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Lecture Comments (13)

1 answer

Last reply by: Dr. William Murray
Fri Oct 16, 2015 2:06 PM

Post by Ahmed Alzayer on October 15, 2015

Greetings Dr. Murray,

If I have the following differential equation, what would be the right guess for the particular solution?

x"+16x = (80t-16)sin(4t)

2 answers

Last reply by: Dr. William Murray
Mon May 18, 2015 12:00 PM

Post by Bahaa Jabbar on May 16, 2015

hi prof., thanks for your great explanation, and I have question! what if I have y"-2y'-3y=3te^(2t)

as I was solving it you way I got to this point where 2Ae^(2t)-3Ate^(2t)=0 (solving for A)
and I don't know how finish this question?

1 answer

Last reply by: Dr. William Murray
Thu Apr 10, 2014 7:53 PM

Post by YILEI GE on April 9, 2014

Hi, prof. Murray. How this problem will solved "y''=t^3"? what is the particular part for this one? Thanks

Lei

3 answers

Last reply by: Dr. William Murray
Tue Jan 14, 2014 11:42 AM

Post by Joel Fredin on January 1, 2014

Murray,

I have a differential equation y''-2y'+2y=e^x*Cos(x). I think the homogeneous one is easy to solve. But for the particular one i have no clue. Can i write y_p=Ae^x(BCos(x)+CSin(x)) and then just take the derivative twice. Plug it into the the equation and then solve it. But if that's the case i don't know which one is equal to which one. It's much easier with plus and minus to right side of the equal sign.

Would be really nice if you could help me out a little. Thanks for your time,

Joel

1 answer

Last reply by: Dr. William Murray
Thu Oct 31, 2013 3:27 PM

Post by michael morris on October 23, 2013

Prof. Murray,

Can you please briefly explain the process of solving systems of linear DE's by elimination. I do not understand the process and i have an exam on fri. There are no videos online to reference.Any help will be appreciated.

Undetermined Coefficients of Inhomogeneous Equations

Undetermined Coefficients of Inhomogeneous Equations (PDF)

Undetermined Coefficients of Inhomogeneous Equations

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 0:11
    • Inhomogeneous Equation Means the Right Hand Side is Not 0 Anymore
    • First Solve the Inhomogeneous Equation
    • Find a Particular Solution to the Inhomogeneous Equation Using Undetermined Coefficients
    • g(t) vs. Guess for ypar
    • If Any Term of Your Guess for ypar Looks Like Any Term of yhom
  • Example 1 7:54
  • Example 2 15:25
  • Example 3 23:45
  • Example 4 33:35
  • Example 5 42:57

Transcription: Undetermined Coefficients of Inhomogeneous Equations

Hi, welcome back to educator.com. I am Will Murray with the differential equations lectures and today, we are going to talk about inhomogeneous equations undetermined coefficients so, let us get started.0000

We are going to be solving linear second-order inhomogeneous constant coefficient differential equations, and the key difference here between these equations and the ones we have been solving before is that they are inhomogeneous which means that the right-hand side is not 0 anymore.0017

We are to have a function G of T in place of the 0 that we had before so, that is the new aspect for these this lecture in the next 1 and is actually 2 methods we are that we are going to be using to solve these.0033

The 1 we are going to learn in this lecture is called undetermined coefficients and in the next lecture, we learned different method called variation of parameters so, there is are 2 main methods this lecture is all about undetermined coefficients and the way it works is you first solve the homogeneous equation .0048

So, you take the same equation and the key difference here is just erase that G of T and replace it by 0 and then you solve the homogeneous equation A Y″ + B Y′ + C Y= 0.0068

So, we studied this type of equation pretty extensively we just of 3 lectures on that so, if you do not remember how to solve those go back and look at the previous 3 lectures where you learn how use the characteristic equation to find the general solution to the homogeneous equation.0083

Remember homogeneous means the right-hand side = 0 and then the inhomogeneous equation is where the right-hand side = G of T so, today we are learning how to solve the inhomogeneous equation but the first step is to actually solve the corresponding homogeneous equation .0102

