WEBVTT mathematics/differential-equations/murray
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Hi welcome back to www.educator.com we are working on differential equations my name is Will Murray and today we are going to look at autonomous equations.
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And we are going to use what is called phase plane analysis to study them so of course i need to tell you what all those words mean so let us go ahead and get started here a little lesson overview.
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Autonomous differential equations the idea there is that you can write the equation in a form y′ is equal to some function just of y that is the real key here is on the right hand side you only see a y and so you do not see the other variable if you are talking about x or t.
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The key thing is that you just have y′ is equal to some function of y, no access or t is on the right hand side.
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And so we have a special way of understanding those which is that we are going to graph y′ versus y so were going to set up a set of axis.
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And we will have y here and y′ here now these graphs are very different from the slope fields that we studied earlier and the solutions that we graph earlier, so do not get these graphs these phase plane graphs mixed up with the slope field graphs.
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Before when we studied slope fields we will have either x or t on the horizontal axis and we will have y on the vertical axis and we are changing this now we have y on the horizontal axis and y′ on the vertical axis.
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So very important difference here, this are totally different graphs from what we have seen before and we are going to study this graphs using what is known as phase plane analysis.
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What we are going to have is a graph of f(y)and so we have something like this when we graph f(y).
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And the point here is that we can look at these and we can see what values of y make y′ be positive or negative or zero and then that can tell us weather our solutions are sloping upwards or downwards or weather they are horizontal so that is the whole point here.
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Let me give you a quick idea of that, in this case ok show this in red down here we have y′ is less than zero, up here we have y′ positive, down here we have y′ less than 0 and up here we have y′ is greater than 0.
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And so if we translate that into the shape of our actual solutions anytime we see y′ less than 0 we know we have solutions that have a negative slopes so their tending downwards.
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Anytime we see y′ greater than 0 we know we have solutions that tend upwards so let me try to draw what these solutions would look like I'm going to make myself a little space here and draw some graphs to illustrate this phase plane analysis so i save 3 places there where that y that f(y) crosses the y axis.
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Now I'm going to put y on the vertical axis and i will put t on the horizontal axis where x if that is another variable so here are those places where y′ where f(y) crossed the 0 mark there.
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And what this tells us is that down here before you get to the first place maybe a column A, B and C.
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Here is A, B and C when y is less than A we see negative values for y′, y′ was less than 0 which means we have curves that are sloping downhill.
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So I'm going to draw negative slopes down here in between A and B we see that y′ is greater than 0 so how curves sloping uphill.
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So there is solutions that slope uphill because i know they have a positive derivative y′ is positive.
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Between B and C, y′ is less than 0 so my solutions between B and C are sloping downhill again.
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I'm making them webel of because I know the solution curves do not cross each other and then bigger than C, I have y′ greater than 0 so I have these solutions that slope up again.
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So that phase plane analysis helps me draw the solution curves in the plain in the y-t plane without even actually solving the different equation.
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So this is a really powerful technique here let us investigate this a little further so let me draw those that phase plane again so you can see how that played out.
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So here is the original craft that we started with and here is the place where it crosses 0 the 3 places where it crosses 0.
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So this was remember graphing y′ versus y and then that translated into a graph of our solution curves when we graph y versus either t or x or whatever the other variable is i call it t.
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And so we have this 3 places corresponding to A, B and C where f(y) is 0 so in between those we figured out that below A and between B and C we have solution curves where y′ is negative so they are is sloping downhill.
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So below A and between B and C they are sloping downhill and then in the other regions between A and B, y′ is positive because f(y) was positive this is y′ greater than 0, y′ is greater than 0 that is bigger than C so solution curves positive here.
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And here y′ is less than 0, y′ is less than 0 that corresponds to the 2 regions over here where the solution curves are going downhill and then right on the boarder line is those places actually at A, B and C here where y′ is equal to 0.
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So we have some stable solutions where y′ is equal to 0 and those are called equilibrium solutions that means the solution just stays completely horizontal it never goes up or down so those are called the equilibrium solutions.
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So in these example we have 3 equilibrium solutions well this are more samples later on with more specific numbers i just kinda wanted to give you the general overview of the possible things that can happen and then will get down to details later on.
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So the question where often ask is if we perturb the equilibrium solutions that means if we started one of this equilibrium solutions and maybe bump it a little bit; bump it up or bump it down which ones will actually return to equilibrium?
