Differential Equations Variation of Parameters for Inhomogeneous Systems

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

My name is Will Murray and today we are still working on systems of differential equations .0004

Were to be studying inhomogeneous systems and said there is really 2 methods to solve inhomogeneous systems that were going to be covering.0009

One is undetermined coefficients and we talked about that E previous lecture so if you are looking for undetermined coefficients just go ahead and get back to the previous lecture .0017

We got a whole lecture on that today we are going to be talking about variation parameters which is a totally different way to solve inhomogeneous systems.0025

So, see how that works out once all the inhomogeneous system X′ = a X + GMT now remember X -prime and X those are short for a column vectors.0032

So, 2 different functions X 1 of T and X2 of T of course when there was X′ E derivatives of both of those and A will be a matrix.0049

Were just cannot do this for 2 by 2 could also do all the same things for 3 by 3s, but that just makes it much more complicated so we will just go over to 2 by 2.0060

Then, what makes it inhomogeneous is this extra term GMT some of that is also a column of functions a column vector functions of the 2 functions in there.0069

And the fact that there is that term GMT it is not just 0 is what makes the system inhomogeneous so the inhomogeneous part refers to the fact that there is an GMT there .0081

If there were no GMT there, it would be a homogeneous system we already had several lectures on how to solve homogeneous systems.0094

Remember , that is where you find the eigenvalues and eigenvectors of a matrix and he do different things depending on whether there is real repeated eigenvalues or complex eigenvalues are real distinct eigenvalues.0101

Forgot lectures on all of those if you do not number how to do those you want to check back on those lectures it is earlier on the same lecture series .0112

The reason you want to be make sure you are up on that is because the first step to solving the inhomogeneous system is to find the corresponding homogeneous system .0119

So that is what this here is all about the first step to solving this inhomogeneous system is to find the solution to the corresponding homogeneous system .0130

What that really means is that, you kind of throw away the GMT see just look at X′ = A X and you want to solve that using the methods of the previous lectures.0142

that is all the business about the real repeated roots E complex roots E real distinct routes where you find the eigenvalues and the eigenvectors and so on so, you go through all that business .0155

You find these 2 solutions to the homogeneous system so, you get were caught X ¹ and X ² those the two solutions to the homogeneous system .0166

Then were going to look for single particular solution which will call X par for particular to the inhomogeneous system and let me show you how that works out.0178

By the way all this so, far is Exactly the same as what we had for undetermined coefficients the difference with variation parameters is how you get that particular solution.0189

So, they both start out with solving the homogeneous system and then both of them variation parameters and undetermined coefficients they both try to find a particular solution.0199

The difference is how you find that particular solution so, let us see how you do it using variation parameters which we are going to guess is you start with that those 2 homogeneous solutions.0210

The X1 and X2 and then you can multiply them called by coefficients and instead just multiplying them by constant C1 and C2 were to guess 2 functions U1 and U 2.0221

Now we do not know what you want and you 2 are yet so, you wanting U2 are those are both functions of T functions of T and we are going to figure them out and teach you how to find those.0233

So, for the time being I am just can say to be determined so, let me show you how you can find that U1 and U2 of T because once we find the 1 U2 see you will be done of the problem.0251

So, what you do is you write the homogeneous solutions into a fundamental matrix say make the 2 homogeneous solutions column vectors and what were calling a fundamental matrix .0263

This matrix we usually denote by the capital Greek letter size that is pronounced sigh capital Greek letter size of C I am going to pronounce it sigh .0275

Then what you do is you find the inverse of size so, you take this matrix and you invert it so, you find sigh in personal video little trick to invert to buy 2 matrices E second so, do not worry about that for the time being.0288

Just know that you have to invert that matrix and then you multiply it by GMT and remember GMT came from the differential equation so, that is already given to you in the differential equation.0303

You will have a differential equation X′ = AX + GMT so, that GMT just illustrate how the differential equation drop it right here and so, you multiply sigh inverse x GMT and what you get is not the U1 U2 directly .0316

You get U1′ and U2′ you get a column vector and the 2 entries are the derivatives of the U .0338

So, what you do with that is you integrate that U1′ to get U1 and you integrate U2′ to get U2 and then just drop it back into that form for the particular solution to drop that you wanting U2 of T back into the form.0346

And we will see how that works out often there is a lot of multiplication and a lot of cancellation that goes on so, this is a kind of a long process .0362

But I think it will make more sense after we work through it with a few examples.0371

So, you take this U1 U2 and plug them back into that form for the particular solution and then you are done.0376

There is quite a bit multiplication involved but it sort of all works out so, you see it is a little bit tedious but will try it out.0385

There is 1 more thing that I want to show you which is the general solution and also, how to invert matrices so, the way to get the general solution once you got that particular solution is you just add on the homogeneous solution which was the 2 solutions that we talked about before on them.0392

And you just add that to the particular solution so, I get the general solution .0413

