Differential Equations Complex Eigenvalues

Hi and welcome back to differential equations lectures here on educator.com.0000

My name is Will Murray and today we are going to be studying systems of differential equations, where the matrix that gives the coefficients for the system turns out to have complex eigenvalues.0004

So we already have a lecture on systems of differential equations, we already saw the basic idea where you find the eigenvalues and eigenvectors the matrix and then drop those into a solution.0015

So if you have not watch that yet we are probably be good to go back and work through that lecture before we learn about the case on complex eigenvalues because this 1 is a little more complicated.0026

So on this case in this lecture I am going to kind of assume that you have already done the 1 with real eigenvalues and we will move on more complicated case of complex eigenvalues.0036

So let us see how that works out.0046

Let me just remind you that to solve the system of linear differential equations X′, this is vector x here and the vector X′ and then this A stands for a matrix.0048

Usually we will solve 2 matrices.0061

The way to solve that to find the eigenvalues and the eigenvectors of the matrix A so that sort of the first step in solving these things.0064

And if the eigenvalues are complex then they will occur in conjugate pairs, meaning you get if A + BI is 1 eigenvalue then A- BI is the other eigenvalue.0073

And that really comes out of the quadratic formula remember the quadratic formula tell you that R = -B + or - the ² root of the ² -4 AC and so all over 2A.0086

And so if you get a - number under the ² root then that is the case where you are going to have complex numbers as your eigenvalues.0100

And so if you got a - number of the ² root then you are really going to get -B + or - that complex value.0108

So to tell you always get these things in conjugate pairs A+ BI and A- BI.0115

So here is how you handle something like that.0122

You choose 1 of the eigenvalues I always choose R = A + BI just because it is 1 less - sin to keep track off.0123

And if you find corresponding eigenvector B, that can be a little challenging to find the eigenvector when using complex numbers but will do some examples here and you will see how it works out.0134

In any form the complex solution it still the same form as before remember our old form was in the RT where R was the eigenvalue x the eigenvector B.0143

So we are doing the same thing here we are forming the complex solution where we put in the eigenvalue as the R.0153

A + BI and those we multiply by the eigenvector B.0160

But the problem is that still has complex numbers in it so let me show you how you can expand that out and get to real solutions.0164

So what you do is you use Euler formula which is E I θ = cosine θ + I sin θ.0171

So to remind you what we had on the previous page we had an E A + BI that was the R x T & we want to expand that out into E AT x E B IT.0187

And the E VAT is no problem we are just gonna leave that the way it is if the B E IT that we are worried about.0210

We think of this as this B E IT B IT as EI x BT and then , we use Euler formula on that so it expands out the BT is playing the role of the θ here.0217

So we have E I x BT and that expands out into cosine θ + I sin θ that cosine of BT + I x sin of BT.0231

So you are gonna expand out that in the B IT part into a cosine BT + I sin of BT and you still have all this time some eigenvector which will itself have complex numbers in it.0242

And so you multiply these complex terms the cosine BT and the i sin of BT you multiply those in eigenvector and so you are going to get some complex numbers inside the eigenvector.0255

Then, what you do is you separate that eigenvector with the E terms inside it into a real part and imaginary part. You are going to have some terms which just a real numbers and some terms which have imaginary have I multiplied by some real numbers.0270

In what you do is you take each 1 of those and you make those into your 2 solutions and you can drop the I now.0287

This first 1 becomes your X1 solution and the second one without the i becomes X2 solution and then you attach constants to each one a C1 and C2 and that becomes your general solution in terms of real numbers.0295

There is no more I after this because you separated out the real part and the imaginary part and then you drop the eye off the imaginary part.0315

We will see how that works out I think I make more sense after reducing examples but that is some kind of the general overview there and let us see how that actually plays out with some specific problems.0323

Let me tell you that we are going to graph these, so in order to grap the solutions the trick here is to ignore the EVA T factor at first.0335

What you have if you ignore that factor is a bunch of terms which have things like cosine of BT and sin of BT and the trick is to pick values of T that are going to make BT equal to some nice common values that it is easy to plug into cosine and sin.0347

Ones I like our 0 π over to π 3 π over 2 and 2 π so for example if B were equal to 3 then what you have is cosine of 3T & the sin of 3T.0368

And so you want to pick values of T that will make 3T a multiple of π over 2 so, I would pick T equal to 0 and then the 3T would be 0 and I probably pick T equal to π over 6 and the point of that is that 3T would be π over 2.0388

So that is an easy value when you put π over 2, when assigning cosine, and it is very easy answers.0409

