 Welcome back, lecture 22, math 241. We are continuing with separable differential equations. If I can say it, that would be good. What we're actually going to do today in class. It's our second day with those. This section is a lot like sections at this point in the book where there's a technique that they talk about solving differential equations if they're separable. Not all of them are separable. They don't have to get real complicated to not be separable. But these are, and then the different special situations, one of which we'll cover today, orthogonal trajectories. Does the word orthogonal ring a bell? What things are orthogonal? What are they? They are perpendicular. So we're going to talk about curves in the plane being mutually perpendicular to one another. And that may sound a little odd since they're curves. How do you get curves to be perpendicular to curves? But there is such a thing. And it does involve obviously separable differential equations because that's where we are. And then the last section, which we're not going to do today, they call it, in this book, mixing problems. Some texts call it change in dilution of a liquid type problem. Or I just call them tank problems because you've got something in some kind of a tank and you're going to add some more bad stuff or maybe some more water to dilute it. And then we'll see after a certain point in time when the dilution in the liquid changes to a point where maybe it's safe or it's drinkable or it's something else, not toxic any longer. So let's continue with separable differential equations. When we have a differential equation that is separable, what is our first step? We did a couple of these yesterday or Wednesday. OK, so we want to separate the x's and the dx's or whatever the letters are in the problem. We'll look at some examples and aren't x's and y's, but maybe it's l's and dl's from u's and du's. So whatever the letters are in the problem. What's the second thing we did to solve a separable differential equation? Integrate both sides. And it might involve different techniques, integration by parts. It might involve the use of a table. But we do want to get hopefully some type of function of y in terms of x. Now, I don't think we did this, but I want to do some examples today that are either a little difficult to solve for y or, in fact, it might be impossible to solve for y, but if we can. So if it's possible, we want to solve for y. So I've got some examples here that are a little bit tougher than what we did the other day. So let's go on with them. So we'll start out kind of tame here with this one. This is one of the problems in the book. It's not one of the examples in the section. Again, if you want good examples that kind of show every step, check the examples in the book. So we could multiply, let me rewrite here, multiply both sides by dx. It kind of tells us where things are going to end up. So now the x's are going to end up on the right. The y's are going to end up on the left. So the square root of x is OK. We need to do what? To get the y's and the dy's separate. Multiply by e to the y. So we have e to the y dy equals the square root of x dx. Next step. Integrate. Integrate both sides. What's the integral of e to the y dy? E to the y. E to the y plus the possibility of a constant which will roll to the other side of the equation. Integral of square root of x integrated with respect to x. x to the 3 halves. Good. 2 thirds x to the 3 halves or x to the 3 halves over 3 halves then division by 3 halves is multiplication by 2 thirds. Put the two constants together into one. If you forget the plus c at this stage, you're not going to be able to successfully retrieve it at the end of the problem. More than likely. Because you can't just, when you're done, say, I forgot the arbitrary constant. I'll just put a plus c at the end of the problem. It's probably not going to be exactly where you want it to be. And in fact, on this problem, if we try to solve this for y, what would we have to do to solve for y? Natural log of both sides. So the natural log of e to the y is y. So we're going to take the natural log of both sides. Natural log of e to the y is y. So we've got the natural log of 2 thirds x to the 3 halves plus c. So there's just a real good example of why you can at the end of the problem say, oh, I forgot my plus c. Let me just tack it on because it's not going to be where we want it to be. More than likely. Because it's inside the parentheses, we have to take the natural log of everything inside the parentheses. Could we distribute the natural log to both of those pieces? OK. I'm glad that you're hesitant there. We cannot distribute the natural log. The natural log of a sum is not the sum of the natural logs. What is the sum of the natural logs? The natural log of a product, right? Is the sum of the natural log. So as tempting as that might look, to distribute that, you cannot distribute that. All right, so we solved it for y. So there's the differential equation