Then 1 more thing that I wanted to mention was that how you find the inverse of a matrix remember were to have this matrix sigh which is the 2 homogeneous solutions and we have to find the inverse of it sort of halfway through solving things by variation parameters.0417

So, here is a trick for doing that if you have a 2 x 2 matrix ABCD the way you can find the inverse is first one you find its determinant which is this AD- BC thing so, that is cross multiplying + on the main diagonal and - - on the back diagonal .0434

So, then you switch the entries around the switch A and D and you put - sins on B and C so, you get - B - C and A in any just divide by the determinant.0454

So, that is a safe way to take the inverse of 2 x 2 matrices does not work for 3 by 3s but it is really nice for 2 by 2 .0468

And you can do that when the entries or numbers are even when their functions and were going to need to do it when the entries are functions.0475

What were doing variation parameters so, keep that formula in mind and will be using that were solving things by variation parameters.0483

Go ahead and see how works out in so, me examples so, the first example here were going to were given a inhomogeneous system and want to find the corresponding homogeneous system and find the fundamental matrix .0491

Remember homogeneous system means you kind of clip off that GMT on the right and that is the GMT part right there that is the GMT .0505

So, when I can worry about that were actually come back in example 2 and saw the inhomogeneous system .0515

For example 1 were just going to solve the corresponding homogeneous system so, what we had here to write it matrix form is X′ = if I write it matrix for my C 114 -2 x X and then remember were clipping off the GMT because were just going to study the homogeneous system .0521

For now come back in example 2 and solve the inhomogeneous system and so, remember the way to solve these homogeneous systems is to find the eigenvalues and the eigenvectors of the coefficient matrix.0546

So, that is this matrix right here and so, A- RI = 1 - R 1 4 and -2 - R just subtracting Rs down the main diagonal and so, I get 1 - R its determinant 1 - R x -2 - R -4 = 0 .0560

If it looks like I got R ² + 2R - R so, that is R ² + R and then -2 and then there is also, a - 4 so, -6 total = 0 .0584

It is nice that factors really easily because that factors into an R+3 x R -2 = 0 as I get my 2 eigenvalues R = 2 and R = -3 so, as my to eigenvalues and now for each 1 have to find an eigenvector so, let me apply in R = to first R + 2 plug that into the A- RI and I get -114 and -2 - 2 is -4.0600

And find the eigenvector R looking for a vector that when I multiply that matrix by it I get 0 0 vector so, mild trick is to plug in 1 for the Y value in and figure out what the X value should be.0634

And I see here that if I make the X value equal to 1 as well then I work that means that the matrix x that vector will be equal to 0 .0649

So, that is my eigenvector corresponding to the eigenvalue R=2 you can also, find these eigenvectors were rigorously using the old linear algebra technique of finding free parameters .0657

You can do that that way if you do not like my sort of off-the-cuff way of just plugging in 1 for the Y value.0673

Let us look at the other eigenvalue which is R= -3 and if we plug-in A- RI that means I am adding 3 along the main diagonal so, get 4 and 1 and 4 -2 +3 is positive 1 .0679

And again I am in a plug-in 1 for the Y value and try to figure out what the X should be to make it equal to 0 4 X + 1 = 0 so, my X would be look like -1/4 so, I plug that in up here -1/4.0699

And I do not really like the fractions in there remember with eigenvectors you can multiply by scalar if you like to make the fractions little easier and so, I think I am going to multiply by - 4.0718

So, my eigenvector corresponding to -3 will be multiply by positive for to get rid of that for the denominators so, 4 x -1/4 on top and 1 the bottom will give me-1 on the bottom and 4 on the top so, now I have an eigenvector corresponding to the eigenvalue -3 .0731

So, when I find my homogeneous solution my homogeneous solution remember, I take each 1 of those eigenvalues and I make those the exponents for E T so, C1 E 2T knows my first eigenvalue and up of the eigenvector as a coefficient 11 x E 2T + C2 and my second eigenvector -14 x E the eigenvalue was - 3T .0756

So, that is my homogeneous solution to that system of differential equations .0796

I have not worried about the inhomogeneous part yet that is the GMT which will solve an example 2 just finding homogeneous solution right now .0804

Then for the fundamental matrix remember you take the 2 homogeneous solutions , you do not need the constants here so, I am in a take those 2 columns 11 x E 2Tis that is E 2T E 2T and then -14 x E -3T .0813

So, - E - 3T and 4 E-3T so, just take those solutions and I make those into the columns of a matrix and that is the fundamental matrix that were going to call sigh .0835

Were to use this E next example in example 2 to find a solution to the inhomogeneous system so, stick around for that but in meantime let me just recap how we found these solutions.0850

So, the key thing here is that even though we been given in inhomogeneous system examples ask us to solve the corresponding homogeneous system which means we really do not pay any attention to this GMT the inhomogeneous terms.0864