And probably take pick 3 equal to π over 3 because then, 3T would be π and then T it would be I would take π over 6 π over 3 π over 2 because then 3T would be 3pie over 2.0415

Finally I would take T = 2 π over 3 so I am going by multiples of π over 6 because π over 6 x 3 is π over 2.0439

Finally I would get 3T would be 2 π.0449

Appointed that is when I saw B was equal to 3, I picked values of T that when I multiplied by 3 are going to give me 0 π over 2PIE 3PIE over 2 or 2PIE.0454

I am going to graph the points that arise very easily.0467

What is going to happen in all of these is you are going to get ellipsis.0473

When you graph out these values of T you are going to get ellipsis.0475

Your basic graph is going to have ellipsis and then you are going to look at this factor of E AT and essentially if A is positive, that means E AT is getting bigger and bigger and bigger so you are gonna draw this ellipsis getting bigger and bigger and bigger.0478

It is going to be spiralling outwards.0496

If a is less than 0 than E AT would go to 0 which means you are going to draw this ellipsis restricting down smaller and smaller and so you can have a pattern that spirals in origin like this.0499

In the A positive case you have a pattern that gets bigger and bigger and so you have like ellipsis but their spiralling bigger and bigger so will see that as we get some examples.0517

Just like all the other problems if you are given initial conditions and you can use those to solve the constant C1 and C2.0529

It is deftly time that we got some examples and you try this out with some actual numbers so worked the first example together.0536

We got X′ = 31 -21 and x X so remember all of the systems problems start out the same way always start by finding the eigenvalues of the matrix and when you do that if you subtract R down the main diagonal.0546

I got 3 - R x 1 - R so 3 - R x 1 - R - -2×1 that is +2 = 0 so if I multiply this out I got R ² - R - 3Rs that is - 4R +3 +2 so we got +5 = 0.0563

That is not a quadratic equation that factors nicely so I am going to go straight quadratic formula.0591

R = - B so positive 4 + or - B ² -4AC B ² is 16 4AC is 4×1×5 so -20 all over to A which is 2 and so that is 4 + or - .0597

Now 16-20 is - 4 so I got the ² root of -4 so that is 2 I all over 2 which is 2 + or - I .0617

So remember, I always I said these roots of the characteristic equation the eigenvalues they always occurring conjugate pairs always have A + BI and A - BI so here we got 2 + I and 2 -I and the nice thing about these is you really only have to work with 1 of them.0627

So we are just going to pick R = 2 + I and will play that back in matrix so 3 - if I put 2+ I in for R that is 3-2 is 1 - I and also have 1 over here -2 over here 1-2+I is -1 - I and then, I want to set this up as to find the eigenvector.0646

So I will find a vectors such that when you multiply that by that vector you get 0 and little trick those works for me is to plug-in Y = 1 and will figure out what the X should be.0682

So let us this trick it is very helpful for finding eigenvectors and I think it is more useful if we look at the bottom row here0694

So I got is -2 x X-1 - I x 1 = 0 and I am gonna solve for X and I see that I got X if I move the -1 -I over other side I get + I over -2 so that is -1/2-1/2 I so my eigenvector would be -1/2-1/2 I A on top and 1 on the bottom.0705

I am really not thrilled with having all those - sins and with having those fractions.0744

Remember you can multiply eigenvector by a scalar if you want to make a little cleaner so that is what i am going to do here so i am going to multiply by -2 and I am going to get that gives me 1 + I in the and on the numerator in the first quarter of the vector and -2 in the second core in a second coordinate.0747

so now I've got my eigenvalue and my eigenvector and I am going to plug that into my generic solution which is number generic solution is E in the RT were R is the eigenvalue x the eigenvector.0766

so E in the RT the R was 2 + I used 2 + IT x the eigenvector is 1 + I20782

I am going to separate out the E 2T so, I will just remind you that E to the 2+ I T = E 2T x E IT and so all right E 2T separately and that is not going to affect the immediate future here.0795

And now E IT I am going to use Euler formula remember, Ei θ = cosine θ + I sin θ so E 2T as our E IT so I had to deal with here that some cosine T + I sin T.0818

And I am still multiplying that by the eigenvector 1 + I and 2 and I am going to multiply that in eigenvector so, E 2T outside so if I multiply that in to each coordinate the eigenvector gets a little bit messy.0845

I get cosine T + I sin T I am going to multiplied it by 1+ I cosine T multiplied the first term by I now I x I so I got -1 sin T and the bottom I get 2 cosine T +2 I sin T0861