that we started with. Here is a family of curves in the plane that are represented by that particular differential equation. If we add more information, which we're going to do those as well today, if we have an initial condition or a certain point that we want to make sure it contains, we can do more. But with this problem, that's the end. Any questions on what we did there? All right, I'm just writing it the way it was written in the book. And it may be, in fact, a web assign question. I don't know. Because a lot of these questions from the book have been coded into web assign. So what do you think if this is our differential equation? It does have y's in it. It does have the derivative of y in it. So it is one of these. y prime is really dy over dx. So that might be helpful since we're separating these things. So we could multiply both sides by dx. And that'll do it, right? We've got the y's and the dy's separate from the x's and the dx's. So it's OK to have numbers added to y's, or in this case, numbers added to the tangent of y. And here, numbers added to x, or in this case, x squared. What really creates a problem is when we have x's added to y's. Because then it becomes difficult to separate them without addition and subtraction. Well, we don't want addition and subtraction of things. We want multiplication and division so we can undo those with the opposite operation. So we've got 1 plus the tangent of y dy. So we're separated. I see one term that's going to be a little stubborn. Trying to integrate the tangent. So let's just talk through these and see if we need to write something before we attempt to integrate all the way through. 1 integrated with respect to y, that's going to be easy. Tangent integrated with respect to y, what is it that has tangent for its derivative? I don't think we have that right at our fingertips right now, so we might have to work with that. I don't know, does somebody have that right at your fingertips? We'll kind of do that off to the side. x squared integrated with respect to x, easy. 1 integrated with respect to x, easy. So let's take this kind of stubborn looking term. Any suggestions or recommendations for how we could find something that has tangent for its derivative? Sine over cosine. Good, OK. Now if that wasn't something you were thinking about, do you see why changing tangent to sine over cosine might get us somewhere? Because they are kind of, in a sense, the derivatives of one another, right? So what would be the take on the next step? U-substitution. OK, u-substitution let u equal? Either one. Yes. Probably if one is the derivative of the other, I mean, think about do we want du over u, or do we want u over du? I don't think we want that one. I don't think we've ever encountered du in the denominator, in this case where we're trying to do an integration problem. So I think we're headed for that. So if that's the case, let u equal cosine y. We've done this problem before, but it's not like a real common antiderivative. Derivative of cosine is negative sine. So it looks like we need a negative one and another negative one, right? One inside, one outside. So when the negative one is in the numerator, this becomes du. This is u. And what is the integral of du over u? Natural. Good. So we've got the negative one that was out in front. And then the stuff in the integrand is du over u. So that's natural log u. And u happens to be cosine of y. Now, if you don't like negative one, natural log of cosine of y, there's another version that is 100% equivalent to that that maybe that might ring about. Natural log of secant, because this negative one could travel back up to the exponent position, right? And cosine of y to the negative first would be secant. So either one of these would work. So we've seen it before, but we don't see it very often, so we kind of have to do that battle again. But I think we're prepared for the battle. So let's go back to this stage of the problem. So the integral of one with respect to y is y, the integral of tangent y with respect to y. I don't care which one we use. That's the one we actually came up with. There's a possibility of a constant here. We'll send it to the right side and put it with the constant over there. Integral of x squared with respect to x, integral of one with respect to x, and then we've got a plus c. I think I want to solve that for y. How fun does that sound? Does that sound fun? Let's not solve that for y. I don't even know that it can be solved for y. So if it's a mass like this, at the end where we've got y's and y squareds and polynomial type expressions, and then some transcendental type stuff with natural logs and secants and cosines, then we'll consider that problem done. At least we have a solution or a family of solutions to the differential equation that we started with. So solve for y is a good recommendation, but it's not gonna work for us on every problem. All right, let's see what this one looks like. This has u's and t's, and we've got a little bit of