 Welcome back. We are at lecture 29. We just took a test over some chapter 7 material. Actually, the chapter 7 material that's in the textbook itself. And now we will switch to the supplement that looks like this. If you have the supplement or if you didn't, I made copies of it and distributed the equivalent of what we're going to do from this course from this supplement to the text. This is material that is the publisher that publishes your textbook. They own the rights to this. It's been determined that some more second-order differential equations are necessary beyond what's in our calculus textbook for some engineering courses and what you're going to encounter pretty quickly in engineering courses. And it kind of makes sense that we've done differential equations. We just extend it a little bit in the second order. And unfortunately, not in the book, but it's in the supplement. So if you didn't bring it, that's fine. But you might want to start bringing it over the next few days of class so you can make notes in your supplement. And that's the type problem we're going to be doing. So second-order linear differential equations. Actually, let's be correct with the type. Second-order linear non-homogeneous differential equations is what we're going to start with. And then we're going to go to non-homogeneous a little bit later which kind of involves the same process and a little bit more. So this is 7.7. It kind of fits in nicely. I think it was actually a chapter from another book and was renumbered so it kind of fits in sequentially with what we're doing. Second-order linear homogeneous differential equations. Second-order because they're going to have a second derivative in there as well as the first derivative as well as the original function. Linear because we don't have a derivative squared or a second derivative squared. So it's always the first power of the second derivative, the first power of the first derivative and so on. And homogeneous because it's going to be equal to zero. So when we move on to non-homogeneous, we've got something else on the right side, some polynomial, sine or cosine or exponential function. We do have a restricted subset when we get further in the supplement. But here's what our equations will look like. So we've got the possibility of some lead coefficient A times the second derivative, some coefficient times the first derivative, some coefficient times the original function. So you can see the second order. Linear because we don't have any derivatives to higher powers and equals zero, which means it's homogeneous. So we've actually examined in section 7.1 when they were throwing a bunch of differential equations at us and they said, what would you do in this setting? And we ask ourselves the question, we're going to throw in, let's actually start at the other end, we're going to throw in some original function, positive amount, negative amount, and the derivative of that original function and the second derivative of that original function and we can have positives, negatives, some of these terms can actually be missing. Can you think of any type of function that we put the original, its derivative, and the second derivative in here and somehow combine them and everything drops out and we get zero? What has for its derivative and its second derivative pretty much the same thing as the original function? Some exponential function. So that's how we start with this now. This will slightly be altered from an exponential function into a trigonometric function, but it does so in a way that we can connect it. But if we start off with a solution that looks something like this, this is what the y things might look like. Some constant, maybe I should use a k, it's not going to make that much difference here, times e to the rx, that way if we use the second derivative and plug it in, what's going to be the derivative of c e to the rx? c comes along for the ride, derivative of e to the rx is e to the rx times r. And the derivative of that, again the c comes along for the ride, r is also a number, so it comes along for the ride. What's the derivative of e to the rx? Right, e to the rx times another r, so we're going to have an r squared. So if we plug these things in and everything is eventually supposed to drop out and we're supposed to get zero, then we ought to have an equation, we ought to have potentially multiple values because we're going to have an equation eventually that has r squared in it, which is what we're looking for, and r in it, and then this term doesn't have an r in it, it just has a constant because we'll be able to factor out e to the rx. In fact in the generic sense, let's see what this is going to look like. So a is going to be some number, for y double prime, let's put in c, r squared, e to the rx. Second term, let's get rid of y prime, and the third term, let's get rid of y. So each term now has an e to the rx, so that can be factored out in front. a is some number that we don't know, c is also some arbitrary number, so a times c is some number, right? Try to be careful here that we use a number that we haven't used yet. So there's an r squared here, we factored out the e to the rx. b times c, b is an arbitrary number, same with c, so we need another number. What can I use here? T times r, and again lower case c and upper case c are both numbers, they're multiplied, we get another number, s, and there aren't any r's here, so we've just already factored out. Sorry, we've got too many letters floating around here, but now we can set each factor equal to zero. So e to the rx equals zero, tell me how