 Hi, I'm Zor. Welcome to Inezor Education. Today I would like to start a new topic, ordinary differential equations. Well, first of all, let me just mention very briefly that if there is anything in the mathematics which you are learning, is really used really widely throughout many, many different subjects and disciplines, that's the differential equations, ordinary as well as partial differential equations, actually partial even more than ordinary. So these will be two topics which I'm going to address. Well, this one is about ordinary differential equations and the next topic will be partial. And I'm sure you actually are familiar with ordinary differential equations from your physics, even such a simple thing as the second Newton's law, for instance, which we all know looks like this, where A is acceleration, M is a mass and F is a force. This is also differential equations because what is acceleration? This is the second derivative of the function which describes the coordinate of the point as a function of t. So this particular lecture, let me just repeat it all the time, I'm kind of repeating this, is the part of the course of advanced mathematics which Unizor.com is offering to students around the world, that's completely free by the way, and it doesn't have advertising so you don't suffer from anything, which disturbs your process of learning. And the site actually contains not only these lectures but also very detailed notes for every lecture and in many cases, exams. If you are watching this lecture from any other source, like directly from the YouTube, for instance, or from any other website, you most likely don't have this additional educational functionality. So that's why I suggest you to go to the website. Okay, first of all, let's start with definition. Now, ordinary differential equation is, first of all, its equation, which means it should be somewhere the equal sign, right? Now, the fact that this is differential equations actually implies that derivatives of the function actually participate. So my most general form of ordinary differential equation is something like this. Function of an argument, the function which is unknown of this argument and its derivatives, maybe only the first derivative or maybe the first and second and third, etc. They are all participating in this equation. So this is the most general kind of expression which describes ordinary differential equation. Now, ordinary means there is only one argument. I was talking about the next topic, which is partial differential equations. That's when we were talking about function of more than one argument, two, three, whatever. For instance, the function of co-ordinate might be actually three different arguments, x, y, and z, right? Okay, so this is the most common representation of equation. And we will never use this type of a common representation. We will definitely use much more narrowly defined expressions, which allow basically to be solved, because obviously we are interested in equations which can be solved. Now, obviously there are equations which cannot be solved. Well, what do people do about this? Well, they solve it numerically using the computers and still get some solution, right? But that's not the subject of our discussion. We are talking about equations which we can actually solve. And many examples will be provided and certain techniques how to solve these differential equations. So again, the purpose is to solve, to find function y of x, which satisfies this particular equation. Now, let me just exemplify the simplest possible differential equation. And it's something like this. So there is some kind of a function y of x, which we have to find. And all I know about this function that its derivative is equal to 2x. Well, obviously we can guess the solution. The solution is this. Well, to be precise, the solution is this, where c is any real constant, real number. Because the derivative doesn't really change if we add the constant. It's very similar to integration in a way. And by the way, if we have an equation with one derivative of the first order, we will always have some kind of a constant. Because whenever we find one particular solution like x square, then we can always add constant and the result would be the same. Now, this is a guess. But not necessarily we are able to always guess that type of thing. So how would you approach this particular very simple problem a little bit more mathematically, a little bit more scientifically if you wish? So here is what one of the methods, which I'm going to talk about, tells you to do. First of all, most likely the first derivative would be better represented by this notation. Which is kind of the same, right? Because this is just different notations for the derivative. Now, whenever we are talking about this notation, we are talking about two infinitesimal variables. And the ratio of these infinitesimal variables is not some kind of an exact number, which is this. It's basically the limit of this ratio is this number. But at any particular moment, when we are talking about two infinitesimal variables, the ratio is some kind of a number plus some kind of infinitesimal error if you wish. So you always can consider this to be like an approximation. And whenever I can write this, it means that for every x, if we are talking about delta, not d, delta y over delta x, that would not be an exact equality. Whereas the delta x is converging to zero, then the limit of this ratio, which we denote as this one, is equal to this. So that's kind of an understanding what is dy over gx. Now, why did I want to explain all this? Well, for a very simple reason. I would like to put this dx to the right and convert it to this. Again, this equality should be understood as a limit. So if we put a delta instead of d here, it would be an approximation. But as delta x converges to zero, then this particular equality becomes more and more precise. And the letter d means that in the limit we are in this particular category. Which implies that right now this very much looks like lots of our examples on indefinite integration. Because if this is equal to this, then I can do indefinite integrals from both sides. And we can actually have the result because the indefinite integral of dy would be equal to y. Plus some kind of a constant. Let's put it c1. And integral of 2x would be x square plus some kind of a constant, right? Well, this is any constant and this is any constant. So I basically can put, combine them together, where c is any constant. So I've just got exactly the same solution as whatever I guessed. But slightly more mathematical way. So what was my mathematical