WEBVTT mathematics/differential-equations/murray
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Hi and welcome back to the differential equations lectures here on educator.com, my name is Will Murray and were working on systems of equations and we already had a couple of lectures where we studied systems of equations where you look at the eigenvalues E matrix.
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We had 1 lecture where you talk about where we talked about 2 distinct eigenvalues and we just had lecture where we talked about complex eigenvalues
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In this lecture, we are going to talk about repeated eigenvalues so, what happens when the eigenvalues, the matrix and you discover that you got a multiple root of the characteristic equation.
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That is what were going to analyze today because it is kind of behaviour from anything we seen before so, it is go ahead and jump right .
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Just remind you where we are to solve the system of linear differential equations X′ = A X .
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A would be a matrix here so, is using it 2 by 2 matrix the way you do that is you find the eigenvalues and the eigenvectors of the matrix A .
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So, will talk about in this lecture is what happens when the characteristic equation has repeated root so, lowers you have a repeated eigenvalues and so, let me show you how to handle that.
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First you find a corresponding eigenvector B so, just like we did before you try to solve A- RI x B = 0.
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That is exactly what we did before then this parts new where to find so, method called generalized eigenvector W which means A- RI W = not 0 but V .
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So, that is the then new element here is that instead of trying to solve A- RI time something= 0 will solve that first will get a V and then we will take that V and will plug that in on the right-hand side and try to find this W such that A- RI VW = V .
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Remember to call that the generalized eigenvector.
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Generalized eigenvector basically when you hit it with the A- RI matrix what you get is the original eigenvector you will get 0 get the original eigenvector .
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So, let us see how that plays E solution reform the 2 principal solutions the first 1 is exactly the same thing we did before.
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It is E RT x B so, that R is the eigenvalue and that V is the eigenvector just like before.
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That is exactly the same as we are been doing it the last couple of lectures here is the new twist.
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The second principle solution is this entire thing T E RT x V+ E the RT x W .
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So, these the same eigenvectors before the W is the generalized eigenvector.
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That whole thing put together is our second solution so, it gets kind of complicated and the general solution is what you get by tacking a constant onto each 1 of those and then adding them up so, I just plugged that in here.
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That C1 x E the RT the plot + C2 x at entire thing E RTV + E RT W and then just like all the other all the other problems use initial conditions to solve for C1 and C2 if there given.
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If you are given initial conditions you just stop your with the general solution if you give an initial conditions you use them at this point the very end to soften the C1 and C2.
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So, there is 1 more thing we have to do with these systems which is graphing the solutions so, the way this works out let me remind you what the general solution looks like it looks like C1 E RT x the first eigenvector V + C2 x this whole quantity T E RT x V + E RT x W .
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So, the first thing to do here is just kind of forget about the second solution just look at C1 E RT x V and when you graph that when you graph that what you get is a sort of straight line of solutions.
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That a straight line of solutions only a line through the origin and then depending on whether R is + or - they E be expanding out a long outline or contracting E origin along that line so, they might be expanding out or this another color they could be contracting in towards the origin depending on whether that Eigen value is + or - .
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Then the other solutions swirl around this line and it kind of depends on whether you looking at + or - values and C2 so, what is going to happen is if C2 is + than they R going to approach + multiples of V and C2 is - then approach - multiples of V .
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Let me try to show you the 2 general pictures and then we will see some more specific examples later.
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So, if C2 is + if R is + then the line is sort of expanding and if if this is V if this is the right here then all the all the other solutions are sort of trying to approach their swirling around and whenever C2 is + they are trying to approach + values of V.
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Whenever C2 is - they are trying to approach - values of V so, get the sort of spiral swirly pattern there is a straight line E middle and everything else is swirling to 1 side or the other .
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So, we will see an example of this so, that is the case when you have a + eigenvalue or rethink sort of expanding as it progresses.
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Let me talk about the case we have a - eigenvalue you will still have that fixed axis defined by the first solution there the first eigenvector but if R is - if R is - then these things will be shrinking down to 0 .
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And so, again when C1 when C2 is + that although be going into 0 but the sort of spiral around and try to approach that first eigenvector so, those were spiralling like this is.
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It does not go round and round it just kind of swirls in and then when C2 is - get that away from the edge of the screen when C2 is - else swirl in to the - multiples of the eigenvector .