extra work at the end of the problem. What is this given to us for? What's the purpose? To find the c or the k or whatever we have at the end of the problem, right? So we're not gonna have a family of curves. We're gonna have a single curve that goes through the point basically zero negative five. When you see a problem like this, where the d u's and the u's are opposite one another and the d t's and t's are opposite one another, you could just in a sense cross multiply, right? That oughta work. So we're gonna have two u, d u, two t. So that separates the u's and the d u's from the t's and the d t's. So we'll integrate. And remember we had d u over d t, so we'll probably want to function for u in terms of t when we're finished, since we've got new letters. The integral of two u, d u is? E squared. E squared plus the possibility of a constant which we'll send to the right side. The integral of two t with respect to t. T squared. T squared, the integral of secant squared. Tangent. So what has secant squared for each derivative? That would be tangent, right? And we'll put our constants together on the right side. So let's solve this for u. I don't know, do you wanna solve it now or do you wanna actually find c now? It doesn't matter. Find c first. So u is equal to negative five when t is equal to zero. So what is c in this problem? Tangent of zero is zero, so everything over here is zero, so c is 25. So let's back up to this version. Sorry. And we wanna solve it for u. Square root, is that gonna be enough? Just the square root. We're taking of the square root ourselves. When we do that, we need to consider plus or minus. Both possibilities. If we're handed a square root, then you don't see a plus or minus there. It's principal square root only, but if we generate this by solving for u by taking the square root of both sides, we'll keep both up. Questions on that? I think I'll forego the last example. Chandler, you said you had a web assignment question, so let's go to that. It's from 7.1. For what values of r, does the function y equal e to the rt satisfy the differential equation? y double prime plus y prime minus 20y equals zero. Okay, y double prime. Minus, no, plus y prime minus 20y equals zero. So there's a second order differential equation, right? It's got second derivatives in it. In fact, it's kind of loaded with derivatives. It's got second derivatives, it's got first derivatives, and it's got the original function in it. So remember, I mentioned the supplement to the book and six or seven of you said you didn't have it, the others have it, so we'll actually be solving problems like this eventually in this chapter, even though the continuation of this chapter is technically not in this book. But we're not supposed to be solving them yet. We're supposed to be verifying a solution. So what is the solution that they propose? Y equals e to the rt. And they want to know the values for r, okay? Actually, this is pretty good, because this kind of leads you into how it is that you would go about solving something like this. Fine, does this say value or values? Values. It should be two values. For r. All right, so if we want to validate a solution, we're going to be able to put this in here for y. We've got some other things to replace in there. One of them is y prime. Well, if this is our y value, what's y prime? Times r, right? Times the derivative of the exponent, r's a number. So it'd be r times e to the rt. And we'll throw that value in there. We also need y double prime. So what's the derivative of r e to the rt? R squared. Good, r squared, because we're going to generate another r when we take the derivative of the exponent. So that times the r that's already there. So we'll throw that in that position. So let's see what we come up with. We're going to have an r squared e to the rt. That's y double prime plus y prime. And then minus 20 of the y things, and they are e to the rt. And we're supposed to get zero. That looks like a mass. So what's going to make that a lot easier to work with? It's equal in e to the rt, right? Okay, they all have e to the rt, so let's take it out front, which leaves r squared plus r. It's looking better. That factors, I hope. Plus five minus four. Thank you. Now don't we have three, the product of three things equal to zero, correct? So we can set each of those three things equal to zero. Now it said solutions, and you said there were like two blanks, right? There should be like two. There should be two, so that one's got a pretty obvious solution. That one's got a pretty obvious solution. What's the solution to this one? It doesn't have a solution because e to any power, right, is not going to be zero. So no solution from that piece, but we do get our two r values. So there are two. So the solution is of this form. So we could have a solution that is e to the negative five t. You could check that. It does work in that equation. We could also have a solution of e to the four t. Actually, if they work, doesn't it also make