I assume that we already solve that and then next thing to do is to find a single what is called a particular solution to the inhomogeneous equation which is where you put the G of T back in .0116

We use a method called undetermined coefficients so, what that means essentially is that you guess function that has the same form as whatever G of T is sitting over there on the right-hand side but you leave it in terms of generic coefficients we put in generic constants for the coefficients .0133

Then whatever your guess is, you plug it into the differential equation and solve for the coefficients so, what I mean by that is if you see on the right-hand side if you see E RT so, E 3T for example maybe multiplied by some constant.0153

What you would guess is the same E RT of putting a generic constant which I am calling A so, you make that guess you plug it work it through when you try to figure out what a should be then so, make sense after we do some examples but I want to get through some of the generic theory before we start.0172

If you see a polynomial on the right-hand side you would guess a generic polynomial for your particular solution so, you guess a T ² + BT + C just leave the coefficients is AB and C and solve for a T for A B and C.0190

I have I am assuming here that it was a quadratic polynomial but basically you just go for the same degree is as whatever G of T is if you see something like a constant time sin of 5T what you would guess is not just sin a 5T but you have to put in cosine term as well so, you would guess a sin 5T + the cosine 5T you plug it in and you try to figure out what the a and B are .0209

Some I just see combinations of the functions that we had above so, you might see something like 3 E 2T + 4 T +7 and what you see here is there is an exponential term and there is a polynomial term so, if you C something like that then your guess for the particular solution while from the 3 E 2T I would guess a generic AE the 2T and for the polynomial I would guess well a B+ C but there is our sorry a T + B but since of I already use a here I would use BT + C .0239

So, that is going to give me the E2T term and this part I would try to figure out what BNC, BNC would have to be to give me the 14+7 term .0285

So, that that is the general idea of what you should be guessing there is 1 more subtlety and this is why you need to solve the homogeneous equation before you start which is that you want to look at your guess for the particular solution and if any term of your guess looks like any term of the homogeneous solution essentially that means it is not to work Because it means when you run it through the differential equation you just can get 0 Because the homogeneous solution was design to give you 0 so, what you do is you multiply your whole guess by T.0302

We will see some example of that in a few minutes couple of points here you will get some you will still have a C1 and C2 from your homogeneous solution you want to solve for those C1 C2 after finding a particular solution make sure you find those constants after finding your particular solution .0337

If you try to solve for those constants before you find your particular solution than the particular solution will screw up the initial conditions and so, you have the wrong value of the constants 1 final note here is you can only use undetermined coefficients for very specific class of functions you can only use it for exponential functions and polynomials and sins and cosine es.0363

Essentially those are functions for which the derivatives look very similar to the original functions so, if that is what you have for G of T then you can use undetermined coefficients and G of T is essentially any other kind of function so, something like natural log of T and T other functions that when you take their derivatives they do not look similar to the original functions anymore .0395

You really can not use undetermined coefficients you have to use variation of parameters which is a completely different technique we are to learn about that E next lecture so, I am not going to say anything more about that right now, but just be aware of the limitations of on its undetermined coefficients and if it is not 1 of these nice functions then you have to watch our next lecture on educator.com and learn how to solve those.0444

So, go ahead and get started with finding the general solution to the differential equation Y″ - 3Y′ +2 Y = 4 E 3T so, remember the idea here is first to solve the homogeneous equation so, for some just can forget about that 4E 3T them a set Y″ - 3Y′ +2 Y = 0 is kind of throw away the right-hand side set = to 0 and now, I am going to solve that as a homogeneous equation.0473

We learn how to solve this in the previous lecture so, if you do not remember maybe go back a couple lectures and look up the first lectures on second-order equations you will C how to solve this the ideas to set of the characteristic equation R² -3 R +2 = 0 and solve that for R this what factors nicely it is R -1 x R-2 = 0 so, R = 1 or 2 and then that tells me my homogeneous solution right away .0508