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So, that is fairly subtler probably be a little easier to understand .
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Check out some examples and do both computations and the graphs and you will see how to make these things on your own .
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So, first example here we find the general solution to the following system X′ is 0 1 -4 and 4 .
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Now remember all of these matrix systems of differential equations it all start out the same way got a find the eigenvalues and the eigenvectors of the matrix so, to find the eigenvalues remember start by subtracting R along the main diagonal .
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So, will get a quadratic equation from that if I cross multiply I get - R x 4 - R - -4×1 so, that is + 4 = 0 so, see about it R ² - 4R + 4 = 0 so, R - 2 quantity ² = 0 so, I get and R I get a double root both of my roots R equal to 2 so, and this repeated eigenvalue situation .
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So, first of all I'm going to find my eigenvector corresponding to R = 2 so, 0 0-2 is - 2 and 1 and 1 and -4 and 4-2 is 2 and I am trying to find an eigenvector that when I multiply that matrix by it I get 0 .
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Remember and easy way to do this is to fill in 1 for Y and see what the X should be in order to make that equal to 0 and I see that it would have to be if I put a 1 half here than -2×1/2 E - 1 + 1 = 0 .
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So, my eigenvector would be 1 half x 1, by the way if you do not like my way of writing in a 1 for the Y you can also, solve this using free parameters .
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So, we practice that back in our review of linear algebra few lectures ago I do not really like this in this fraction in here and remembered eigenvector you can multiply by a number by scalar and it will still be in eigenvectors.
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To get 1 and 2 for my eigenvector and that is it for my eigenvector but now I have to find this generalized eigenvector will remind you how that works you try to solve A- R I x W = V where V is the eigenvector you just found.
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So, I am in a try to solve that now for W A- R I is the same matrix is before so, you write down what I am doing here trying to find the generalized eigenvector.
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So, I got to the same matrix as before is it is the same A- R I -21 -42 x my W which I will does call X and Y for now X and Y = the eigenvector none 0 anymore but 1 2 .
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So, right this out I got - 2X + Y = 1 -4 X +2 Y = 2 .
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Those equations are redundant and you can get 1 from the other 1 and other that is a general principle here that that ought to be true if not then we would have made some mistakes some risk across out that second equation and look for any X and Y that satisfy the first equation.
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The easiest 1 I can see there would be X and Y = 0 and 1 if Y is 1 in X0 then that will be a solution to that equation.
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By the way I mentioned when you are finding eigenvectors which can multiply it by a scalar if you like to make the numbers nicer that is right in here multiply that V by 2 in order to make the numbers nicer that is not true with the generalized eigenvector.
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You cannot multiply by scalar's anymore because it would no longer be a solution to this equation so, if you do not like the numbers you get for the generalized eigenvector you kind of stuck with them you have to keep going with that generalized eigenvector .
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Now that I found the eigenvector and the generalized eigenvector let me go and plug goes E general form of the solution which was all remind you what it was in general.
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It was C1 E RT x V + C2 x ERT x V + E RT x the generalized eigenvector W .
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So, I figured out all the quantities you are just need to drop them in and I will get my solution so, in this case my C1 and do not know what that is just I just leave it there my R is 2 came from over here so, E 2T now my V is 1 2+ C2 x T E 2T my V still 12+ E RT E 2T and my W is 01.
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And that is my general solution to the system .
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So, let me show you a quick review of how that worked out we always start by finding the eigenvalues in eigenvectors of the matrix they subtract R down the main diagonal cross multiply here + - that is when doing here across multiplying - 4 give me a +4 get a quadratic equation turns out has double root which means I am in this situation repeated roots .
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So, I take that is that the 1 word that we got and plug it back E matrix's meaning subtracted down the main diagonal lines I had a - 2 and 2 that A- R I .
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And I try to find the eigenvector which means installing for a vector sets it when you multiply that matrix by that vector you get 0 start out by putting a Y in his 1 might figured out that X had to be 1 half to make that work .
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You could also, do that with free parameters if you are more comfortable with that method from linear algebra so, i got this eigenvector 1 half 1 did not like the fractions I multiplied by 2 to get 1 2 .
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Those my eigenvector and if generalized eigenvector in solving this equation A- R IW = V so, I A- R I is the same matrix there is my W on trying to find that and there is that V that comes from the eigenvector I just found .
