 Welcome back. We're at lecture 34. We're in the supplement to the textbook, which is second order differential equations. We're actually kind of scheduled to start a new section, but we've had a couple of web assign situations that might need to be addressed. One of them is one that I actually had as an example to show what could happen with the particular solution. And the other one, let's just take a look at that, at least enough of it where you feel like you can take it from that point. One of the web assign problems from 7.8 looks like this. Does anybody have that printout so you know which one it is? Number two. OK. They give you the complementary solution, or what we have called the homogeneous solution. We could figure that out. It'd probably take less than a minute to get the characteristic equation r squared plus 2r plus 1. If you factor it, that would be r plus 1, the quantity squared, double root of negative 1. So if we have a double root of negative 1, then we ought to have c1 e to the negative x plus c2 x e to the negative x. That look right? For that part, I think that's given to you, but certainly have the ability to come up with that very quickly. That's part of our answer. I think it's our job in this problem to find the particular solution. So we've got a generic linear term there. So if we wanted to kind of characterize what that is, that would be something of the form ax plus b would generate, have the potential to generate something like that. We're going to change this, but we could also kind of generalize e to the negative x. If we want to end up with an e to the negative x, we can start with a certain amount of e to the negative x's. I think we had a problem like this two days ago in class where the d could be distributed to these. We don't know what they are anyway, so it's really kind of redundant to have that. So we can kind of dispense with that. If you were to distribute this solution, or the proposed particular solution, you'd have ax e to the negative x plus b e to the negative x. Anybody see a problem? OK, aren't both of those up here? This is really this. We don't know what a is. We don't know what c is. So they're both the same. So that's really not going to be the particular solution. This solution is right here. So both of these pieces, it seems to not be enough, or we're not going to need an ax e to the negative x term. We might as well dispense with that completely. We're not going to need a constant times e to the negative x either. So we kind of throw that out at this point and try again. Well, if the x e to the negative x we don't need, and the constant e to the negative x, what do we do from that point? Multiply by x squared. So here's what we don't need to do, which is kind of typical to what we do need to do on some other problems. This is not going to be correct. We don't need to say, OK, now we failed with this linear expression times e to the negative x. Let's come up with some arbitrary quadratic. Well, we don't need this, and we don't need this. We've already ruled those out. Those don't help us. Those are pieces to the homogeneous solution. So we could go ax squared e to the negative x. If I remember this problem correctly, that also does not work. And that's really frustrating, kind of annoying, borderline annoying. Now, it actually crosses over. It is annoying. So this doesn't work. You can't generate x e to the negative x with that as well, even though it's not part of the particular solution. So this kind of came up the other day in class, too. If we determined that this, and some of you had success on this problem, you determined that this was not the particular solution, how would you determine that? Make them. Say that again, Jacob. Maple. Maple? Oh, OK. Well, that could be a determining factor if you had accessibility to maple, which we don't have on a test. I can try to do it with something weird, like 2a equals 1 and 2a equals 0 or something. I remember that trying to plug this in, this is kind of bizarre, and we do not get a solution for a. So we rule that out after some effort. You do this, and you do this, and you plug them into the original equation, and you're going to try to generate and 1 x e to the negative x, and it's impossible to do that with this problem, with this particular solution. So we scrap that one as well, and then we go on to a x cubed e to the negative x. And from what I remember about this problem, you can successfully get a solution for a, starting with this as your particular solution. Yes, product rule. So there's the first term of your product, there's the second term of your product. Would you have to, because you didn't really determine that a x squared e to the negative x wasn't, did we determine that that wasn't part of the solution, like would you not have to make it a quadratic, including? Like I'm sorry, like in the first part we did, we crossed out in the second try, we didn't use, when we did a x squared plus b x plus c, we didn't use b x plus c because they're part of the homogenous one, they'd be equal. Okay, so your question is do we need to include the quadratic in this one? Yeah. The quadratic did something bizarre, I can't remember what happened, but it, it created a problem and it very well might be part of the homogenous solution. We only found two terms that caused this to be zero with this shortcut, there may in fact be more, in fact maybe this is one of them. So I think once we rule it out, we don't need it anymore. So when we realized that it failed here, the quadratic term failed, we can just go on up and as it says in the recommendation, if I can find it, that it says if you fail with one, keep multiplying it by x, modification, I think they call it a modification principle or whatever in the reading, but when we fail with this, when we multiply it by x and try again, yeah, that does work, a x cubed e to the negative x does work in, when you do the problem, you can find a value for a, so you're not putting in a into your answer, you'll know a number for a that works in this problem. So hopefully that'll jump start you on that problem if that would cause some difficulty. The other problem that was recommended that we take a brief look at, so we have a little extra information