 We are at lecture 30 today probably makes pretty good sense because this course when it's all said and done will be Somewhere in the low 60s as far as lectures or DVD type lessons So we are I think exactly at the halfway point of the semester with spring break pending So if we look a little sparse today in the classroom, I guess they just didn't read their Information properly and they thought that spring break started this morning But it actually starts this evening at the end of all the classes. So I'm sure that they would have been here But it's a clarification error So we will kind of finish up 7.7 today, I think It's a battle to get this complex Roots situation covered because some of the ammunition we need we haven't had yet So we have to kind of jump ahead and pick a couple of things out of chapter 8 and use them There's a chance you've used them before in another math or math related class But I want to take a look at them Kind of validate that they do in fact work and then use them in this particular case So here's where we are in Terms of the cases that we've covered We didn't really handle it this way But if the value under the discriminant is positive The square root of b squared minus 4ac you actually get two distinct real roots So they're real and distinct So we know what the solution looks like it looks something like this. We've handled that we've done examples with that We've even done a boundary condition problem with that. So that's first case Get the characteristic equation get the roots go directly to the solution and then find the C values if in fact we have additional information Case two is where the value under the discriminant is zero. So you don't get two roots you get a double root and So the common value for r1 and r2 it is the same So we kind of upgrade the solution a little bit and we Validated that something like that has a chance of working because of the nature of the product rule in the terms that you create When you look at the product rule. So here's our third case So we've got a negative value under the discriminant If we have real coefficients in our characteristic equation, which will always be true for this class We will have two roots that are Complex conjugates of one another So there's the nature of our two roots that we're going to deal with today So there's case three now, there's some more to that but it kind of goes jumps directly to the answer and That's too big of a leap as far as I'm concerned at this point in time So we're still in seven point seven which is in the supplement to the text We're looking at the third case of second-order linear Homogeneous differential equations. So the when I say complex roots Complex roots to the characteristic equation. So there's the nature of our two roots We've got a real part and an imaginary part you should expect something to be very different from Real numbers because these numbers are Not real at all and they're very strange when you start dealing with square roots of negative numbers especially in the exponent position because the solution Is going to look something like this and since they are distinct we can allow for The two roots in this fashion as opposed to the double root So look at the ugly thing that's going to occupy R1 and The equally ugly thing that's going to occupy the position of R2. So it turns out that it Looks like an exponential type function because we've got the base e but when all is said and done and all the simplification is done, we're going to turn out with a Trigonometric result. So that's our goal today is to take this and Simplify it in some way shape or form. So we'll come back to this But there's a couple things I want us to do that actually kind of rate a couple topics from Chapter 8 Get them enough not to get full understanding of them, but just enough to use them so that we can simplify this So here's a couple things that possibly you've seen before in another math class Or math related class that we now need to use these are Taylor series have you dealt with Taylor series before and other math classes or Maclaurin series power series that Express what e to the x is in terms of the thing that's occupying the x position So it's 1 plus x plus 1 squared over 2 factorial x squared over 2 factorial x cubed over 3 factorial and so on and this is an infinite series But you can see that it would converge if let's just take an example Let's say we want e to the I Don't know keep it simple. We could do e to the first or we could do e squared Let's do e to the first and you can see Please make this series believable. This is how your calculator operates by the way your calculator doesn't know how to do e to the x problems, but it converts this to Probably a 11th or 13th Taylor series and it computes based on what you're trying to find E to a certain power so if this is true This is e to the x then everywhere. I see an x out of be able to replace it with a 1. Is that correct? This is a series for e to the x. We now want e to the 1 so x is equal to 1 so 1 plus here's an x That'd be a 1 We'd have 1 squared over 2 factorial 1 cubed over 3 factorial 1 to the 4th over 4 factorial. Let's get one more 1 to the 5th Over 5 factorial so it's a really easy series To write out so if we wanted to e cubed everywhere we put a 1 we'd put a 3 But we know what e is approximately so it's probably makes sense for our first validation to put e to the 1 So I'll put a question mark there since we're trying to validate it so the 1 plus the 1 is 2 well we're On our way 1 over 2 factorial is 1 half. That's pretty darn close already, right? We're not very far along in the series 1 over 3 factorial. What's 3 factorial? 