 Welcome back to our lecture series Math 1220, Calculus II for students at Southern Utah University. As usual of your professor today, Dr. Andrew Missildine. Now you might've been thrown off by the title here if you're reading it, Math 2280, what is this? What we're gonna be doing is starting chapter nine in James Stewart's Calculus textbook, which is based upon the topic of differential equations. A differential equation is an equation that involves a derivative of some kind. That is, we'll have an equation that involves some unknown quantity, I should say unknown function, y equals f of x, and the equation's gonna involve a finite number of its derivatives. So we have an equation that could be involving the first derivative, the second derivative, third derivative, et cetera. We'll come and talk about that a little bit more detail in just a second. Now, the differential equations turn out to be very, very important in not just mathematics, but in broad science as well, as one can use differential equations to approach many, many, many real life applications. And so one cannot undersell how important the study of differential equations, how important that study is. Now, it is not our desire to do an extensive coverage of differential equations. Instead, we plan to do just an extended trailer. Most universities will have at least an introductory course to differential equations at Southern Utah University, that class is math 2280. And of course, other programs might have extensive courses in differential equations, many reaching into a graduate program. Differential equations is pretty important, and the prerequisite at a university is typically gonna be calculus two with maybe some linear algebra, because there's a lot of linear algebra in differential equations. Some universities will require it, some don't. It depends on how they approach it, of course. So let's talk about differential equations. Differential equations, like I said, there are equations that involve functions, which typically we'll call that function Y. The solution to a differential equation will be a function. It'll be a function F, right? So what I want you to do is to try to compare differential equations with algebraic equations you've seen in the past, where in algebraic equations, you have some algebraic combination of the variable X for which the variable X is typically a number, a quantity of some kind. With a differential equation, we're gonna have these combinations of the function Y, we're gonna hear Y as a function. Now in an algebraic equation, you'll have like X squared, X cubed, X to the fourth. You have these powers of X showing up. In a differential equation, the powers of Y are actually gonna be the first derivative, the second derivative, the third derivative, et cetera. And so solving a differential equation means looking for a function Y which satisfies the differential relationship. The order of a differential equation is gonna be the highest derivative present in the differential equation. In this course, we'll be mostly interested in first order differential equations. You might see some in the homework questions, involving second order or higher, but for the most part, we're gonna be focused on solving special types of first order differential equations. Let's give you an example of one. So this right here is a first order differential equation. Y prime is equal to X times Y. So how you wanna interpret this equation is the following. We're looking for a function Y whose derivative is the same thing as just multiplication by the variable X. Because again, Y is a function, it's a function of X. So can we find a function whose derivative is no different than just multiplying by X? You could see how that might actually be sort of a useful information. It's like, okay, derivatives can be complicated times with the chain rule, the product rule, the quotient rule, et cetera. If I knew the derivative was just multiply by X, you could use that for efficiency of some kind. Now, in this video, we're not gonna really talk about how one finds a solution of differential equation. I mostly just wanna convince you of what it means to be a solution, right? So what we're gonna do is we're gonna take as our candidate F of X equals E to the X squared over two, power. And I claim this is a solution. This is a solution to the differential equation. Well, how am I gonna check that? Well, when it comes to an equation, you wanna take your variable assignment and you're gonna wanna plug this into the left side of the equation and the right-hand side of the equation. So if we consider the left-hand side, we wanna take the derivative of our substitution Y here. So what is the derivative of F of X? By the usual rules of derivatives. Because we have an exponential expression, the outer derivative will just fit back out itself, E to the X squared over two. But then we have to multiply it by the inner derivative, X squared over two. That by the power rule is gonna give us a two X over two. You'll notice of course that the two's here cancel and we end up with simply just X times E to the X squared over two. And when you compare this to the right-hand side, the right-hand side is supposed to be X times Y, which if you take X times F of X here, you're gonna end up with X times E to the X squared over two. And Bob's your uncle, you see that the left-hand side and the right-hand side are actually equal to each other. So in fact, this right here is, where did it go? This is a solution to our differential equation. But one thing we're gonna see about differential equations is that solutions are not necessarily unique. Take, as another example, take Y to equal the function G of X, which is given as three times E to the X squared over two. Now, when you look at this one, take the left-hand side here, we're supposed to take the derivative of G. Well, the derivative of G is gonna be three times E to the X squared over two times X. Well, that's just the same thing as three X E to the X squared over two. But if you switch things up a little bit, you get X times three E to the X squared over two. And that's the same thing as X times G of X. So you can see that in this situation, we have a second solution to a differential equation. Now, this might not be any surprise, given that algebraic equations can have multiple