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So let me give you an example of this if we start at B here and we suddenly bump up a little bit notice that we get forced back down to B or if we bump it down little bit we get forced back up to B.
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So that means B is known as a stable equilibrium and the way you can identify that looking back at the phase plane is that y′ is going downwards because when your below it y′ is positive when your above the y′ is negative.
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That means either way you get bumped you are going to go back towards that equilibrium solution at y = B.
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Now on the other side of that look is at what happens if we start at C and then we got bumped a little all of the sudden we get knock down to the solution to that tree they goes far away from C or if we get bumped a little bit down we get knock down to the solution to that tree that again goes far away from C.
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So C is considered to be an unstable equilibrium even though if you stay exactly at C you will never deviate from that path if you get bumped a little bit away from C you get bumped to an solution which takes you farther and farther away from C so that is considered to be an unstable equilibrium.
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And we could have identified that back in the phase plane by seeing that y′ is going upwards that means that if you are a little bit smaller than C the curves are going down away from C if you are little bit bigger than C the curves are going up away from C.
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So notice that A is the same way y′ is going upwards to that 0 at A so if again if you start at A and you get bumped a little bit you are going to go far away from A.
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The solution curves are tending to run away from A so A is an unstable equilibrium and there is one sort of mix case here which is semi stable equilibrium.
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i do not have an example of that in my picture here but a semi stable equilibrium is one where there is stable on one side so the solution curves are approaching it and they are unstable on the other side
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So if it gets bumped, if it gets bumped one way say down here it would return to the equilibrium but if it gets bumped up it would go far away from the equilibrium.
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So that is called a semi stable equilibrium and we will see an example of that later as we get in to some more specifics so i definitely think we are overdue to see an example here and actually work it out with some numbers.
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So starting with the differential equation y′ is equal to y x y - 1 x 2 - y so what makes this autonomous differential equation is that there is no other variable on the right hand side except for y so in particular there is no x or t on the right hand side there is just y's.
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So we are going to draw a graph of y′ versus y we are going to draw our phase plane and then we are going to the identify the equilibrium solutions.
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So to draw our graph of y′ versus y that just means we are trying to graph this polynomial y x y - 1 x 2 - y so here is y and here is y′ and that polynomial is already nicely factored here.
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So we know that if 0 when y is 0 and 1 and 2 that is because this tells you that if 0 and y is 0 this tells you that if 0 when y is 1 and this tells you that if 0 when y is 2 and notice that if you multiply this all together what you get is a cubic equation because we got 3 powers of y here.
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So you get a cubic equation and the first coefficient will be negative that is because of this negative right here would make that first coefficient of the y cube be negative.
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So there is one of 2 general shapes for cubic equation it can have this upwards curve or it can have it downwards curve and because of this - we know we got a downwards curve here.
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So i am going to draw my f(y) as a downwards cubic there and of course if you want to confirm this it is easy to pull out a graphing calculator and just graph.
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Just graph (f)y = y x y - 1 x 2 - y and you will see that it does not did look like this curve and then from that we want to identify the equilibrium solutions.
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So that means the solutions where y′ is equal to 0 so those equilibrium solutions are exactly where this curve crosses 0 and so there they are at y = 0, 1 and 2 and so if you wanna graph those solutions.
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Remember now this is a graph of y versus t it is not y′ versus y anymore we have here graphing y versus t so the equilibrium solutions come when y = 0 or 1 or 2 so i draw in my equilibrium solutions there to 1 and 0.
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So those are my 3 equilibrium solutions we got those by looking at the graph of y′ versus y and figuring out where it crosses 0 each time.
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Those tell us where y′ is 0 which means that the solutions for y will be completely horizontal their slope will be 0.
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So I'm going to keep working with this example with this equation and the following example, so we are going to going with this.
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We have the same differential equation y′ = y x y - 1 x 2 - y and we have 3 tasks here we are going to sketch other solutions.
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We are going to label each one of the equilibrium solutions as stable and semi stable or unstable and then we are going to predict y of infinity predict a long term behaviour of this solutions for a bunch of real life initial conditions.
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So we will see what would happen if we started at each of these initial points so I'm going to go to a new slide we will have more space but i will keep the equations here.
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So this is the same differential equation y′ = y x y - 1 x 2 - y so let me just remind you how the phase plane analysis work for that.01642 We graphed y′ versus y here so there is y, 0, 1, and 2 and we graph y′ here so that was a downward cubic curve.