1 over 6 1 over 4 factorial 1 over 24 And 1 over 5 factorial 5 factorials 5 times 4 times 3 times 2 times 1 Okay, sorry. I didn't see that look on anybody's face that you know, what in the world's a factorial but Got that clarified that's 5 4 3 2 1 120. Is that right? Somebody that has their calculator out and working what's 2 plus a half plus a sixth plus 1 over 24 plus 1 over 120 Of course to get e to the first we need to let this thing run indefinitely, right? See how close we are at this point 2.716 1 6. Okay, so we're not there yet So you can envision the rest of these terms being added in and this is a legitimate Power series progression for e to the x So why do we need that? We have e to some ugly power We want to be able to write it out almost in polynomial type form So this is powers of x x squared x cubed So it's kind of an infinite termed polynomial if that makes any sense We're also going to need in The simplification. I know it doesn't look a whole lot like simplification yet I think it's going to get worse before it gets better But we're also going to need this same kind of a series expansion for the sine function Because we're going to actually end up with the sine of something and we're also going to end up with the cosine of something One way to remember this is that the sine is an odd function Not like strange or mysterious odd because the f of Negative x is the negative of the f of x Notice the powers they're all odd and notice the factorials. They're all odd So that's a handy way to remember this. Let's just try something here. Let's try the sine of pi over 6 So if this particular series works for the sine of something again This is how your calculator finds the sine of something. It doesn't know the sine function So if this is the sine of x everywhere I see an x I should be able to replace it with a pi over 6 Now let's say we stop there. We're not very far along in this infinite series, right? We got the first term Are we reasonably close? What is the sine of pi over 6? That's a fair question for a Friday When yeah, what is the number pi divided by 6? Somebody take the number pi and divide it by 6 Point five two three six We're doing pretty well already aren't we this is a half and we're at point five two three six Here's how these things plot along which will explore these in depth in chapter 8 We've got too much right? We need a half. We've got point five two three six So we're going to subtract some away. We're going to subtract in this case pi over six cubed over three factorial Guess what's going to happen? We're going to subtract away too much So what are we going to have to do we're going to have to add some back in what's going to happen? We're going to add too much back in what's the next term. It's subtracted. Oh, we subtract too much So that's how it goes it kind of converges eventually on this value point five. So let's get a Couple of terms here and see what we get Let's do these and see how close we come to the value of One half, which is what the sine of pi over six is supposed to be So what's pi over six the quantity cubed all over three factorial anybody? Didn't we subtract away a little too much right? We needed to subtract out point oh two three six But aren't we getting closer to a half? Which is what we're supposed to get closer to now. We're under one half. We have to add something back in What's pi over six raised to the fifth power? Divided by five factorial. It's not going to be very large Negative fourth So three zeros three two eight So if we were to add these up knowing that we're not even close to the entire infinite series But we've picked off the first three terms. That's pretty close right to the value of one half This thing minus point oh two three nine and add this back in. What is it point five zero zero zero two eight? Is that right? Zero zero two one three okay pretty close to one half So if we kept going we'd get each term we add in and subtract add and subtract We get closer and closer and closer to this value that we know is the sine of pi over six Doesn't prove it by the way. We're not proving these but we're at least Seeing if they're plausible see if they're believable so it seems to get us what we need to get for the sine of pi over six I don't know that we need to take our time to do the cosine as well Because it works just the same way The more terms you pick off the closer you get to the actual value that you want Notice cosine is this is an odd function. We've got odd powers and odd factorials cosine is an Even function right because the f of x is equal to the f of negative x which makes it even all the powers are even all The factorials are even so that's those are handy Memory devices to remember the power series. You don't have to remember them now that go along with sine and cosine All right, let's use this one first we've got e to a power so we want to Go back to our complex solutions to our characteristic equation and then eventually we're going to use all three of these But let's use this one first so our solution at this point just plugging in r1 By the way, if you just plug in the complex roots Then you don't really have the answer because you don't have the function what it really looks like So this is why we have to Do what we're doing here Since I've got a sun I could distribute the x so x times the alpha and x times the beta times i then I Could take what is a sum in the exponent position and kind of work backward algebraically Into a product is that correct is that equivalent to the one that's above it? We've done that enough. I don't think I need to break that up into a separate step Same thing here. I could distribute the x x times the alpha that's here and x times the negative beta times i So that's going to be e to the alpha x times e to the negative i times beta x Anything gained by doing that right there you can combine some of the okay, what what do both terms have we've got two things added together Do they both have something either they both have e to the alpha x right? So by taking that sum in the exponent position Translating it backward algebraically to the product of two things with like bases We're able to now not that it's anything major, but it's Ugly enough, so let's try to farm some of that stuff out in front if we can I'm going to call this i times negative beta x is that okay negative i beta x So here's what we off to the side we have to get Something that's going to substitute for i times Something in this case that something is beta x We'll