at the end, what purpose does this serve, why would, what additionally are we gonna be able to do by the fact that we know the y of zero is one or the y prime of zero is zero? Find our c values. Our c values from the. Homogeneous. From the homogeneous part of the solution. Is this one where they gave you the complementary part or the homogeneous part? Okay, so r squared plus r minus two, characteristic equation. So we need factors of negative two that'll give us plus one in the middle, plus two minus one, does that work? So c one e to the negative two x plus c two e to the x. The c values now. Okay, that's a good question. I was gonna ask that, but since, that I decided not to, but since you did, let's go ahead and. Finding the c values now, so should we bring this into play and this into play now? Because that is the final y, that's not just y sub h of zero equals one. So that's the final y of zero equals one. So we've got some work to do. If this were a homogeneous equation, then we can opt right now to solve for c one and c two, but we've got some other work to do before we do that, which does affect the y of zero being one potentially. You could do this in pieces or you could do this all at once. The only way you're gonna be able to generate an x term on the right side is to start with what? x plus b. Do you see any reason why we would have to kind of get rid of any part of that solution based on what we have thus far in the homogeneous solution? Don't see any. Double roots or duplicity of roots or anything, so I think that should work. We should be able to go through and find out a and b so that we could generate one x and then the second piece, I'm just trying to get it to the point where you feel like you could take it over. Now we've gotta generate one sine of two x. How could we generate one sine of two x? Okay, give me some more other than that. Right, we don't know how many sines to start with and we don't know how many cosines but you've got the right terms. We need a sine, let's say c cosine two x plus d sine two x and then you can go through and that's the only way you're gonna generate that kind of term is with one or the other of those. Do we probably need them both? What happened in a problem Wednesday in here? One of them can't take that. Yeah, we didn't really need them both because we only had the original function and its second derivative which were the same, right? If you start with a sine and you take the derivative twice, aren't you back to a sine? We need them both here because we've got the original function and the first derivative and the second derivative. We're hoping to keep the sines involved in the problem but we eventually want the cosines to drop out but we need to start with both of them. For p one, when you take the second derivative of that, it's just zero so is that supposed to happen? But the issue is can we generate one x by plugging in something like this for the particular solution and yes we can with that term because the y term has an x in it. The derivative term doesn't have an x in it but that's okay, it's present here. So any issues about what to do to find a and b from this point? I'm not gonna go all the way through it because I know what else we have to do today. Everybody okay with that? Take the first derivative, second derivative, plug it into here and try to generate one x. See what a and b have to be in order for that to happen. Start again, find the first derivative, second derivative with the second piece of the particular solution, plug them in here on the left side and try to generate one sine of two x and solve for c and d in order to do that. This kind of lengthy problem because you're not done at that point in time. Once you find a and b, if in fact they're both present which I know a is, I don't know if b is or not. And c and d, then you come up with the final solution. Does anybody happen to have the a and b and c and d values for this problem? Andy? So this won't be enough, this will be a number. I'm gonna have an a here but you'll have a number by this point in time in the problem. You'll have a number there. I used a c, I probably shouldn't have. So let's call this d, you'll have a number there and you'll have a number there. So the only unknowns at this point in time in the problem are c one and c two. Some of these might be gone. I mean it's possible that I don't, I think you're gonna need them both but it is possible that d or e could be zero. Then you would need to bring these into play, right? Take the original equation for y plug in one every time you see an x plug in a zero and then see what value for c one or c two would give you a y of zero equals one. Probably gonna have an equation with c one and c two both in it, right? And then take the derivative of this long mess that we have now plug in zero for x and it should kick out zero for y prime which will be on this side. So I think you're gonna have an equation with c one and c two here and equation with c one and c two here. Put those two equations with the two unknowns together substitution, linear combination, something to get the two solution. That's a lengthy problem. Probably an unfair test question. It just takes too long. That would be the only question on the test then it'd probably be fair but that's probably, chances of that happening are, well there's two chances of that happening. Slim and not, okay? So you're gonna have more than one problem on the test, Nicole. I told P one. Yes. But you just said I, I just. You need the derivative to plug in there which is what? A, you need the second derivative to plug in up there which is zero. So when you plug those things in, let's go ahead and do that. Second derivative is zero plus the first derivative which is a, sorry, minus two of the original y things and we wanna somehow figure out what it takes to generate the x term, the linear term. We're not gonna generate the sign of two x with this stuff. So when you equate the coefficient of x on the left side which is gonna be what? What's coefficient of x on the left side? Negative two a, it better be equal to one. So that's the only x term on this side. Everything else would be a constant. So a and what, minus two b. So the coefficient of x here better