Remember those become the exponents of the coefficients of the exponents E T + C2 E 2T so, that is my homogeneous solution and now, I want to guess a single particular solution to the inhomogeneous equation so, what I do is I look at that 4 E 3T and I guess something generic that resembles it .0549

I am going to guess my particular solution a E 3T and the whole point here is I do not know, what that a is right now, that is to be determined the 3 at I take right from that exponent that that always stays fixed but my a is generic and I do not know, what that is yet so, I am not plug that into the differential equation which means I need to know, it is derivative and the second derivative suits first derivatives 3a E 3T and its second derivative is 9 a U 3T .0582

So, plugged those all into the differential equation and Y″ is 9 a E 3T - 3Y′ is 3×3 a U 3T +2 Y is 2 a E 3T and that is = to 4 E 3T so, this actually that E 3T cancels everywhere so, cancel all that out and I get 9 a -9 a + 2 a = 4 the 9 A cancel I get 2a = 4 so, a =2 and so, my particular solution is a particular solution is 2E 3T .0621

That 2 back into my guess for the particular solution and the general solution to this differential equation is I add the homogeneous part + a particular part and so, my general solution where is my homogeneous part that is over here C 1 E T + C 2 E 2T +2 E 3T and that is the general solution to the inhomogeneous differential equation.0686

So, that is the end of that example if we had some initial conditions we be using them right now, to solve for the C common mistake students make is the use the initial conditions here to solve for the C that is incorrect because if you did then the subsequent edition of 2 E 3T would screw up the initial conditions so, use the initial conditions if you are given initial conditions.0727

In this case we we are given any so, we would not worry about it but if there given use the initial conditions to find C1 and C2 after finding the particular solution and adding it to the homogeneous solution to get the general solution so, use the initial conditions at the very end.0773

So, what am I that the let me recap what was going on here in fact we are going to be using this same differential equation all the way through this lecture for the further example so, you might want to keep this homogeneous solution in mind because that will be the same every time and then will experiment using different right-hand sides here. 0805

So, the initial work will be the same for every example so, we did there was we took that homogeneous equation we set the right-hand side = to 0 found the characteristic equation factored it down to get the R and then plugged those in as the exponents to get C 1 E T + C 2 E 2T that is the same in every example in this lecture.0830

And then we looked at the right-hand side the G of T and we guess something like it but we have generic coefficient now, that is that a E 3T and then to plug that in we have to take the first derivative the second derivative we plugged those into the differential equation the inhomogeneous differential equation including the right-hand side now, .0854

That simplifies down and we figure out that = 2 so, we plugged that 2 back in for a here we get a particular solution and we had a particular solution to the homogeneous solution to get the general solution .0880

That is how that all worked out if we had initial conditions we would use them at this point to find the C1 and C2 we do not have initial conditions in this problem so, we are just can a stop with the general solution so, remember this homogeneous solution to the corresponding homogeneous equation because we are to be using that for every example from now on .0898

So, keep moving to the next example to find the general solution to the differential equation Y″ - 3Y′ +2 Y = 4T ² so, remember this is the same equation that we had in the previous example at least on the left-hand side so, the homogeneous equation is Y″ - 3Y′ +2 Y = 0 .0923

From example 1 we figured out that the general solution for the homogeneous solution there was C 1 E T + C 2E 2T work that out you can check back in example 1 if you have not watch that recently and now, are going to guess a solution to the inhomogeneous equation which means we are going to look at that 4 T ² .0952

That is a polynomial a quadratic polynomial so, I am going to guess my particular solution I guess a generic quadratic polynomial I cannot just say a T ² that is enough that is not enough have to go BT + C sum up find the first and second derivatives plug that in and then try to figure out what those coefficients A B and C should be.0986

So, Y′ would be 2A T + B and Y″ would just be 2a and now, I am going to plugged those into the differential equations so, Y″ is 2 a - 3Y′ so, -3 x 2 a T + B +2 Y so, 2 x a T ² + BT + C = 4T ² and sort this out according to powers of T so, looking for T ² here.1015