pick the problem up back here when we get what we need to simplify this So let's say we have something e to the i Theta just so we don't have a beta and an x there yet, so we have something occupying that position According to our power series For e to the x whatever is occupying that position where x is we can put it here Put it here and put it wherever x is in that infinite Series that infinite power series So for e to the i theta it ought to be one Plus now normally it's x So now it's going to be i theta Then it's normally x squared over two factorial, so it's going to be i theta squared over two factorial i theta cubed Over three factorial, so we're using that e to the x expansion Everywhere there's an x we're putting in a an i theta Let's get one more This by the way is in the supplement they kind of make the leap to Here's what e to the i theta is but it was my guess and I think it's true that we We don't have all the background that's necessary for this, so we're trying to fill in a little bit of the gaps here So we're going to square the i. What do you get when you square i? Negative one So it's going to be negative theta squared over two factorial Next term what do you get when you take i and cube it? What's i cubed? negative i and i to the fourth what's i to the fourth is One and then things are going to start to repeat right? Because then we're going to have an i to the fifth and i to the fifth is back to i Right, so am I being redundant here? Is this something you've done in? Another math class or math related you've done this anybody you've done it, okay? But it looks like several of you haven't done this I don't I don't like mystery formulas to all the sudden appear So let's just a couple steps and we'll be able to validate this The grouping that's going to happen from this point forward is we have some terms that have i in them Every other term will have an i in it. Is that correct? And then we have terms that don't have an i in it So let's group together the terms that don't have i in them So there's one actually it is one minus theta squared over two factorial plus theta to the fourth Over four factorial is that enough of a pattern to know what's the next term is going to be and what it's s i g n is going to be Next term although we don't have it would be what? negative Theta to the sixth over six factorial and so on right we're going to have a bunch of those Every term that I've underlined Has an i in it. So let's factor the i out in front and if we do that we get a theta Then we get a theta cube sorry a minus theta cubed Over three factorial. We don't have the next term. Is that enough to figure out what the pattern is and what the next term would be? It's going to be plus Theta to the fifth over Five factorial the next term is going to be minus and it would go on forever now We need to does that look familiar at all to something I had up here five minutes ago One minus something squared over two factorial plus something to the fourth over four factorial Is that cosine? Here's here's what I had up here a while ago Does that look like what we have the terms? Excuse me that did not have an i One minus something squared over two factorial plus something to the fourth over four factorial isn't that exactly what we have? So this thing we can call the cosine of theta right because on the green paper It was cosine of x the variable was an x in the expansion. So that's a cosine of theta How about this thing that is multiplied by i does that look familiar? That should be our sine Taylor series expansion or power series expansion Sine of x is x minus x cubed over three factorial x to the fifth over five factorial and so on We don't have x's we have Theta's So Theta minus Theta cubed over three factorial that should be Sine of Theta is that okay? So those are the three things that we haven't really had in this book yet that that we need to make that leap So that's what e to the i Theta is we have e to the i Something else so everywhere. We see a Theta. We're going to have to put in a beta x and over here every time We see a Theta. We're going to have to put in a negative beta x So here's where we are. I'll put the beta x in parentheses And we are going to use That that we just kind of developed in a way So e to the i beta x I want to put a beta x everywhere. There's a Theta So that's going to be what cosine said a cosine Theta. It's going to be cosine beta x plus i sine Beta x I know you don't think it's getting better But it is getting a little bit better because we'll get be able to consolidate some terms So we made a substitution for e to the i beta x and we plug beta x in here in here That's what we've done. We need to do a similar Thing over here. We've got i times something. What's the something it's negative beta x So where there's a Theta in this expansion. We need to put in a negative beta x parenthesis So it's probably a good thing We were able to get e to the alpha x out in front because it's already a mess in there We don't need to make it any messier Let's see if we can distribute and combine like terms all in one step So we're going to have a c1 that gets sent to both of these We're going to have a c2 that gets sent to both of these We need to do one more thing before we do that Cosine beta x and cosine negative beta x are they the same thing is the cosine of something Exactly the same as the cosine Of the negative of that That's true right because it's an even function so because cosine is even Cosine beta x and cosine negative beta x are the same if they're the same I'm going to get rid of this and Just call it cosine of beta x is The sine of beta x is it the same thing as the sine of negative beta x It's the negative of it. Why because the sine is an odd function, right? So sine of negative beta x is the same thing as negative sine of Beta x So we'll get rid of that. So now everything is in terms of beta x So we're going to have a and you're not responsible for this Reproducing this development. Isn't that nice news today? That's the best news yet That now it gets a little emotion out of you So c1 is