sense that their sum would also work? Their sum would also work, and also some other variations of their sum. So we'll do more with this later in this chapter, but here's what the solution looks like, plug it in, plug the first derivative in, plug the second derivative in, and then solve the thing that we can solve is this quadratic equation. Have any of you worked with second order differential equations before prior to this class? Wes, you have? No, no, no, no. No, okay. Because this has a name, I was gonna ask what the name of this is. It's called Steve. No, it's not. What would be the name of this? It's called either an auxiliary equation or a characteristic equation. That's the quadratic equation that goes along with solving the second order differential equation. But we'll revisit this in detail later. We'll see Steve a little bit later. Anything else from WebAssign while we are there? Okay, another kind of subsection in this portion of the book on separable differential equations would be orthogonal trajectories. So they are sets of points in the plane, lines, curves, whatever they might be, that are mutually perpendicular. Now if it bothers you that curves could meet at right angles, it's actually not the curves. It's the tangent lines to the curves. So the tangent lines are perpendicular to the two curves at the point of tangency. So there's a couple pictures in your book and we're gonna come up with some other pictures of our own here on these two examples that I have. But for orthogonal trajectories or curves that are mutually orthogonal, there are a couple pictures that are in your book. So these actually are lines. So one family of curves would be lines that go through the origin. What would they look like? What would their equation look like? So these are all a family of curves. Y equals CX or MX or something, X, right? And it looks like the others are concentric circles that have centers at the origin. What would they look like? R squared equals X squared plus Y squared. Okay, and we would say R squared but R squared is just a number, right? So we could say C or K or whatever. So here's a family of curves, they're not curves, they're lines through the origin. Here's a family of curves, concentric circles of varying radii depending on the value of K, they are mutually perpendicular. So they are orthogonal trajectories. So the slope of the tangent to the curve at the point of intersection looks like that and it should form a right angle with the other curve while it's not a curve, it's a line. So that one really kind of gets easier than most. The slope of the tangent line and the other curve, in this case is linear so its tangent line is coinciding with the curve itself. But if you check these out at any point, they're gonna form right angles. So again, we don't have specific curves here but the slope of the tangent here and the slope of the tangent form right angles. So it's the tangent lines at the point of tangency that are mutually perpendicular. So I had a couple of examples picked out, I scrapped them. So we're going to example three and four. They have good examples in the book, make sure you take a look at those. Here are some other examples. So here's the typical orthogonal trajectories problem. So we are given a family of curves. Here's what we're given on our first example. What is that? What are these curves? Now if it were x squared plus y squared equals k squared we've already seen that. That's a set of concentric circles all centered at the origin, right? What are these? Ellipses. Now if you wanna put it in a kind of a standard form for an ellipse we could divide through everything by two to get the left side equal to what we normally like for it to look like. And if we divided by two we'd have k squared over two. We really like for it to be one, right? On the right side to be the standard form of an ellipse. But the left side certainly has that look of an ellipse. So we've got a bunch of ellipses, a family of ellipses, center at the origin. How do you know the center's at the origin? Because there's no x minus h the quantity squared, y minus k the quantity squared. So we're centered at the origin, major axis on the x-axis or on the y-axis? What's indicative of the major axis? The larger denominator, right? So on an ellipse the larger denominator in this case is two so the major axis would be on or parallel to the x-axis. Use your imagination here. So we've got a family of ellipses all centered at the origin having major axis on the x-axis. So let's go back to the original equation. Well, let's think about the picture. We want the slopes of the tangent, excuse me, the tangent lines themselves to be perpendicular to one another. So the slopes of the tangent lines tell me how they should be related to one another. If the lines themselves are perpendicular, what about the slopes? They are negative reciprocals. OK. So we want to find dy over dx on this first set of curves, first family of curves, flip that, negate that, and then use