I see it 2a T ² and I am looking for T so, I get my T terms will be -6 a and here is a T term right here + 2b and now, my constant terms I see it 2a here -3 B + 2C from there is 4T ² so, I am sorting it out by powers of T and I think of this as 4T ² + 0T +0 and I got to polynomials = to each other that means they must have the same coefficients term by term so, from my T ² coefficients.1066

I get 2 a = 4 that is from there and there from T term I get -6 a + 2B = 0 and from my constants I get 2a -3 B + 2C = 0 from this I get a =2 from this I get 2b = 6 a so, B= 3 a so, B would be 6 and if I plugged this in I get 2x 2-3 B is 6 + 2C = 0 so, that is 4-18 so, -14+ 2C = 0 .1123

If I move that over to the other side 2C = 14 so, C = 7 .1200

So, figure out what is what B is what C is plugged that back into my original guess I get Y particular solution is a T ² + BT + C that is 2T ² +60+7 that was a particular solution the general solution remember is always the homogeneous solution + the particular solution so, in this case the homogeneous solution was C 1 E T + C2 E 2T + the particular solution 2T ² +60+7.1208

Clean things up a little bit here and again I do not have any initial conditions so, I do not need to go any farther than this if I did have initial conditions now, is the time when I would be using the B plugging them figuring out the values of those constant C1 and C2 .1251

But I do not have those so, I am done with this problem we just recap quickly we used our first first we erase the 4T ² we said = to 0 to get the corresponding homogeneous equation and that is the same homogeneous equation we had back in example 1 so, we will remember how we solve that just check back in example 1 and you will see that homogeneous equation is C1 E T + C 2 E 2T.1295

For inhomogeneous equation we look at that right-hand side 4T ² and we guess a generic polynomial of the same degree is that is a T ² + BT + C take its first derivative and second derivative plug them back into the equation so, plug it all back in.1324

We get all this stuff = to 4T ² and then we score this kind of sort things out according to powers of T so, I collected all the T ² stuff all the terms with just a single T and all the constants here and then I set like terms = to like terms are I said looked at the coefficients.1347

So, the coefficients of T ² on the left was 2a on the right was 4 so, I got a = 2 coefficient of T on the left was -6 A+ 2b on the right was 0 so, I solve that out was what I know, from a and I get the = 6 the coefficient of a constant was 2 A- 3 B+ 2C on the right it was 0 I plug-in what I knew for a and B and I figured out C was 7 plugged those back and for A B and C to the particular solution that is coming in here.1370

I got this particular solution added back onto the homogeneous solution and I got my general solution so, that is kind of very common way that undetermined coefficients works out let us explore that some more with another example of sins and cosine , the sin we have, Y″ - 3Y′ +2 Y = 5 cosine 2T again remember the first step is to replace the right-hand side was 0 and solve the homogeneous equation .1407

This time again we have the same homogeneous equation that we had in example 1 so, we get our homogeneous solution C 1 E T + C 2E 2T that work we did back in example 1C can check back in example 1 if you do not remember how to do that and now, are looking at the possible solutions to the inhomogeneous equation so, looking at 5 cosine 2T and I am going to guess something that looks like that.1441

A cosine 2T remember we have a sin or cosine we also, have to include the other 1 it is because when you take derivatives of sins and cosine es you get the other 1 as well so, throwing a generic sin term as well B sin 2T and now, I am going to take my derivatives so, that I can plug it into the differential equation Y′ derivative of cosine is - sin and there is also, a 2 so, - 2 a sin 2T derivative of sin is cosine and 2 will pop out.1480

Let us chain rule there 2b cosine 2T got a find the second derivative here Y″ is through the sin is cosine so, - 4a that before that is not in a right that a little more clearly 4 a cosine 2T through cosine is - sins of - 4B sin of 2T, take these 3 pieces and plug them into the differential equation so, Y″ is - 4a cosine 2T -4 B sin .1518