We just kind of throw them away for now these later in example 2 so, we write this in matrix form there is her matrix A and want to find the eigenvalues and eigenvectors of a .0879

So, I am subtracting R down the main diagonal and taking its determinant and I get a polynomial which I saw for R get to values for R each 1 of those I plug them back into A- RI .0890

Then I try to multiply it by a vector to get 0 and each time there I started with putting the 1 E Y coordinate and the just kind of figuring out what the x-coordinate should be.0904

So, that is where I got 11 for the first vector for second vector I got 1 and then a -1/4 E the top position.0916

And I did not like that because of the fractions so, because it is an eigenvector I can multiply it by scalars that is what I am doing here multiplied by 4 to get rid of the fraction and I got -14 .0925

Then our homogeneous solution just take the eigenvalues and you make them the exponents of E and then you take the eigenvectors and make those the coefficients and tack on a C1 and C2 to each part.0937

For the fundamental matrix what you do is you make each 1 of those homogeneous solutions into a column vector E individual column E fundamental matrix .0950

This E 2T E 2T that comes from the 11 E 2T and then - E- 3T and 4E - 3T that comes from that second homogeneous solution .0961

So, hang on to this matrix because we are going to use it to find the inhomogeneous solution in the next example .0975

So, example 2 being asked to solve this inhomogeneous system and what you notice here is that this is the same inhomogeneous system that we started working on an example 1.0983

So, this is the same as example 1 and we already started to working on that so, we already found our fundamental matrix back in example 1 and I am just going to copy what we got from there .0994

We found the eigenvalues the eigenvectors and so, our sigh from back from example 1 just a copy that matrix over from the previous slide was E 2T E 2T -E -3T and 4E - 3T .1015

So, we worked that out in example 1 if you did not just watch example 1 might want to go back and re-watch example 1 we will see where that matrix comes from.1040

What I need to do remember for the key equation that were to solve for a variation of parameters is to find the derivatives of U1 U2 we find sigh inverse x GMT.1049

So, we need to find the inverse of sigh and part of finding the inverse of the matrix ABCD few will find the inverse of that matrix.1066

First thing we do is divide by the determinate AD - BC so, figures out what that is for this matrix right here the AD - BC for this matrix is E 2T x 4E - 3T .1079

Without before now E 2T x E - 3T is just E - T - - E - 3T x E 2T so, that is + because it is - - so, - in a - 3T x E 2T is E - T .1096

And so, we get 5 E the - T there and so, sigh inverse is 1 over the determinate so, 1 over 5 E - T and let me finish my formula for the determinate for the inverse here.1119

Remember U2 switch the D and A and you put - on the other entries - B and - C so, they fill that in here to switch the A and B entries so, 4E - 3T and E 2T.1139

Switch those and I negate the other 2 entries so, I got positive E - 3T and - E 2T.1158

And so, I want to multiply in that external factor and number 1 over E - T is the same as multiplying by E T so, I am multiplying by 1/5 E T and so, on get on the top left 4/5 x E T x E - 3T is E - 2T E - 3T so, 1/5 E - 3T x E T is E - 2T .1168

Remember and dividing by E - T that is the same as multiplying by either the T and then -1/5 E 2T x E T is E 3T and 1/5 E 2T x E 1/5 E 2T x E T is E 3T.1206

And I want to figure out sigh inverse G I am writing a little G everywhere here so, G and a copy E G which was the inhomogeneous terms from the original equation.1230

So, that is these 2 terms here got include the - sin so, E -2T and - 2ET so, if I work that through then I have to multiply these 2 matrices so, when I am going to get is 4/5 E - 4 T E - 4 T -2/5 E - 2T E T is E - T and then on the second row on multiplying the second row by the column on the right so, -1/5 E 3T E - 2T is E T now x - -2/5 E 3T x E T is E 4T .1249

So, that is my U' , what that means is that top row is U1′ and that bottom row is U2′ have to do is integrate each 1 of those to get the U1 and U2.1320

Slowly go ahead and fill that in U1 I integrate is the integral of U1′ DT so, me integrate that top row there 4/5 E - 4T the integral of that is just what you divide by - 4 so, it is -1/5 E - 4T .1338

And then I divide the next term by -1 because it is E - T so, I get +2/5 E - T and then U2 is the integral of U2′ DT so, -1/5 E T because I just integrates to itself .1368

Now I divide by 4 so, -2/5÷4 is -1/10 E 4T .1396

So, that is my U1 and U2 repackage them together as as into a column vector if you like and we will see what to do with next just a copy those over onto before I get rid of the slide .1406

Let me show you where each step came from so, were working with this general formula U′ is = sigh inverse G that is it absolute to a permanent formula for variation parameters but that sigh in that formula that Greek letter sigh comes from the 2 homogeneous solutions.1417

So, we found that back in example 1 you can look up an example 1 where this matrix came from so, there is that sigh to find the inverse of it we need to find its determinate first sigh found A/D - BC cross multiply their A/D - BC .1439

