Differential Equations Distinct Real Eigenvalues

Hi there and welcome back to www.educator.com and these are the differential equations lectures, my name is Will Murray.0000

Today we are going to talk about systems of equations and the way you solve systems of equations is essentially by finding the eigenvalues of a matrix, there is a sort of three different things that can happen.0007

It depends on what kind of eigenvalues you get when you solve for those eigenvalues in the matrix, sometimes you get real eigenvalues but they are repeated and sometimes you get complex eigenvalues.0019

What we are going to do is we will have a separate lecture for distinct real eigenvalues when you get different real eigenvalues and then we will have another lecture for complex eigenvalues and then we will have another one for repeated real eigenvalues.0033

Each one of those kind of has its own separate behavior, what we are talking about today is distinct real eigenvalues, if you are working on some homework problems and you are finding that you are getting complex eigenvalues or repeated eigenvalues.0046

Then we got separate lectures to cover those cases, you might want to check out those lectures instead, today we are talking about distinct real eigenvalues, let us jump in and see how that works out.0061

We are going to solve the system of linear differential equations x1(t) is equal to a1-1 x1 + a1,2 x2 and x2′(t)=2,1 x1, a2, 2 x2(t) that is the system we are trying to solve.0072

Right away we are going to try to write that in matrix form, first of all I'm going to drop off the t's on the left hand side and I'm going to try to simplify this equation, I simplify it into that form that really has not changed much.0095

Then I'm going to try write it in matrix form which means I'm just going to extract the coefficients (a1,1), (a1,2) (a2,1), (a2,2) and write those in matrix and then I'm going to write x1 x2 as a column vector and the derivatives of those as a column vector on the left.0110

The shorthand version of this whole thing is to write this 2 x 2 matrix as a and then this x stands for the column vector x1 x2 and the x′ stands for its derivatives, that is bold vector x standing for (x1 x2).0130

Bold vector x′ standing for its derivative, this X′ equals (ax) in bold form that very small equation there is just short hand for this entire system and we are going to figure out how to solve that.0151

Let us go ahead and see how that works out, the secret to solving that is to look at the matrix a and find its eigenvalues and their corresponding eigenvectors.0168

It all comes down to the eigenvalues and eigenvectors, now if you are a little rusty on your linear algebra, we had of lecture here on the differential equation series which was a review of linear algebra.0179

We practice finding eigenvalues and eigenvectors, if you are little rusty on that, if you do not remember that, why don't you go back and look at the review of linear algebra around the lecture, it was the previous one here in the differential equation series.0190

You will get some practice calculating eigenvalues and eigenvectors then come back to this lecture, we are going to use it to solve differential equations, we are going to find the eigenvalues, we are going to call those r1 and r2.0204

Each one of those is going to come along with the corresponding eigenvector and we are learning a new Greek letter here, this is called xi 1, it is hard to say, it sounds like KSI, KSI is the way you pronounce that.0217

We write it as a backwards three, it is a complicated but with that is the Greek letter that we are going to use for the eigenvector, fortunately you would not have to write it very often.0235

But we will call the eigenvector xi 1 and xi 2, xi 1 is the eigenvector corresponding to the eigenvalue r1 and xi 2 is the eigenvector corresponding to the eigenvalue r2.0249

Once you find those eigenvalues and eigenvectors you can write down the general solution to the differential equation right away, the general solution is just a constant times that first eigenvectors c1 times e^r1(t)remember r1 was the eigenvalue.0265

And then a constant times the second eigenvector times e^r2(t), basically it is just a matter of finding the eigenvalues and eigenvectors and then just dropping them into the solution formula.0283

The initial conditions there is a small mistake here on my print out here, the initial conditions are something we would use to find the c1 and c2, we use the initial conditions that were given to find c1 and c2 here.0298

Let us see what else we want to do with those, the next step you take with that would be to graph the solutions and let me give you a quick idea of that and I'm probably make more sense after we study some examples.0318

What we will have is a set of axis, this is X1 in this direction and x2 in this direction, and we will have 2 vectors c1 and c2, first I want to graph those vectors c1 and c2.0339

Maybe you have a vector there which is c1, it got a little messy there and then maybe in another direction you will have c2, and what you want to do is to set up axis defined by those lines .0355

I'm going to set my axis up, define by those lines by the eigenvectors and the solution trajectories, now remember that they all had this term of e^r1(t) and e^r2(t).0385