- 3Y′ so, -3 x - 2 a sin + 2B cosine +2 Y +2 x a cosine + B sin = 5 cosine go ahead and say 0 sin so, now, let me sort things out in terms of sins and cosine and set the 2 sides = to each other.1569

For the cosine I see - 4 a -6 B from this term here so, -6 B and + 2 a - 4a -6 B+ 2a those are all my coefficients cosine + coefficients of sin a b - 4b +6 a + 2b +6 a + 2B sin = 5 cosine +0 sin and just like the polynomials of the cosine and sins the coefficients of cosine is sin = to each other - 4a -6 b+ 2 a this the cosine coefficients = 5 and the sin coefficients give me - 4B +6 a + 2B = 0 .1611

I'm going to work on the second want to looks little easier so, I see - 2B +6 a = 0 and that tells me 6 a = 2b and so, B = 3 a and now, I am going to work with the first 1 so, I see - 2a -6 B = 5 this is just 2 equations and 2 unknowns by the way and is a lot of different ways you can solve this depending on what you learned your out the class and what comfortable with.1705

If you use a different technique from the one, I am using it does not really matter there is linear algebra techniques, there is substitution techniques, there is linear combinations, there is matrix techniques.1748

It is all different ways of solving 2 equations and 2 unknowns but what I'm going to use is a substitution here when it put B = 3A in here so, I get - 2a -6×3 a = 5 - 2 a -18 a = 5 and finally -20 a = 5 .1760

So, a is -5/20 which is -1/4 and if I plug that back in here I get B is 3A B is - 1/4.1785

If I plugged those back into my generic guess back here,I will get my particular solution is my a was -1/4 cosine 2T + B is - B was 3 A is that should have been -3/4 -3/4 sin of 2T.1800

That is my particular solution the general solution says you add your homogeneous solution to your particular solution so, the homogeneous part + the particulars solution.1826

Now, the homogeneous part we already had back from example 1, C 1 E^ T by E a little more clear there, + C2 E^2T + the particular solution which is X and has - coefficients so, -1/4 cosine of 2T -3/4 sin of 2T so, that is my general solution.1849

As usual if we had if we had some initial conditions we can use them at this stage to find the 2 constant C1 and C2 and we use it now, we would use them earlier but we do not have initial conditions so, that means we stop here with the general solution .1881

So, let me remind you what we did here we already had the homogeneous solution that means is that the right-hand side = to 0.1908

We figure that out back in example 1 see can check back there if you do not where that comes from.1916

The particular solution we get something that looks like the right-hand side except we put a generic coefficient there and if you have a sin or cosine, you always have to include the other one as well.1921

We take its derivatives we take it second derivative we plugged these back into the differential equation so, we plug everything into the differential equation and I stopped writing the 2Ts here because it was so long.1933

But basically your sorting out a whole bunch of sins and cosine of both sides so, sort out all the cosine terms sort out all a sin terms set the cosine terms, the coefficients on both sides equal to each other set the sin coefficients on both sides equal to each other.1948

And that really gives you 2 equations and 2 unknowns so, at this point it becomes and high school algebra problem to solve these 2 equations and 2 unknowns .1965

The way I solve that was to solve for BN terms of a in the second equation plug that into the first equation and then figure out what a was and then go back and solve for what B was.1976

Then I find my A and B I plugged those back in to the particular solution for a and B here so, that gives me my particular solution that is my a that is my B at that onto the homogeneous solution and I get the general solution.1993

Let us keep going with another differential equation here we find the general solution to the differential equation Y″ - 3Y′ +2 Y = 5 E T.2011

And as I mentioned before that is the same left-hand side as we have been seeing in every example so, we originally solve that in example 1, we solve the homogeneous equation and we found the homogeneous solution Y homogeneous is C1 E^T + C2 E^2T .2027

That is been the same in every example so far in this lecture and we look at that right-hand side and we make a guess about a particular solution, my guess would be my particular solution would be a x generic a generic coefficient x ET .2051