be equal to the coefficient of x here and this constant on the left side better equal the constant on the right side while the constant on the right side is zero. So a is what, negative one half which makes b what? Negative one fourth. So back to what I wrote out earlier. Once you write out the particular solution I wrote out a, x plus b. We're gonna know what a is. A is negative a half. B is negative a fourth. So when you write this part out you just put the a and b back into your y of b. Like the values that you found for a. Yes. So the first piece of the particular solution instead of just being a, x plus b is negative a half x minus one fourth. Now you go on to the second particular solution and find, I wrote it c and d probably would have been better written d and e. If you have your supplement with you or the copy that I made of the supplement there are a couple of diagrams that I think are helpful to look at as we go into section 7.9. Not a big deal is made about some things on the picture but I think we wanna point out some things. Now it seems awkward that the page number is 1,172 but the first page on 7.9 you'll see a couple diagrams and let me back up a half a step. So we're gonna do some applications now some of which we've actually kind of started a little bit when we looked at how it is we might need to use or apply second order differential equations one of which was a spring. We talked about the Hooke's law and the restoring force and mass times acceleration those kinds of things. They've already been mentioned in here one time but the diagrams that you see on the left side of the page you see the spring with the mass here at the bottom of the spring. Notice the scale out to the right that if this is in equilibrium position it shouldn't be a mystery that the equilibrium position is zero but look what happens when you go down the scale. It's a subtle way of telling you that in this particular model down is positive, this diagram. So as we go let's say we've got this mass at the end of the spring it sits here at its equilibrium position in order to start the mass in motion let's say we pull the mass down and release it well we need to prepare ourselves for the fact that pulling the mass down and releasing it down is positive on this particular picture. Now the other picture it's not a stretch for you to think of we've got a spring here they've got a scale down here on the other direction where the equilibrium position is here and they move it out to the right and it's not a stretch probably shouldn't use that word it's not a leap to think that moving it out to the right beyond its equilibrium is positive that seems positive but this one doesn't really seem positive down we normally think as negative but it's positive here that kind of starts the oscillation and you say in this particular model if there isn't any friction which we start the first model we look at doesn't have any friction at all I know that's not good but then we'll see what happens when we throw friction into the picture. So if you take this mass that's at the bottom of the spring and you pull it down positive beyond equilibrium and then you release it it's gonna oscillate so this spring the position of the spring over time it's gonna go back up beyond its equilibrium and then come back down and if there isn't any friction it's gonna oscillate indefinitely so that makes sense this spring once we set it in motion is just gonna stay in motion forever there's no friction to stop it so it looks a little bit redundant when you say it goes back up here comes back down to here well let's get a picture that kinda scrolls for time so if we let T be along this axis where is it? What's it do? Well we started it up here as you scroll time it comes back down here and then it goes back up to where it was and then it comes back down here and then it goes back up to where it was so we've got this nice little oscillatory model in reality it's going kind of moving in a straight line but if you let time scroll along the axis and show how it moves what would you say that type of oscillatory model is? Sine, cosine or sine plus cosine? Because if it is a sine plus a cosine we could, and we've already done this once we could put that in the form of a sine only or a cosine only and this might be a good time to take an answer that we had earlier in this class four or five days ago we graphed this, I'm not gonna graph the whole thing again but we graphed the sine, we graphed the cosine we added them up on the coordinate plane and we ended up with a function that itself was oscillatory we went back and revisited that and we gave it an amplitude of square root of two square root of two, is that ringing a bell? The final graph that was sine plus cosine we went back and revisited that another day and we said if the period of this is two pi and the period of this is two pi when you add them together they have the same period which was also two pi so we left the coefficient of x one and then we figured out kind of backed it up a little bit and figured out what the phase shift was I'm not getting any good visual prompts from you is this ringing a distant bell? Yes, okay and what was that phase shift to the left? Good, okay. Thought maybe I was talking Spanish there for a while I do that from time to time so this sine graph with a changed amplitude up the square root of two units down the square root of two units period of two pi and a phase shift of pi over four to the left is exactly the same graph as sine x plus cosine x. I wasn't gonna do this but I think it might actually be worthwhile to do this. We drew pictures and I don't have the pictures here right now so since we don't have the picture let's see if you remember this it might not be a fair question. The sine of the sum of two things. What is that? Sine of... How is that? Does that look familiar? I guess it was a fair question. Thank you. Now let's, what are our two, what are our a and b values? Well a is just x we're not gonna be able to do much with that but b is pi over four so let's see what we get from the sine of x plus pi over four. So it should be the sine of the first cosine of the second plus cosine of the first sine of the second. Everybody okay? That this equation