The solution trajectories will either go to zero or infinity depending on whether the r1 and r2 are positive or negative.0407

If r1 is positive then we will have trajectories on the c1 axis going out to infinity, if r1 is negative then we would reverse these arrows and have them going in toward 0.0415

For r2 again if it positive and xi2 axis we would have solution trajectories going out towards infinity, if it is negative I will graph it as if they are negative then these solution trajectories would tend in toward zero.0422

We also have solution trajectories in between these two axis and what happens with the ones in between is they tend toward the axis spanned by the eigenvector that corresponds to the larger eigenvalue.0447

You look at r1 and r2 and you see which one is bigger and then you have the solution trajectories tending towards that axis and the way I graphed it right here, I graphed it as if r1 is positive because those are going off to infinity.0461

I graphed it as if r2 was negative because I graphed them going into zero, what that would mean is that the solution trajectories in between these axis would tend towards the axis defined by the r1 because that is the larger one.0477

They follow these these curves that would follow the other axis for little way but then get drawn off in the direction of the r1 axis, you see how all these red curves are tending towards this axis determined by the first eigenvector there.0495

All these axis, all these solution trajectories are tending towards this southwest to northeast axis, they do not cross each other by the way, we never have solution trajectories crossing each other.0520

We will get some practice of that, I think it is probably hard to see in the abstract and when practice several examples you will start get the hang of it, let us go ahead and start our first example here.0535

We have to solve the following system, this is actually an initial value problem solution, there are going to be two parts of this, first is solving the differential equation which is of the form X′ = ax.0548

We will have some initial conditions which we are not going to worry about until later, let us set up the the differential equation first, now remember that the solution to all of these things depends on finding the eigenvalues and eigenvectors of the matrix.0561

Let us find the eigenvalues and eigenvectors of the matrix, A-R(i) , this is how you find the eigenvalues is 6 - R - 2, 9 - R and we want to take the determinant of that.0582

The determinant we cross multiply is 6 - R x 9 - R - 2 x -2, that is - + 4, -4, we will get R^2 here - 9R - 6R, -15R + 54 - 4 is + 50=0.0600

That factors really nicely, that factors into R -5 times R -10 is equal to 0, we get our eigenvalues our R is 5 and 10, two eigenvalues for each one of those, we are going to throw those back in to the matrix and find an eigenvector.0632

For our R=5 I will do A- R(i) A-5(i), I will subtract 5 from the main diagonal, 6-5 is 1, -2 -2, 9-5 is 4 and I can simplify that matrix using my row operations, going through the linear algebra kind of quickly.0654

If you are little rusty on this linear algebra we do have a whole set of lectures on linear algebra here on www.educator.com and I also gave one quick review of linear algebra lecture just before this one.0677

You might want to go back and check on that if you want to get a quick refresher for linear algebra if you are really rusty and you want to do it from scratch.0689

We got a whole set of different of linear algebra lectures that you can check those out and get a lot of practice on finding eigenvalues and eigenvectors.0697

What I'm going to do here is do some row reduction, (1, -2) and then I can use that row to cancel out the second row of 0's here and I want to find a vector that will give me 0 when I multiply that matrix by it.0706

My standard trick and I talked about this last time in the review of linear algebra lecture is to put a 1 in for the z component or for the Y component and then figure out what I should have for the X component.0724

It looks to me like I have -2 is equal to 0, that first entry that will be a 2, if you are not comfortable with that method you can use the free parameter method, maybe I will just show that quickly.0738

We got x -2y=0, we use a free parameter for y, we use y=t and then x would have to be 2t, xy would be 2t and t, if we factor out the t we would have 2 and 1 x t.0750

That is another way to get that factor but I usually use my short hand trick of just throwing a 1 in a y position and then figuring out what the exposition would have to be so that they multiply to be 0.0771

That is my little shortcut of finding an eigenvectors, we still have to find the eigenvector corresponding to R=10, what we just did was fighting eigenvector corresponding to R=5.0783

Let us do the same thing for R=10, we subtract 10 down the main diagonal A-10(i) is 6-10 is =4, -2 -2, 9 - 10 is -1 and if we will reduce that then that row reduces into, I could bring the bottom row up to top and maybe multiply it by -1 to get rid of negatives.0795

(2, 1) and I can use that to simplify the first row as well, I will simplify that into 0 and if I do my normal trick of putting one in a y position, 2 x 1 you would just give me 1/2 in the x position and I do not really like that.0828