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And then that translated into finding the equilibrium solutions on the graph of y versus t be very careful to keep these two graph straight it is really easy to confuse them.
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One is y′ versus y so y is on the horizontal axis and then over here we have y versus t so y on the vertical axis.
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So these equilibrium solutions at 0, 1, and 2 on the y axis those turn into values on the vertical axis over here and those give us horizontal equilibrium solutions.
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So what we have to do now is sketch other solutions by looking back at the phase plane we can figure out here y′ is less sorry greater than 0 because it is above the horizontal axis here.
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That tells us that when back here in this range where y is greater than 0 let me put some scales here 0, 1, and 2 when y is less than 0 that is down here we are going to see positive slopes.
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So I'm going to draw some solution curves with positive slopes here in between 0 and 1 when y is in between 0 and 1 we see y′ is less than 0 so I'm going to draw some curves with negative slopes here.
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Negative slopes in between y = 1 and y = 2 we see y′ is positive again so i will draw in some solution curves in between 1 and 2 with positive slopes.
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And then when y gets up to be bigger than 2 we see that y′ < 0 again so i will draw a curves with negative slopes.0
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So from looking at the way the other solution curves behave we can tell which of our equilibrium solutions are stable and unstable.
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if we look at y = 2 here then the solution curves are all tending towards y = 2 which means if you are at y = 2 and you got perturb a little bit if you got bump a little bit draw a little bump here you will and up going to y = 2 so that is a stable equilibrium.
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Now here at the equilibrium here y = 1 and you got bump a little bit you would become part of the solution curve which goes away from y = 1 so that is an unstable equilibrium.
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And finally if you are at y = 0 and maybe you got bump a little bit you are going to end up going back towards y = 0 so that is also a stable equilibrium.
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And remember we said we could look at the phase plane analysis and have predicted that anytime the curve is going downwards through the axis it is a stable equilibrium.
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So that means y = 0 and y = 2 are both stable equilibria and then when it is going upwards through the axis like here y = 1 it is an unstable equilibrium.
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Now last we are supposed to do with this problem is to identify the long term behaviour if we started this different initial conditions.
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So let me label this 1, 2, 3, and 4 so y have 2 = 5 that means we should have started the point at t = 2 and y = 5.
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So i did not put a scale on my t axis for it does not really matter because the important thing here is that y is starting somewhere way up here at 5 let me put this in blue.
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And you can see that any solution up here is going to tend downwards towards y approaching 2 so let me write that down.
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So as y gets very large here y would as t goes to infinity y would approach 2 so that is that first initial condition.
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The second initial condition says y(1) = 3 halfs so that means we are starting at 3 t = 1 and 3 halfses between 1 and 2 so maybe somewhere here.
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And so we can see that this solution curve also goes up and approaches the equilibrium solutions at y = 2 so y(t) goes to infinity while very large values of t there will also go to 2.
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Now so that was the 2nd initial condition we are given y(3) = 1 that means t = 3 we are starting exactly at y = 1 now that would mean we will continue exactly along that equilibrium solution.
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So y(t) going to infinity would approach 1 now that is a sort of a theoretical answer and i have to qualify a little bit remember that 1 is unstable.
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So in real life it is essentially never going to happen that anything will come to rest that are unstable equilibrium.
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Everything is always getting bumped or jazzled or perturbed a little bit so in real life you would not ever expect to see a solution come to rest at an unstable equilibrium.
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in real life it would at some point get bumped or jazzled or perturbed and it would go either or way up to 2 or a way down to 1 it would approach, I'm sorry, down to 0. It would approach the stable equilibria at 0 or at 2.
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So in sort of a theoretical abstract perfect mathematical world it would keep going forever along this equilibrium at 1 in real life when everything gets bumped or jazzled a little bit by some kind of external forces it would actually tend away from 1.
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And it would tend towards the stable equilibrium at 2 or the stable equilibrium down at 0 so that is kind of the most complicated case there.
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Last case here is y(2) = -6 so that means t is 2 and y is somewhere down here at negative 6 and we can see that all the solution to that trace at this starting down here tend upwards and then they level off in your t = 0 or in your y = 0 so y as t goes to infinity would approach 0.
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So let us just recap what we did there first we graphed y′ versus y we got this phase plane analysis which was very instrumental on sort of understanding everything that came later on.
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So we graph this function y x y 1 x 2 - y and we got this downward cubic and then we used that to get all the information about the solution to the differential equation.