kind of morphs into this equation with this sine of a sum identity. What's an exact value for the cosine of pi over four? What's an exact value for the sine of pi over four? Same thing, right? Sine of pi over four and cosine of pi over four are the same. All right, let's distribute the square root of two to both terms. So what's the square root of two times the sine of x times the square root of two over two? Sine x, is that right? What's the square root of two times the cosine of x times the square root of two over two? Just cosine of x. Is that what we started with? Sine x plus cosine of x was our original. So they are the same. We're not validating that by a picture. We're validating that kind of algebraically with some trig identities thrown in. So you can take this thing that is sum of sines and cosines and write it as a single sine function. This is called a phase amplitude form. In your supplement, on the top of page 1173, you will see kind of some conversion, some interrelationships between something given to us in this form and how to convert that into the so-called phase amplitude form. I'm gonna actually give you something slightly different from this that I think simplifies it a little better. So that if we have an answer that is the sum of sines and cosines, so this is a unit circle and it's a unit circle centered at the origin, so the equation is x squared plus y squared equals one. Here is a point on that unit circle. This angle phi or phi is actually this angle right here. So it's from the zero gradient or zero degree mark rotated through positive phi or phi radians which takes us to this place in the plane. So here is a point on the unit circle. Typically when you see a point on the unit circle the x value I'm hoping that some of this rings about because we haven't really looked at any of this in this course, the x value is the cosine of the angle that brought you to that point. The y value is the sine of the angle that brought you to that place in the plane. So from this angle phi or phi we have the cosine of phi is a over the square root of a squared plus b squared. The sine of phi is b over the square root of a squared plus b squared. So those things are stated right down here. Do you remember that from a unit circle that the x value is the cosine of what got you there and the y value is the sine of what got you there? I guess we could validate that by saying the x value is here and the y value is here and the hypotenuse is one right and you could say sine is opposite over hypotenuse cosine is adjacent over hypotenuse and you could come up with these identities. If that's distant for you then mess with that a little bit till you feel assured that this is legitimate information. So far we haven't deviated a whole lot. I guess we're going to deviate from this one that's in the supplement because they convert it to a cosine and we're going to convert it to a sine. Maybe one more thing before we completely leave this diagram alone. This is supposed to be true. If you plug in the x value you get the cosine of the angle cosine squared plug in the y value you get sine squared certainly cosine squared plus sine squared is equal to one. You could also plug in the coordinates if you take the cosine or in this case the x value and square it plug in the sine value and square it or call it the y value that's the same thing. You can check that out to verify that you do in fact get one when you square those and add them together. Square the numerator you get a squared square the denominator you drop the square root symbol square the numerator you get b squared square the denominator you drop the square root symbol these have the same denominator which is what a squared plus b squared if you add the numerators what do you get a squared plus b squared you do get one so it does satisfy the unit circle equation. This is going to look familiar but we had specific numbers earlier. I'm going to propose that the amplitude in the phase amplitude form is going to be the square root of a squared plus b squared. Let me see in the what do they call this? Yeah they do call it an omega a lower case omega so I'm going to use the same looks like a w in this phase shift angle the angle that we're talking about on the unit circle diagram is the angle phi p h i so that is let's say that's what now we had this in terms of x this particular example has it in terms of t same thing if we were to use that sign of a sum again it would be sign of the first cosine of the second plus cosine of the first sign of the second from the unit circle what was the cosine of phi it was the x value of that point which was what and what was the sign the sign was the y coordinate of that point on the unit circle if we distributed this thing out in front like we did before it was a number it was the square root of two now it's the square root of a squared plus p squared what is the first term would there be an a left? because the numerator and denominator would knock out so you'd have a sign of omega or w t and if you do that same thing to the second term don't you have a b cosine of omega times t isn't that the form that we're gonna have when something oscillates right? sign plus cosine can we take something in this form this is kind of in reverse but hopefully it justifies if it works this way coming down this page couldn't we start here and work our way back up to this equation if it's an equation as we work down the page we should be able to basically say they're equations also as we work up the page so we're gonna have something like this and we want to put it into this form so how do we find the amplitude? well we're gonna have a number here potentially and we're gonna have a number here so we are in which we haven't done this yet we picked it off of a diagram didn't we? we added the square root of two over two to the square root of two over two and we got the square root of two on the diagram now we know with numbers whatever the numbers happen to be we'll take a which is the coefficient of sign and b which is the coefficient of cosine we'll square them and add them together take the square root that's gonna be our amplitude once we put it in this form let's check it we started with this one should the amplitude