Now, at this point there is an issue coming out that we have not seen so far in this lecture except for the overview that I mentioned quite a while ago.2074

You might not remember it but what you have to do is look at this particular solution and then compare it to the homogeneous solution and see if any term of the particular solution matches any term of the homogeneous solution .2084

Of course if you look at this particular solution then it matches perfectly one term of the homogeneous solution what that means is that if you plug-in this particular solution into the differential equation.2102

If you plug-in this particular solution to the differential equation what you are going to get on the right-hand side is 0. 2124

Because the homogeneous solution was exactly what makes the differential equation come out to be 0 so, that means that this different that this particular solution is also, a this and threatening red color is doomed to fail.2130

That particular solution is guaranteed not to work as a particular solution to our inhomogeneous equation.2155

Because it already was a solution to the homogeneous equation so, how do we fix that we are going to write a new particular solution and we are to multiply it by T, which is what I mentioned in the lesson overview at the beginning.2164

When you have part of your particular solution looking like your homogeneous solution, you replace the particular solution with, or multiply your particular solution by T and then it all works as you will see when we go ahead and figure out the derivatives and plug it so, let us do that now.2182

Y′ would then B okay the derivative of this is harder now, because we have a product so, let me use the product rule it is a x now, T x the derivative of U T.2206

That is UT + E T x the derivative of T which is 1 that is E T there and Y″ still have a x again I have to use the product rule.2221

T x derivative of E T + E T x the derivative of T + this E T it is derivative is another E T so, that is a x T E T + 2 copies of E T so, that is what I get for Y′ and Y″ and I will plugged those into my differential equation.2236

Y″ is a x T E T +2 E T - 3Y′ so, 3A x T E T + E T +2 Y so, 2 a x T E T .2265

This is all supposed to be = to 5 E^T so, let us expand things out and sort things by terms I see an a .2288

I try to isolate everything it has a TU^ET so, from here I get a from here I get -3 a and from here I get a + 2 a and now, my ET terms I get 2a that is coming from here here -3 a and that is = to 5 ET .2300

Now, a -3 A+ 2 a that gives me 0 so, that term completely drops out and I get here - a E T = 5 E T.2334

To make that work my a would have to be -5 so, that is all for that coefficient, I plug that back in no not there that was my initial failed guess for Y particular plug that back in here.2351

My particular solution is -5 TE T and my general solution is the homogeneous solution + the particular solution.2371

Still have that general solution from example 1 C 1 E T + C2 E 2T now, my particular solution -5 remember you do not put arbitrary constants on the particular solution because we took a lot of care to get that -5 just right so, -5 TE T .2385

That is my general solution if I had initial conditions I would use them now, to find the C1 and C2 use the initial conditions to find C1 and C2, but in this problem we have not been given initial conditions.2412

There is no C1 and C2 or there is no specific information about C1 and C2 to be found.2436

We just leave them generic and we stop with the general solution so, that means we are done with this one, except for got just going back and doing recap there.2441

We started out by finding the homogeneous solution which I we had already found back in example 1, you can check back earlier in the lecture look at example 1 to see where that comes from .2451

We wrote down our particular solution by looking at this right-hand side and guessing something generic that looks like the right-hand side, that would be a E^T.2462

However at that point, we noticed that a E^T looks like the homogeneous solution which means if we plug it into the differential equation we are going to get 0 on the right-hand side.2473

In other words we cannot get 5 E T so, that is why, I said that a E T is doomed to fail so, instead we bump it up by a power of T and we go with a particular solution a TE the T.2488

And we are going to work through with that we still know, what a is we found its derivatives first and second derivative the heavy use of the product rule that of course comes from calculus 1 the product rule .2503

Once we found its derivatives we plugged them back into the differential equation , set it equal to 5 E^T, what we did on this line was we sorted out everything that had at TE the T and everything that was just E^T and E^TE the T terms dropped out very nicely just leaving us with a E T terms and that let us figure out that a was -5 plug at -5 back in to our guest for the particular solution our new guess not our original failed guess.2516