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We never actually solve the differential equation here and often times with autonomous differential equation you do not have to.
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Because you can learn what you need to just from looking at the phase plane so here by looking at this phase plane we see the places where it crosses 0 and we know that those correspond to this equilibrium solutions all did actual differential equation.
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So these 3 places where it crosses 0 correspond to horizontal equilibrium solutions to the differential equation and then when we look at a little more detail we can see where y′ is positive and negative and positive and negative.
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And we can translate that into the solution through that trace is positive, negative, positive and negative meaning that they are slope uphill or downhill, uphill or downhill.
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And so those give us the solution through that trace to predict the sort of qualitative behaviour starting at any initial condition and so finally what we did here was we took 4 initial conditions to 5 and one 3 halfs and three 1 and two -6
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And then we started at each of those initial conditions and we saw where the solutions through that trace ended up.
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So the one starting up here would go down and level of in your y = 2 this one down here would approach upwards y = 2 and then the most interesting one or the most complicated one is the that starts y = 1.
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Because in the mathematically perfect world it would just stay at y=1 but in real life what would happen is it will get perturbed a little bit up or down and so it will end up at one of these stable equilibria you would never stay at an unstable equilibrium.
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And finally y(2) = -6 that was the one that started down here and so we can see to that grows and approaches the stable equilibrium at y=0.
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So we are going to practice this a few more times with some more examples we have a differential equation for example 3 here is y - 1² x y - 2 x y - 3 and again we are going to start out by drawing a graph of y′ versus y.
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And then we are going to try to identify the equilibrium solutions so i will start out with y′ versus y and here is y on the horizontal axis and where else put y on the horizontal axis for this phase plane analysis.
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And i see that i have a fourth degree equation here because it is y² here and then 2 more powers of y and i see that has zeros at 1 and 2 and 3.
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And i see that that is going to be a positively oriented fourth degree equation it is going to be a positive value of y to the 4 which means it is going to be going to infinity.
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Let me draw those in blue it is going to be going to an to positive infinity as y goes to positive or negative infinity so i know it is going up as it goes to positive or negative infinity y is 3 when y is 3 it crosses through the axis.
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So there it is crossing there i will make route a little more visible there it crosses again at 2 and at 1 we have a double route which means it just barely touches the axis it is just tangent right there.
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So that is how we can graph y′ versus y of course if you wanted to use graphing calculator that is no problem you just take this equation and throw that into your calculator and you will generate the same graph.
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And then we want to identify the equilibrium solutions remember the equilibrium solutions are wherever this function crosses the y axis so here we got 3 equilibrium solutions at y = 1 at y = 2 and at y = 3 those are the values of y.
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Then if you plug them in would give you′ is equal to 0 so when we draw the actual solutions on the next page those will be a horizontal lines.
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The next example is actually a follow up with the same equation so I'm going to go ahead and keep moving with next example we will draw the equilibrium solutions and we will see how the other solutions behave.
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So example 4 here same equation y′ is equal to y - 1² x y - 2 x y - 3 while we are being ask to do a sketch other solutions label each of those equilibrium solutions remember we had already identified those as y = 1, 2, and 3.
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We are going to label each one of those a stable, semi stable or unstable and then we are going to predict the long term behaviour y of infinity for a couple of initial conditions here.
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So i have got this information copy on to the next slide same differential equation here and we have already draw the graph of the phase plane here.
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So let me recap that for you remember the phase plane has y′ on the vertical axis and y on the horizontal axis and we know that this one crosses 0 at 1, 2, and 3 those are the equilibrium solutions.
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And we figured out on the previous examples you can go back and check that out if you do not remember how that works that it just barely touches 0 at y = 1 goes up comes down again and it goes of to positive infinity so we got these 3 equilibrium solutions.
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And let me draw those and then we will try to figure out the behaviour of other solutions so now I'm actually drawing the solutions this is not the phase plane anymore so notice that y is on the vertical axis now back on the phase plane the y is on the horizontal axis.
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So now my t is on my horizontal axis and I'm going to draw those equilibrium solutions at y = 1, 2, and 3 that was supposed to be a 3 so each one of those is going to be an equilibrium solution let me draw those in 1, 2, 3.
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Equilibrium solution means y′ is 0 means your slope is 0 that is why you get horizontal lines thats what an equilibrium solution means and then we wanna identify how the other solutions play in between those solutions.