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So, that turns into 2 equations but 1 of them is redundant so, I got rid of the second 1 and I just try to find any X and Y that works and I got X and Y to be 0 and 1.
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Again you could solve that with free parameters if you like but it is really such an easy 1 that I do not think it is necessary.
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So, that I took all that information I plug it into my generic general solution for so, had this back at the beginning of the lecture in the overview got this generic general solution for so, plug-in R = my eigenvalue my eigenvector my eigenvalue again my eigenvector again it was that eigenvector.
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That eigenvalue, eigenvalue again and then finally my generalized eigenvector of plug that in and I get my general solution .
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So, hang on to this week is really use this exact 1 E next example were to learn how to graph this.
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So, for the second example here were the graphs and solution trajectories to the previous system of equations so, I got written down here is the exactly the general solution to example 1 so, the escorts R looks like a big mystery then go back and watch example 1 because that is where we figured out this general solution here .
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We start with the matrix we found the eigenvalues in eigenvectors and we eventually came up with this general solution .
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Let me show you how the graph plays out .
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These graphs are little bit complicated does take some practice and working through some examples before you get comfortable with him .
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So, let me put a scale on here and it is all sort of motivated by the eigenvector and the eigenvalues that is really what you want to focus on is the eigenvector E eigenvalue .
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What I am first going to do is just graph this first solution C112 x E 2T so, 12 x E 2t to go to vector 12 that is 1 of the horizontal direction and 2 E vertical direction and what that gives me is a fixed axis of solutions.
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So, make my axis there and I notice that 2 is + which means that these solutions are growing there getting bigger and bigger and bigger.
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and in the - direction that would correspond so, this would be C1 being + in the + direction the kind of going off to - infinity to getting bigger E -direction and that is really because the eigenvalue is + .
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So, now I want to look at what happens when I consider the second set of solutions , these over here and so, if I start out with if I take for example C2 = 1 and C 1= 0 C2 is equal 1 at T = 0 I am going to be at 01 because the T = 0 will make the 1 2-term dropout.
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That means I will start at 01 so, my 01 is here and it is going to grow from there what's can happen is as a grows the dominant term is the TE the 2T.
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So, it is going to pull towards the 1 2 direction so, it is cannot pull towards the 1 2 direction .
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So, as this grows it is going to pull out in the 1 2 direction number that was 1 2 right there their first eigenvector right there.
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And that is can happen any time, C2 is greater than 0, anytime C2 is going great is greater than 0 it is been a start of the top left side of this first axis and its could end up pulling out E + direction .
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So, even though the curses are down here are good and up swirling around and attracted back to that + direction.
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So, you get these this swirling effect all of these repeated eigenvalue graphs have the swirling effect they always swirl E out if the eigenvalue is + or inward if the eigenvalue is - .
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Now down here were looking at - values of C2 which means for large values of time they are going to be going to - when T is very large they are going to - values of 12.
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So, that means there swirling down to the - part of the plane here.
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Slowly draw some curves there swirling to the - playing here everything on the side of the major axis of the axis defined by the eigenvector is swirling down to - multiples of the eigenvector.
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So, even though they start out sort of parallel to the major axis they swirl around and try to go to the - side of the major axis .
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So, that is what the graph would look like let me again remind you how you would know this you would start out so, so, for so, you start out with the equation and we figured out the equation in example 1
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I did not do that right here but you can go back and look at example 1 if you do not remember where the general solution came from.
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I started out by looking at this first solution 12 x E 2T and that is I got this major axis with the initial eigenvector and then as T gets bigger it grows and grows and grows and a C1 were - that is kind of the - of that vector and there it is going down into - territory so, there is-1 and 2 .
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Now the trickier part is when you look at solutions with C2 we notice when T = 0 that term drops out and so, they just start at 01 so, that is that term right there that 01.
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But when T gets bigger this 1 2-term starts to dominate and he gets pulled over E 1 2 direction and anything on this side of the the major axis has the same effect so, anything on the site has C2 greater than 0 and so, it is can end up getting pulled towards the + 1 2 direction.
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Anything on the other side gets pulled down towards the -1 direction and that is why everything down below is swirling in that -12 direction .
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So, all of these graphs they all have this sort of major axis swirling on each side swirling in opposite directions on opposite sides the only difference is if it is a + eigenvalue like we have here there swirling outwards if it is a - eigenvalue were to see that E next example there swirling inwards.