Now I said that the goal here is to separate this out into a real part and imaginary parts so, each 1 will have the E 2T in the 2T if I just collect all in real terms over here and I will collect my imaginary terms over here and we will see what we get for each 1.0887

For the real terms I see at the top I have a cosine T - sin T bottom I got 2 cosine T and now , the imaginary terms as you have it I x the sin T + a cosine T so feel that in sin T + a cosine T and 2 sin T on the bottom.0908

So, those are my real and imaginary part of the solution remember I said you can take each 1 of those and make those into separate solutions that is going to be our first solution our X1 and that is can be our second solution X2 .0933

And then just take each 1 of those attaches a constant to each 1 and you can write down the general solution right away.0953

So, our general solution is just C1 x that first solution C1 x E 2T x cosine T - sin T and something is bothering me a little that I see what is bothering me I have a -2 here and I forgot to bring that -2 along that - sin along for the ride so they fill that up here.0962

There is - here - that should be - and this should be - so all my 2s through there should be - so we go ahead and fix that.0994

-2 cosine T sorry about that little mistake there + C 2 E 2T x the other solution sin T + cosine T all over -2 sin T .1006

And that is our general solution to that system of equations or to that matrix differential equation, if you prefer that for so let's start recap how we got that.1027

First thing for any of these problems is to find the eigenvalues and eigenvectors the matrix that is related here.1047

Subtracted are the main diagonal, I took the determinant , I cross multiplied to get the determinant, I got this quadratic equation for R which did not factor nicely so I had to use the quadratic formula and simplify down to 2 + or - I.1056

Remember these complex 1s you only have to take the first route .1069

I always was to go for though the 1 with + just to make me have to do with fewer - sins later so I used R = 2+ I plugged that back in the matrix for this R here I got 2+I in there as I gotten 1 - I and -1 - I .1074

On the main diagonal here and try to find the eigenvector here so putting a Y 1 for the Y and I try to figure out what action be correspondingly so looking at the second equation here that is -2 X -1 -I =0 .1090

And I solve that out for X plugged that back into my eigenvector and I see I got some nasty fractions but member it is okay to multiply eigenvectors by scalars if necessary to clear some nasty fractions .1107

Since of this -2 just multiply -2 in top and the bottom and so, I get a new eigenvector which does not have any bad fractions anymore and out with his in vector I plug at back into my generic form for the solution E of the eigenvalue x T x the eigenvector.1123

In E + 2 +E IT I expanded that out in E 2T and in the E IT just say the E 2T on the sides not really get into anything more .1148

The E IT expands out by Euler formula that was Euler formula up there so expands out into cosine T + I sin T .1156

And then we multiply that in eigenvector which requires keeping track of all the complex terms, it is a little messy in the top here we multiplied it through the -2 bottom and then we look through all those terms segregate out the real terms and the imaginary terms.1162

So here, I've segregated all the real terms and here I segregated away all the imaginary terms and if I factored out and I from all the imaginary terms is all the terms it had and I affected out group them altogether.1184

And it turns out that each 1 of those gives you a fundamental solution to your original system so you do not have to consider the I anymore you just take each 1 of these and you form a solution to original system .1202

Since I d1 that here I took just copied those 2 solutions down C1 and C2 onto each 1 and then put them together to get my general solution.1219

So you want to hang on this general solution we we're d1 with all the competitions what we will be working in the next example is taking a look at the graph of this and plug in some values for T and see how it plays out on some actual axis.1228

So in the second example, here we are going to graph the solution trajectories the previous system of equations .1243

I got here is the solution that we derived in the previous example so, if you are not sure where this equation came from then you might want to go back and rewatch example 1 because it was example 1 were we derived the solution1257

What really doing now in example 2 is working with this equation and trying to graphs and solution trajectories so, let me show you how that plays out the setup some axis here.1272

The nice scale so 1 2 on my X 1 axis here's X2 1 2 there .1284

So remember the trick here is you do not worry about the E the 2T term at the beginning so let me not worry about in E 2T .1307

What I am going to do is just going to look at this part to start with just a graph 1 of the solutions because I am going to take the thing that is taken C1 = 1 and C2 = 0 and I am also going to ignore the E2T at the beginning .1315

We will see how that what role that plays in a moment so, I am gonna plug in some values of T into that solution and I am going to see what kinds of numbers I get1333

So, I am going to take T = 0 I want to take the values that are and can have me plugging in multiples of π over 2 so T is equal π over to T = 3 π over 2 and T = 2 π .1344

So, think that will be enough values of T to get an idea of what the shape is the plug-in T 0 plug-in and here we get cosine of 0 that is 1 - sin of T that 0 and - 2cosine T that is -2 so that gives me 1 -2 .1367