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So in order to do that let us go back and look at phase plane remember we are not solving the differential equations everything we are doing in these lecture has to do with the phase plane analysis and just using that phase plane to figured out the shape of solutions without solving the differential equations.
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So if go back and look at this i see the curve is positive here y′ is greater than 0, here y′ is still positive, here y′ is negative and here y′ is positive again.
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So what that means is that when y is less than 1 over here I'm going to see positive slopes so over in the actual graph of the solutions less than 1 I'm going to see positive slopes here.
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So I'm going to draw solution curves with positive slopes and now when y is between 1 and 2 i see that y′ is still positive so I'm going to draw some more positive slopes in between 1 and 2.
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So y′ is positive derivative is positive means you are going to see uphill curves here between 1 and 2 now between 2 and 3 when y is between 2 and 3 i see that y′ is less than 0 it is negative which means I'm going to see downward slopes here.
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And then finally when y is bigger than 3 I'm going to see positive slopes so i will some positive slopes here and then this is the only informations i need to identify which of the equilibrium equations are stable and unstable.0
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So if i look at this, if you look at this solution at y = 3 you see that the solution curves are tending away from it that means if you bumped in a little bit you would and up running far away from that solution so y = 3 is an unstable equilibrium there.
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Y = 2 if you bumped a little bit away from that one the solution curves are coming back to get closer to it so the solution curves are approaching to it as a stable equilibrium.
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And then at y = 1 thats sort of the most interesting case here because the solution curves on one side are tending away from it.
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So if you bumped it a little bit up you will run away from this solution but if you bumped it a little bit down you would tend back towards the solution so that is a semi stable equilibrium.
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So what we have done here is figured out how each one of these equilibria behave in the sense of if you bumped it a little bit will you go back towards the equilibrium or run far away from the equilibrium.
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And as before we could have notice that just by looking back at the phase plane let me start with 2 here we see that y′ is going downwards through the axis so that was a stable equilibrium.
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Over here at y = 3 it is going upwards through the axis so that is an unstable equilibrium and at y = 1 it is sort of a not doing either 1 it is that special case that goes down and just barely touches the axis and then rans away again so that is a semi stable equilibrium.
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So finally we are ask about a couple of initial conditions here y(0) = 4 that means you would start at 0 and 4 and wanna figured when you had and up so let me look at the point 0, 4.
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So that is somewhere up about here and we can see that up here whenever y is bigger that 3 we are in the region of positive slopes, we are in the region of sloping uphill so this first initial condition here would have, would give us the solutions that slopes upwards to positive infinity.
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This second solution y(0) = 1/2 that means we start at t = 0 y is equal to 1/2 so 0 1/2 would be right here you see that this curves around there slope upwards and so for that one as t goes to infinity it slopes upwards towards the equilibrium solution at y = 1.
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in fact since that is only a semi stable equilibrium what would probably happen in the real world is at some point the solution curves will get bumped a little bit and it would been head up and level up at y = 2.
00:39:59.000 --> 00:40:28.000
So i will just say in the real world in which all solutions are constantly getting bumped or perturbed or you will always introducing small external factors into a system we would expect at some point that solution curve would get bumped a little bit about 2.
00:40:28.000 --> 00:40:58.000
Above one and at that point it would immediately latch on to an upper trajectory and head up towards the stable solution at y = 2 so sort of as mathematical extraction we know that y would approach one but in practice in a long term we would expect to see y jumping pass the solution at 1 and approaching the stable solution at y = 2.
00:40:58.000 --> 00:41:13.000
So just a recap there what we did was first we graphed this equation in y′ versus y so we graphed that and that is how we got this blue curve here.
00:41:13.000 --> 00:41:29.000
And then by looking at that we could read off a lot information about the solutions, we could read off at our equilibrium solutions at y = 1, 2, and 3 so that is how we got these three solutions at y = 1, 2, 3 that is three horizontal lines there.
00:41:29.000 --> 00:41:48.000
We could also read off information about when y′ is positive and negative from that graph and then we use those to draw those solution curves in between y′ is positive, positive, negative and positive.
00:41:48.000 --> 00:41:59.000
And so once we have that we could identify which of the equilibrium where unstable, stable and semi stable.
00:41:59.000 --> 00:42:22.000
Of course we could have figured that out just by looking at the original phase plane analysis because we know that stable solutions have f(y) going downwards through the axis and unstable solutions have f(y) going upwards through the axis and then this is sort of the special case.