that to help us generate the orthogonal trajectories because we want to force the slopes of the tangents to be negative reciprocals of one another. So when we're done with this problem, we're in doing this problem, we want the slopes of the tangent lines to be negative reciprocals of one another. That forces them to be perpendicular, which then when we get back to the curves that represent each slope, their derivatives, differential equations, the curves themselves would be orthogonal to one another. So let's take the derivative of our given curves, our given family of curves. And let's derive with respect to x, derivative of x squared with respect to x. 2x, derivative of 2y squared with respect to x. 4y, dy over dx, and the derivative of some number with respect to x. Zero. So if we solve this for dy over dx, we get we move the 2x over here to the other side, negative 2x, and then we will have divided by 4y. Is that correct? Which is negative x over 2y. Now, this is probably a good first example because this is easier than our next example, which at this stage of the game, we don't have any k's in our equation. So our next example, we're going to have a k in here and we're going to have to get rid of k based on what k is in the original equation. We don't have that here. So we've got x's and y's only. Now, these are the slopes of the tangent lines on any one of the ellipses that we want to talk about. It's always negative of the x value over twice the y value. But I want to know the slope at this point right here. I know it's the negative of the x value divided by twice the y value for any point on any of these ellipses. So now we want to switch to the orthogonal trajectories. Well, if that's the slope on this set of curves, what do we want the slope to be on the orthogonal trajectories? We want it to be the negative reciprocal. So we want dy over dx on our new set of points, our orthogonal set, to be the negative reciprocal, which would be 2y over x. Something's not seeming to be right to me right now. OK, let's keep going. All right, so now we have the slopes of the orthogonal set. Maybe it's just not looking quite right at this moment. So we want to take this differential equation and see what family of curves are represented by this differential equation, the new one, the one we came up with. So we would multiply both sides by dx. What else? Divide by 2y. Divide by either 2y or just y. The 2 is not a problem. But if you want to multiply or divide by 2y, let's just divide by y. I wasn't thinking this was going to happen on this problem. OK, let's keep going. I think I've made an error somewhere. Does anybody see it to this point? I know what the answer is supposed to be. And I'm not thinking I'm going to get there with what I have right now. All right, let's integrate both sides. We might be able to get there. What's the integral of 1 over y dy? Natural log y plus a possible constant, which we'll send to the right side. What's the integral of 2 over x integrated with respect to x? 2. Natural log x. We've got a possibility of a constant here. Yeah, I think we're going to be OK. We're going to get there. Natural log. What's that word? Exponentiate. Let's exponentiate both sides. So e to the natural log of y, that's y. e to the, I'll go ahead and split that piece up right now, e to the sum in the exponent position goes backward algebraically to the product of two things with like faces. So e to the 2 natural log x times e to the c is equivalent to the statement that's above. Two natural log of x. I want to put that 2 somewhere else, other than out in front. Where else could that go? x squared. Good. That's the same thing as e to the natural log of x squared. If you had this, you'd say, well, I want to bring that 2 out in front, which is certainly legal. Or if the 2 is out front, you can kind of put it back, if that's what you want. And e to the c, e's a number, c's a number. What's e to the c? It's a number. There's no variable quantity in this, so it's just a number. e to the natural log of x squared. x squared. That'd be x squared. I knew that's what we had to get, but it didn't look like we were going to be able to get there. But we did. So describe to me what this family of curves is. Parabolos. That's a bunch of parabolas, right? And these are the parabolas that are functions, right? y equals something in terms of x squared. Could they open up or they could open down, right? Depending on the sign of v. So we've got a whole bunch of parabolas that could be real fat or real skinny. They could open up or they could open down. So let's go back to our original picture, which is a bunch of concentric ellipses centered at the origin that have major axis on the x-axis. Now we've got, and I need to be a little careful here when I draw these. And the reason I need to be a little careful is because where they meet, the tangent lines should be perpendicular, right? So the slope of the tangent here and the slope of the tangent here, if my picture is accurate, those