Remember you can scale eigenvectors if you want and they will still be eigenvectors, I'm going to scale that eigenvector by a factor of 2 and get (1, 2) and I see I made a small mistake there, that 1/2 should have been negative.0847

The whole point is that when we multiply this matrix by this vector, this should be 0, that 1/2 should have been negative in order to make that come out to be 0.0868

When we scale it up by, I will multiply that by 2 just get rid of the fraction, I will put a 2 in the y component and a -1 in the x component, that is my eigenvector corresponding to the eigenvalue R=10.0879

Remember once you have the eigenvalues and eigenvectors there is a generic solution that always works for the general solution to these differential equations, I will remind you what that was.0900

We had that in the beginning of the lecture but it is c1 x e^r1(t), first eigenvector x e^eigenvalue(t) + c2 x c2 x e^ r2(t), what I'm going to do is just drop those eigenvalues and eigenvectors that I figured out into this generic solution.0910

My generic solution here is c1, now my first eigenvector was (2, 1) x e^r1(t) that is e^5t + c2 x my second eigenvector is (-1, 2) x e^10t that is my general solution.0941

If I did not have initial conditions this is where I would stop and say that is the complete solution to the problem, but now I'm going to use the initial conditions which is the x(0)=(0, 5).0970

I have not use that for anything yet, I'm going to incorporate that, that means I plug in 0 for t, I get c1 x (2, 1), e^5 x 0, I'm plugging in e^0 here.0982

I also plug it in over there later, e^5 x 0 is just e^0, that is just 1, c1 x (2, 1) + c2 x (-1, 2), again e^10 x 0 is just 1, that is supposed to be equal to 0× 5.1005

I'm going to separate that out and get two equations and 2 unknowns for C1 and C2, it looks like from the first row I have 2c1 - c2 is equal to 0 and from the second row I have c1 +2c2 is equal to 5.1026

Now you can solve this using whatever technique you like best from algebra, let me use substitution, it looks like c2 is equal to 2c1, I will plug that in and I get c1 + 2 x 2c1 is equal to 5.1048

It looks like I get 5c1 =5, c1 = 1 and from c2 =2c1, I get c2 is equal to 2, I will take those two constants and plug those back into my general solution, I will get my particular solution x=1, that is my c1 (2, 1) e^5 + 2 x -(1, 2) x e^10t.1067

If you wanted you can bring that to into the vector, multiply it by each of the components, does not really make much difference so I'm not going to do that.1107

Now I have finished solving the initial value problem but let me go back and show you what we did with each step there.1120

The whole key to solving this systems of differential equations is to start with the matrix and find the eigenvalues and eigenvectors, that is what we are doing here, we found A- R(i), that means you subtract R along the main diagonal.1127

Take the determinant, that means you cross multiply and we set that equal to 0, we get polynomial which is this one turns to be easy to factor and we got 2 values for R.1140

For each one, you plug those back in, you subtract them back along the main diagonal that one came from a 6 - 5 and that 4 right there came from that 9 - 5, that is using the eigenvalue R=5.1157

We will row reduce that matrix using the techniques we learned in linear algebra for row reduction and I just guess the vector, I started out with the 1 in the y component and then I figured out that the x component would have to be a 2.1169

If you do not like that technique you could also use free parameters where you start out with y=t and you go back to x=2t and then you figure out again that your eigenvectors (2, 1) do the same kind of thing for R=10, the other eigenvalue.1186

Plug it in on the main diagonal, reduce the matrix and then try to find an eigenvector where when you multiply the matrix by that eigenvector you will get zero, our eigenvector there is (-1 and 2).1202

We just drop those eigenvalues and eigenvectors into our general solution, those are the eigenvectors and those are the eigenvalues right there and that is our general solution, we will be done except the we also have to satisfy this initial condition.1216

The initial condition we use it right here and that means we are plugging in t=0 which makes the E terms dropout, you just get e^0 is 1 and we end up getting 2 equations for 2 unknowns in the c's which is pretty easy to solve.1234

It is just kind of high school algebra there to solve the 2 equations and 2 unknowns, get values for c, plug those back in there is c1 and there is c2, we got 1 and 2 and we have our particular solution to the differential equation.1249

In the next example we are to going to practice graphing some solution trajectories to the general system to the general solution for the previous system, let us see how that works out.1267