00:42:22.000 --> 00:42:35.000
So from there we got which solutions are stable, unstable and semi stable and then we look at these two initial conditions that we were given and we look at with those fell on the graph.
00:42:35.000 --> 00:43:01.000
And we identified where they will go as y as t goes to infinity where the y values would go so here we saw that would the y values will go to infinity here we saw curve that would sort of in theory we go up to one but in practice it would probably get bump up to y = 2.
00:43:01.000 --> 00:43:21.000
So lets try another another example here we got a failing messy differential equation here y′ is equal to y² - 4y³ + 5y - 2 and so again we wanna start by drawing a graph of y′ versus y and we to identify the equilibrium solutions.
00:43:21.000 --> 00:43:39.000
So this one is a little tricky because it has not been factored for us and so we have to factor a cubic here and really if you do not know how to do that a good place to look is in the educator lectures for algebra so i will cover that in detail here.
00:43:39.000 --> 00:43:51.000
That i will show you that in order to factor this i wanna check for roots factors of the right hand term, the constant term here.
00:43:51.000 --> 00:44:12.000
And I'm going to use synthetic substitution so on factors of 2 and since i already plan this out I'm going to try y = 2 first so I'm going to use 2 as my potential route and then I'm going to write down this coefficient 1, -4, 5, and -2.
00:44:12.000 --> 00:44:24.000
So 1, -4, 5, and -2 and I'm going to do some synthetic solution here there is something you can learn about reviewing your algebra from a few years back.
00:44:24.000 --> 00:44:52.000
1 i just bring that down and i get 1 multiply that by 2 and i get 2 at -4 to 2 and i get -2 multiply that by 2 and i get -4 at 5 and -4 i get 1 multiply 1 by 2 and i get 2.
00:44:52.000 --> 00:45:19.000
And at -2 and 2 i get 0 what that means is that 2 is a root of that cubic and moreover it means that when i factor y - 2 from this the coefficient of the reduced equation are given exactly by 1 -2 and 1.
00:45:19.000 --> 00:45:39.000
So this factors into y² - 2y + 1 so that is really useful because it means i can keep factoring here y - 2 and then y² - 2y = 1 is y - 1².
00:45:39.000 --> 00:45:56.000
So that is just a little algebra that we have to do in order to give us some meaningful equation to solve here so now we have that we can graph this equation so I'm going to graph y′ versus y remember that is how you do that phase plane analysis y′ versus y.
00:45:56.000 --> 00:46:30.000
Y′ versus y and looking at this i see that it is going to be 0 when y is 1 or 2 so there is y = 1 and there is y = 2 now this is a positive cubic curve so i know it is going to have a general shape like this.
00:46:30.000 --> 00:46:54.000
And i see that y = 1 is a double root that is because of the y - 1³ there, it is a double root which it is just going to be tangent to the axis there and then y=2 is just a single root so that is just a regular root there.
00:46:54.000 --> 00:47:16.000
And now we have to identify the equilibrium solutions well that just means where y′ is equal to 0 so wherever this curve touches the horizontal axis that is where we get the equilibrium solutions so the equilibrium solutions there y = 1 and y = 2.
00:47:16.000 --> 00:47:36.000
So where done with everything this example asks us because we graphed y′ as a function of y that mend factoring y in this cubic equation and of course factoring takes us all the way back to algebra so we mend to through synthetic substitution here.
00:47:36.000 --> 00:47:44.000
We got 0 at the end which meant 2 is really a root of course i tried 2 first because i had already kinda plan this out i already knew 2 is already going to work .
00:47:44.000 --> 00:47:57.000
if you did not know that you that you would have probably tried 1 and -1 and 2 and -2 and you are kept trying this numbers until got something that would give you a 0 at the end there.
00:47:57.000 --> 00:48:17.000
But once you do get 0 then it is very nice that we get these reduced equation 1, -1, -2, 1 that tells us these coefficients here 1, -2, 1 and so that lets us factor the equation on down to y - 2 x y - 1³.
00:48:17.000 --> 00:48:40.000
And then that gives us a graph of course if you a graphing calculator and you are allowed to use it then you could have skip all that algebra and just strong as equation into your graphing calculator and of course it would give you this same kind of graph here of y′ is equal to f(y).
00:48:40.000 --> 00:48:50.000
And then we identify the equilibrium solutions those are just the places where this crosses through the y axis or the horizontal axis those were the equilibrium solutions.