solutions, right? Quadratic equations typically have two solutions. Maybe that's no surprise here. Now, in some respect, we do wanna consider these two solutions one and the same thing. What I mean is in the following context, if we take Y to equal C times E to the X squared over two, notice that Y prime is gonna equal C, X E to the X squared over two, which is the same thing as X times C E to the X squared over two. This is just X times Y. And so this right here, Y equals C times E to the X squared over two or C is just any real number. This actually turns out to be a more general solution to this differential equation. And the example F and G that we had before are just specific instances of this example right here. The first one where we take C to equal one and the second one where we take C to be equal to three. We could choose any real number. We're gonna come back and talk about this a little bit more in just a second. So in the meanwhile, I wanna take a look at another example of this. So let's take the differential equation Y prime equals one half Y squared minus Y. This is a first order differential equation. What we're claiming is that the, we wanna find a function whose derivative is equal to just a quadratic expression of the original function, one half Y squared minus one. Now it might, again, at this early stage, it might not exactly be obvious to the viewer why would someone want to solve such a differential equation. But, and we're not even gonna talk about how. So we're skipping the Y, we're skipping the how. All we're doing right now is verifying that this right here represents a family of functions because you take any parameter C that you want right here. You can pick any real number C you want and this would be a solution to this differential equation. Now to do that, we check the left-hand side and the right-hand side. The left-hand side, we wanna check the derivative of the proposed function. So taking its derivative, we get one plus C e to the t over one, or one minus C e to the t. In this situation, our variables in play is C here. That's no big deal. Take the derivative, by the quotient rule, we're gonna get low d high. Take the derivative of the top, we're gonna get C e to the t. And then we subtract high d low. The derivative of the bottom can be a negative C e to the t, like so. Square the bottom, here we go. Just doing our little poem right here. We're gonna leave the denominator factored. There's usually never a benefit of multiplying out the denominator when it comes to a fraction like this. Distributing and combining like terms the numerator though actually will prove to be profitable. Distribute the C e to the t right there. Since we have a negative here and a negative here, that's actually a double negative, like so distribute this C e to the t right here. Actually now that I mentioned it, there's a C e to the t common to both. Let's factor this out. Again, this is a positive there. Factor out the C e to the t. What's left over is one minus C e to the t and one plus C e to the t. All over one minus C e to the t squared. You're gonna see that the C e to the t is in the numerator will cancel. The ones will combine to give us a two. And so the derivative of our functions in these two C e to the t over one minus C e to the t quantity squared. That's what we get for the derivative of this function. Let's compare this to the right-hand side. The right-hand side of the function, sorry, of the equation, that should be one half y squared minus one. So taking the function we had here, let's plug it in one half times. Well, what was the function again? You can see it, we can still see it right here. We're gonna take one plus C e to the t over one minus C e to the t, quantity squared minus one. And so what we wanna do is we wanna show that this is, we wanna check, is this equal to the derivative that we found earlier? That's a parenthesis right there. So we wanna check that. So it's gonna take a little bit of work to do. So let us begin by trying to combine some like terms right there. We'll find a common fraction. Now notice we have a one half in front. We'll just leave that alone for right now. If you're squaring a fraction, that means you square the top and you square the denominator. But again, we're gonna leave the denominator factored. In order to combine the fractions, I need to write one with a common denominator, one minus C e to the t quantity squared. As it's one, that'll be the numerator as well. Now in order to combine some like terms, and actually if you're times a by one half, I can stick that in the denominator. So we're gonna get two times one minus C e to the t squared. Now in the numerator, we need to foil some things out. If you foil the one plus C e to the t squared, you're gonna get one plus two C e to the t plus C squared e to the t, like so. That's the first bit. And then we're gonna subtract from it a one, a minus two C e to the t, and then a plus C squared e to the two t. Sorry, kind of ran out of space there. But there will be some terms that cancel. There's gonna be a one minus one, they get each other. There's a C squared e to t, they get each other. And then the other ones that are left over, this is actually a double negative again, you're gonna get some positives. And so as you add those together, we end up with four C e to the t all over two times one minus C e to the t squared. Of course, two goes into four two times. Since you're left with a two C e to the t over one minus C e to the t squared. And so this is what our right-hand side turned out to be if we plug in this function. And if we come back and compare, where do we put it? Oh, it's right here. Two C e to the t over one minus C e to the t squared. So you can see that those two expressions are in fact one of the same thing. So this shows us that every function, every function of this form is a solution to this differential equation right here, where it didn't actually matter what the parameter C was, it's gonna turn out to be the same thing. So this shows us how one checks if a function is a solution to a differential equation and not again, I haven't yet talked about how one finds that we'll get to that in the future. We just wanna clarify that we understand what a differential equation means and how one can verify whether we have a solution or not a solution.