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So, practice 1 like that and then you will really be aware of all the possibilities since try out that next example.
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So, this example we got a find a general solution to ex-' = -41 -40 and remember all start with finding the eigenvalues in eigenvectors.
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So, find the eigenvalues subtract the R and I will get -4 - R x - R - -4 is the same as +4 = 0 .
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So, R ² - R ² + 4R + 4= 0 and we want to solve that of course that just factors into R +2 ² = 0 so, we get a double root at R = - 2 and so, we only have 1eigenvalue with repeated eigenvalue and so, plug that in a find the eigenvector .
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So, that means -4 - R is - 4+2 so, - 2 1 -4 and 0 - -2 is 2 and I will find eigenvector that if you multiply that by that matrix you get 0 .
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To do my normal trick plug in a Y = 1 here feel like that trick you can always do this using free parameters for - 2X + 1 = 0 if I make that to a 1 half there then that would work nicely goes I would get -1+1 = 0 .
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So, my eigenvector is 1 half 1 but I do not like this fraction so, multiplied by 2 and I will get 1 2 for my eigenvector .
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Now, here's where it starts to diverge from the previous examples of systems of linear equations difference here is because we got this repeated roots we cannot find another eigenvalue eigenvector combination.
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Instead, we find the generalized eigenvector which is what we do that as we solve A- R I x W = V and more that V is the eigenvector that we party found the W is what were trying to solve for.
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So, try to solve-21 -42 x my W which for now call XY and do not what is yet 12 and so, get - 2X+ Y= 1-4 X+ 2 Y = I see that those equations are just multiples of each other so, there is really no use in hanging on to the second 1.
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I cannot so, I am looking for any X and Y that will solve the first 1 and again I can see that and easy XY that will work would be 01 if Y is 1 and X is 0 then not that would certainly work there.
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Some use that as my W my generalized eigenvector and let me remind you cannot go multiplying generalized eigenvectors by scalars appear we took we had an eigenvector and we multiplied by a scalar to get rid of some nasty fractions you can do that with eigenvectors but not so, with generalized eigenvectors they are very specific things so, you cannot just multiply them by scalars.
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Having found that generalized eigenvector of all set to go with my general solution for me remind you the generic form for the general solution we had this at the beginning of the lecture was C1 in the RT R the eigenvalue x V + C2 V the eigenvector C2 now gets confiscated T RT x V+ E RT x W .
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So, W is the generalized eigenvector so, figured out all those quantities except for the C C1 each of it now my R is my R as was - 2 so, E - 2T x V is where my V is 1 2+ C2 x T RT R still - 2T x V is 1 2 + E RT's that is each of the same thing E - 2T now my W is 01 .
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That is my complete general solution so, that is the end of that problem but before we move on to be moving on a drawing a graph of that but let me show you again out what the steps were solving that.
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So, we start out by finding eigenvalues in eigenvectors so, that means to find the eigenvalues you subtract R off the main diagonal finding A- R I there find its determinant C Cross multiply there and that is what I am doing with this equation right here finding its determinant there is-4 but that is on the - diagonals.
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That is why that turned into a + 4, their + 4 and that simplifies into a quadratic equation which is pretty easy to solve, but we get a double root it - 2 .
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Go ahead of plug at - to backend find the eigenvector and since I had some fractions that I did not like there I multiplied by 2 which is okay for an eigenvector is not okay for generalized eigenvector.
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Which is what were to find next so, got to make that via vector there it is a vector so, finding the generalized eigenvector you again look at A- R I the difference, though is that on the right-hand side he open 0 you put the original eigenvector and then you try to solve this generalized eigenvector W here .
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So, got a couple equations for the X and Y and W Little is redundant so, we got rid of it and then we figured out a X and Y quickly that would work there and so, really use that as our generalized eigenvector.
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And then our general solution this is something that gave you at the very beginning of lectures on just quoting it at this point go back and looking up at the beginning of lecture.
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That is where this comes from and then I just plug in my eigenvalue - 2 my eigenvector eigenvalue eigenvector eigenvalue and finally the generalized eigenvector so, that is where the solution comes from now you want to hang onto the solution were to be going farther with this really graphing it that is the next examples .
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Just keep track of the solution and will use it right away in the next example so, example 4 graphing some solution trajectories to the previous system of equations so, I got here is the solution that we derive in previous example .