have been the square root of two? yes one here and one here one squared added to one squared is two square root of two that would be the amplitude uh... I don't want to skip over this this got mentioned but let's the omega value is here and here it will also be the same up here we want the periods of the individual sign and cosine functions to be the same as the period of the so-called phase amplitude form we want them to all have the same period because they're gonna all repeat in the same cycle so omega is unchanged from the original sine plus cosine equation to the phase amplitude version how do we get the angle phi see if this makes sense from our diagram back to the unit circle diagram if the x value is the cosine and the y value is the sine or if you just wanted to use the right triangle here tangent should be what sine over cosine so if we put this thing over this thing what would you get b over a so if you want to find the so-called uh... phase shift or the phase angle then take b now look at what we have for b actually doesn't that look really good no it's not on that page so on this page what is b b is the coefficient of cosine sometimes we write that first we write the cosine term first and the sine term second so just remember that b the coefficient of the cosine term is in the numerator a is the coefficient of the sine term that's in the denominator and if we do an inverse tangent problem then we can get the angle back to our example we had sine of x plus cosine of x if we did tangent of phi we would get b over a which they're both one coefficient of cosine is one coefficient of sine is one what angle has one for its tangent five or four right and that's not what we got this way but we got it another way so this way we don't have to draw them figure out what we think is the the phase shift involved in that graph we can do this inverse tangent problem so in our example we got pi over four so is it possible without drawing a diagram and i'm gonna change to x is now to go directly from this version to a single sine or cosine we've got sine so what's the amplitude the amplitude is the square root don't need a graph for that here's a one here's a one square add them together and take the square root square root of two sine well the coefficient of x here is one so our omega or w value is one same thing here so we're gonna have one x again tangent of this angle is the b value over the a value one over one what angle has one for its tangent pi over four now by adding pi over four the net effect is to shift the graph what? left so it's actually going left pi over four i think we found that to be true on our picture which we brought over here and saw that it actually was shifted by pi over four units to the left so you can go from this version to this version without graphing it and trying to figure out from the graph what's the phase shift what's the amplitude the periods will always be the same where did you get the x? this is this one? the x in here whatever the coefficient of x is in the sine and the cosine it's going to be the same here because the period of this sum it's got to be the period of the net single sine or single cosine graph so that's slightly different from what's in your book but i think getting the amplitude there and getting the phase angle here is going to be a little more straightforward than what is illustrated in the in the book itself alright let's set up a problem and that's probably all we're going to get is the setup alright thank you hook's law which we have already encountered a couple of times in this class says that the force required to stretch or compress the spring is directly proportional to the elongation of the spring itself how long how far it's already been stretched or compressed so we've also used this at the restoring force then if the force required to stretch or compress it is k times x that's hook's law directly proportional to how long it's been stretched the restoring force is that opposite motion so the restoring force is going to be mass times acceleration and the forces involved in this motion of this spring with a mass at the end of it they've got to be equal so the restoring force basically given to us by hook's law and force being mass times acceleration these forces have to be equal you put everything on the same side you get this and if you want to make that look a little more familiar to what we've been doing in this material m is the mass it's a number second derivative of x which is really just acceleration but second derivative of x we'll call x double prime isn't this a second-order linear homogeneous differential equation it is there's nothing in this problem about friction or any kind of a damping force at all when that is thrown into the equation so here we've got a second-order differential equation when we throw in the uh... damping force or the friction into the problem we're going to have this and then we'll get this written down and then we'll stop at that point uh... i don't want to write it like this first time let me scrap that so here's our mass times acceleration we've got our kind of hooks law restoring force over here now let me read this next we consider the motion of a spring that is subject to a frictional force in the case of the horizontal spring or a damping force in the case where a vertical spring moves through a fluid an example of the damping force supplied by a shock by a shock absorber in a car or a bicycle okay here's the statement that's going to get us what we need out of this equation we assume that the damping force is proportional to the velocity of the mass and acts in the direction opposite to the motion velocity well if this is acceleration the second derivative of x what's velocity first derivative of x so there's our damping force there's our friction so now this is what i was going to write a minute ago but i wanted to write it in this form first it'd be nice if i could actually write it i want to bring these terms over to this side x prime is really dx dt now we have a second order linear homogeneous equation but it has all three terms there's our second derivative term there's our first derivative term there's our original x term and we are equal to zero we'll pick up at that point