Remember, there is a sort of several stages here, you start out by graphing the eigenvectors, I'm going to make a big graph here.1280

First I will graph the eigenvectors and we will see what the axis are defined by those eigenvectors, let me put a scale on here.1292

That first eigenvector is (2,1) and remember you think of that as being X and Y are X1 and x2, your x1 is 2, x2 is 1 or x is 2 and y is 1, I'm going to go over to (2, 1).1313

There it is right there, there is the vector (2, 1), 2 horizontally and 1 vertically and I will graph negative multiples of that and I'm going to set up an axis defined by that eigenvector, I will do that in blue.1331

I see that I have a positive eigenvalue there, the eigenvalues 5 that is positive which means solutions on that axis are going to go off to infinity in the positive direction and negative infinity in the negative direction.1353

That really depends on whether C1 is positive or negative, this is kind of C1 is positive, this C1 is negative that tells me whether solutions go to positive infinity or negative infinity.1370

I'm going to set up the other eigenvector as (-1, 2) so -1 on the horizontal axis, 2 on the vertical axis, there is (-1, 2 ) and let me set up my axis, I wanted to do that in blue.1382

I see also that I have a positive eigenvalue there, R+10 my solutions are also going to infinity in the positive direction along that axis, that is c2 greater than zero because that is (-1, 2), this is c2 less than 0.1415

All of my solutions seem to be going to infinity that is because I have 2 positive eigenvalues but now let me look at solutions in between there, I know that all the solutions go to infinity but if I mix these two solutions what I notice is that the e^10t is the dominant term.1437

That is because that is the larger eigenvalues this is really key here, 10 is bigger than 5 which means that mix solutions in between these axis are going to tend towards the c2 axis.1457

The axis defined by the larger eigenvalues, if I start in between these solutions my other solution trajectories are going to bend over and I'm going to pull towards the axis defined by the larger eigenvalue.1475

That is why all of these solutions in between are going to approach the direction defined by the larger eigenvalue that e^10t, they all bend towards the axis defined by the vector with the e^10t.1495

Let us recap what we did there, we got a nice graph of solutions but how do we get there, the first thing was to graph the two eigenvectors and set up axis there, (2, 1) that was my vector right here, there is (2, 1).1523

Here is (-1, 2) and each one of those gives us an axis of solutions and both of those have positive eigenvalues which means we are going to infinity along both of those axis, that is why these arrows here go outwards.1540

And why these arrows go outwards and then when we look at solutions in between, they want to pull towards the axis determined by the larger eigenvalue, let me just remind you this was the eigenvalue R=10 this is the eigenvalue R=5.1559

All the mix solutions are going to pull towards the R=10 axis and that is why all these curves tend towards the c2 axis, the axis defined by that eigenvalue of R=10.1578

In example 3 we are going to solve another system of linear differential equations, again remember the whole key to solving these things is to find the eigenvalues and the eigenvectors of the matrix.1599

Let us jump right in and find the eigenvalues of this matrix, we want to find A- R(i), we are going to subtract R's down the main diagonal, 3 - R2 (-3 and -4 - R), I will multiply that out, 3 - R x -4 - R, - -3 x 2, that is + 6 = 0.1615

If I work that out I will get r^2 + 4R -3r, r^2 + r and then -12 + 6, that is -6 is equal to 0 and that is a nice format factors very nicely into R +3 and R -2=0.1646

I get my eigenvalues very nice, R=2 and -3, that is the first step is getting the eigenvalues but then remember for each one of those you are going to find an eigenvector, let me start out with R=2 and find my eigenvectors there.1672

That means if I plug in A- R(i), I get 3 - 2 is 1, 2, -3 - 4 -2 is -6 and I can quickly row reduce that, the 1, 2 will cancel out the -3 -6 I get 1, 2,0, 0 and I'm going to use my trick of putting a 1 and Y component.1688

I see that I would have to put a -2 in the X component in order to get zero multiplying those out, if you do not like that trick, you can use free parameters to get the eigenvector but I can see there that my eigenvector is (-2, 1).1716

For R=-3, that was my other eigenvalue, again I'm going to find A- R(i) that means I'm going to add 3 along the main diagonal, 3 + 3 is 6, 2 - 3 now -4 + 3 is -1 .1736

I see that I can simplify that a bit, I can divide the top row by 3 and then add it to the bottom row to cancel out the bottom row and I see that if I put a 1 in the y component, then my X component, in order to make it cancel would have to be -1/3.1756