Basically I am just looking at the cosine terms drop out if I plug in T = π over 2, I see well, now for π cosine is 0 in the sin is 1 so going to get the - 1 on the top and 0 on the bottom .1385

For T = π the cosine is -1 and the sin is 0 so I get - 1 on top and then -2 x -1 so, positive 2 on the bottom 3 π / T cosine a 0 sin is -1 so I got positive 1 and 0 and finally T = 2 π .1403

Remember, I am plugging them all into this first solution right here T= 2 π is the same as T = 0 so I get positive 1 on top from the cosine and -2 on the bottom .1423

Slowly plot those out when T = 0 I see I got 1 -2 so 1 on the horizontal axis -2 on the vertical there is T = 0 and T = π over 2 got -1 0 -1 0's there is π over 2 and T = π I see I have -1 2's that goes up here there is π and T = 3 π over 2 I see 1 0 so there is 3 π over2 and T = 2 π is back to 1 -2 .1439

So that is back the same place we started, it is T = 0 and if I connect these up it is pretty clear that I have an ellipse itself is a circling around forever but, it is it is not a circle it is an ellipse so it is the ellipse in around forever .1479

And we know it is going in that direction in the clockwise direction so that is that sort of the basic shape of the graph .1501

However there was something I was not considering yet which was that there is a E2T in there as well .1513

So, what happens is, the E2T course that goes to infinity so what that does is it sort of makes this ellipse get bigger and bigger and bigger each time you circle around.1521

So, this makes it spiral outwards because basically that is because the eigenvalue 2 is positive there so what I am going to do is say this blue ellipse is some sort of a first draft that is not the actual graph.1533

I will make the actual graph will make it red and what I want to do is have that ellipse Getting bigger and bigger as it goes around and around.1560

I will draw few more trajectories here basically, they're all ellipsis Except that they are getting bigger and bigger every time they go around.1576

And even down here in the origin they're getting bigger and bigger following the same general shape is that ellipse .1585

But getting bigger and bigger each time so really the blue graph here is not the final graph that is not the answer, it is these red spirals that are the true shape of the solution trajectories.1605

So, let me show you how we derived that I remember we only need to look at 1 essentially if you look at the other at the other solution here, you are just going to end up with the same points just all shifted around by π over 2 .1620

So you would end up drawing the same thing if you looked the other 1 so we just looked at the first 1 the cosine T - sin T on top and then -2 cosine T on the bottom .1635

And what we did was we plug-in select values of T to give us a shape of the graph C = 0 π over 2 π 3 π over 2 and 2 pies.1645

We plugged those in and we got different points year 1 -2 -1 0 -1 2 1 0 and went back start 1 -2 .1656

So, I grasped each 1 of those points is what I did here and I got an ellipse and I also got the direction of travel of the ellipse so just by graphing those is how I got this blue ellipse .1666

Kind of going around clockwise but then I noticed that I also had this term of E 2T and that is going to be magnifying the size of the ellipse each time it travels around .1682

So a set of a pure ellipse, what I actually have is an elliptical spiral that is getting bigger and bigger each time he travels around .1694

So that is what you see here with the red curves the red curves are actually the 1 that are actually giving me the solution to the problem .1702

So for example 3 were to find the general solution following system .1712

3 -5 5 -5 so, just like all the others here the first very first step in all of these is to find the eigenvalues and the eigenvectors of the matrix.1719

So, remember the way you do that is you subtract R from the main diagonal and then you get yourself a quadratic equation for R, 2901, were to cross multiply here so got 3 - R x -5 - R -5 x -5 so +25 = 0+25 = 0 and let's see how that plays out I see them and have a positive R ² +5 R -3 R so +2 R and now got -15+25 so that is +10 = 0 .1731

Again that is not a polynomial that factors nicely so in order to solve that I have to use a quadratic equation R = =B 2 + or - now be ² is 4 -4 AC , AC is 1×10 so -40 all over 2A so that is - 2 + or - 4-40 is -36 so I am going to get complex solutions the ² root of -36 is 6I still divided by 2 so I get -1+ or -3 I .1780

So again I get 2complex roots remember they always occurring conjugate pairs -1+3 I and -1 - 3I1822

The nice thing about solving systems with complex eigenvalues is you only ever have to treat 1 of the of the eigenvalues so, we are going to take the 1 with the R -1+3, I re-plug that back in matrix for that R on the main diagonal so what we will get is 3 - R 3 - -1 is 4 -3 I -5 and 5 and -5 - -1 is -5+1 it is -4 -3 I .1834

