 Welcome back after a spring break, so it's been a while since we've been here, so it's probably going to take a few minutes for us to get back in sync with COP2 things. We are in the supplement. We have spent a couple days in the supplement already. In the bulk of this, we will finish, I hope, this week, possibly a little bit of it carrying over into next week. We have the second order linear homogenous differential equations. We've got the three basic cases. I think it would probably be wise today to take a look at each of the three cases because you're probably remembering that there were three cases but not necessarily remembering what those three cases are at this very moment. So we'll take a review of each of the cases. One of the problems that I had some flawed information on that we were solving for, I think we solved for K1 and then we tried to solve for K2 with some other data and K2 was eliminated because it had a coefficient of zero. I think I didn't actually find the problem, my source of the problem but I think I've got a way to make that work out so we can go back and revisit that. Usually when we cover this information and we use this equation which we had to use, we had to use that equation in the third case where we had complex roots to the characteristic equation. So this was the third case. We had complex roots to the characteristic equation and we put those complex roots up in the exponent position and they had an i in them so we had to simplify and then one of the formulas we used was this one. Characteristic equation, let's make sure everything that gets mentioned today we adequately review. Where's that coming from? Characteristic or as it's called in the little supplement it's called an auxiliary equation. So if we have an equation y double prime minus 4y prime what's the characteristic equation that goes along with that? R squared minus 4r plus 11. R squared minus 4r plus 11. And that's where we're going to get our roots, right? So it's all based on the kind of roots that we get. Do we have two complex roots? Then we fall into case three. We'll do an example here in just a minute. If we have a double root, remember that, that was case two. We had a certain type of solution based on the solutions to the characteristic equation and if we had two distinct real roots we had another kind of equation. So we'll refresh those in a minute. But there's our characteristic equation. If the roots were complex we had to use this thing right here to get it simplified so we had this new function in terms of sines and cosines. I probably should not let that go without looking at this equation. Let's look at e to the i pi. That's three pretty strange numbers put together all in the same equation. What do you think e to the i pi might be? Something really strange, right? e is strange, i is very strange, pi is kind of strange. Let's see what that is. So according to this equation that we used the other day e to the i theta would be cosine theta so theta is pi so it should be cosine pi. So everywhere there's a theta we replace it with a pi. Got a pretty ugly mess there on the left side but the right side is actually pretty easily simplifiable. What's the cosine of pi? Negative one. It's been spring break, right? Reading about that stuff and thinking about it over spring break a whole lot I can tell that in your faces. Then i times the sine of pi. Zero. So if that's zero that's gone. So e is a pretty strange number, i is a strange number, pi is a strange number but when you combine them in this fashion e to the i pi is negative one. That's kind of bizarre. Now the equation that you sometimes see as a result of this e to the i pi equals negative one since e to the i pi is negative one not often but I mean that's one of the more strange equations in all of mathematics because it's got some of the more important numbers that go all the way across all disciplines of mathematics. E, i and pi, one is a pretty important number because that's how we progress through our number system and zero. There's a pretty unique equation which I failed to use the other day but I happen to think about that. This tells you the difference between where you are at your age and where I am at my age. I think about stuff like this when we're not in class or when I'm walking to and from class in my office and I should have done something with that last week and I failed to do it so there's that one. Now I've actually seen this, I don't know if I'd go this far but this particular equation is one of the most beautiful in mathematics because of all the numbers that are in it. I see those faces too like not beautiful but it is pretty neat to have all those strange numbers in that equation. Alright let's look at some examples of these cases. This brought a sheet of problems. We'll put one up here I don't even know necessarily when we start what case that is. Can we prop that door open? I don't know if it's just me but it seems a little hot in here. Is that it? Yes. Okay thank you. Okay let's say we had this second order differential equation and we're trying to get ourselves reviewed a little bit. So I said we'd review just about everything today. Second order, why is this second order? Second derivatives? Okay linear. Why is it linear? You don't really see anything squared. We don't have a second derivative squared, a first derivative squared, the original function squared so it's all linear. Homogeneous, where'd that word come from? It's all y. It's all in terms of y's and y' and y' double primes but we're going to have starting tomorrow we're going to have non-homogeneous equation. It's equal to zero. So the fact that it's equal to zero means