going to be distributed to this and c2 is going to be distributed to this So don't we have c1 plus c2 Cosine beta x is that correct c1 is going to be sent to this C2 is going to be sent to that. So if we factor out cosine beta x we'd have c1 plus c2 Well, these are constants. So the sum of two constants is just another constant. So we'll be able to simplify that C1 is going to also be sent to this and C2 is also going to be sent to this so we've got some sine of beta x's and Here we've got negative which let's associate that with the c2 So we've got some more sine of beta x's. How many sine of beta x's do we have? C1 minus c2 That's how many sine of Beta x's we're going to have And that takes care of everything done it didn't we use all the terms c1 they got sent to both of these terms C2 they got sent to both of these terms This is not much of a stretch at all to say that some arbitrary constant added to another arbitrary constant is a constant So we'll call that k1. This is a little bit of a stretch So we have to allow for our constants in this particular problem. I'm sorry. I forgot something didn't I? This had this had an I and this had an I right so there should be an I here In that second term So if we take c1 minus c2, that's not a stretch. That's a constant But that constant is multiplied by I if we allow for our k values to be Non real numbers then we can go ahead and say for c1 minus c2 times I We're going to call that k2. We're just opening it up and saying it's not just some normal number It might in fact be complex so we have taken a Solution that looked very exponential Because we started with e to some ugly power and e to some other ugly power and we've converted that into something that's very trigonometric and You'll see when we get to applied problems why we want a solution that is Oscillatory which this is an oscillatory function That can be affected by this coefficient out in front because something is going to have a Path or a description of its motion that is Oscillating so we want the model to match what the motion is actually going to look like so if we have two complex roots that are in the form of Alpha plus beta I In alpha minus beta I There is what our solution is going to look like the problems are going to be very easy now that we've done this battle We've identified what alpha is it's very clearly identified in the solution We've identified what we do with beta once we identify beta beta is very clearly identified in our solution So back to this sheet and let me unfold what the author says our solution should look like in case 3 So if there's a root alpha plus Beta times I and alpha minus beta times I Here's what our solution is going to look like e to the alpha x. We got that right they have c1 We had k1 same kind of deal Cosine beta x plus c2 sine beta x so it is the sum of sines and cosines And you'll see that we're going to need an oscillatory model and this is exactly what we want in the third case So let's do our first example of this type and then we'll add some boundary conditions time permitting Which I think we're going to have time to do this all the way through Let's go from the second order linear homogeneous equation To the characteristic equation that skips a couple steps What is the characteristic equation that goes along with this? That 41 is not looking too good is it we're not liking the 41 But let's see what happens Probably not going to factor. I think we can concede that fact because 41 is kind of stubborn So let's see what the quadratic formula is going to yield We know what it's going to yield because it's an example of this type So it's going to have a negative under the radical. It's not going to be a surprise. Let's just validate it negative B plus or minus B squared Minus four times a is one times c is 41 all over twice a So we've got a hundred four times one times 41 is what 164 So we've got a square root of negative 64 What are we going to call the square root of negative 64? Let's call it Steve now Ed Ed Let's call it Ed now. What would we call it? Eight I would be a better name and the fact that everything is divided by two Let me tell you what I see occasionally from intelligent young people such as yourselves somebody will make a just a radical Departure from reality and reduce ten and the two Okay, what's wrong with that? Can I just reduce the ten and call it a five and reduce the two and call it a one? So the answer is five plus or minus eight. I I'm not going to write it down because it's not correct How can I reduce numerator and denominator? It's got to be a factor of the numerator, right the entire numerator. Don't we have two terms in the numerator? So in order to reduce a two in the numerator and the denominator I've got to have it as a factor not just one of them. It's got to be a factor of both of them And now that it's a factor of the numerator and a factor of the denominator We have our two solutions. I know nobody's going to admit that you have ever actually done that Reduced the ten and the two but I've seen it enough in the 30-plus years. I've been teaching mathematics, too I almost believed that it's true But then reality kicks in and I know it's not true So we have alpha equals five Beta equals and we are always in our problems going to have complex conjugates of one another Because the coefficients of our characteristic equation will be real and that guarantees this kind of solution Beta is four you could say beta is four or negative four. It's not going to affect the solution So let's choose the easier of the two. So there's what our solution is supposed to look like in this particular category So we know what to sub in alpha and beta e to the 5x So since we've done the battle and come up with this It makes the getting from our solutions complex solutions to the characteristic equation Getting to a final solution a pretty easy process now. There is some memorization So you're probably going to have to memorize this But it's worth memorizing as opposed to trying to develop it. I think we'd get a hundred percent agreement on that Some things it's