And again , we want to find eigenvector corresponding to that eigenvalues set th= 0 , and as usual I am going to put a 1 in here and I am in a try to figure out what action be to make it work and I am going to look at the second equation that is often useful in these complex in these matrices with complex eigenvalues .1875

So looking at 5X -4-3 I = 0 using that second equation there so if I solve for X I move the 4+3 I over the other side X = 4+3 I over 5 .1896

So my Eigen vector is then read that is 4/5 + 3I +3/5I on top and 1 on the bottom .1921

Again I do not like these fractions and because it is an eigenvector it is okay to multiply the whole thing by a scaler in order to eliminate whatever fractions you might want to get rid off.1932

So to multiply this whole thing by 5 and I will get on the top there i will get 4+3 I and on the bottom of this get 5 so it is a much nicer eigenvector so now have an eigenvalue and an eigenvector .1941

So I am going to plugged this and I got remember my generic formula is in E RT x V where R is the eigenvalue and V is the eigenvector.1948

So in this case my Eigen value is -1+3 I all of that x T and now my eigenvector is 4+3 I / 51970

The point is I like to sort that out into real parts and imaginary parts, I may be using Euler formula and but First I am going to right E -1+3 I see the separate that out as E -1T x 3T .1989

Of course that should have been an I so E -1 is E 3 IT remember this Euler formula was E I θ = cosine θ + I sin θ .2009

And the key part here is that , θ in question is multiplied by I that is going to be 3T in our case that is our θ right there the 3 and the T right there .2029

So let me expand this out this is E -T x now cosine θ + I sin θ so that is cosine of 3T + I sin of 3T x my eigenvector which is 4+3 I x 5.2043

Now , I will multiply the real and complex parts the real and imaginary parts in eigenvector so i am just going to preserve this in the - T it is not really doing anything for the near future.2065

Multiply everything in eigenvector so I get 4 cosine 3T + 4 I sin of 3T is + 4 I sin of 3T +3 I cosine of 3T +3 I x I which is -1 so - 3 sin of 3T .2078

So that is the top part is a hard 1 in the room bottom I get 5 cosine 3T +5 I sin of 3T .2109

So, if I expand that out into a real part and imaginary part of E the - T x both of them in real terms over here + I E - T and all collect all my imaginary terms over here wasn't E - 2 in my T got run together a little bit there that looks little better .2121

So let me collect all the real terms on the left so 4 cosine of 3T and I see I have a - 3 sin of 3T in the bottom for the real side I have five cosine 3T .2147

Now , on the compliment on the imaginary side, Ii am going to collect all the terms that have I in them remember with the eye outsides would only to write an I inside 4 sin of 3T and 3 I cosine of 3 T I outsides up at 3 cosine of 3T .2165

On the bottom I see I have a 5 sin of 3T because I can go outside.2187

So remember, the way this works once you find your real solution in your imaginary solution you can click the I off the imaginary solution and then each 1 of those is going to be 1 of the terms of your general solutions .2193

That is X1 and that is X2, so you just put those together with constants to form your general solution.2209

Slowly go to do that X is C1 x E to the - T x that first solution 4 cosine of 3T -3 sin of 3T and 5 cosine of 3T + C2 x that second solution which was E - T .2217

Notice I am not writing the I anymore you do not need the I resolve this into real solutions 4 sin of 3T +3 cosine of 3T and then five sin of 3T on the bottom .2245

So that entire thing is our general solution we got 2 specific solutions but constants multiplied by each 1 and put them together to get your general solution.2264

Since the end of that problem in the next example , were to go through it and working to graph this see how this looks on an actual set of axis but before we do that I like to recap how we got there because the next problem were just in a start with this general solution will say where it is coming from.2279

So what we started out by doing was to find the eigenvalues and eigenvectors , the matrix the system from always E way, you always find eigenvalues and eigenvectors.2297

So that means, subtracting R from the main diagonal of the matrix cross multiplying cross those and subtract them which is how I got this quadratic equation, subtracting -5×5 some subtracting it turns into a positive 25 .2307

Simplify that and run the quadratic formula on it to get my 2 eigenvectors or much eigenvalues which are always conjugates of each other so they always look like a + or - B I so -1+ or + or - 3 I .2326

I pick the positive 1 -1+3, I plugged it back in for the R so plug-in -1+3 I in the R in the matrix as I got that matrix there and I want to find the eigenvector and my trick for doing that is always a start with the 1 in the y-coordinate.2341