it's a homogeneous, now we're going to use that, we're not going to drop that and it's going to fall off the map starting tomorrow. We're still going to use the homogeneous part of a solution but if it's something that's not equal to zero we're going to have to kind of think about how it is that we get those things that aren't equal to zero. Right now we don't have to worry about them, we want to make them zero. So we go to the characteristic equation, probably ought to review this part too. What are these y things that have a chance of working in a second-order linear homogeneous differential equation? E to the rt. E to the rt or e to the r, we don't know what the other variable is, let's call it x here. Why do they have a chance of working because they are their own derivatives with some different coefficients in front as well as their own second derivatives, again with different coefficients out in front? What is y prime? And y double prime. And if we put them in the equation, which you don't have to do this but we're trying to kind of get reviewed on these things, doesn't have to be every problem where you do this. Now the fact that it is homogeneous, we're actually going to be able to get a solution we can factor out. E to the rx, remember this? Stuff from before the break, I hope. There's something on the test that I need to address too before we do a second review problem. Since we have the product of two things equals zero, that equals zero. Well, we're not going to get any solutions out of that. E to something is never zero. So there's our characteristic equation. It has a whole lot of similar features to what we started with. So we'll go about solving that and then we'll jump to the solution. So what are the r values that we could actually plug into that position for this particular example? R minus six, r plus one. All right, well I heard one and six and three and two. So let's see which... Minus three and minus two. Minus three minus two is going to work. So that gives us the minus five in the middle and their product is the six. Now if you use six and one, you'd have to have one positive and one negative, right? Which won't get us plus six. So we could get negative five in the middle but we can't get plus six unless they both have the same sign. So r one, if we want to call it that, is three. R two is two. So what's the nature of the solution here? What are the y? And it should be a family of curves unless we know some additional information which we don't on this problem. Interesting looks this morning. So while we're contemplating and pondering on this problem, I got an interesting email from a student that I taught in 1979 and he was going across the cable TV numbers and happened to stumble on channel 18, which this broadcast is on. And he goes, I found myself looking away from the TV screen so you wouldn't call on me just like I did in 1979, which is kind of funny that he was quite a character. I got a kick out of that. If you knew this kid too, you'd probably get his forties now but pretty funny. All right, that's plenty of time. A little off to the side there. Plenty of time for you to think about this. C1, I like that. E to the 3x. Go ahead. Plus C2x. No, that'd be a double root. E to the 2. Yeah. So this is case one, right? Two distinct real roots. So it's, in general, when you have two distinct real roots, that's what the solution is going to look like. So you put one of those real roots up here, the other real root here. If you don't have any additional information that's going to help us find C1 or C2, then we're done. That's it for that type. Questions about any piece of that before we go forward? Look familiar? I hope. Okay, let's see what happens. This one doesn't look very kind. So I'm thinking with that 13 there that this is not going to factor when we get to the characteristic equation. But let's see what happens. Okay, what's characteristic equation? Plus 13 equals 0. Thank you. Don't want to waste a whole lot of time trying to factor this one, right? Factors of 13. Not a lot there to work with. Certainly not going to give us negative 4 in the middle. So quadratic equation solves for the variable that's being squared, which in this case is r. Tell me what to write down. Plus minus square root 16 minus... 4ac, right? B squared minus 4ac under the radical. In fact that's how the authors of this supplement that you have, that's how they categorize the solutions based on what's happening with the B squared minus 4ac term. So if it's negative, which I think is going to be the case here, then we're into case 3 all over twice A. All right, what do we get under the radical? Negative 36. Negative 36. Negative means we're dealing with imaginary solutions or complex solutions. So 4 plus or minus square root of negative 36. 6i. And everything in the numerator can be reduced by 2, as well as the denominator so the end result is... So let's jump from here to the solution. I don't want to go to that one with this ugly stuff in the exponent position. We've already fought that battle. We've got another type of solution when we have two complex solutions that are conjugates of one another. What's that look like? 2 plus 3i? We could, but we had another version of that after two or three pages of squaring and cubing and grouping together like terms and putting in k1s and k2s. e to the alpha root. Well, we're going to get an alpha and a beta out of this solution, solutions to the characteristic equation. Alpha is 2, beta is 3. So what's the solution in terms of alpha and beta? e to the 2x. So e to the 2x, so e to the alpha x? In principle, c1 cosine of k1. Doesn't matter, c1 or k1 doesn't matter. 