And I came out to be 5 E - T the inverse is 1 divided by that and then you mix up the entries in the middle this is D A and - C that is where each 1 of those came from where the original entries are A B C and D .1455

If you multiply that through then it is like dividing by E the - T but that is the same as multiplying by E T .2443 So, multiplying everything by 1/5 E T sets we got these entries here right each 1 of the exponents I bumped up by 1 so, -3 you - 3T x he is seeking the even - 2T and so, on there.1475

Now we want to multiply that by G from this formula up here so, this is our GMT and were getting that from the original set of differential equations so, multiplying this 2 by 2matrix by this column vector and it works out to be this not so, pleasant expression.1497

But each 1 of those tells us remember that U′ so, that U1′ and U2′ with integrate each 1 of those to get back to U1 U2.1518

So, I integrated the top row that vector on the right that is what I got when I integrated the bottom row that vector on the right that is what I got through.1530

Just to take this over and really use this on the next side would see how that works out.1539

So, here is the U1′ that we just worked out on the previous side there is the U2′ that we just worked out on the previous side and so, what to do is room multiply U1 x X1 + U2 x X2 .1544

So, just copy that down that is -1/5 of the little messy -1/5 E - 4T +2/5 E-T x my X1 is my first homogeneous solution E 2T E 2T now + -1/5 uses this is U2 that I am copying now used T -1/10 E 4T x my second homogeneous solution the X2 is - E - 3T and 4E - 3T .1564

So, multiply that through to get a little complicated on the top I see you have a -1/5 E -4+2 is - 2T and +2/5 E - T x E 2T is E T .1605

Go ahead and combine this with what I get from the second solution that the first row the second solution.1628

So, I have -1/5 x - here I get +1/5 E T E - 3T that gives me E - 2T and -1/10 x -1 gives me a +1/10 E4T E3T gives ET .1635

In the bottom row here I see have got while the first 2 terms of the same soldiers copy those quickly -1/5 E - 2T +2/5 E T.1659

Now the second on a second set of terms here I got -1/5 x E T so, x 4 E - 3T so, -4/5-4/5 E - 2T now -1/10×4 is +2/5 E 4T x E - 3T is E T.1672

So, let us see if that simplifies at all us looks at a should okay I see I have a -1/5 E - 2T and a 1/5 E - 2Tis nice and then I see have 2/5 E T +1/10 E T that is 4x +1/10/5/10 that is 1/2 E T .1706

On the bottom row I see I have got 2/5 looks like I may error down the bottom row that + should been a - that is coming from that -1/10 right there and that is nice that it works out that way because that means that 2/5 and that 2/5 E T they cancel each other out as I got -1/5 E - 2T and a -4/5 E - 2T.1726

Put those together you get -5/5 is - E - 2T.1754

What I just found there was my particular solution to the inhomogeneous system of the differential equations.1764

By the way there is feature here that is very common to variation parameters which is lots of cancellation at the end see how all these terms canceled at the end and that sort of all par for the course for variation parameters.1770

You expect that to happen and you often get very very complicated solutions that sort of gradually counts cancel down and turned the things that are fairly simple .1782

Let us put these together and get our general solution for general solution is the homogeneous solution + the particular solution .1793

And so, in this case our homogeneous solution we work this out back in example 1 is C1 x 11 x E 2T + C2 x -14 x E - 3T i am just copying this back from what we learned in example 1.1804

It had on the particular solution +1/2 E T and -E- 2T E - 2T so, that is my general solution we go ahead and box that often offer that is a formal solution there.1826

Alright let us recap how we found that a lot of the work was done the previous slide and we are over that we found the middle of mistake here and calling that U1' and U2′ .1851

This is actually U1 itself when we found and U2 itself on the previous side we found U1′ and U2′.1874

But we are integrated on the previous sites this is really U1 and U2 that we found here and so, we do is we take that U1 U2 and we multiply those buyer to homogeneous solutions.1881

The X1 there and the X2 so, after we multiply everything in and combine it into a single vector that a lot of cancellation which is kind of par for the course for variation parameters.1893

Simplifies down to this particular solution 1 half E T and - E - 2T and so, that is her particular solution to get the general solution we just add that onto the homogeneous solution which remember, we already found that much of the answer back in example 1.1907

So, homogeneous solution we found that in example 1 we just take that homogeneous solution and we add on the particular solution that we just found to find the general solution.1926

It is the end of the that example let us go ahead and try another 1.1937

So, example 3 were start out just by finding the eigenvalues and eigenvectors of the matrix 2 -5 1 -2 of course really use this later on E examples 4 and 5 directly solve inhomogeneous system of differential equations or just a start out by finding the eigenvalues and eigenvectors.1943

Remember , the way do that is you subtract Rs off the main diagonal so, that subtract those Rs C2 - R take the determinant now -2 - R - -5×1 +5 = 0 and so, I get R ² about a + to R - 2R and then -4+ 5 R ² +1 = 0 .1961