Then you had to figure out what the X should be and it is usually useful to use the bottom line here so this is multiplying the bottom line x that vector 5X -4-3 I = 0 .2359

Solve for X which is pretty easy, plug it back in for the X values the eigenvector many notice all got these bad fractions like to remove the denominators but since it is an eigenvector can multiplied by a scalar and you can clear the denominators away.2372

Since I did hear multiplied by 5 and I got the eigenvector 4 +3 I over 5 .2389

So then, I plugged that back into my generic solution E RT x V so there is the eigenvector and there is the eigenvalue right there .2395

I expanded that into E - T which does not really do anything for a while and E3 IT which I used Euler formula up here to expand that out into cosine of 3T plus I sin of 3T.2406

And I multiplied that back in eigenvector so , we multiplied this in top and bottom here we got 4 terms in the top because it is a binomial x a binomial 2 terms in the bottom and then we segregate out the real parts of the solution and the imaginary parts .2424

Imaginary parts of the 1s multiplied by I, so we segregated those out we sorted them out into everything real and everything imaginary.2446

The imaginary we factor out and I and then each 1 of those becomes 1 of our solutions that were going to use for general solution so, that is the real 1 over there at the imaginary 1 over there except that we do not need an I anymore we just attach arbitrary constants C1 and C2 to it.2454

So that is how we derived that solution in the next example you want to hang on solution because we’re going to start with this equation and were going to go graph were going to see what it looks like .2473

So, let us go ahead and see how that 1 plays out so next example this is the exact I copy down the general solution from before but in bother writing down the C1 and C2 .2484

So, got E - T x that first vector E - T x that second vector .2496

If you just tuning in if you have lost the previous example for while, these 2 vectors they came out of the answers from the previous example2506

So, you are wondering where these came from what we did in example 3 was we solve the matrix we found the eigenvalues and eigenvectors and then we went through some algebra to resolve the solutions in to these 2 pieces.2511

So if you have a just watch example might go back and check it out and see where the solutions come from .2529

Example 4 what were going to be doing is graphing it so I am a set up some axis here vertical axis here very vertical to see if I can know little better than that.2535

Little hard to draw vertical lines on these tablets but I will do the best I can do is a little bit better alright and looks like I am be going up to five some scale on there.2553

12345, now the key thing here is you want to be plugging in values of T that are going to give you multiples of π over 2 .2571

So , I am seeing in my solutions here I got a bunch of 3 T so far I want to get 3T to be a multiple of π over 2 then I see I got a make T is a multiple of π over 6 .2595

So, I am going to use here multiples of π/6 and so that will give me nice multiples of π/2 in the sin cosines which will be very easy values to plug-in and find nice numbers for.2610

So, multiples of π over 6 start out with 0 C = π over 6 T = 3 π over 6 which is the same as π over 3 and finally T = 3 π over 6 which is the same as to π over 2 and finally T = 4 π over 6 which is the same as 4 π over 3 .2625

And the point of doing that as a something I mentioned briefly at the beginning of the lecture but in the absence of an example it might not of made sense back then now it is supposed to make sense .2651

The point of doing that is that 3T would be in this case 0 here 3T will be π over 2 so very easy value to plug-in 3T will be π 3T will be 3 π over 2 and finally down here 3T will be 2 π .2662

So since I am going to be plugging in 3T , I already notice that here that means I have some very nice values here nice multiples of π over 2 to plug-in so get nice easy answers.2686

So far plug-in 3T = 0 I see I got 4cosine of 0 so 4 in the top and 5 in the bottom but im not to worry about each of the - T yet so I will just put that aside for now will figure out how that affects the solution later .2698

So you have 4 5 and the first 1 by playing π over 2 that means sin is 1 cosine of 0 so I just gets -3 and 0 here for π over 3 that means 3T is equal π so cosine is -1 & sin is 0 so -4 -5 for 3 π over 2 my cosine a 0 my sin is - 1 I get positive 3 and 0 and finally for 2 π I am back to where cosine is 1 and sin is 0 so I get 4 and five here .2717

That gives me a nice set of points to start plotting so, let me go ahead and plot those I see a start out T = 0 on it 4 5 so there is 4 and 5 is up there C = π over 6 assume it -3 0 so this is We flag this as T = 0 π over 6 is at -3 and 0 so there is π over 6 .2757

π over 3 is at -4 -5 so, there is -4 -5 there is where we are at π over 3 and π over 2 were 30 and at 2 π over 3 we come back to where we started to π over 3 come back to where we started more starting to repeat ourselves.2787

So I can see a nice ellipse forming here we see if I can connect these points up to make a nice ellipse .2820