I do not really like having a fraction in there, it is a little easier to have whole numbers, I multiplied both entries there by 3, get rid of that 3 in the denominator and you can do that with eigenvectors.1777

Remember any scaler multiple of an eigenvector is still an eigenvector, then I will get -3, 1, I'm sorry (-1, 3) that is my eigenvector for corresponding to R= -3 and now I can find my general solution.1795

Remember my general solution that is always the same, we saw the general form of the solution at the beginning of the lecture in the overview, maybe go back check that out if you are not sure where this is coming from.1817

But it C1 times the first eigenvector xi1 e^r1(t), r1 is the first eigenvalue plus a constant times a second eigenvector xi2 times e^r2(t), R2 being the second eigenvalue, my general solution in this case is C1 first eigenvector is (-2, 1).1831

e^first eigenvalue is 2t + c2, second eigenvector is (-1, 3) and e^-3t, that is my general solution but I also got this initial condition that I want to satisfy, I want to plug that in, I will plug in t=0 then I'm going get e^0 twice here.1860

Both of those terms turn into 1 and they dropout, I will get C1 x (-2, 1) + c2 x (-1, 3) = it is supposed to be (5, 0), let me set that up.1888

That will give me a system of 2 equations and 2 unknown, it should not be too bad, -2 c1 - c2= 5 and c1 + 3c2 = 0, we can solve this out using whatever algebraic system you are comfortable with.1910

But I'm going to use a substitution, I see I'm going to use the C1 as -3 C2 and if I plug that into the first equation I'm going to get 6C2 because I'm plugging them in right here, -3 x -2 is 6, 6c2 - c2 is 5.1932

5c2=5, c2=1 if I plug that back and I get C1 i= -3, I can drop those two constants into my general solution, now my general solution is c1 is -3 x where was my eigenvector (-2, 1) times e^2t + C2 that is just 1 x my eigenvector is (-1, 3) x e^-3t.1957

That is not a general solution anymore because we already found the constants, that is my solution to the initial value problem.2007

That is the end of that problem but let us recap and see the steps that went into solving that, the first step to solving systems of differential equations is to find the eigenvalues and the eigenvectors of the corresponding matrix.2025

That is what we are doing here, we are finding A- R(i), we are subtracting R's down the main diagonal, take the determinant that means cross multiply pluses and minuses here and we get a fairly simple quadratic equation for R which solve quickly into (2, -3).2039

That means those are the two eigenvalues, but for each one of those we have to find an eigenvector, we plug the first one back in R=2, we are subtracting -R down the main diagonal, that is where we get a 1 here and a -6 here.2057

By row operations that simplifies down to 1, 2, 0, 0 and we are looking for vector that matrix would kill and if we just start with a 1 and a y component, then it is pretty easy to see the X has to be a -2.2076

You can also use a free parameters to solve that if you are more comfortable with that technique, then we try to do the same thing with R= -3, that means we are adding 3 to the main diagonal of the original matrix.2090

And (3, 1), if we start out with a 1 and Y component we get -1/3 in the X component, which I do not like very much because of the fraction, you can scale up eigenvectors if you like,I'm going to scale that up to (-1, 3).2105

We drop those eigenvalues and eigenvectors into the general solution, the eigenvalues are (2, -3), the eigenvectors are (-2, 1) and (-1, 3) and that would be the end of finding the general solution but we also got this initial condition here.2119

Using the initial condition means we are plug in T=0, the E terms you get e^0, those turn into1 and then we have C1 times a vector plus C2 times a vector is equal to this given vector (5, 0).2138

That just resolves into a system of two equations and two unknowns which you can solve however using whatever technique your most comfortable with, you can use substitution to get a value of C1 and C2.2154

And I drop them back into my equations my C1 was -3 and my C2 just came in there as a 1 and I had my complete solution there, hang on to this general solution.2170

We are going to come back and look at that general solution in the next example and see how that would graph.2188

Let us go ahead and move on to the next example and try to draw some solution trajectories for that general solution. In example 4 we are going to graph some solution trajectories to the previous system of equations.2194

We already found the general solution back in example 3, this came from example 3, if you did not just watch example 3, you might want go back and look at example 3 and you will see where this solution came from.2208

What we are going to do in this problem is we are going to see what kind of solution trajectories that would give us, I'm going to set up some axis here and we are going to draw some solution trajectories and try to get a feel for the behavior of some of those solution.2223

