculate something.
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But they are the most important numbers in the matrix. They're hidden, but they are the things that control how this system behaves.
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Those are called the eigenvalues. Now, they're various purists. They're a fair number of them in the world who do not like this word because it begins German and ends English.
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The eigenvalues are first introduced by German mathematician. The matrices came into being in 1880 or so. A little while after, eigenvalues came into being two. Since all this happened in Germany, they were named eigenvalues in German, which begins eigen and ends the value.
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But people who don't like that call them the characteristic values. It takes, unfortunately, it's two words and it takes a lot more space to write out.
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An older generation even calls them something different, which you are not so likely to see nowadays, but you will in slightly older books.
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You can also call them the proper values. Proper is a standard characteristic, not a translation of eigen, but proper is.
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But it means it in a funny sense, which is almost disappeared nowadays. It means proper in the sense of belonging to.
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The only example I could think of it is the word property. Property is something that belongs to you. That's the use of the word proper that's something that belongs to the matrix. The matrix has its proper values.
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It does not mean proper as in the sense of fitting and proper. I hope you'll behave properly. We'll go to antagithism, something like that.
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But as I say, by far the most popular thing, slowly the word eigenvalues is pretty much taking over the literature. Just because it's one word, that's a tremendous advantage.
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Now, what now is still to be done? Well, there are those vectors to be found. So you have to then solve. So the very last step would be to solve the system to find the vectors a1 and a2.
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So for each lambda i, find the associated vector. The vector will call it alpha i. That's the a1 and a2. Of course it's going to be indexed by, you've got to put another subscript on it because there are two of them.
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So the a1 and a2 is stretched a little too far. By solving the system, the system will be the system which I'll write this way, a-lambda, b, c, d-lambda.
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So the system that was over there, but I'll recopy it, a-lambda, a1, a2 equals 0, 0.
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And these are called the eigenvectors. Each of these is called the eigenvector associated with belonging to, again, in that sense of property.
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So the eigenvectors associated with, let's say, belonging to, I see that a little more frequently, belonging to lambda i.
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So we've got the eigenvectors and of course the people who call them characteristic values, also call them characters. These guys characteristic vectors.
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I think I've ever seen proper vectors, but that's because I'm not old enough. I think that's what they used to be called long time ago, but not anymore.
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And then finally, then the general solution will be by the superposition principle, x, y equals the arbitrary constant times the first eigenvector
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times the eigenvalue times the e to the corresponding eigenvalue, and then the same thing for the second one. A1, A2, but now the second index will be 2 to indicate that it goes with the eigenvalue lambda 2, e to the lambda 2t.
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Okay, I've done that twice, and now in the remaining five minutes I will do it a third time, because it's possible to write this, it's still a more condensed form.
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And the advantage of the more condensed form is, A, it takes only that much space to write, and B, it applies to systems, not just the 2 by 2 systems, but to end by n systems. The method is exactly the same.
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So let's write it out, let's write it out as it would apply to n by n systems. Okay, what's the vector I start with is x, y, and so on, but I'll simply abbreviate this as is done in 1802 by x with an arrow over it.
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So I'll abbreviate this as is, and then the system looks like x prime is equal to x prime is what? A, x, that's all there is to it.
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So there's a green system. Now notice in this form I don't even tell you how many, whether this is a 2 by 2 matrix or an n by n, and in this condensed form it will look the same no matter how many equations you have.
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Your book deals from the beginning with n by n systems, that's one of its, in my view is one of its weaknesses, because I don't think most students start with 2 by 2. Fortunately, the book double talks.
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The theory is n by n, but all the examples are 2 by 2. So just read the examples, read the notes instead which just do 2 by 2 to start out with. Okay, the trial solution is x equals what?
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The known vector alpha times e to the lambda t. Alpha is what we called a 1 and a 2 before. Plug this into there and cancel the e to the lambda t's. What do you get? Well, this is lambda alpha.
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So it's lambda alpha, well, I put it in for the moment, e to the lambda t equals a alpha e to the lambda t. Cancels these two cancel and we get the system to be solved is a alpha equals lambda alpha.
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Now the question is how do you solve that system? Well, you can tell if a book is written by a scoundrel or not by where how they go a book which is, in my opinion, completely scoundrally.
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Simply says you subtract one from the other and without further ado, write a alpha a minus lambda. They tuck a little i in there and write alpha equals 0.
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Why is the i put in there? Well, this is what you like to write. What's wrong with this equation? This is not a valid matrix equation because that's a square n by n matrix, a square 2 by 2 matrix, if you like.
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This is a scalar. You cannot subtract the scalar for the matrix. It's not an operation. To subtract matrices, they have to be the same size, the same shape.
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So what's done is you make this a 2 by 2 matrix. This is a 2 by 2 matrix where lambda is down the main diagonal and i elsewhere. And the justification is that lambda alpha is the same thing as lambda i times alpha because i is an identity matrix.
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Now, in fact, jumping from here to here is not something that would occur to anybody. The way it should occur to you to do this is, you do this, you write that, you realize it doesn't work, and then you say to yourself, I don't understand what these matrices are all about.
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I think I'd better write it all out. Then you would write it all out and you'd write that equation on the left hand board there. Now, I see what it should look like. I should subtract lambda from the main diagonal. That's the way it will come out.
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So, hey, the way to say lambda from the main diagonal is put an identity matrix. That will do it for me. So, in other words, there's a little detour that goes from here to here. One of the ways I judge books is by how well they explain the passage from this to that.
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If they don't explain it at all, but just write it down. They never talk to students. They've just written books.
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So, where we get finally here. So, the characteristic equation from that, I forget what color that's in salmon. The characteristic equation then is going to be the thing which says that the determinant of that, the determinant is zero, that's the circumstances under which it's solvable.
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So, in general, this is the way the characteristic equation looks. And it's root once again, or the eigenvalues. And from them, you calculate the corresponding eigenvectors.
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Okay, go home and practice. In recitation, you'll practice on both 2x2 and 3x3 cases, and we'll talk more next time.