many solutions it has. None. e to the rx will never be zero, e to the exponential value to some power is never zero, it's asymptotic to zero if you graph it, but we're never going to get a solution from this piece, so the solution we get is going to be from this piece right here. Now in the supplement, this is called the auxiliary equation, that's probably the only time I'll ever refer to this as the auxiliary equation, I'll probably refer to it as the characteristic equation. More often than not, but it is in the book called the auxiliary equation. So it's an equation that we're going to solve because when we start with our value for r, there's going to be a c in our answer anyway, we're not going to know what c is. e to the rx, what we want to find out is what is the r value that helps us solve our particular differential equation. So r is what we're looking for, does that look like an equation, this auxiliary equation or characteristic equation that we can solve for r? It might factor, right? It's a quadratic equation, so there's going to be two potential solutions for r, and then we'll have to figure out how we handle that too when we have two solutions for r. They might be the same solution, so it might be a double root, so either two distinct real roots, and you'll see the categories that we're going to have as we progress. Two real roots, but they're really both the same number, so that would be a double root. In the third category, you tell me, what other kind of solution could we have to this quadratic characteristic equation? Non-real. Non-real, okay? Two complex roots, if you want to call them that, so roots that have some imaginary value in them. So that's the third case, and that's probably the most stubborn. I don't know that we'll get to that today, but hopefully we'll do the other two cases. So this characteristic equation is this equation that once we solve for r could have, let's cover the first category, two distinct real roots. Now, it'd be nice if they're also rational, so we've just got these numbers that we're used to working with, or even nicer, it'd be nice if they're integers, but at least they're real. So an example, probably should do an intermediate example, but we'll just do all that stuff that I was going to do before we get to this example. We'll do it with this example. So there's our equation. So we said we're going to have some numbers, potentially in front of the y double prime. Easy version, we've got a one, a number in front of the y prime, and a number in front of the y, and we know we're going to be equal to zero. So this process that we're going to do right now, on this, our first real example, you don't have to do what we're doing now. You're going to be able to skip this process, but let's do it on the first example. So let's say that our solution is something like this. It's actually going to be bigger than this, but we'll get to that in this problem. So that's the kind of thing that has a chance of working. Second derivatives, first derivatives, original functions, put them all together, a certain amount of them. They all drop each other out, and we end up with zero. We'll check the solution on this problem, too. So this is the same thing we did on that kind of generic problem. Now we're going to make it specific to this one, and we know the a and the b and the c that are the coefficients. So for y double prime, minus three of the y primes, there's the y primes, minus 10 of the ys, and we want everything on the left side to knock each other out, and we're left with zero. So it's a homogeneous equation. Factor out e to the rx. Actually, we can factor out the c also, right? Is that correct? Everything has a c. What's left? Minus 10. So we could set each factor, c is a factor, c equals zero, that's pretty absurd. Unless c were happened to be zero, then it's not a very interesting problem. e to the rx, I'll go ahead and put those together, equals zero, it never is. So we're going to get our solutions just like we thought we would from this, from this characteristic equation. Now, can you see how easy it's going to be to jump from this equation to this equation? Do you see the similarity in the two equations? Coefficient is one here, the coefficients minus three, it's minus three, coefficients minus ten, same thing here. Here we've got y double prime, we end up with an r squared. Here we've got y prime, we end up with an r, and here we've got minus ten y, which just becomes the zero degree term, the constant term in the characteristic equation. So is it permissible to jump from here to here? Yes. To go right directly to the characteristic equation. But it's the first example, so we went through the whole process. So we're not going to get a solution here. Let's go ahead and solve our characteristic equation. Factors, I hope. I think I chose an example that the first time through, factors, factors of minus ten that would somehow be combined to get us a minus three in the middle. Good. So as we would expect in a quadratic equation, two, and I said this would be two distinct real roots, so r could be five, and r could be negative two. So we really have two solutions. So when we started, we said that the solution was going to look something like this. So we have, since we're going to have two solutions, I'm going to give it one of them a C one, and the other one a C two. Again, it's the first example, so let's kind of labor through. It's not that bad. Let's verify that both of these are solutions. So back to the original equation, which is this. So if we're going to use this first solution and verify that it