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Lecture Comments (2)

1 answer

Last reply by: Dr. William Murray
Fri Oct 24, 2014 2:36 PM

Post by Paul Carrera on October 22, 2014

Hi Dr. Murray:

I'm not sure if you explained this in a previous lecture, but will using your short hand method always yield an Eigen vector? It is very quick and easy to use in comparison to doing row operations. Can I always use this method.  

Undetermined Coefficients for Inhomogeneous Systems

Undetermined Coefficients for Inhomogeneous Systems (PDF)

Undetermined Coefficients for Inhomogeneous Systems

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Lesson Overview 0:35
    • First Solve the Corresponding Homogeneous System x'=Ax
  • Solving the Inhomogeneous System 2:32
    • Look for a Single Particular Solution xpar to the Inhomogeneous System
    • Plug the Guess Into the System and Solve for the Coefficients
    • Add the Homogeneous Solution and the Particular Solution to Get the General Solution
  • Example 1 4:49
  • Example 2 9:30
  • Example 3 15:54
  • Example 4 20:39
  • Example 5 29:43
  • Example 6 37:41

Transcription: Undetermined Coefficients for Inhomogeneous Systems

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

My name is Will Murray and today we are going to talk about inhomogeneous systems and we are going to study 2 methods of solution for inhomogeneous systems.0005

Today we are going to talk about undetermined coefficients there is another method called variation parameters, we are going to talk about in the next lecture So, if you are looking for a determined coefficients you are in the right place.0015

If you are looking for solving things by variation parameters then, just wait to the next lecture because that is a totally different method of solution.0026

So, let us take a look at what we’re talking about here.0032

I will also, solve the inhomogeneous system of differential equations vector X′ is = a x vector X + GMT.0036

So, the idea here is this this a would be a matrix So, usually be a user 2 by 2 matrix of Constance vector X is the functions that we are solving for that is that is a column vector X1 and X2 .0047

Then this vector Dt would be some extra functions that we did not have before this what makes it inhomogeneous this GMT So, that is that is really what inhomogeneous means is that we have that GMT .0062

So, some extra functions will be added on the outside there and today we are to learn how to solve something with of that form .0077

So, the idea here is that first you want to solve the corresponding homogeneous system X′ = a X and you can do that by the methods of the previous lectures So, you can get a homogeneous solution C1 x vector X1 and C2 x vector X 2 .0085

So, it had several lectures now on how you solve that member that was the whole business of finding the eigenvalues and the eigenvectors of the matrix A and then plugging them back in and you get C1 E RT x the first eigenvector C2 in the second RT x the second eigenvector so, we had several lectures on that.0103

What you do when you have real distinct eigenvalues what you do when you have complex eigenvalues and what you do when you have real repeated eigenvalues.0125

So, if you do not remember that, maybe check back in those lectures because we’re going to be using that material sort of as a first step today to solving these inhomogeneous systems, the very first step is to solve the corresponding homogeneous system just as we did before.0134

So, see what happens after that for the inhomogeneous system we are to look for a single is called a particular solution which call X R to the inhomogeneous system and the idea is you look at this GMT look at the extra functions that are given in your equation and you guess something of the same form but with generic coefficients.0151

So, for example if GMT we are say 3 E to the 5T and 2 E to the 5T will be of you would be a vector there E to the 5T So, GMT is always a vector because these are these are systems of equations So, he would think of that is being vector 32 x E 5T.0171

So, that when the GMT that you we are given you would guess something that is just like that except you would be specific about the 3 2 so, you would guess you would guess something of the form of multiple of E 5T except that you would guess generic coefficients A and B instead of 3 2.0198

So, then what you would do and that is you plug it back in and try and solve for the A and B, will see some examples of it and you will get the hang of it once you figure out what your particular solution as once you plug it back in and figured out what those A and B are then you are going to take that particular solution and plug it right back in here, which means you can add it on to the general solution and you are going to get I should not have labelled this homogeneous session of labelled this the general solution So, fix that right there.0220

Next the inhomogeneous solution So, the general inhomogeneous solution is C1X1 + C-2 X2 + the particular solution .0259

The realized that new part of this lecture is really making that guess right here and solving for those undetermined coefficients and that really will make much more sense when we do some examples .0275

So, separated to some examples for the following inhomogeneous system for similar to do is just solve the corresponding homogeneous system so, what that means is the when resolve the corresponding homogeneous system and means you ignore the inhomogeneous part of the equation .0288