This represents X1 and X2 or you can think of them as X and Y, it does not really matter and I'm putting a scaler here.2242

Now remember the first step here is to graph the eigenvectors and to set up 2 axis corresponding to the eigenvectors, our first eigenvector is (-2, 1), I'm going to go over to -2 and then up one unit and there is my eigenvector (-2, 1).2262

I'm going to set up an axis spanned by that eigenvector, let me see where to be on the negative side, right there, there is my axis defined by the first eigenvector.2282

My second eigenvector is (-1, 3), there is -1 on the horizontal axis, 3 on the vertical axis, it is like it is up there and its counterpart in the negative direction will be down there, I'm going to set up an axis corresponding to that.2302

Those are my two axis corresponding to the two eigenvectors, you can think of this as being the the C1 axis corresponding to the first solution c1 and there is the C2 axis corresponding to the second solution there.2328

I will fill in my axis a little bit and now let us try and figure out whether those solutions are going to zero or infinity and that really comes down to the sign of the eigenvalue, if the eigenvalue is positive then it goes to infinity.2346

If it is negative it goes to zero, let us look right here, we got a positive eigenvalue for C1, that means the solutions on that axis are going to infinity, I'm going to draw these axis going out to infinity.2362

On the second axis, we got a negative eigenvalue that means those solutions are dying down to zero, that tells us which direction everything is going along those two axis.2378

Finally our mix solutions will be in between these two axes and I always approach the axis defined by the larger eigenvalue.2399

Here are two eigenvalues are 2 and -3, what we see is the 2 is the larger eigenvalue right there, that means the mix solutions will approach the C1 axis, which is this axis right here.2410

The mix solutions are all going to tend asymptotically towards the C1 axis, they might follow the C2 axis for a while but they all end up approaching the C1 axis.2431

That is because we got a larger eigenvalue for that solution, you see how I'm kind of drawing the arrows, drawing the lines to make them parallel to the arrows on the two axis defined by the eigenvalue.2451

But they are always heading towards the one defined by the larger eigenvalue which is the C1 axis because it had an eigenvalue 2, the other one was negative.2470

Let me recap how we solve that one, we got a nice graph of solutions here but the way we did it was by starting out by graphing the eigenvectors, I graphed that first eigenvector it is right there is (-2, 1).2485

-2 horizontally and 1 vertically, then I graph that second eigenvector was (-1, 3), there it is right there, it is that black dot right here is (-1, 3) and then I set up axis based on those two eigenvectors.2500

I drew my blue lines here, those are my axis based on those two eigenvectors and then I looked at whether the eigenvalues were positive or negative to see whether solutions are expanding or contracting along those axis.2516

What I see is that the first axis is positive, it has a positive eigenvalue, that means solutions are expanding along this axis, C1 I drew my arrows outwards, all those arrows are going outwards along the C1 axis because the solutions are expanding.2529

On C2 I got a negative eigenvalue which means the solutions are contracting along that axis, that is why you see these arrows are going in towards the origin we are on that axis.2546

I want to draw my mix solutions, what I do is I looked at which of the eigenvalues is bigger, well 2 is definitely bigger the -3, that means the C1 axis is going to dominate, which means all these mix solutions tend towards the C1 axis there.2559

They are parallel to the C2 axis may be at the beginning but then they end up tending towards the C1 axis, because C1 had the larger eigenvalue.2578

That is how you can fill in solution trajectories to cover the whole plane, remember they never cross each other which you can see what the behavior of any solution trajectories starting at any point in the plane would be.2591

That is the end of that example, let us move on to example 5 where we are going to solve a system that looks a bit like one we solved earlier today but it got some important sign differences, let us see how that works out.2604

We are going to solve the system (-6, 2), (2, -9) as our matrix, remember on all of these the first thing we do is find the eigenvalues and the eigenvectors of the matrix, let us jump right in there.2616

A- R(i) means I will subtract R down the main diagonal, -6 - R2, 2 -9 - R and I want to take the determinant of that, I will cross multiply -6 -R x -9 -R, -2 x 2 is -4 and I want to that equal to 0, if I expand out the algebra I see I get a negative on both places.2630

That will give me a positive R^2 -6 x -R that is + 6R, same with a 9, so + 9R + 15R and then -6 and -9 multiply to positive 54 but -4 gives me a + 50 is equal to 0 and that factors really nicely into R +5 and R +10.2660