really does work, all this stuff on the left side combines and drops out and gives us zero on the right side, we need to know y prime, which is what? Five C one, e to the five x, and y double prime is 25, C one, e to the five x. Let's plug them in and see that it, in fact, works. So for y double prime, we've got that. We're going to subtract three y primes, and we're going to subtract ten of the original y's. Do we get zero? We better if it's a solution. So we've got a bunch of things that we're talking about. We're talking about some of these C one, e to the five x's. We've got 25 of them here. We've got what? Minus 15 of them here, and we've got minus 10 of them here. So are these C one, e to the five x's, are they going to drop out? They will. So it is a solution. You don't have to check it each time. That's the first problem, so we're kind of validating that what we think is going to work out really did, in fact, work out. Now let's check this one. It's also supposed to be a solution. So if y is C two, e to the negative two x, what is y prime? And y double prime, and we're out of room here, so let me move to the next page. So our equation is this. We already have one solution. We're now trying to validate another solution. For y double prime, we put that in, minus three y primes. You can get caught up in the algebra and the stuff that's going on to solve the problem. We don't have any idea right now what kind of problems we're going to be able to solve with these techniques. So right now we're going to get the techniques, and then later get some problems that turn out to be second-order linear differential equations, and then we'll see how to, we'll already know how to solve them. So right now it's just the kind of the mechanics behind solving this kind of problem. So the y thing, the second y thing, is that. So we've got some of these things, these C two, e to the negative two x's. We've got four of them here. We've got six of them here. And we've got minus 10 of them here. So do they all knock each other out on the left side? So both of these solutions work. So we just say that it has two solutions. No, we say that it has a solution. And in fact, the solution is their sum. If they work individually, isn't it going to be true that their sum, if I put this in by itself, it gives zero. If I put this in by itself, it gives zero. And we're adding these things together. If I put them in together, added together, aren't we going to get zero again? So I don't think we need to go through the fact that the sum of the two individual solutions is also a solution. So every time we have a characteristic equation that has two distinct real roots, it will always have this for its solution. So we just put the two r values up in the exponent and we move on to the next problem. So it really can be nice and quick. You can go from the original equation to the characteristic equation, get the two values for r that solve it, write down the answer, and you're ready to either move on to the non-homogeneous part if it's that kind of problem, or you're done with the problem. So two distinct real roots, that's what the solution looks like all the time. We're convinced that this solution does in fact give you zero on the left side. Take the derivative of the sum, take the second derivative of the sum, plug them in, and you'll see that it in fact gives zero. Alright, so let's take another example, and this time we won't go through every possible step that it takes to solve the problem. Second order, linear, homogeneous differential equation. So let's skip the formalities and the pleasantries here, and let's go right to the characteristic equation, which would be what? That helps. Save some time, energy, space. Can we factor this? We could try six r and r, or three r and two r, any recommendations. Six r and r. We've got to work with factors of two to get negative seven in the middle. I don't know, maybe it doesn't factor. We can still solve it if it doesn't factor. I don't know what works. I don't have the answer written down. I'm hoping that it has two distinct real roots because I wrote it under that heading as an example. Yeah, three r, two, three r minus two, two r minus one. So we need a two here. Two. That'll give us four r and a one here. And these both have to have the same sign, right, to get plus two. So it's going to be minus and minus. So there's our minus four r, there's our minus three r, so we've got minus seven r in the middle. So our solutions, one of the solutions is what? One-half-thirds. Two-thirds? Two-thirds and one-half. And one-half. So our final solution, so you can see how quick these problems can be but you don't have to do all the intermediate steps. C1 e to the two-thirds x plus C2 e to the one-half x. Is that all right? Now we don't have an example here, but let's suppose that this doesn't factor quadratic formula. So depending on the variable, in this case it's r that's being squared, so r would equal what? Quick review on quadratic formula. Negative three plus or minus b squared minus four x e over two a. All over two a, right? Now real, this is two distinct real roots. There's an example in your book and without going through the whole problem, they end up with r equals negative one plus or minus the square root of 13. That's a delightful answer. There's our two r values. So r1 is that value, r2 is that value. Are those two distinct real roots? Is this the same category that we're working in right now? They're real. What are numbers that are not real? Okay, imagine here. Not real is imaginary and imaginary would have a square root of negative