You ignore this part will just ignore that and will solve the corresponding homogeneous system which means really use the techniques we learn several lectures ago so, if you do not remember how to solve this, go back and check a couple lectures ago when we first started solving systems of differential equations.0304

The way you do it is you find the eigenvalues and eigenvectors of the matrix So, what we will going to do 1 - R and 1 - R doing A- R I your subtracting RI down the main diagonal of the matrix .0322

So, now I got 1 - R ² 1 - R x 1 - R -4×1 is = 0 and that is quadratic equation to get R ² -2 R +1-4-3 is = 0 this with factors nicely it is R -3 x R +1 is = 0 So, R is that 3 or -1 those my 2 eigenvalues .0336

For each 1 of you find an eigenvector by plugging that value of R back into A- R I So, R is = 3, plug that back in 1 - R is - 2 41 -2 and look for an eigenvector there so, my eigenvector I plug and Y = 1 and try to make that = 0 that is my little shortcut way of finding eigenvectors .0375

If you do not like it then there is a way to solve this using free parameters as a way learned your algebra signal and use that if you like it better, but I see all gets I want to see what the first component should be to make it = 0 and I see if I put the 2 there then when I multiply that matrix by this eigenvector I will get 0 .0403

So, that is my eigenvector corresponding to R = 3 for R = - 1 that means I am adding 1 of down the main diagonal so, get 2 4 and 1 and 2 get them to put a - 41 down the Y component and I see that I would have to put a - 2 in the X component to make that come out to be 0.0423

So, now got my eigenvalues and my eigenvectors and I just drop those into my generic formula for the general solution for the general homogeneous solution it is C1 x E RT in the R1T x the 1+ C-2 x ER 2T x V2.0446

We saw this a few lectures ago and so, this is C1 well , my first eigenvalue -1 is the first 1 So, E - T x -21 and its + C2 x E 3T x 21 and that is my homogeneous solution .0471

We use that to add on to the inhomogeneous part in the next example but let us recap what happened with this example .0497

It says solve the corresponding homogeneous system which means we are going to ignore the inhomogeneous part for now what will taken it in taken into account in the next example So, working to find the eigenvalues and eigenvectors of this matrix .0509

Subtract R down the main diagonal and then take the determinant So, cross multiply get a quadratic equation there which factors easily get 2 eigenvalues for each 1 we plug it back in the matrix and solve A- R I x the eigenvector is = 0 So, my shorthand way of doing that is to fill in 1 for the Y component and figure out what the X component should be .0525

I get my 2 eigenvectors and I just drop them in of the eigenvalues as exponents of E and the eigenvectors in as coefficients and that is my general homogeneous solution .0550

I saw to solve the inhomogeneous system that is what we are going to do in the next example .0565

So, the next example we are going to solve the inhomogeneous system which means we are going to guess something of the form that is comparable to the inhomogeneous part of the equation .0569

So, that that right there we are going to guess something that looks like that but has generic coefficients sort in a guess X is = AB x Et .0585

Plug that back into the system but will need to know what ex-' is the derivative Et is just Et was just going to be A x AB x E T so, plug that back in plug into those into the differential equation So, I get AB x E T is = 1411 x AB E T + -12 -12 x ET T .0600

I see right away every things got E T So, cancel those out right away and get AB is = now 14 xAB is = a + 4B and A+ B that is 1 x AB + -1 and 2 is going to turn into a pretty easy system of equations right away.0642

A is = a + 4B - 1 B is = A+ B +2 .0666

So, this is a system of equations 2 equations 2unknowns So, you can really solve this using any method you like from high school algebra for this 1 turned out to be really easy we just get the A's cancel there So, 4B -1 is = 0 that tells me that 4B is 1 So, be is 1 4th .0677

Then, on the second 1, that these cancel and get A+ 2 is = 0 So, a is = - 2 and so, my particular solution I guess AB even now I know that a is - 2 B is 1 4th and I have E T there and Ss, my general solution is the homogeneous part X homogeneous + the particular part that I just figured out.0699

Now the homogeneous part I figure this out on the previous example so, if you did not just watch the previous example go back and look at example 1 and we will see where this comes from.0737

You will see that for example 1 what we got was C1 x E - T x the vector -21+ C2 x E 3T x 21 now + the particular solution.0749