Of course that is easy to solve, my two eigenvalues are R= -5 and R= -10 and if that means for each one of those I'm going to plug it back in and find the corresponding eigenvectors, for R= -5 I'm going to start with that one, R=-5.2694

My A- R(i) is the same as adding 5 to each entry on the main diagonal, -6 + 5 is -1, I will put a 2 there, -9 + 5 is -4 and if I do a little matrix reduction, I'm going to take that first row and multiply it by -1, I get (1, -2) in the first row.2714

I can use that to completely wipe out the second row, I will just get 0 there and to find the eigenvector I have this little trick of writing a and y component and then seeing what the x component would have to be.2744

If they multiply to be zero, I see in this case the x component would have to be 2 to multiply the 0, if you are not comfortable with that trick just use free parameters to solve this, the same way you learned how to do a linear algebra class.2759

If you want some review on that I have a quick review on linear algebra in the differential equation series, it is a lecture right before this one and you can get a longer review on linear algebra with the whole separate linear algebra series here on www.educator.com.2772

My eigenvector(2, 1) right there and now let us find one for R= -10, that means I'm going to add 10 to each element on the main diagonal to get A- R(i) is -6 + 10 is (4, 2), 2, -9 + 10 is 1, I can row reduce that matrix.2787

I'm going to chop the first row down by a factor of 2, I get (2, 1) and then I will use that to wipe out the second row, I get (0, 0) and again I'm going to put a 1 in the y component.2819

That looks like I will have to put a -1/2 in the x component to make that come out to be 0, I'm not that comfortable with a fraction but remember you can multiply eigenvectors by a scaler and you will still get eigenvectors.2831

I'm going to pick 2 for my scaler and I will get that -1 and 2 as my eigenvector corresponding to the eigenvalue R= -10 and remember my general solution gave this at the beginning of the lecture in the lesson overview it is always the same for the systems.2849

It is always C1 times your first eigenvector xi1 times e^ri(t), r1 is the first eigenvalue + C2 x C2 x e^r2(t) and I drop in the xi1 and xi2, those are my eigenvalues and eigenvectors.2855

My general solution is C1, my first eigenvector was (2, 1) and my first eigenvalue was -5t and my second eigenvector was (-1, 2) x e^-10t and ordinarily at this point if I had some initial conditions I will be using them right now to figure out what my C1 and C2 are.2900

But I was not given any initial conditions in this problem, I'm just going to stop there and say that my general solution is the best that I can do, I will present that as my solution there.2931

That is the end of that one, what we are going to do in the next example is come back and graph some solution trajectories based on this general solution but before we go ahead and do that, let us recap what we did to get to this general solution.2946

We started with the matrix and we got find the eigenvalues and eigenvectors, to find the eigenvalues you subtract R off the main diagonal and then you take the determinant by cross multiplying positive in that direction and negative in the back direction.2959

That gives you this quadratic polynomial which factor really easily, R was -5 and 10, if you plug those values back in R= -5 means you are adding 5 to the main diagonal, that -6 + 5 that is how I got this -1 right here and -9 + 5 + 5 is how I got this -4.2975

I row reduced that, which meant I could wipe out the bottom row, simplify the first row and then I was looking for an eigenvector which means I am looking for X and Y that are killed by that matrix.3001

By starting out with a y=1 I was able to guess fairly easily that x had to be 2. If you do not like that kind of guessing technique, use free parameters. It is a very systematic technique and it will get you to the same answer.3014

That was my eigenvector corresponding to R= -5 I did the same thing with R= -10, I added 10 to the main diagonal, simplify down the matrix and I got an eigenvector, I did not like that one because it had a fraction it so I multiply that by 2.3030

You can do that with an eigenvector, you can scale it by something if it makes it more convenient, I was trying to get rid of that fraction and I got this new eigenvector, I took my two answers and I drop my eigenvectors into my general solution.3048

I drop my eigenvalues the -5 -10 into the general solution and I got my general solution and for this problem that is as far as we can take it because we do not have an initial condition.3062

There is no way to solve for the C1 and C2, that is the end of this problem but hang on to this general solution, we are going to use it in the next example to draw some solution trajectories for this.3073

You will be wanting to remember this general solution, we are going to use it right away in the next example, in example 6 we are going to graph some solution trajectories to the previous system of equations.3087