3x plus k2 sine of 3x. Does that look familiar? So it's e to the alpha times c1 cosine beta x plus c2 sine of beta x? Yes. You can use Caesar k's. I think we ended up with k's because we had some c1's and c2's and i's and we got rid of them and put in k's. So that was our case 3, right? Two complex roots. So you can go right to the characteristic equation, get the solutions, identify alpha and beta and go right to the solution. If we don't have any additional information, then we just leave it here. So I know it's supposed to be exponential type functions. That's what we started with when we were kind of building the characteristic equation. And then we found y prime and we found y double prime and so on. But we now have a way of converting something that's in this form into something that isn't just exponential in nature, but in a metric in nature. So that's why we end up with cosines and sines because of the equation that we had earlier in the class and before we went on break. So that's case 3. Two complex roots to the characteristic equation. So we're missing case 2. What was that? A double root. So real roots, but they are the same. R1 and R2 are the same. Let's see if I can find one of those real quickly. Characteristic equation? R squared minus 4r plus 1. Factored? R minus 2 minus 2. So R1 and R2 are the same. So this is case 2, double root to the characteristic equation. What did that solution type look like? C1 e to the 2x plus C2x. Look right. So that was case 2, double root. Both roots are real. So this is kind of the answer we would normally come up with if we had a C2 and we didn't put the x there. This term and this term would be the same. So they're redundant. You really don't need both of them. This actually has a chance of working and in fact does work because of the terms that are produced when you're finding y prime and y double prime because of the product rule. So because of how the product rule works, we can generate some more e to the 2x terms as well as x e to the 2x terms. They all have a chance of knocking each other out and you can end up with all this stuff on the left side dropping out and just having zero on the right side. All right, so there's the three cases. One of the things I noticed on the test, let's see what problem that was. This is one of the parts and this was problem 4. Yes, problem 4 had two parts. One where you approximated the solution using Euler's method and the other where you found the exact solution. So this was a separable differential equation. So I saw this error and if you see a big blue arrow in a capital no underlined with some exclamation marks after it, you're one of the guilty parties here and that's why I'm addressing this particular situation. If each problem is worth 16 points, actually this is part of the problem, so 4a was worth 8 points and 4b was worth 8 points. So 4b being worth 8 points, if I had my druthers, I would take off 30 points on the 8 point problem for this mistake. But I didn't do that, that would be unfair. But this is a mistake we need to, from this point forward, we need to avoid. So we want to separate. This wasn't the problem. We need to multiply by 1 over y squared. This is fine, everybody, that got to this point successfully. Now here's where the error came, where we integrated both sides. What you see on the left side never has been a natural odd. It's not today a natural odd and will never be a natural odd. For something to be a natural odd, we have to have that to the what power in the denominator, to the first power. This is to the second power. Has never been a natural odd, is not today a natural odd, and will never be a natural odd. So you can't develop this mindset anytime you see 1 over something, like 1 over secant, that's natural log secant, or 1 over Steve, well that's natural log of Steve. It doesn't work that way. The only time we have a natural log is when we have 1 over y, and we also have dy, 1 over u, and we also have du, 1 over t, and we also have dt. So when you see that y squared there, that takes it out of that situation. In fact, 1 over y squared is really y to the negative second, so some of you said to the negative first so that you were thinking correctly, if you saw negative first, and you also had derivative of that, then that's a natural odd, but this isn't. What is the integral of y to the negative second dy? Negative. It's a power rule, right? Don't we add 1 to the exponent? Divide by the new exponent. So if we add 1 to the exponent and divide by that new exponent, that's what we get, which drastically changes the nature of the solution from this point. If you add a natural log, you're going to exponentiate both sides. If you didn't have a natural log, which this is not, then you're not going to exponentiate both sides. So the solution is just completely different from this point. I think it's probably worth a couple of minutes to finish this one because of what some of you did. You did some flipping, which is fine, as long as you completely take the equation left side, flip it, right side, flip it. There could be a constant which will fuse into the right side. What's the integral of negative x dx? Negative x. And I'll put that constant there. That was a mistake that some people made. So there's our equation. We separated, we integrated properly on the left side. I hope. So why do the negative first over negative one is the same thing as some of you, which I would join in on that particular bandwagon. I would, seeing a negative here and a negative here in a constant, doesn't really matter. It's positive or negative. You could multiply through the whole equation by negative one. You don't have to. But if you do