one in it, right? Somehow, some i in it. So this is real, square root of 13 is not a nice number, not a very kind number, but it's real. That we're square root of negative 13 and we'd have something that falls into a completely different category. So you can envision what your answer would look like when you have ugly values like this, but it still is the category two distinct real roots to the characteristic equation. So we plug them up there in the r1 and the r2 position. Still is this category, two distinct real roots. So if it doesn't factor, it's a little bit annoying, but as long as they're real, it falls into this category. Questions about this? All right, second classification and there are three. Three types of second order linear or homogeneous differential equation solutions. So this would be the roots are real, however they're not different from one another. r1 and r2 are exactly the same. So we should expect a different category, a different classification because if we use the same, look what it would look like. So we would start out saying there's our solution, but in reality, r1 and r2 are the same, so I'm going to sub in r1 for r2. Aren't these terms really the same? So couldn't we factor out e to the r1x and we'd get c1 plus c2? Well, what is c1 plus c2? It's a constant. So really, this is redundant. If we say this is a solution and this is a solution, we're really talking about the same solution. But there is a second. So that is going to be part of our solution, but it's not going to be both parts of the solution. So there is another type of term and I know what your question is going to be because it was my question the first time I saw this. So that will be one of our solutions. There's the other one. And I see that same question that I had the first. How do you know that? Where did that come from? And the answer is, I don't really know. No, I do know. Somebody a whole lot smarter than us decided that in addition to this kind of obvious solution based on what we've done to this point so far, there is the potential for another solution. And you'll see as we do problems that when you kind of upgrade a solution, this is the way that you come up with the next most logical solution. This really probably was discovered or came about from somebody verifying or validating solutions using the product rule. So in using the product rule because there's a variable quantity and there's another variable quantity, it just so happens that when you take the first derivative of this kind of term and the second derivative of this kind of term, you end up generating terms that have the potential to knock each other out. That's probably not a real good explanation. But when we use, on our first example, we'll show that this kind of solution really does have merit and it has the potential to drop everything out on the left side. And it's the product rule that kind of causes that stuff to happen. So our example here, so it is a second order linear homogeneous differential equation. Let's skip to the characteristic equation. We know that exponential things are going to solve this. So let's just skip that because it's going to be the same on each problem. What's the characteristic equation that goes along with this? Factors into the same factor. So r1 and r2 are the same and they're each three. Since we already know what the solution is going to be, I don't think we need to mess with this. I think we've kind of validated that in our first example, that stuff like this is actually going to work. The other part of the solution, why don't we take it through the derivative process and the second derivative process and show that it actually does work and why something like that has the potential to work. So let's isolate on this. So let's say that our solution is something like that. So I'm leaving off the first term because we know it works. So we've got a product rule going. There's the first term of the product. There's the second term of the product. What is the derivative of the thing that we think has a chance of working for why? First times derivative of second. What's derivative of second? Plus second times derivative of first. C2. You can kind of begin to see now how it has a chance of working because when we take the derivative of something that is x, e to the 3x, aren't we going to generate some x, e to the 3x terms in addition to just good old fashioned e to the 3x terms? We're going to generate both up. So they have the potential of kind of knocking each other out all along the way. And it gets, luckily, we only have to go to the second derivative because it gets kind of progressively uglier here as we go. Let's call this the first term. We could combine these, but let's just do them one at a time. There's the first term. There's the second term. So let's use the product rule on that. First, which is what? 3c2x times derivative of the second plus second times derivative of the first would be what? 3c2. So there's our first product rule on this product. And then we also have to take the derivative of e to the 3x times c2. What's that? That's just a number, right? Just bring it along. Derivative of e to the 3x is 3e to the 3x. This is going to be delightful. So we're going to plug this in for y double prime. We're going to plug this in for y prime, and we're going to plug this in for y. Just to show that, in fact, you can start with a solution that looks like that. Take the original function, take the derivative, take the second derivative, and it is possible for everything on the left side to knock out every other term, and we end up with 