Now, this does not get multiplied by a constant this is very important particular solutions to inhomogeneous equations you can just multiply them by arbitrary constant So, you know I am not So, thrilled about that 1 4th right there really like to maybe multiply by something in clear denominators I cannot do that because that would make it not be a solution anymore.0777

So, I am stuck with that 4th -2 1/4 x E T So, really emphasize there is no arbitrary constant there those - that -2 1 4th are absolutely fixed, I cannot change them and I cannot multiply by anything to clear that denominator .0797

But having done that I am now finished with that example I have got my complete solution to the homogeneous system So, recap how that worked out.0825

For solving the inhomogeneous system really the first thing to do is to solve the corresponding homogeneous system that is what we did back in example 1 so, really the first Here was to find a homogeneous system .0842

To find the C1 E - T + C-2 E 3T x 21 you will see the derivations for that back in example 1 what we are doing here in example 2 was finding the particular solution to the inhomogeneous equation.0853

Which meant we guessed which I saw over here and E to T So, I mimicked it here for generic coefficients undetermined coefficients a and B and then when I take the derivative E T is just itself I get the same thing .0870

When I plug that in X′ plug of this and for X, and I try to simplify my equation right away the E T's go way I multiply this matrix x this vector which is how I get this column vector and then I separate that out into a system of linear equations.0886

Now , you can solve the system of linear equations using several different methods that you learned in high school algebra this 1 solve really easily because we had several variables cancelling right away.0908

So, got a value for A and B and then I plugged those back into my particular solution that was coming from the original guess but now I have determine what my coefficients would be .0921

That is my particular solution , I cannot multiply that by arbitrary constant am forced to just take it and I plug it back in here not there but over here is where the particular solution got plug-in add that onto my homogeneous solution and I get my complete answer there.0931

So, let us keep going with another example here we got the following inhomogeneous system and we are being asked to solve the corresponding homogeneous system with that really means is we are not can worry about the inhomogeneous part for now .0952

The next couple of examples we are to come back and we are in a solve the inhomogeneous system So, we will see that in examples 4 and 5 but we do not want me to worry about that right now we just need solve the corresponding homogeneous system .0967

Remember those all work the same way your finding eigenvalues and eigenvectors subtracts and find A- R I will take the determinant there So, I get 5 - R x 5 - R - -3 x -3 that is -9 is = 0 .0987

And that is R ² -10 R now +25-9 is +16 = 0 factors nicely R = to only show the factorization R -2 x R -8 is = 0 So, R is = 2 or 8 .1009

So, for those of the 2 eigenvalues each 1 of those we are in a plugged back in for R and get it corresponding eigenvector So, in R = 2 I am doing 5 - R main diagonal assuming it 3 -3 -3 5-2 is 3 and 1 multiply that by some eigenvector and get 0 .1038

Do my usual trickle plug in a 1 for the Y you do not like that you can use free parameters to solve the system but plenty and 1 for the Y works very well and I can see that I would have to have A1 for the X to make that work 1 for the X to make that work.1062

So, got my eigenvector corresponding to R = 2 R = 8 then I subtract R down the main diagonal there which is getting to me -3 -3 - 3s everywhere 5-8 is -3 and the put a 1 in for the Y .1080

I see that I would have to to make that work out since I got done - 3 x something -3 is = 0 that for something to be -1/2 in my eigenvalues and eigenvectors and I am in a drop those into my generic form to get my homogeneous solution .1103

Can be C1 x E of the first eigenvalue x it corresponding eigenvector 11+ C2 x E of the second eigenvalue x its corresponding eigenvector -11 .1122

That does it for the homogeneous solution remember when I can worry about the inhomogeneous part of this equation until the next couple of examples So, what I did there was I kind of ignored the inhomogeneous part.1138

I just focused on the homogeneous parts which means I need to solve for the eigenvalues and eigenvectors major subtracting R down the main diagonal and then cross multiplying to get the determinant so, that is how I got this quadratic equation here.1157

Which factored really easily and I got at R= 2 or 8, and then for each 1 of those a plugged in and I found A- R I .1173

So, plugged the 2 back in 5-2 is 3 that is we are that 3 and that 3 came from then I try to guess a vector that would make it come out to be 0s I plugged in Y= 1 and I figured out that X had to be = 1 in order to make that come out to 0 sending articles 8 5-8 is -3 So, that is for those come from .1183

And again, I plugged in my = 1 and then I realized that if I put at X = - 1 that would give me 0s and confirm that that is an eigenvector that I just dropped my eigenvalues and my eigenvectors into the generic form for the solution and I get my homogeneous solution .1205