Now that does not factor of the real numbers you can use the quadratic formula we could just knows that this is R ² = -1 so, Rs equal to ± I .1993

So, got complex eigenvalues here so, let us go ahead and figure out the corresponding eigenvalues eigenvectors we should expect complex numbers to appear .2005

So, me plug-in articles I into A- RI and so, I get 2- I -5 and 1 -2 - I and I want that to multiply by an eigenvector and give me 0 .2015

Now my favourite trick here is to plug in 1 for the Y value and it was try to figure out what the X should be.2037

If I look at that second row there I see I get X -2 - I multiplied by 1 = 0 and so, X = 2 + I and so, my eigenvector is then 2+ I and 1 and so, they asked that the eigenvector corresponding to the eigenvalues R = I.2045

Now very nice property of matrices of real matrices with complex eigenvalues is that you only need to find 1 of the eigenvectors because what happens is the other eigenvalue hellos be a conjugate of the first eigenvalue and the other eigenvector will be a conjugate of the first eigenvector.2071

So, what that means is for R = - I don’t have to go through all this work I just know it can be the conjugate of the first eigenvectors can be 2- I and that is a 1 there.2089

Were conjugate just means you change a + π to a - π and so, those that is my Eigen value and corresponding eigenvector for the first 1 and then that is my second eigenvalue and corresponding eigenvector so, were done with that for the problem.2105

What we will do with the next problem is use these eigenvalues and eigenvectors to solve the differential of a system of differential equations.2126

Let me recap quickly how this worked out, we want to find A- RI want to find eigenvalues and eigenvectors so, you subtract R down the main diagonal and then cross multiply take the determinant get R ² = -1 .2135

You could use the quadratic formula that would work out very nicely on that but I did not think I needed to is I know how to solve R ² = -1 is just R = ± I.2152

Plug-in plugged that back in logic 1 of those articles I plugged it back into A- RI and so, I subtracted either on the main diagonal .2161

Looking for eigenvector and my favorite way of doing that is to put 1 in the y-coordinate and then just try to figure out what the X coordinates should be.2173

If you do not like doing that, you can solve this by linear combinations and free parameters the same way you learned linear algebra so, if you are not a call if you are not comfortable that then there is certainly other ways to do it.2182

But my way works okay and then when you look at the second equation we get X -2 - I x 1 = 0 as we solve X = 2+ I is I put plugged that 2+ back in for X and I got my eigenvalue and my eigenvector corresponding to I .2194

And the nice thing is that when you find the conjugate eigenvalue you can just take the conjugate eigenvector so, for articles - I just with the switch that + to a - and I get 2 - I and 1 for the corresponding eigenvector there.2215

So, in the next part of it examples 4 and 5, were going to do is use this matrix to solve for to solve the system of differential equations so, see how that plays out2233

So, in example 4 what were going to do is solve the following inhomogeneous system actually on the inhomogeneous system that were just being asked to solve the corresponding homogeneous system and then find the fundamental matrix and invert the matrix .2252

Were not really worry about the inhomogeneous system that were just going to try to resolve the homogeneous system and then ask the put these together and solve the the inhomogeneous system were put that off until example 5 .2272

You have a small typo here I wanted to make that a + cosine T a arithmetic works out better later so, were to change that to a + cosine T so, make little change there let us go ahead and work that out solve the homogeneous system first and that is nice because we have gotten started on this.2290

Remember in example 3 we already found the eigenvalues and eigenvectors of the coefficient matrix here 2 -5 and 1 -2 so, we found the eigenvalues and eigenvectors of the coefficient matrix there we remind you what 1 of them was.2312

The first 1 was R = I R = I, and the corresponding Eigen vector was 2 + I and 1 so, use that to solve the homogeneous system corresponding to this the inhomogeneous system that we were given.2330

So, what that means is remember the solution is E RT x the eigenvector V in this case the R is the complex number I and the eigenvector is 2 + I and 1 .2351

Remember, the way resolve such things we talked about this way the inner previous lecture where we had the complex eigenvalues so, if you do not remember this maybe go back and check that previous lecture relearned how to handle complex eigenvalues.2372

But the way to so, rt it out was we use this this identity E I θ = cosine θ + I x sin of θ .2385

So, that is what were going to invoke here this is E IT so, it is cosine T + I x sin of T and were to multiply that by 2+ I and 1.2398

And so, if I multiply that in I see I got 2 cosine T +2 I sine T and now the I terms + I cosine T I x I is - 1 a - sin of T and then on the bottom I just have 1x cosine T + I sine T2419

And separate out a real part and a purely imaginary part for that so, separate out all the real terms and all the terms that have an I on them.2443

Look for the real terms I got it 2 cosine T - sin T and on the bottom of got a cosine T the all the purely imaginary terms I see I have a cosine T +2 sin T .2459