not that big a deal You know if you don't remember is it one plus tangent squared that secant squared not that big of a deal because you can take sign squared plus cosine squared equals one and 12 seconds later you can have the tangent squared secant squared identity But I think it took more than 12 seconds to develop this. So that's probably one you want to remember Let's add to this problem Some initial conditions or some boundary conditions. Let's say that we also know or we want y of zero to be one and The y of pi to be two So you know what's going to Result from this we're going to be able to solve more than likely for k one and k two So we'll have a solution as opposed to a family of solutions So y of zero equals one So x equals zero y equals one, right? So one equals e to the five zero. I never did address the test I didn't have any grading time yesterday. I am taking over a class for a teacher who's going on family medical leave So I kind of was bringing myself up to speed to take over that class for her so I'll be doing that the rest of the semester, so No grading time yesterday. I apologize for that. I do want to grade them all before I decide what to do anyway as far as just Scaling it by a certain amount of points or throwing out everybody's worst problem. So I Knew I wasn't going to have them all graded So e to the five zero is e to the zero, which is one Cosine of four times zero. That's the cosine of zero, which is one and sine of four times zero is sine of zero, which is zero we go ahead and Eat of the five zero is e to the zero, which is one so we end up with a solution k one is one so we also know that Y of pi equals two so when x is pi Y is two K one we already know so I just put a one in there not good That's what I was thinking about when I was writing down the other one I'm not liking the way the second piece of this is going to work out What happens? Why is it not good? Okay, we've got sine of four pi which is zero so we lose our K2 so we don't know what our K2 is right So I've got flawed data here. Let me see if I can find that where I got this Maybe I got it from the list of problems at the end. Maybe I can make the correction. I'm not seeing it here In this list of problems But we're not going to get a solution with this right I've got Y of Pi Well, I wrote that down wrong. Maybe it's Y prime of pi Is equal to two? Let me see what that would correct things Yeah, I'll get it corrected and we'll look at the boundary value problem But that's not going to get us that needs to change to something else because we lose our K2 and that's What we're looking for out of this equation. So something happened. I wrote down the wrong Conditions All right, I'll get that corrected and we'll look at that At another time we have one more boundary condition problem to look at and Then we'll be right where we need to be In this section so let's suppose we had this problem that bothers me that I wrote that down like that That's not good. This Doesn't necessarily have to fit in the same category. We just finished it could fit in any category since we're now done with all three cases so the characteristic equation is Can you go through the mechanism? Say that again, please. Can you go through the mechanism for that the how we get from here to here? Yeah, okay Let me get this part done. We'll get our two R values and then I'll kind of go backward That's right you weren't It's okay to revisit that but I'm going to go a little quicker since Our square. Thank you. That's what she said, but I didn't write that down. So we've got a plus four and a plus four Isn't that what you said? Yeah At this point in time in this class today, I probably trust you more than I trust myself with values since I've got a flawed Conditions in that one problem So we have a double root, okay? Jacob wanted to know where how we get from here to here. So this is going to be quick But we know that something some original function plus its derivative Times the number plus its second derivative possibly times the number a Function that has a chance of working there is the exponential function And we could have a C in front of it, but I'll forgo that just to make this a little bit simpler So since this is the nature of what y might eventually be We need to be able to plug in its derivative and its second derivative. So that are in the next one to be our square Okay, good, and then you can factor out an e to the rx, which is never equal to zero Therefore, this is the only realistic part that has a chance of being zero All right double root. So let's jump to the solution. This is case two when we have a double root What does the solution look like? That work we will see similar Situations when we have to kind of keep going with this Philosophy of adding an x till we get a solution that actually works Hopefully my data here is good So let's see Y is two and Y prime is one. I bet you that's where my mistake was on the other problem So here's some initial conditions or boundary conditions so when x is zero y is two and When x is zero Y prime is one. So we do need to take the derivative. Unfortunately, it's got a product rule, but it should be fairly quick So when x is zero y is two So that term is zero We've got e to the negative four zero which is e to the zero which is one so C One is two Y prime. All right, here's our y value. So we need our y prime. What's the derivative of C1 e to the negative 4x? 4 Negative 4 C1 Okay, and now we have a product rule first times derivative of second plus second times derivative of first and And we want when x is zero we want y prime to be one. So y prime is one So there's a zero as a factor. So that term is gone Negative four. Don't we already know that C1 is two. So let's plug that in e to the zero is one That whole term is gone. So we've got C2 e to the zero. So that's just C2 So we should be able to solve for C2 So it looks like C2 is nine So our final solution C1 is two C2 is nine So there's our final answer to that with the given boundary conditions. Have a wonderful spring break. I will see you in a little over a week