should be perpendicular. We kind of made that happen. We forced that to happen. The slope of the tangent right here on the ellipse and the slope of the tangent on one of these parabolas, they should be the negative reciprocals of one another. They should meet at right angles. So these two sets of curves are orthogonal trajectories of one another, Chris. Doesn't matter what the constant is for either of the. That's correct. So the differing constant, whatever you put in for k here, it will generate a new one of these ellipses. And every one of these ellipses, even the one that we put in k equals 30, that ellipse would be perpendicular to every one of the members of the other family. So that's why I was trying to be careful, because I want it to look like that tangent line and that tangent line. Because everywhere they meet on any one of the two that we draw, they're mutually perpendicular, regardless of the k or the c that's chosen. All right, let's start the next one, simply because it has a situation where it has a k in the derivative and we have to get rid of that. What is this family of curves? Think of just put a 1 in where k is. y equals 1 over x. Do you know what that looks like? It's the thing with the asymptote in it. OK, it's got lines that are asymptotes. If you graphed y equals 1 over x, it's going to look like that. What is that? That's a hyperbola. Now, it's not the kind of hyperbola that has its axis on or parallel to the x or the y axis. It's rotated, in this case, 45 degrees in the plane. But if we had, let's say, negative 1 over x, it would look like this. So we've got a bunch of hyperbolas. It's a family of hyperbolas situated in the plane in this fashion. So we want to come up with the orthogonal trajectories. I don't know if you could visualize what they would look like. But let's see what their equations are and then we can know what they are by the basis of their equation. So we need dy over dx. So what's the derivative of k times 1 over x? k times negative 1 over x squared. Is that correct? So the derivative of x to the negative first is negative 1, x to the negative second, which is that. So this example, we've got a k in the derivative. So we have to go back to the original equation. What is k in terms of the original x and y values? So k equals xy. xy? I heard a couple people say that. So let's throw that into this position of the equation. And if we forget to do that, this problem becomes a nightmare. Because of the k's that are there, we don't really know what the k's represent from the original equation in terms of the interrelationship of x and y. So it really messes things up. So we need to know what is k in terms of x and y from the original equation. Pardon? Can you take out an x? Right. So we're going to have what when we reduce? Negative y over x. Is that correct? Yes. So that's the slope of the tangent line of every point here. So if we want the slope of the tangent right there, it's the negative of the y value divided by the x value, regardless of what k is. So now we want to go on and find the orthogonal trajectories. So if these are the slopes of tangent lines to every one of these hyperbolas, we want to negate them and flip them so we can see what the slopes look like on the orthogonal set. So the negative reciprocal is x over y, which is the negative reciprocal of this. And then we want to solve for y possible. So we could, in fact, cross multiply here. Is that right? Separate the x's and dx's from the y's and dy's. Integrate both sides. y squared over 2 plus a possible constant equals x squared over 2. And then we'll throw the two constants together over here. So if you don't like the division by two part, take the whole equation and multiply everything by 2. That would be twice c, which we can give that another name, a something. And how would we normally see that? We would normally see this as y squared minus x squared equals a number. And in fact, y squared is the same thing as y squared over 1. x squared is the same thing as x squared over 1. I'm trying to make it look like it's standard form. What is that? That's a bunch of hyperbolas. So we started with a family of hyperbolas. And now we have another set of hyperbolas that are mutually perpendicular. So we've got about 30 seconds. So let's doctor up our picture here. These hyperbolas that we're talking about are hyperbolas that have their axis on the x or the y axis. So they're going to look something like this. And I'm trying to make it so that the slopes of their tangent lines are negative reciprocals of each other. They're mutually perpendicular. So we've got a family of hyperbolas kind of altered in the plane by a degree factor of 45 degrees. And the orthogonal set that is paired with that would be another set of hyperbolas that are actually, we could come over here too as well, can we? So we've got a whole bunch of hyperbolas. They're all perpendicular to one another in the two sets. All right, we're out of time. Have a great weekend. See you on Monday.