I hang onto that will be using that later but in the meantime in the next examples we are to try to solve for the particular solutions to the inhomogeneous problem So, stick around for that.1224

So, example 4 , you find a particular solution to the inhomogeneous system is the same 1 we had in example 3 So, you might go back and check example 3 see how we solve the homogeneous problem.1237

But now, resolve the inhomogeneous system but I see when I look at this I got some terms with E to T and I have also, got some terms with E 3T and he should think of those beings we are fundamentally different .1251

So, what we are going to do is in this example we are just going to try to solve for the E T terms So, let me separate this into an E 3T term and and E T term .1263

we are just going to worry about the E T terms as he was in the 3 T OF 0 and 5 and 4 in the T I would have 2 and -5 E T .1279

So, you should think of those beings we are fundamentally different functions there and So, what you do is to solve for particular solution corresponding to each 1 separately.1292

So, in this example we are just going to solve the ET terms and then example 5 that is what we are going to handle the E 3T terms .1304

That is going to be example 5, and what we will do right now is to solve for the ET terms so, see how that works out we are going to guess an X = something of the same forms on right AB E T but I do not know what those A and D are yet I leave those coefficients generic .1314

That is why it is called undetermined coefficients to find my X′ that is the derivative of that but ET its derivative is just self So, A.B.1335

I am in a plugged each 1 of those into the differential equation X′ and X So, I get A B this is the X′ of writing down right now is = the matrix 5 -3 -35 x my X is A.B. E T + remember of just trying to solve for the terms we are E T so , put 2 -5 will count will account for the other terms later E T .1347

I see now that it really I have has ET cancels off if you do that by multiplying by E - T if you liked and now I see that that matrix I can write it as 5 A- 3B and -3 A+ 5 B .1381

So, now I can sort this whole thing out into a system of equations on the top I got a is = 5 a -3 B+ 2 and B is = - 3A + 5B -5 .1403

So, let me try to simplify that a little bit if I move that a over to that side and move the 2 over to the other side I get 4 A - 3B is = now -2 the multiplex I moved it over to the other side if I bring that be over and move that -5 back I will get - 3A + 4B is = 5 .1422

So, I got this system of equations and indexing not a very nice 1 in incidents go work out my 6th practice it before but it is it is a little ugly kind of along the way but basically it is a system of 2 equations into unknowns So, you can use whatever technique you learn in high school algebra to solid there is lots of different techniques.1463

Leave substitution but there is other techniques you might use that would debate basically they all work to solve this.1487

So, what I am going to do , is take that first equation and I am going to say 4A is = move the 3B over the other side 3 B- 2 and so, a is = 3 4ths B / 4-1/2 and then I am going to substitute that into the second equations to get a little messier.1495

Warning you in advance, it could be a little ugly So, -3×3/4 B -1/2 of substituting in for a + 4B is = 5 and So, that is - now -9/4 B +3/2+ 4B is = 5 now 4B- 9/4 that is 16/4-9/4 that is 7/4 B and if I move that 3/2 over the other side 5-3/2 is 10 half -3/2 7/2 and that actually is a very nicely because it turns out that B is = 2.1518

So, kind of ugly in the meantime, in the intermediate steps but it worked out nicely and then remember we have 4A is = 3 B- 2 So, I use that So, 4A is = 3 B- 2 3 x 2-2 which is 6-2 is 4 So, 4A is 4 and so, A is = 1 and So, my X let me remind you was A.B. x E 3T now third out what A and D areA is 1 and B is 2 .1574

We still have not E T there So, that is my particular solution on the subscript on there to remind you that that is the particular solution corresponding which would if we plugged it into the equation would give us the E T part of the inhomogeneous terms there.1615

So, it is it is 12 x E T. Let me emphasize here there is no extra constants here there is no C1 or C2 So, no more constants that 1 and 2 are very much fixed you are not allowed to multiply this thing by any arbitrary constants because it would ruin the fact that it is a solution there.1637

So, that is the end of that example we are going to revisit this later and solve that last term the 05 E 3T.1660

Let me remind you what we did here we guessed we we are trying to solve for this term the 2 -5 E T So, we guess something that looked just like it A.B. in the 3T except we left the coefficients generic the A.B. and set up to a -5 .1668