On the bottom sine T so, remember you do not ever have to worry about the second eigenvalue and eigenvector with these complex systems is you is you get it separated out like this then each 1 of these is 1 of your homogeneous solutions.2478

Slowly put these together and I will get the homogeneous set us the homogeneous solution my X homogeneous is C1 x that first solution there 2 cosine T - T over cosine T + C2 x that second solution cosine T +2 sin T over sin T.2501

So, that is my homogeneous solution we have elected that inhomogeneous part were just solving the homogeneous equations or not worrying about that that homogeneous part yet.2534

To find the fundamental matrix from the fundamental matrix, that just means you take the 2 homogeneous solutions and a packaging together as the 2 columns of a matrix so, we put those together.2549

My sigh will be understood for those 2 those 2 solutions as a columns of a matrix to cosine T - T and cosine of T down here and then cosine of T +2 sin T and sin of T down here .2562

So, that is my side and I will find the inverse of that we remind you of our formula for inverting a 2by 2 matrix ABCD inverse of a 2 x 2 matrix you divide by the determinant A D- BC .2586

Then you switch to the A and D supposed that a down there and you put - sins on the B C .2605

So, that means you got a find the determinant there so, to find the determinant size and sets part of my solution the determinant if I cross multiply there I get 2cosine sin can write the T 2 cosine sin - sin square that is multiplying that way and that if I multiply that way which I need to subtract all those terms.2620

- cosine ² - oh second got another 2 cosine sin and so, the 2 cosine sins cancel each other out and -sin ² - cosine ² is -1 so, sigh inverse is 1 over the determinants that is 1 of her -1 which is - .2652

Now just can go ahead and write my generic form D- B and a front for the might - C up here - C and A .2680

And so, that is - D positive B positive C and - A down here.2694

So, let us figure out what that is the context of this matrix number that is a ABC and D so, top left corner for sigh inverse of got a - D - sin T top right of got B so, that is just cosine T +2 sin T cosine T +2 sin T .2704

Bottom left of got a C that is cosine T and bottom right I got - A so, - A is the - of this so, that is sin of T positive sin T -2 cosine T.2725

So, that is my sigh inverse me go ahead rewrite the letter sigh inverse we will be using all this when we solve the inhomogeneous system in the next example.2746

But before we do that let me really use all these answers without going over them again so, they recap right now in show you where all these answers came from the first thing to notice here of the problem is asking us to solve the homogeneous system and so, what I am doing here is a really ignoring the inhomogeneous part of the of the system .2762

So, I am just kind of ignoring that for now looking at the homogeneous system so, that coefficient matrix which is 2 -5 1 -2 2 -5 1 -2 and we already found the eigenvalues and the eigenvectors there back in example 1.2787

In example 3 so, if you do not member how we got those eigenvalues and eigenvectors go back and check example 3 see how we got 2R = I for the eigenvalue 2and 2 + I over 1 for the eigenvector.2803

So, I plugged those in here E IT that is my eigenvalue right there that is the R and then there is the eigenvector and I expanded E IT using our old formula that we learned in the lecture on complex eigenvalues.2820

So, expanded that out of the cosine T + I sine T I multiplied that into the eigenvector multiplied into the top and multiplied into the bottom.2838

Gets kind of messy but we can so rt out the real terms of terms without an I and E terms with an I and remember what we learned about solutions for complex eigenvalues which do is you take each 1 of those and you make those each 1 of those columns 1 of your homogeneous solutions.2849

So, my general homogeneous solution is just those 2 columns multiplied by so, me arbitrary constant C1 and C2 and in fundamental matrix means that you just take those and make those the 2 columns of a square matrix.2866

So, there is my fundamental matrix and the last thing he asked us to do in this problem is to find the inverse of that so, I am using this well first I do find the determinant.2883

So, I cross multiplied here and it really worked out nicely, supply down into - 1 for my determinant so, I found the inverse that - right there that is really doing 1 over AD- BC that is why have that - on the outside .2894

Then I switched around the entries on the inside and when I multiply the - 1 and I got - DBC - A and I just read off my ABC and D from the fundamental matrix plugged those in and I got the inverse to the fundamental matrix .2909

So, use all this to solve the inhomogeneous system on the next slide see how that works out.2928

So, on example 5 here were to find a particular solution to the inhomogeneous system and remember there is a little typo there that should have been a + original had and solve it with a + there.2935

It is really messy if you work it out with a -sin solve it with a + and the way were going to attack this is really use the information that we had from the previous that previous examples.2951

A lot of the work has already been done in example 3 in example 4 so, me just remind you of what we figured out in example 3 in example 4.2966

We already figured out the fundamental matrix and we found its inverse and were ready to go to find the solutions to the inhomogeneous system.2975

So, what we learned at the beginning of this lecture is that you can find your derivatives of the U by doing the inverse of the fundamental matrix x G where G is the stuff right here.2986

That is our GMT it is the inhomogeneous terms of the differential equation so, we go ahead and write those out.3003