00:48:50.000 --> 00:49:00.000
So we are going to be using this phase plane analysis in the next example to actually draw out the various kinds of solution to the differential equations.
00:49:00.000 --> 00:49:13.000
So let us go ahead with that we are asked this is the same differential equation let me emphasise y² - 4y³ + 5y - 2 last just sketch the other solution.
00:49:13.000 --> 00:49:30.000
We are going to label the equilibrium solutions as stable, semi stable or unstable and then we are going to describe what ranges of values for an initial condition would lead to what limiting behaviour for the solution.
00:49:30.000 --> 00:49:56.000
So in order to understand that let me redraw my phase plane for you remember we factor this down into y - 1² x y - 2 and so when we graph that we are graphing y′ versus y that is the phase plane there.
00:49:56.000 --> 00:50:19.000
We had roots at y = 1 and y = 2 and then we have a curve that just goes up and just barely touch this at y = 1 and then it goes back down again and then it goes up through y = 2.
00:50:19.000 --> 00:50:37.000
And we going to translate that into a graph of solutions in terms of y and t so remember when you are graphing the solutions y is the vertical axis when you are graphing the phase plane analysis y is the horizontal axis.
00:50:37.000 --> 00:50:44.000
So that is always a little confusing but you wanna keep be very clear to keep track of which craft is which and do not mix them up.
00:50:44.000 --> 00:51:00.000
So here I'm going to start by graphing y = 1 and y = 2 and those correspond to equilibrium solutions those are places where y′ is equal to 0.
00:51:00.000 --> 00:51:13.000
So i will graph a couple of horizontal lines there those are my 2 equilibrium solutions and then i wanna figured out what the other solutions do.
00:51:13.000 --> 00:51:26.000
So i will go back and look at the phase plane analysis if i look down here when y is less than 1 i see that y′ is less than 0 because the curve is down here below the axis.
00:51:26.000 --> 00:51:35.000
When y is between 1 and 2 i steel see that y′ is less than 0 and then when y is bigger than 2 i see that y′ is greater than 0.
00:51:35.000 --> 00:51:48.000
So I'm going to used that to draw my solution curves when y is less than 1 i have solution curves that slope downwards so i will draw these curves sloping downwards.
00:51:48.000 --> 00:52:06.000
When y is between 1 and 2 i still have y′ less than 0 so I'm going to draw my curves still sloping downwards, levelling off when they get to the equilibrium solutions.
00:52:06.000 --> 00:52:23.000
And when y is bigger than 2 up in this region that corresponds to this region here i see that y′ is positive so I'm going to draw my solution curves positive.
00:52:23.000 --> 00:52:33.000
So on the strength of that having drawn these other solution curves i can label which of my equilibrium solutions are stable or unstable or semi stable.
00:52:33.000 --> 00:52:49.000
So if i look at y′ to start with sorry or y = 1 to start with that is this the solution curve here i see that if i got bumped a little bit if you got bumped a little bit down then solution curve will go far away.
00:52:49.000 --> 00:53:16.000
if you got bumped a little bit up then the solution curve would come back and approach the y = 1 again so what i have there is that y = 1 is a semi stable it depends on which direction you bumped it it is a semi stable equilibrium.
00:53:16.000 --> 00:53:30.000
Y = 1 is a semi stable equilibrium because if you bump it a little bit down and then the solution curves goes far away but if you bump it far up a little bit up then the solutions come back to down to y = 1.
00:53:30.000 --> 00:53:59.000
Y = 2 if I'm on that solution that equilibrium solution that equilibrium solution and i get bumped a little bit it is going to go far away and that is true it need a direction it is going to go far away so y = 2 is an unstable equilibrium because even the slightest bump will send you to a solution curve that goes far away from the equilibrium solution.
00:53:59.000 --> 00:54:15.000
So even though it is an equilibrium solution it is unstable you would not expect to see it happen in real life because any slight perturbation will take the solutions far away from that.
00:54:15.000 --> 00:54:23.000
So what we done is identify our two equilibrium solutions one was semi stable and one was stable we did not stable equilibria there.
00:54:23.000 --> 00:54:43.000
And finally the question prompt asked us to identify which ranges of initial values will lead to which limiting behaviour for the solution so let us go back look at our solution curves.
00:54:43.000 --> 00:55:18.000
We see that if we are in this region that is if y is less than if y 0 if we started the point when y 0 is less than 1 then these solution curves go down to negative infinity so y goes to as t goes to infinity as t grows the y is trending downwards to negative infinity.