We took its derivative which because it was just E T did not change anything we plugged it back in there is the X′ there is the vector X and I plugged it back into the ' there the plugged it back in the equation just including the E T partners.1683

we are solve in E 3T part later on and then I took this matrix multiplied by A.B. and expanded that out into a vector and this gave me 2 equations and 2 unknowns, A.B. is = these equations. 2840 Now, that was a little bit of an ugly system of equations there quite aware to solve those but was nothing that you have not done in high school algebra so whatever methods used in high school algebra you can use those to reduce and soften the A.B.1702

I use substitution in the first equation of the first equation I saw for A and terms of the third is right there plugged it back in to the second equation .1734

Then I we are that go came over here and salt for B and I finally plugged that back into the first equation and solve for A.1745

So, little messy there but it be actual values of A.B. came out nicely and I drop them back into my initial guess and I got my particular solution which remember you do not want to multiply by any arbitrary constants.1755

So, the next example what we are going to do is where the try to account for this other turn this in the 3T term and R finally in example 6 will put them all together and get our complete solution to this system.1767

So, let us go ahead and look example 5 here also, example 5 this is the same system that we have been working on for the past couple of examples the difference is that we are now going to try to account for the E 3 term in E 3T term on the right-hand side .1781

Let me remind you that this E 3T term that this right-hand side this inhomogeneous part where you write this as 05 x E 3T +2 -5 x E T .1799

Separate out the ET stuff from the E the 3T stuff anything of those is completely different problems in fact E T we solved that in example 4 single back and check that out if you do not member how we did that.1818

What we are going to do right now is we are going to try to solve for this term the E 3T and so, the way we do that is we make a guess for a particular solution our X is going to be something like that So, we have E 3T .1833

Really generic coefficients A.B. and we want to plug that back in know the derivative X′ now E3T its derivative begin extra 3 there So, it is 3 x A.B. x E 3T I just got an extra 3 there from taking the derivative of E 3T.1849

Plugged each 1 of those back into X′ and X and remember also, going to include that inhomogeneous term of E 3T So, get 3 x A.B. E 3T is = our matrix 5 -3 -35 x our X is A.B. E 3T we are to try to solve for A.B. now + our inhomogeneous term 05 E 3T .1873

Do not worry about the rest of the inhomogeneous terms where he found a particular solution back in example 4 corresponding to that So, I see now in my equation got a E to E 3T everywhere.1909

So, many cancel that out in the 3T there there there the way to do that if you wanted to be really formal about it is your multiplying both sides by E - 3T.1921

Cancels that out and now I see that this matrix will turn into if I multiplied by the vector AB I get 58 - 3B and -3 A+ 5B and So, I see that I got a system of equations here write that out.1930

3A is = 5 A- 3B +0 there So, going right +0 and 3B is = - 3A + 5B +5 at 5 is coming from that term that inhomogeneous term over at the end there .1952

So, I have again 2 equations and 2 unknowns is a matter of high school algebra to try and solve these things using really whatever method you like also, simplify them a bit first .1976

If I subtract that 3A from the 5A, I get 2 A- 3B is = 0 and if I subtract that 3B over there and maybe plugged move that -5 over to the other side - 3A + 2B is = -5 So, I think I am many you substitution.1989

So, right over here , I see that 2A is = 3B So, a is = 3/2 B gets a little bit messy but I think I plan this 1 So, that the numbers will work out nicely in the end .2019

We play that back in the second equation I get -3×3/2 B + 2B is -5 and So, -9/2 B + 2B is -5 now 2B is 4/2 B So, for has -5/2 is -5/2 B is = -5 So, I see the B is = - 2 no positive 2 to make that.2036

Work that I plugged that back into a = 3 has been get a is = 3/2 x 2 which is just 3 and So, I figured out my in my B and now I just have to drop them back into my initial guess my particular solution right there was A.B. 3 in the 3T So, my is 3 my B is to x E3T.2073

Kind of looking at this guess over here and that is my particular solution corresponding to that, that if you plug it and would produce this part of the inhomogeneous to the inhomogeneous term there would produce the 05 it would not solve the 2 -5 problem but that member we solved in example 4.2102

So, let me recap what we are doing here we got this complicated inhomogeneous differential equation we solve the homogeneous part back in example I think was 3 and in example 4 we solved the ET part .2126

Now in example 5 what we are trying to do is to solve E 3T part and So, the way we do that as we make this generic guess E 3T to mimic what we are trying to find but we leave under the coefficients of determine their the A and B .2142

