Now we're ready to look at non-homogeneous equations that have a forcing function in them. For first-order problems, we started out with variation of parameters. That is still an option for second-order problems. But it's a long formula with a lot of minerals, it's not desirable. Instead, we're going to talk about the easier method, the method of undetermined coefficients. Of course it only works if f is exponential, polynomial sine or cosine or some combination of them. Fortunately, it follows the same rules that we learned before. Really there's nothing new to learn, It's just new examples. Our first case study is when we have an exponential forcing term. This is the easiest one. The particular solution is a constant times that same exponential. We just have to insert that into the differential equation. Each derivative is another factor of negative one. And exponentials are never 0. So we can divide through in all the terms. That takes the t dependence out completely. And we get a very simple equation to solve for that constant a. Our next case study has a polynomial forcing function. Since it's a quadratic polynomial. That means that the particular solution is also a quadratic polynomial. Whose coefficients we have to find. It's useful to write down the first second derivatives. And then those get inserted into the differential equation. Now I'm going to match like powers of t on both sides of the equation. So I'm going to collect powers of t on the left side. For this type of problem, you always want to work from the highest degree down because it makes it easy to solve for the coefficients. It's not going to match powers of t squared. Obviously get that a equals 3. Then I'll match powers of t. Since I already know what a is, I can put it right in. And I find that B equals 0. Finally, I match coefficients of t to the 0. The right hand side doesn't have that term, so it's coefficient is 0. Then since I know a and b, I can find, see, the linear system that we get for a, B and C is automatically triangular, so it's easy to solve. Here's the third major type of example. This time with a sign forcing. Whenever you have sine or cosine or both at a particular frequency. Xp always needs both terms. And it's best to write out the derivatives so that you don't make mistakes and what's coming up? I'm going to plug this into the differential equation. And I know I'm going to have cosine of three t terms and sine of 32 in turn. So I'm going to gather those, buy it from the beginning. From x double prime, I get minus 9 8 times cosine and minus nine times sine. From minus x prime, I get negative three b cosine and positive 3 a sign. And then from 3 x, I get two more contributions. Thou has to equal 30 sine three key. Implicitly there's 0 times cosine three t. So the coefficients of cosine have to match and the coefficients of sine have to match. From the cosine. I get negative six, a minus three, b equals 0. From the sign, I get three a minus 6, b equals 30. You know all about solving this two-by-two systems. I'll just write down the answer. So tell us what x sub p should be. Now, as with first-order problems, I haven't told you all the rules of this method, and so sometimes it can fail. The only case we care about is pure oscillation with forcing at the natural frequency. So if I try the standard method as we've talked about it to this point, then I can collect sines and cosines. When I plug this into the ODE from x double prime, I get minus omega squared a times cosine minus Omega 0 squared b times sine. From the other term, I get plus omega 0 squared a and plus omega 0 squared b. That's all supposed to equal sine of omega 0 times t. But when you look, these coefficients are both 0. And I have a contradiction. 0 equals sine omega 0 t. That can't be satisfied. The root cause of the problem is that when you look at the homogeneous equation, that solution is also a combination of cosine omega 0 t and sine omega 0 t. It's essentially the same as the x sub p that we tried. So it's not surprising that it canceled out and give us 0. Well, the fix in this case is to just multiply x sub P by t. So it no longer looks like x sub h. That turns out to be the right thing to do. Now I've got product rule terms and the derivatives. And I don't want to have to write all this out. You'd probably don't want to watch me write it out. Bottom line when I plug it into that left side. The t times cosine, t times sine terms. Now those are the ones that cancel out. Which is good because I have nothing to match them on the right hand side. But now I do have leftover cosine and sine that I can use. So from here we can see that b has to equal 0. Then a has to equal negative one over two omega 0.
IV.7 Method of undetermined coefficients
From Tobin Driscoll March 29, 2021
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