My U′ is there U′ is my inverse x G now inverse. We figured this out in example 4 so, we are not going to work it out from scratch again.3011

Example 4, that is what I am using right here to get my sigh inverse of just a copy that down from what we had is that the answer to example 4.3027

So, - sin T cosine T and then cosine T +2 sin T +2 sin T and sine T -2 cosine T that was the inverse aside.3038

Copy my GMT , that is coming straight from the differential equation here so, this cosine T & sin T and I need to multiply those through so, I see on the first row Ihave got - sinT x cosine T and then I got a cosine Tx sin T .3056

So, cancel it will just get 2sin ² T So, got that was I multiplied this row by this column and I saw that I had the 2 terms canceling the term of cosine T sin T in as occurring with the - and with the positives that is why cancel out those out right away.3081

Now we multiply the second row by that same columns I see about cosine ² + sin² set the 1 -2 sin T cosine T 2 sin T cosine T.3103

And I think that remember that is U's what I am going to do is use so, me trigger metric identities here geometric identity and many used to make it easier to integrate is +sin ² of T = 1/2×1 - cosine 2T .3122

That is a trigonometric identity and if you do not remember that we got so, me lectures on trigonometry here on that educator and we also, use that pretty heavily in calculus 2 lectures here on educator .3141

Calculus 2 learn how to integrate trigonometric functions and 1 of the tricks there was when you have a sin ² cosine ² you want to use this double angle identity so, that is what I am kind of recalling that from 2.3154

And so, if I write this out I get a get well 2 sin ² T would just be 1 - cosine 2T and then I saw the bottom I have 1 -2 sin T cosine T .3171

So, I really want to think of those as that U′ that U1 U2′ so, U1 is the integral of that first row 1 - cosine 2T DT and so, by integrate that I just get T.3189

Now the integral cosine assine but since it is cosine 2T it is 1 half sin of 2T.3211

You do not have to include the constant when youare finding this U1 U2 if you did include the constant it would just give you more multiples of the homogeneous solution so, you do not have to worry about the constant here.3219

The U2 work a little more with U1 I am going to use another trig identity there which is that the sin of 2T = 2 sin T x cosine T and so, T - ½ 2T is just T -1/2 of what I just wrote.3234

So, T - sin T cosine T to go to be a little easier to work with later on, go ahead and find U2 that is the integral this was U2′ the top row was U1′ so, to integrate U2′ to get U2 .3260

So, 1-2 sin T cosine T that was a lot of ways you could integrate this but I think the 1 that is going to work the best is to do a little substitution here for cosine T and I like using U substitution but some are using the variable U for something else.3278

I am in a use of different variable interviews W W = cosine T and so, DW = - sin T DT and so that will give me if I plug those and now do maybe integral of + 2W because the - is included E DW + 2W DW .3299

So, the integral 2W is just W ² so, that it be very easy so, when I integrate this the 1 just integrates to T the 2 sin T cosine T integrates to W ² so, that is + cosine ² T and so, I have got my U2.3325

Were to figure out how to use that on the next slide but let me just quickly recap what we did here we already found sigh inverse back in example 4 so, our sigh inverse that comes from example 4 if you do not remember how he got that just go back and read over example 4 you will see where it came from.3348

And our G is coming right here from the inhomogeneous terms in the differential equation so, plug those in right there I multiplied to the 2 rows there and there was so, me cancellation going on and a simple fight a bit into 2 sin ² and 1-2 sin T cosine T .3364

Now my sin ² T I use this old trigonometric identity a lot and calculus 2 to convert that into 1 - cosine of 2T .3388

So, then what I have here is U1′ and U2′ and I integrated each 1 of those little substitution here turn it into his ½ of 2T and then I used this old trigonometric identity to simplify that into sin T x cosine of T .3398

Then the second 1 I use a little I wanted to call you substitution but so, me are using you for something else I call it W so, W = cosine T and then DW = - sin T .3417

So, had 2 W and that integrates to W ² so, we get T + cosine ² T that T came from that 1 right there and then need the rest of the game in a cosine ² T .3432

Since my U1 U2 keep going with that on the next slide so, in our next slide we have our U1 here that is actually not quite what we had for U so, on the there is there is a couple extra terms your little typo here slowly fix that quickly.3445

The U that we had was T - sin T cosine T and bottom part erase that in fix that quickly in the bottom we had 2+ cosine ² T .3471

So, that is the answers that we got from the previous sides go back in and wants at previous side if you do not remember where they came from and so, now let me show you how you use these to write down the completes solutions of the problem.3492

So, far particular solution is let us see it is a U 1 x our X1 + U 2 xX2 .3506

So, that is can be a little elaborate and complicated might right that out there U1 is T - sinT cosine T and X 1 we get from this first column of the matrix so, that is 2cosine T - sin T and cosine T in bottom.3527