This is the general solution that we figured out in the previous system, if you are not sure where this comes from, go back and rewatch example 5 if that was where we figured out this general solution, this is coming straight from example 5.3099

We started out with a matrix there and we found its eigenvalues and eigenvectors and that is how we got this general solution, what we are doing in this example is we are going to try to graph this.3115

I'm going to set up some axis here and you want to start with your eigenvalues and eigenvectors, here is x1 and x2, I'm going to give myself some scale.3127

You want to start out by graphing your eigenvalues and eigenvectors, first eigenvector I see is (2, 1), I'm going to graph that right here horizontal 2, vertical 1 on the other direction and I'm to go ahead and set up an axis there.3156

Corresponding to the first solution there, that is my C1 axis corresponding to that first solution, I'm going to look at the second eigenvector which is (-1, 2) .3177

-1 in the horizontal direction, 2 in the vertical direction, there it is right there, there it is negative, let me set up an axis coming down through there.3192

I want to figure out which direction the solutions are traveling along those axis and the way I do that is by checking the eigenvalues and seeing whether positive or negative.3214

I see on my C1 solution I got a negative eigenvalues that means that the solutions are shrinking there, they are getting smaller and smaller, all along this axis I'm going to show my solutions gradually drifting into 0.3229

My second one, I see the eigenvalue there is negative 10, that means the solutions are also getting smaller on the second axis, all of these solutions are drifting into 0 and I want to think about my mixed solutions where I start mixing the C1 solutions and the C2 solutions.3248

The mixed solutions are always dominated by the larger eigenvalue, the mixed solutions I'm going to look for the larger eigenvalue, and my 2 eigenvalues are -5 and -10, you want to be really careful here.3270

-5 is larger than -10 because it is bigger on the number line than -10, it is -5, that means that is the dominant solution here, the dominant axis is the one defined by C1, that means all my solutions in between these 2 blue axis will try to tend towards the axis defined by C1.3284

I'm going to draw these solutions in between here, they are all going to 0 because everything here goes to 0, but they are going to 0 along the axis defined by C1.3323

All of these things are going to try to follow the axis defined by C1, even the ones that start near C2 are ultimately head over and try to follow the solution defined by C1.3336

The reason for that is because C1 has the larger eigenvalue even though -5 and -10 they are both negative, they are both very small but you still want to find the larger one and -5 is larger than -10, let me emphasize that.3365

We are using the fact that -5 is larger the -10, -5 gets to call the shots on the other solutions, all the other solutions will be attracted towards that larger eigenvalue, it is essentially means that e^-10t dies out quickly and the e^-5t is the dominant term there.3382

That is our graph of solutions. Let me recap the steps to doing that. First of all we took the general solution and we took that back from example 5.3406

If you are wondering where this line comes from, maybe go back and watch example 5 and you will see we did all the arithmetic to work out those eigenvalues and eigenvectors.3415

This problem we are just graphing them, we start out with the first eigenvector (2, 1), we graph (2, 1) right there that is 2 on the horizontal, 1 on the vertical that is where we got that first black dot there.3423

We looked at (-1, 2) and we graph that -1 horizontal, 2 vertical, graph that point as that second black dot and then we use those vectors to define our 2 axis, that is where we got these two blue lines.3439

We had to figure out are the solutions along those lines expanding or contracting and we figure that out is we look at the eigenvalues, here we see a negative eigenvalue which means our solutions are contracting.3457

Which is why we had these arrows going into zero all along that line, here we have also a negative eigenvalue again we have solutions contracting, they are going into 0 all along that line.3473

Our solutions in between, we figure out that when they want to follow the dominant eigenvalue which is the larger of the two numbers even if there both negative, we figured out that -5 is bigger the -10.3488

That means all the solutions in between are trying to approach this C1, this dominant axis, it is dominant because it has a larger eigenvalue and that is why all these red curves you see are approaching this dominant axis.3502

They might follow the other axis parallel for a little while but ultimately they all end up getting close to the dominant axis and asymptotically approaching it as they all dropped to 0.3521

That is the end of our lectures on real distinct eigenvalues, this is part of the systems of linear differential equations chapter, we are going to have a couple more lectures, one for complex eigenvalues.3534

We will get totally different pictures there and one for real repeated eigenvalues that is all part of the systems of differential equations chapters, I hope you will stick around for that.3547

This is all part of the differential equations lectures series here on www.educator.com. My name is Will Murray and I really want to thank you for watching, bye.3557