We had to take its derivative sorting the derivative X′ the 3 props out and get an extra factor of 3 out there and we plug it in for X′ and for X and So, for X′ got 3 A.B. E 3T for X we got to A.B. E 3T .2159

We noticeably got in the 3T everywhere So, those all go away those cancel away but this matrix x this vector expands out into 5 A- 3B -3 A+ 5B and So, on the left-hand side we got 3 A3 B on the right-hand side we got those terms + the 05 those are coming in here and here .2180

It turns into to equations into unknowns not too bad you can use whatever you learned in high school algebra to solve these substitution linear combinations determines any of those work.2202

I use substitution so, I solve for A in terms of B here it is right there that was in the first equation than I plugged it back into the second equation for a .2215

That gave me a nice equation for B I got B = 2 plug that back in here and got a = 3 that took the whole thing and plug them back into my initial guess for the particular solution 3 and 2 x E 3T .2223

So, my particular solution corresponding for E 3T terms on the right-hand side So, finally an example 6 over to do is to put all these different results back together and get a complete solution to that inhomogeneous system of differential equations.2242

So, we will see how that plays out example 6 is the same system we been studying and examples 3 4 and 5 but we want to put all those answers back together and find out what the complete solution will be.2260

So, redo all the work of the answers that we figured out in her previous examples So, an example example 3 we solve the homogeneous system which means we completely ignored all the inhomogeneous terms always in the T terms .2278

We got X homogeneous was C1 x 11 x E 2T + C2 x -11 x E AT.2294

That was in example 3 go back and watch example 3 that came from an example 4 we focused on the E T terms of right-hand side .2312

We looked at that and that and we figured out a particular solution that produced those 2 terms in that particular solution was 12 Et that was using undetermined coefficients.2327

Very important here there is no C here there is no C1 C2 C3 anything like that So, there is no constants there is no C because that is a particular solution if you change the constants you are going to be changing the answer and it will work out.2342

In example 5, that was where we found a particular solution to account for the E 3T term and that turned out to be it was 3 2 E3T and again there is no constants involved there you do not multiply that by an arbitrary constant you just leave that.2359

We very carefully found those coefficients 3 and 2 and you cannot multiply them by anything without messing up your solution and so, what we want to do is just add all the solutions together and will get the general solution to the problem.2389

So, all I am going to do is just add up those 3 parts of the solution I still have the C1 and C2 for it from the homogeneous part E 2T + C2 -11 E 8T and now minute add-on each 1 of those X particular.2405

So, configure those of being the first particular solution in the second particular solution 12 E T and for the second 1 32 E 3T so, to the 3T So, we add all those together.2432

That is the complete solution to that inhomogeneous system.2460

So, let us recap everything I did in this example which actually was not much work it off because we did all the work in the previous example So, if you have not watched this previous examples that is where I am getting all these individual solutions for .2467

We did all the work back there in this example we are just kind of assembling them together so, we first looked at the homogeneous solution and then back in example 3 we found that the homogeneous solution had this form right here .2485

An example 4 , we looked at this part of the ET terms of the inhomogeneous solution we found a particular solution using undetermined coefficients that will produce that part of the inhomogeneous of the inhomogeneous terms .2501

Finally an example 5 , we looked at E 3T and begin using undetermined coefficients we found a particular solution that would produce that part of the inhomogeneous terms .2518

We do not multiply E 1 of those by C1 or C2 we do have a C1 and C2 inhomogeneous solution essentially that is because the homogeneous solution your setting something = 0 .2532

So, if you multiply it by constants it will still work out to be = 0 but with the inhomogeneous solution you got some terms that are none 0 and So, you very carefully arrange your solutions and if you multiply them by something that we multiplying a none number by something and So, would not be come out to work after you did that.2544

So, after we do all that all this was in the previous examples we can go back and check those out to see where those numbers came from.2563

All we are doing in this example is just adding those together just assembling the homogeneous solution the first particular solution and the second particular solution into 1 big solution to the inhomogeneous system.2573

So, that kind of wraps up our lecture on the inhomogeneous systems using undetermined coefficients, got another method called variation parameters which is a totally different method for solving inhomogeneous systems.2586

So, hope you will stick around and watch that lecture it is the very next lecture we will seem to learn about variation parameters.2600

In the meantime you been watching the differential equations lecture series here on .2607

My name is Will Murray and I really appreciate your watching, thanks very much, bye bye.2612