Now + U2 so, that is T + cosine ² T + cosine ² T and X2 to get that from the second column of the fundamental matrix so, cosine T +2 sin T & sin T on the bottom .3557

Now this gets to be a little ugly incompetent but it is going to simplify nicely after we plow through all the mess so, let me out work through that first row here I see about 2 T cosine T - T sin T -2 sin T cosine ² T + sin ² T cosine T .3583

And to go ahead and add on the first row of the second term there so, + T cosine T + 2T sin T + a cosine 2T +2 cosine ² T sin T .3611

So, very long and messy but I am hoping there is going to be so, me good Cancellation coming here.3637

In the bottom we get T cosine T - sin T cosine ² T cosine ² T + T sin T + cosine ² T sin T3642

So, I am really hoping so, me will simplify and it is customary when the new variation parameters to get very long complicated solutions and then they do somewhat simplified.3661

So, if I look at this top term here I see I have got 2T cosine T and I got another T cosine T so, and put those together get 3T cosine T now - T sin T + 2T sin T so, - T + 2T so, just + 1 after all that simplifies now -2 sin T cosine ² T and +2 cosine ² T sin T .3672

Those 2 terms just cancel each other out that is a really nice and I see got 2 other terms here + a sin ² T cosine T forgot my T cosine T + cosine cube T .3715

Now the bottom I see I got T cosine T + T sin T and then I got a -sin T cosine ² T + cosine ² decided those to cancel so, really did cancel quite nicely and it actually is no work out even better because if you look at these 2 terms I can factor out a cosine T x sin ² T + cosine ² T .3733

Of course cosine ² + ² is just 1 so, those 2 terms R to combining just give me a cosine T and so, the thing we will do is factor out at T whatever terms I cannot get T x 3 cosine T + sin T .3762

And then the bottom I got cosine T + sinT and I still have this cosine T so, let me not forget that is just E top so, I am in a factor that out on the right get a cosine T and other say I have 10 because I had that term only appears on the top and everything else canceled on the bottom.3786

I think on this 1 were just asked to find a particular solution so, that is my particular solution so, let me recap and show you all the steps that went into that.3811

This sigh first of all came from solving the homogeneous system we did that back in example 4is that is where we found that sigh that is example for coming in and being useful here.3826

Watch example 4 recently so, back and check example 4 you will see exactly where that side came from .3840

These 2 entries for U that was U1 that was the U2 that we found on the previous page originally wrote it wrong in my in my sigh so, I just corrected it right there that that is the T + cosine ² T which we found the previous page .3848

Then our general form for the particular solution is U1 x X1 + U2 x X2 so, copied my U1 down there copied my X 1 that was the first column of sigh there is my U2 and then I copy my X 2 that is the second column of sigh X2.3866

So, we always have this form U1 X1 + U2 X2 every time you variation parameters you got that form and then multiplying this through the first row that gave me these first f4R terms there.3889

Multiplying this through the first row gave me the next 4 terms of very ugly expression there.3902

The multiplying this through the second row gave me those 2 terms and then multiplying this to the second row gave me those 2 terms.3912

So, get this really horrible expression but it is nice a lot of things cancel the these 2 terms cancel this 1 cancels with this 1 this term combines with this term the T sin T combines with the 2T sin T .3923

So, we end up with 3T cosine T T sin T and these other 2 terms sin ² T cosine T + cosine QT and those 2 we can factor out and simplify that 1 just down into cosine Tof this whole thing is turned in a cosine T .3943

In the bottom we get terms canceling those big terms cancel we just get T cosine T + T sin T and then what I see now basically I am done I just want to factor the T out make him look a little nicer so, I factored out that T from the first 4 terms factor that out here .3963

That left me with a 3 cosine T + T and a cosine T + T and I still had that 1 terminal cosine T can factor T out of that side to write it separately so, there it is 10 x cosine T .3982

Sets the particular solution to that inhomogeneous system that we start with way back in example 4 then had again in example 5 .3998

If you want to find the general solution it would really be no more work because what you do is you would just take the homogeneous solution and then add on the particular solution.4006

We partly found both of those copy them again because it is kind along and there is really nothing to be gained here but we found the homogeneous solution in example 4 .4019

Then this particular solution is what we just found right here so, if you want to you can just put those together and get the general solution to that inhomogeneous system of differential equations.4030

So, that is the end of our lecture on solving inhomogeneous systems of differential equations .4043

Remember we had 2 methods there was a bit undetermined coefficients that was the previous lecture and then variation parameters that was this lecture .4051

So, 2 completely different ways of finding this particular solution but then they both work and they both end up giving you solutions to inhomogeneous systems of differential equations.4059

That actually wraps of this chapter on systems of differential equations.4072

Our next chapter is going to be on numerical method so, totally different stuff I hope you will stick around and watch that chapter as well.4076

In the meantime, you are watching the differential equation series here on educator.com. My name is Will Murray and thank you very much for watching, bye bye.4083