And I also know the direction of travel because it started out T equal to 0 and it goes down Southwest here.2827

So that is heading down and spirals back up back to where it started so, that is kind of the basic shape of the graph but there is 1 element that I did not incorporate here which it well actually there is I didn't look at the other solution.2843

If you graph this solution here you just can get the same graph sort of rotated around by a little bit so , go ahead and try it if you are instead plug in some values of T and you will see that you end up getting the same ellipse that is kind of rotated around.2860

So let me now incorporate the other missing element which is this, E in - T and so E - T goes to 0 when T gets large so that makes the this ellipse Smaller and small smaller so it makes it instead of following up your ellipse an elliptical spiral getting smaller and smaller and makes it spiral in origin .2876

I will do my best to draw that for you mix a spiral in origin so instead of just following the ellipse purely getting smaller and smaller as it goes around. 4844 Still following the same basic shape of the track but it is getting smaller and smaller as it spirals in in towards the origin.2907

There be bigger versions to that or they are all kind of following this general elliptical shape but they are all getting hoops are all getting smaller and smaller and spiraling in.2937

So to draw a few more those spiraling in all going counterclockwise and getting smaller and smaller as they do it .2951

Slowly recap how we figure that out first thing I just picked 1 of the solutions you could pick E 1 I just picked the first 1 but you do not need to look at both that is a nice part and then I noticed that I had that 3T in there and I wanted to make that the multiples of π over 2 .2968

Because I know I have multiples of π over 2 fit into sin and cosine them and get very easy values to plot.2985

So, that is why I set 3T's π over to will and T would have to be π over 6 so that is why, I pick multiples of P over 6 down here is my x and then that meant that when I multiplied each 1 by 3 I got multiples of π over 2.2992

And so when I plug-in those multiples of π over 2 π into my cosine and sin, I got these points for 5 -3 0 -4 -4 5 +3 0 and then asserts repeat itself it for 5.3007

Sets was graphing when I got that point that point this point in this point and it start repeating itself over back beginning .3023

So, I connected those up I got ellipse which is the blue graph right here now that blue graph is not the final graph that is not the final shape of the solution trajectories because, we have this other piece we have to incorporate this in the - T .3032

Everything is getting multiplied by E - T as it as it circles around here and that E- T is getting smaller and smaller so, it is kind of dragging the whole thing down the 0 which means a set of a single ellipse.3048

You got it getting this shrinking and shrinking as it goes around and around which really means you have this elliptical spiral in origin.3063

That is what I was drawing with these red curves here is getting smaller and smaller and it is shrinking down in origin.3070

Such a happy without 1 in our next example here were going to solve all the following system and were going to graph the solution trajectory.3078

This is X = X′ = 3 -5 5 -5 you will notice that that is the exact same 1 as before sort of start with her general solution from before .3092

But now we also got this initial condition , X0 = 615 so, we are going to use that initial condition to find the values of the constants in the general solution from before .3102

So, we write down what the general solution was the general solution I am just quoting this from my think was example 3 where we solve this so if you just tuned and for example five here you want to go back and look at example 3 and you'll see where we solve the same matrix.3114

We did not have the initial condition but we just use the matrix to get C1 E - T x 4 cosine of 3T -3 sin of 3T /5 cosine of 3T + C2 E - T x 3 cosine of 3T +4 sin of 3T /5 sin of 3T.3134

So that was our general solution from example 3 using the same matrix of that star general solution.3171

I do not want to rehash all that you can go back and look at the video for example 3 what were going to do is work in a plug-in T = 0 for any use that value of T = 0.3180

So everywhere through here in a plug-in T = 0 and were these the fact that E 0 is 1 cosine of 0 is 1 & sin of 0 is 0 so all the sin terms are to drop out all the cosine terms or disbelievers with 1 so exit 0 plug-in 0 everywhere is C1 x in the 0s that is just 1 for cosine that is 4-3 sin which is 0 5 cosine is 5+ C2 x 3 in the 0 again drops out since 1 3 cosine is 3+4 sin is 0 5 sin a 0.3190

Okay so, that is supposed to be equal to 615 and what that really is doing is it is encoding a system of 2 equations and 2 unknowns.3238

So, if I write out the first line there I get 4 C1 +3 C2 = six bottom line says 5C1 +0 C2 = 15 .3249