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So, fairly complicated solution here but we worked it out in previous example example 3 so, having just watched example 3 and go back take a look at example 3 and that is where this complicated solution came from .
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Now I want to try to graph it which is no small task set up some axis for myself here .
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So, were to start with the simpler of the 2 solutions which is the C1 solution we start with the simpler of the 2 solutions , I see 12 x E - 2T so, there 1 2 right there.
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There is 1 2 and E - 2T is going make that thing just steadily shrink down towards the origin that E - 2T just makes it gets smaller and smaller and smaller .
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It is basically because the eigenvalue is - because that and any - eigenvalue is can have that affects only show all multiples of this 1 R just getting smaller and smaller and smaller make that and blacks with little easier to there.
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I will go ahead and show the - side of that as well so, these R - multiples of that initial solution there is the - of 12 and then these R getting smaller and smaller there gradually drifting in towards the origin but I will always on a straight line.
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That is our major axis and effects can be our only axis because remember we said for all of these repeated eigenvalue problems they always swirl around that major axis 1 way or another so, we just have to figure out which way this is swirling around .
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So, the way we do that is we look at this second solution here so, let me now take C1 equal to 0 get rid of the first solution C2 = 1 and so, that means my solution is 1 2x T E - 2T +01 x E - 2T for plug-in T = 0 in that first part will drop out and I will just get 01 so, little start up here at 01.
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But when I let T get bigger and bigger all these terms are going to go to 0 but I see that the dominant term here is going to be the 1 2 so, as to when affinity it is going to approach 0 but along a trajectory defined by 12.
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So, it is going to kind of swirl in and approach 0 along that trajectory defined by 12 and that is kind of going to kind of be a general pattern whenever I have C2 bigger than 0 it is going to try to approach the + multiples of 12.
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So, we have all of the all these curves swirling around and approaching the + multiples of 1 2 .
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Now the exact opposite thing happens when C2 is less than 0 were to start on the lower side of the major axis were going to approach the - multiples of 1 2so, everything's going to swirl in an approach from the - side of 12 .
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Easier for me to draw those backwards that look like it goes more towards the origin there we go.
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So, everything kind of swirl and approaches from the - side of 12 so, this is the the there is always a picture like this where there is a major axis things are either expanding or contracting along the major axis and that depends on whether the eigenvalue is + or - .
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Here, our eigenvalue was - 2 which means that things are contracting along that major axis and then outside of that major axis things are always swirling around that major axis in 1 way or another and is just all you have to do is figure out which way their swirling and also, whether their expanding or contracting as their swirling.
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So, in this case we have a - eigenvalue which means are all contracting as their swirling so, swirling in the origin around that major axis.
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So, just to recap the individual steps there we got the solution from example 3 this was kind of this kind came straight out of example 3 so, I did not work that out now if you were that comes from example 3 is the place to look.
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But then we started out by graphing that first solution 12 x E- 2T so, there is1 to right there and were taking different multiples of that which is a whole line of that E - 2T being - makes everything along that line drift in towards the origin along that line.
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That gives us fixed axis and then with C2 to graph the C2 solution wrote down what we would get if we just plug in C2 = 1 which means when T = 0 is first-term drops out and we just get 0 1s that is why were starting right here at 01.
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But when T gets bigger the dominant term here is the 1 2 is not the 01 anymore , which is why it drifts into 0 along that 1 to direction and so, that is true anytime you are on this side of the major axis or the northwest side of the major axis and time C2 is + you can adrift in 0 along + multiples of 12 anytime you are on the lower side and means you have a - value of C2 which means you can adrift E 0 along - multiples of 1 2.
00:37:52.000 --> 00:38:15.000
See every following this 1 right here so, that is that is how you work out that swirling out 1 E next 1 were going to see what happens when we form some what when we use some initial conditions and see how that affects things and how we can get some solutions with specific values of C1 and C2 .
00:38:15.000 --> 00:38:27.000
So, see how that works out so, example 5 resolve the initial value problem -41 -40 and X0 = 38 .
00:38:27.000 --> 00:38:48.000
So, let us see how that affects things let me remind you what our solution was to the previous problem because this is the same matrix that we had actually was in example 3 and example 3 we solve this in general we found a general solution .