00:55:18.000 --> 00:55:49.000
if we start at 1 then we are just going to stay at 1 and if we start anywhere in between 1 and 2 then we are going to thrift down to 1 so i will say if y if our initial value of y is 1 or anything between 1 and 2 then y of large values will tend down towards the solution at 1.
00:55:49.000 --> 00:56:18.000
if we start at 2 if y not the initial value of y is 2 then will stay exactly on that solution it will be unstable so in the real world you would not expect to see this happen but we would see for large values of time we would still see y right there there on that equilibrium solution.
00:56:18.000 --> 00:56:43.000
And if we start anywhere up here if y not is anything bigger than 2 then we see that the solution curves go up to infinity so as we plug in larger and larger values for t we see that y would increase up to infinity.
00:56:43.000 --> 00:57:00.000
So now if we i identify all the possible things that could happen depending on which values of y not you start at we can tell you where the ultimate behaviour will be for any given value of y not.
00:57:00.000 --> 00:57:26.000
So let me just recap what we did there we started with the differential equation we had to do a lot of algebra back on the previous example to factor down into y - 1³ x y - 2 once we had that we drew this phase plane analysis where we drew the graph of y′ versus y.
00:57:26.000 --> 00:57:47.000
So we are graphing a polynomial there of course you can check that on your calculator and from there we are able to identify the equilibrium solutions at y = 1 and y = 2 and then those translated into our equilibrium solutions over here y = 1 and y = 2 that is how we got these 2 horizontal lines here.
00:57:47.000 --> 00:58:13.000
Looking at it in a little more detail we see that y is negative, sorry y′ is negative when y is less than 1 y′ is negative when y is between 1 and 2 and y′ is positive when y is bigger than 2 so that helps us draw these solution curves sloping downhill when y is less than 1.
00:58:13.000 --> 00:58:29.000
Still sloping downhill when y is between 1 and 2 and then sloping uphill when y is bigger than 2 so we are able to draw these solution curves and then we can look at those and identify which of our equilibria are stable or unstable.
00:58:29.000 --> 00:58:40.000
Turned out that y = 1 was semi stable because solution curves sort of approaching on one side, on one side they want to tend towards that solution.
00:58:40.000 --> 00:58:58.000
On the other side they run away from that solution so it is stable on one side, unstable on the other side we call it semi stable y = 2 was an unstable equilibrium because the solution curves are drifting away from it.
00:58:58.000 --> 00:59:12.000
Of course we could have also notice that by looking at the original phase plane graph since at y = 2 the graph is, the curve is going upwards across the horizontal axis.
00:59:12.000 --> 00:59:26.000
We know that is an unstable equilibrium and at y = 1 since it is just barely touching it is tangent to the horizontal axis we know that is a semi stable equilibrium.
00:59:26.000 --> 00:59:45.000
And finally we are able to look at the solution curves and identify for any possible value of y not weather be less than 1 between 1 and 2 equal 2 or others are small mistake here i should have said y not greater than 2.
00:59:45.000 --> 00:59:58.000
That corresponds to these values here when y not is greater than 2 so we are able to look at those and identify for any possible range where the solution curves will end up.
00:59:58.000 --> 01:00:10.000
if you are less than 1 you are going to be going down to negative infinity, if you are between 1 and 2 you are going to be force down to this solution at y = 1.
01:00:10.000 --> 01:00:23.000
if you are at 2 then you are just going to keep going exactly at y = 2 but if you are any think bigger than 2 then you are going to be going up to positive infinity.
01:00:23.000 --> 01:00:34.000
So the whole point of this study of autonomous equations and phase plane analysis is that nowhere in any of this did we actually solve the differential equations.
01:00:34.000 --> 01:00:43.000
instead we have these autonomous equation where we have no x's or t's on the right hand side just y's.
01:00:43.000 --> 01:00:57.000
And so from that we are able produce a graph of y′ versus y and from there we are able to really get a lot informations about the actual solutions y versus t.
01:00:57.000 --> 01:01:05.000
We are able to figure out the equilibrium solutions weather they are stable or unstable where the other solutions go and where you are going from the initial point.
01:01:05.000 --> 01:01:20.000
So i hope you had fun practicing some of those on your own that wraps up this lecture on autonomous equations and this is part of differential equations lecture series my name is Will Murray and you are watchIng educator.com, thanks.