So, that is obviously a pretty easy system to solve the bottom line right there tells me that C1 = 3 if I plug that in up here I get to 12+3 C2 = six and so I would get C2 = - 2 .3264

so it gives me my C1 and C2 Upland is back in general solution .3281

Is my C1 was 3 E - T x 4 cosine of 3T -3 sin of 3T over 5 cosine of 3T.3288

Now my C2 is - 2 so, -2 in the - T x 3 cosine of 3T +4 sin of 3T over five sin of 3T and I think I can combine those together so I will have an E - T everywhere and if I look at the top I see I got 3 x 4 cosine.3305

So that is a 12 cosine over here, I've got -6 cosine so that please me and with a net of six cosine 3T now -9 sin and -8 sin over here because of the - 2outside so -9-8 is -17 sin of 3T .3338

And then in the bottom I see I got 15 cosine 3T -10 sin of 3T.3367

So that is my solution and I am in a try to graph that and remember you want to graph it where you plug in multiples of π over 2 in sining cosine .3383

So discuss a 3T is 0 3T is π over 2 3T is π 3T is 3 π over 2 and 3T is 2 π , you see what happens with each 1 when 3T is 0 the sins dropout I get the cosine 5708 I get six and 15 right that is a fraction it is really a vector they're so 615 π over 2 the sins give me 1 so -17 -10 π will give me the cosine give me-1 so -6 -15 and the 3T over that the 3 π over 2 give me the sin give me -1 so 17 and 10 .3398

And the 2 π will give me the same as if we're plug-in 0 so be six and 15 again.3470

So, let me set up some axis and try to graph that will substrate it all.3479

That looks bit better so all graph this by 3s, 3 6 9 12 15 18, 3 6 9 12 15, 3 6 9 12 15 18 , 3 6 9 12 15 18.3488

So, start out at 6 15 20 graph 615 six down six over horizontally there is my starting point there now -17 10 90 17 just short of -18 10 is just up from nine that I go to -6 -15 so and up down there .3521

And finally, -17 -10 is -17 that should have been -10 so that takes me down right there and 17 10 is your 17 and here is 10 .3548

And so I seem to have basically an elliptical shape because we know that it starts repeating after that.3576

however there is 1 aspect of this that I have not taken into account which is that there is an E - T so that E - T what happens as it takes that ellipse And it depresses it as it spirals around .3588

so each each time his browse around it is going to get smaller and smaller.3606

So let me draw the true graph and read it does start there is T = 0 there but it gets smaller and smaller each time it travels around the origin.3611

So that red graph is this true solution trajectory system .3626

So it is go back and see what we did there we started out with the solution matrix equation which is something really figured out an example 3 so, he do not know where that came from and go back and look at example 3 and you will see where we got this general solution.3642

So I just copy that over from example 3 and we plug-in T = 0 because that was the initial condition were given which met the E- T .3658

Those alternative to 1 the cosine the 3 T those alternative to 1 and so those are my thoughts right at the 4 and 5 here and that is forgot the 3 here.3666

And that all the sin term sin is 0, 0 so those all kind of dropped away when we plug-in T = 0 .3679

So I got C1 x 4 5+ C2 x 30 = 615 , and that resolves into 2 equations in C1 and C2 which are very easy to solve C1 right away plug that back in plug back in plug C1 back in first equation we get C2 so I put the C1 and general solution and the C2 in general solution.3686

And then I gathered like terms I got all my cosine's together all my sins together which is where I got the 6 in the 17 that six for example came from 3 x for which is 12-2×3 so 12-6 give me that 6.3712

That 17 came from 3 x -3 and -2×4 so that is how I got -9 -18 -17 that is where that came from .3727

Then the 15 cosine 3T and 10 sin 3T that just came from the bottom of each 1 of those vectors that was my actual solution and then I graph of plug-in values of T that would make 3T equal to multiples of π over 2 .3740

So 0 π over 2 π 3 π over 2 and 2 π of plug-in those values of T, and I got specific points here and I plotted each 1 of those points there is the first 1 there is the second 1 there is the third 1 here's the 4th 1 and then it started to repeat itself after the 4th 1 .3757

I drew my first blue ellipse Which is what the solution would look like if there were no E the - T there but since there is even - T there , I know that ellipse is getting smaller and smaller as it travels around so it is not a true ellipse it is an elliptic spiral it is getting smaller and smaller and spiralling in towards the origin .3780

So that red curve there is our actual solution for which describes the solution trajectory system .3801

So that is the end of our lecturer on complex eigenvalues and next lecture were to talk about what to do when you have repeated real eigenvalues.3809

This is all part of the series on systems of differential equations and in general this is part of our series of lectures on differential equations here on educator.com .3820

My name is Will Murray, thanks for watching and will see in the next lecture, bye bye.3831