0. Alright, so for y double prime, we're going to have 9c2x e to the 3x. We're going to have 3c2e to the 3x. And we're going to have another 3c2e to the 3x. So there's y double prime. We could put that together, which we will as we go forward. From that, we're going to subtract 6y prime. So minus 6, let's see what y prime is. 3c2x e to the 3x plus c2e to the 3x. And the last term is plus 9y. So the question is, do we get 0? The potential is there, right? Because we have some x e to the 3x terms. Hopefully they'll all knock each other out. And we have some e to the 3x terms. Hopefully they'll all knock each other out. Let's just track the x e to the 3x terms. Let's just go down the line. Here we have what, 9c2 of them? Here we have what? Negative 18c2 of them. And here we have 9c2. So did we lose the x e to the 3x terms on the left side? They all knocked each other out. Because the net result here is that we have 0 of them. 9 of them, sorry, 9c2 of them, 9 more c2 of them, and we lose all 18c2. So the other term is e to the 3x. So here's 3c2 and 3c2 more. So there's 6c2. Here is what? Minus 6c2. And we don't have any of them here. So they all knock each other out here too. So this term has a chance of working, and in fact in this example it did work. So we allow for that term kind of trusting that somebody else has done the leg work behind us and potentially because of the product rule generating x e to the 3x terms as well as e to the 3x terms they will all knock each other out. So for this equation the solution of it can be actually this quick. Go from here to the characteristic equation factor certainly quicker now because we've already done it and go right from there to the solution because every time we have a double root to the characteristic equation the solution will look like this. We're not done for the day, we're done with this problem. Some of you got a little too excited when I said we're done. Saw you moving toward the door. Alright. We can get one kind of preliminary thing out of the way. Actually I'm not going to go that way. Let's do this. So I'm going to save the whole Category 3 stuff till tomorrow which is when we have two complex roots. So both of you that are here with me tomorrow you might want to take good notes so you can give them to your classmates who might not be in attendance tomorrow, last day before spring break. So we've got a second order linear homogeneous differential equation but we've also got a little extra baggage. So hopefully this can be solved with one of the two categories that we already have and then what do you think this is for? What, how will that come into play on our final solution? So we can find the C value. So we won't have a C1 and a C2 we'll have specific information that's relevant to the fact that y is 0 when x is 0 and y prime is 2 when x is 0. So let's go to the characteristic equation which is factors into minus 2 minus 1. So our two solutions are 2 and 1. Two distinct real roots, that's our first category. So y is C1 e to the 2x and if you change this around and your C1 is somebody else's C2 and vice versa, then you'll have values that will be flip-flop. So when it's all said and done, if this final answer is if the coefficient of e to the 2x is supposed to be 5, you'll both get 5 even though you called it C1 and the other person called it C2. C2 e to the x, right? So let's incorporate this additional information. So it might be written like this y of 0 is 0 and y prime of 0 is 2. So the key thing is x is 0 in both cases. y is 0 in the first case y prime is 2 in the other initial condition that we're given. So let's take y of 0 equals 0. That's our answer in terms of y. So for y, let's put in 0 and for x, let's also put in 0. When x is 0, y is 0. Not that we're going to get a solution for C1 and C2 but we're going to get potentially something that's going to help us get a solution for C1 and C2. So C1 times e to the 0 is just C1, right? And C2 times e to the 0 is C2. So not a solution yet but something that's going to help us. Now let's go on to our second fact. So we need y prime of 0. We need the derivative, here's y. So what's y prime? What's the derivative of C1 e to the 2x? And what's the derivative of C2 e to the x? So now we have an equation for y prime. We know that when we put in 0 for x, y prime is going to be equal to 2. So for y prime we put in 2. Very thankful in this particular problem that x is 0 because it gets considerably ugly when x is not 0 because e to the 0 very conveniently is a nice real number. So we end up with 2 equals what? 2 C1 plus C2. Is that right? So we have an equation that has C1 and C2 in it. And we now have a second equation that has C1 and C2. Two equations and two variables. There's a variety of techniques that you can use to solve. Somebody give me one of those techniques. Okay. So C1 is equal to? Negative C2. Negative C2. And what do you want to do with that? Plug it in for C1. Okay. If that's equal to C1, let's plug it in for C1 in the second equation. So instead of 2C1 it's 2 times negative C2. And we end up with what? Two of these. I'm sorry. Negative two of these. And then we add one back so we have negative one. So C2 is negative 2. What do we do to C2 to get C1? Negate it. Negate it. Right? So it's 2. So our final solution that shouldn't have any C's in it at all, where was our solution? C1 was the coefficient of e to the 2x. And C2 was the coefficient of e to the x. So when you have boundary conditions or initial conditions, then we're going to be able to find specific values for C1 and C2. All right. Good stopping place. I will see you tomorrow.