We get -5 TE the T and then our general solution is whatever you get from adding the homogeneous solution to the particular solution so, our general solution is that homogeneous solution and then our particular solution is -5 TE the T.2553

That is our general solution there and we are completely done with that differential equation.2571

For our last example, we want to give an appropriate form for the particular solution to the differential equation Y″ - 3Y′ +2 Y is T the 4th +7 E the 3T now, remember on all these equations you have to start by solving the corresponding homogeneous differential equation.2575

So, in this case we solve the homogeneous equation back in example 1, you can check back there if you do not remember how we solve that but it was using the characteristic equation.2601

We figured out that the homogeneous solution was C 1 E^T + C2 E^2T so, that was the homogeneous equation and then the way you guess a solution to the inhomogeneous equation is you use functions that look like the right-hand side.2614

In this case we got a kind of combination right-hand side, there is a T the 4th term and there is E 3 term is 3T term, we are going to treat those totally separately and solve for those completely separately.2636

We are going to guess for the first term there will make our first particular solution.2654

I see a T the 4th which means I need a 4th degree polynomial generic I can just say a T the 4th, I got a say a T the 4th + BT cubed + CT ² + DT + E and we would take that guess.2666

We would solve Y″ - 3Y′ +2 Y = T to the 4th we would solve that for the undetermined coefficients A B C D and E .2695

So, that would be the first part of solving this problem, by way we are not actually going again to go through in solve probably coefficients here be really messy the idea of this problem is just to figure out what the general form of the guess would be so, kind of understand the ideas without getting too bogged down into the nitty-gritty details.2713

So, after you solve that part then we have to deal with this 7 E 3T so, we guess a new particular solution Y P2 now, I'm going to guess something of the form E the 3T.2742

My guess generically why do not we use a again because I already use a above so, used F, E^3T and then you would find the first derivative second derivative plug-in and solve Y″ - 3Y′ +2 Y = 7 E 3T.2754

We solve that for capital F so, that would give us a particular solution that matched up the EU the 3T part of the equation and then once we found all those coefficients ABC DEF we would use our Y particular we would just add those 2 solutions.2784

So, YP1 + Y P2 or once we found those coefficients a T the 4th + BT cubed + CT ² + DT + E + F E 3T that would be our final particular solution.2813

And if you want to find the general solution as usual you take the homogeneous solution and tack on the particular solution that you found above by filling in all those constants.2853

So, the key idea of this example just to recap here is that when you have a combination right-hand side like this what you want to do is split it up into the 2 different types of functions.2867

We have a polynomial part and an exponential part and sort of solve for particular solution separately.2883

This is all of course after finding the homogeneous solution so, your first particular solution is to match the polynomial part, you guess a generic polynomial of the same degree.2891

The degree was at T the 4th so, you got... it's I guess a generic…a 4th degree polynomial. Plug it in, find its first and second derivatives. Solve for all those coefficients; it would be kind of messy which is why I don't work it out .2904

Then you would find the exponential part by guessing and X a generic exponential term it does not match anything in homogeneous terms, we do not have to bump it up by T like we did in the previous example.2918

You just find its first and second derivatives, plug it in and solve for F and then those 2 particular solutions that you found one for the polynomial part, one for the exponential part, you add them together and get your combines your combines general solution there we clean that up little bit .2934

That would be as your combined particular solution and then you would add that onto your homogeneous solution to get your general solution .2956

So, that is the end of our lecture on undetermined coefficients basically a way to solve inhomogeneous equations if you have a nice functional on the right-hand side. 2968

In our next lecture we are going to study variation parameters which is a way to solve inhomogeneous equations when you have a nastier function on the right-hand side so, to completely new technique it will be fun to work that out.2981

In the meantime I am Will Murray, and you have been watching the differential equations lectures here on educator.com. Thanks for joining us. Bye bye.2993