that, that's positive. This is now positive. And this is now negative, so you can just change it to another letter. It's an unknown constant. It's the negative of what we started with. You can still call it C if you want to. This is where if you which you don't really have to, to plug in the values and get a solution, but normally we like to solve things for y. So that's easily done on the left side by just flipping it. Now the right side, you can't just flip the two pieces. You got to flip the whole thing. So the right side is really this over one. Right? So we've got one over y. We've got a single fraction here. We've got this ugly numerator. Over one. Now if you flip both sides, you've got that. y over one is just y. And I think we know a point, right? That this goes through. Don't we have a starting point in problem four? Yes. When x is zero, y is one. So that'll help you find b. I think b in this problem is one. So if you plug in zero for x and one for y, it turns out that b is one. So you have this equation. And we want to know what is the y value. And in part a, we did the Weiler's method and we incrementally went by what? A half each time? No. A quarter? What was it? Went from zero to two, right? Our increment was a half each time. Our delta x. So when x is two, what is the y value? So there's a two plus one, which is a third, which ought to fit in with part a in the problem if you did that properly, which most of you did. You got point three zero for the approximation, which seems reasonable here that the exact value is a third. We approximated it using the tangent plane to the curve and we got point three zero. So seems like it's reasonable. So when you flip one side, you have to flip the entire other side. You can't just flip each piece. All right, let's get one of these and then we'll kind of see what it is that we're going to encounter in the next section and we'll kind of dive into the nuts and bolts of that tomorrow. That's a different look, right? Second derivative, first derivative, original y thing. So we can go directly to the characteristic equation, which is factor? Probably not. So what? Four plus or minus four minus four times one times five. Isn't it two? Oh, two, negative B. Thank you. Just testing you to see if you're awake here after the vacation. I guess I'm not a wing. So B squared minus four A, C, all over two A. Under the radical, we've got a four minus 20, negative 16, two plus or minus four I, all over two. So the final solution of the characteristic equation, so alpha's one, beta is two. We have y's and x's so we do want y in terms of x. Tell me what the solution is and then we'll continue with additional information. E to the x. E to the one x. C1 cosine 2x. C2 sine two. Okay. Got agreement on that? So we've got this information. Let's put it to use which will probably allow us to solve for C1 and C2. When x is zero, y is zero let's see what happens. So does anything disappear here? Sine of two zero would be sine of zero which is zero so that eliminates C2. Cosine of two zero is cosine of zero which is one. So we have C1 times one C2 times zero and what's E to the zero? Is one. So C2 times zero that's gone. Oh this is kind of good this hasn't happened on a problem yet but obviously it happens from time to time because it happened in this one. So we end up with C1 equals zero. So what's that say? Go ahead. Cosine that whole piece cancels out. That's right. There isn't one. We allowed for the possibility of C1 cosine two x but apparently in this particular problem C1 is zero and that term doesn't appear in the final answer. So C1 is zero. So at this point our solution normally we would have a C1 cosine two x here C1 is zero. So we just have what? C2 sine two x? Is that right? We might have the time today to do something. This isn't that different from the original solution. I mean it looks like there's a term that's gone. It's going to be drastically different. It's not very different from the solution that we had earlier. Alright so we found C1 it's zero. That term's gone. Now we need to find C2 we've got additional information and that is up here when x is zero the derivative is one. So now we need to take this equation right here and take the derivative. That's a little bit easier now that C1 was zero. So it's gone from the equation. What's the derivative? First times derivative of second. Derivative sine of two x is two times two cosine two x. Is that right? Derivative of sine of u is cosine of u du. So there's the du. So there's first times derivative of second plus second times derivative of first. Derivative of e to the x is e to the x. So there's a product rule made a little bit simpler because C1 was zero. So what is supposed to be the case when x is zero dy over dx is one. So for y prime we're going to put in one everywhere we see an x we're going to put in zero. What about that second term? It's the gone. Right? Because sine of zero is zero and it's multiplied by two other things so that's all zero. E to the zero is one. Cosine of zero is one. So we've got one times C2 times two times one, which is what? Two C2 C2 is one half. So our final answer e to the one x, e to the x C2 which is one half sine of two x. Let's analyze this just slightly and then I want to adapt what could potentially be a solution and then see what you think the graph would look like. So if we were to ignore this e to the x what would this thing look like? One half sine of two x. What's the amplitude? Right. How far up down it goes from its axis? Be a half, right? Amplitude is a half and the period is it's affected inversely by the two so it's half of what it normally is. So normally the period of sine is