00:38:48.000 --> 00:39:26.000
So, if you kind of just and for example 5 go back and look at example 3 and see that we found a general solution X general was C1 x E - 2T x 12+ C2 x the quantity T E - 2T x 12+ E - 2T x 01 .
00:39:26.000 --> 00:39:35.000
So, that was which we did some work in example 3 to get to that point so, like a repeat that now but if you do not remember that just go back and check example 3 see it all worked out.
00:39:35.000 --> 00:40:10.000
Now let us see how this works with our initial condition because this is the new part of this problem X0 = 3 8 so, i am going to plug-in T = 0 here so, get C1 E 0 is just that 1 2 + C2 x now 0 E 0×12+ E 0 x 01 and so, a lot of these terms drop out.
00:40:10.000 --> 00:40:35.000
There E 0s are all to me once I get C1 x 12+ C2 now this this term completely drops out because multiplied by 0 and then I just have C2 x 01 because E 0 is still 1 and that is supposed to be equal to 3H so, I can expand this out into 2 equations if I distribute the CE vectors.
00:40:35.000 --> 00:41:01.000
Looks like I got C1 +0 C2 = 3 and 2 C1 + C2 +1 C2 = 8 right away I can figure out the C1 = 3 and if I plug-in 3 there got to 6+ C2 = 8 so, C2 would have to be 2 .
00:41:01.000 --> 00:41:36.000
So, I got my 3 and my 2there so, to plug those in for C1 and C2 so, I have got to see C1 was 3 E - 2T x 12+ C2 is 2 x T E- 2T x 12+ E - 2T in this copying the general solution now 01 filled in C1 and C2 .
00:41:36.000 --> 00:41:55.000
I see that I could simplify a little bit because this to this 2E- 2T to combine it with this 3 E- 2T that the T′ might have to write separately so, this is what we expand out maybe this right-hand set of terms .
00:41:55.000 --> 00:42:24.000
This is 2T E- 2T x 12+2 E - 2T x 12 x 01 and then I can combine my light term E - 2T gets E - 2T x 3×13×1 and 2×0 so, 3 there .
00:42:24.000 --> 00:42:40.000
Then on the bottom I got 3×2 and 2×1 6+2 is 8 + T E- 2T if I multiply that this 2in with this vector I get 24 there .
00:42:40.000 --> 00:42:46.000
And that is a simple as it is going to get.
00:42:46.000 --> 00:43:08.000
By the way notice that it is very easy to check now if I plug-in T = 0 the right-hand side would all go to 0 and the left-hand term would give me 38 so, that is what it was supposed to be so, I can check that I got it right that is very easy.
00:43:08.000 --> 00:43:21.000
So, recap what we did there start out with a general solution that I worked out in example 3 so, that general solution was a complete district you just go back and watch example 3 see it .
00:43:21.000 --> 00:43:39.000
Then I plug-in T = 0 here they so, my initial conditions on plug-in T = 0 so, were sought T through E 0 so, all my E 2Ts become 1 and the T E 2T actually drops out become 0 because of that 0 .
00:43:39.000 --> 00:44:08.000
So, I just end up with C1 x 1 2+ C2 x 01 and that resolve itself into two equations, in two unknowns are very easy equations I get my solutionsC1 = 3 C2 =2 so, plot those back in here C1 is 3 C2 is 2 plot those back in here and it is a little bit messy so, I want to maybe expand out this 2.
00:44:08.000 --> 00:44:34.000
Distribute it to here and here and I see of got E 2T term here and used 2T term here so, I combined those had to distribute the 3 into the 1 and 2's that turn into 3 and 636+210 010236+02 gave me 38 .
00:44:34.000 --> 00:44:49.000
Then, this to distributing E 1 and 2 gave me that 2 and 4 that is my final form of the solution there but it is also, easy to check if I plug-in T = 0 in here this term would drop out because of the T outside that T outside this may turn into 1 .
00:44:49.000 --> 00:44:54.000
I get 3 8 so, it does check that I match the initial condition that I was given.
00:44:54.000 --> 00:45:08.000
Since our last example and that wraps up our lecture on repeated eigenvalues so, this is the systems of differential equations where we had repeated eigenvalues.
00:45:08.000 --> 00:45:13.000
This is all part of a larger lecture series on differential equations here on educator.com .
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My name is Will Murray and I thank you very much for watching, bye bye.