two pi, now the period is pi. So we've got this oscillatory function, if I ignore the e to the x, we've got this function that it's a sine so it goes up to a half down to negative half, up to a half down to negative half and it goes through that full cycle every pi radiance because of the coefficient two. Now what would that do to it? Can't we kind of loop that together in a way with this one half and say that's now the amplitude, right? e to the x each time is going to be a number as x gets larger, e to the x gets larger so this would have a varying increasing amplitude, right? As x gets larger e to the x gets larger so if we had an increasing amplitude this is kind of unrealistic that's why I want to change the example so our first amplitude is a half and then the next time maybe we're down here and the next time we're up here. I think if you were designing a bridge this would be a problem if you were a bridge designer and the amplitude as the force kind of increases on the bridge the bridge begins to kind of move a little bit and then as time goes by it begins to move even further, not a good bridge right? That bridge in fact will not be a bridge it's pretty soon down the road here so you want what to happen with vibrations like this. You would want a lead coefficient instead of being e to the x you would want it to be e to the negative x right? If you had a lead coefficient what would that do to the oscillations or the vibrations? Wouldn't they get smaller as x gets larger? Doesn't this get smaller as x gets larger? So if your first one is up here your next one is here the period stays the same by the way this is why when the battery watch goes down it doesn't do a bad job keeping time it still keeps time it just kind of gets to the point where it doesn't move things anymore so the period stays the same that's what we would like to happen with vibrations for them eventually to diminish where they dissipate and they're gone we don't want this to happen this is trouble this is expected so if we had something out in front that would cause the amplitude to diminish that would be a good thing here's the amplitude there it is there it is and so on so it's getting less as we go let's finish up today with this and we can just start completely new in 7.8 tomorrow with the non-homogeneous let's say we had an equation let me try to keep this pretty simple so we've got some sines added to some cosines and of course now we've got some coefficients out in front but let's say that coefficient is zero or is one if it's zero then we can just go home now because it's not very interesting but if it's one now sine plus cosine I don't know if you've ever used this technique but sometimes it comes in handy let's split this up into two pieces one piece is the sine the other piece is the cosine because we know each of those pretty quickly and then we'll just add them up on the graph so let's take it through one cycle and hopefully we'll see enough of it to at least say that it's believable so let's do the sine of x sine of zero is zero pi over two is one pi we're back here to zero three pi over two alright let's do cosine cosine of zero is one then it's at pi over two it's down to zero at pi it's negative one then back to zero and then here now we've got the sum of these two can't we just add them up graphically at individual values the answer to that is yes we can so at x equals zero let's add them up where are the two individual pieces one and zero sum is one zero plus one added together is one let's actually do this point right here at pi over four the sine of pi over four is square root of two over two the cosine of pi over four that was kind of funny the cosine of pi over four is even funnier it's also square root of two over two what do you get when you add half the square root of two to half the square root of two you get the square root of two so it's this distance added to itself which would be roughly up there alright let's move on to pi over two we're adding zero to one which is one hope my graph's decent here I don't know here we're adding positive square root of two to negative square root of two what's there sum would be zero, one's positive one's negative so if you add them together you get zero zero plus negative one is one here's negative square root of two over two added to itself which is negative square root of two here's zero added to one positive square root of two over two negative square root of two over two we'll add those together and zero added to one So the black marks with the X on them should be sine plus cosine, I just added them up graphically. So we added two functions that individually are oscillatory functions. Is there some also an oscillatory function? Is it oscillating with the same kind of peaks and coming the same period each time we go forward to the right? It is also oscillatory. So it's not that different from a sine or a cosine. In fact, we don't have time for this today, but we'll get an equation tomorrow of what sine plus cosine actually looks like. What's the period by the way? Does it start to repeat itself in exactly the same space that the sine and cosine repeated themselves? Sure, it kind of almost has to, right? Aren't we repeating the same values here that we were here? So we're going to start the same cycle all over again. So our final graph is oscillatory also, just like the two components of it. So if you eliminate one of the components, it doesn't make it that much different than if you had both of them present. So we'll do slightly, slightly a little bit more with this, but that'll at least make it believable that we can eliminate C1 or C2 and get the same answer. See you tomorrow.