way? I separated dy from dx into two different sides and separated y also to different sides. So on the right side I had only function of x and dx. On the left side I had only function of y and dy. And then I can integrate this of dy and this is by dx. And got something which is resembling the function. So that's one of the ways you can solve certain differential equations which allow this type of things. And this methodology is a separation. So differential equations which allow this type of approach to solve them are called separable. Okay, so we talked about separable differential equations. And separation is actually, I would say more frequently occurred way to solve the equations. At least within the realm of the problems which you will be dealing in school, in physics, whatever. Okay, now let's make some examples. Example number one. I have the following equation. x square times y derivative is equal to y. Well, obviously this can be separated, right? How dy by dx is equal to y over x square, right? x square to here. Or I can separate dy by dy is equal to dx by x square. Right? Now I can integrate both sides. Now what is integral of logarithm y? Now in theory it's logarithm of absolute value of y. But let's just restrict ourselves. That would be much easier for educational purposes. So we consider that we are talking about function which is only allowed to have positive values. So we can divide by x square. You see, that's the problem. Whenever you go into idea of how to solve, this is basically the idea. If you're talking about much more, again, mathematically rigorous way, you really have to think about can I divide by x square, can x be equal to 0, etc. So this is, I would say, more on the idea side. And that's okay for now. So I will use the simplest methodology possible. So the integral from 1 over y is equal to logarithm y, right? Because the derivative from logarithm y is 1 over y. And derivative plus some kind of a constant, c1. And derivative of which function gives me 1 over x square? Well, 1 over x gives me minus 1 over x square as a derivative, right? So I have to put minus 1 over x and derivative of this would be 1 over x square. So I have integrated plus c2, which we can rewrite as logarithm y is y is equal to minus 1 over x plus some kind of a free constant, c. Now this is, again, any constant. Again, what I will do is I will make a little simplification now. And from this, if I will raise e to the power of left and the right, e to the power logarithm y would be y. So it's equal to e to the power of minus 1 over x plus c. Or e to the power of 1 minus x times e to the power of c, right? And this is, again, a constant. So I can always say that it's equal to some kind of a constant. I'll use the same c, but it's different. e to the power of 1 minus x. So this is what came to me, came to, basically, as a solution to this. Well, let's try to check it out. Y derivative is equal to, this is a constant, times derivative of exponent, it's this, times derivative of internal inner function, right? It's a composition, it's a chain rule, right? So derivative of minus 1 over x is 1 over x square, right? And this is the same y, right? So if we will put x square here, we will get this. So it checks. Again, I did not pay a lot of attention what if x is equal to 0, etc. Obviously, there are certain, I would say, more rigid considerations. It has a general idea of how to approach this particular differential equations are presented. And by the way, integration has been involved, obviously, right? So in many cases, to solve differential equation, people are using the word to integrate differential equation, which means basically to solve, since integration becomes the major tool of solving. Okay, next example. Tangent x, con times y derivative equals to y square. Now, again, this is a kind of a separable differential equation, and here is how. So y derivative is dy by dx. It's equal to y times, now, tangent is sine over cosine. If I will move it to the right, it will be cosine divided by sine, right? Now, y can be moved here, dx can be moved there. So I have dy divided by y square equals cosine x dx divided by sine of dx. So now I'm integrating. Now I have separately function of y and function of x. Here I will have minus 1 over y, because the derivative of this is equal to that. And this is kind of easy, because what is cosine x dx? It's basically d of sine x divided by sine x, right? So the solution to this would be logarithm of sine of x. Again, some kind of a free constant is always present. So that's what I've got here. And to get the result, I have to flip this thing. So it would be 1 over minus logarithm sine of x plus c. Or again, we can always consider that this is not just a constant c. We can always replace c with a logarithm of c. Again, forgetting for a while that in this particular case c is any constant. If I will put logarithm of c, it would be the positive only. And if this is a logarithm c, I can say that this would be minus logarithm of c times sine of x, right? Because logarithm of product is equal to sum of two logarithm, logarithm c and logarithm of sine of x. And minus is here. So that's the final solution. Again, I kind of shortcut a little bit all these negative positive verifications. I just did it very, very briefly, idea-oriented. And let's just check if I will take why derivative would be what? First of all, it's 1 over something. It's minus 1 over something. It would be 1 over logarithm square of c times sine of x times inner function. Inner function is logarithm. Inner function is logarithm. So the derivative would be 1 over c sine of x. And derivative of inner function would be c cosine of x. Am I right? I don't know. Well, c is cancelling. That's fine. That's OK. Now this is y square, right? This is y. This is y square. And this cosine over sine, if I will transfer it this way to the left, I will have sine of x divided by cosine of x, y. And it's equal to y square. And this is the tangent. All right, so it checks. So we should not really forget all these three constants, any constants which are supposed to be added during the process of integration. But now that means that we have an infinite number of solutions, right? So going back to the first example, first example was this. And the solution was c times e to the power equal 1 minus x. So it's infinite number of solutions because the c is any constant, right? And all of them satisfy this equation. So does it mean that we cannot really find a real solution? Well, no. It means that the real solution is infinite number of solutions. That's fine. However, if we are interested in a particular solution, so this is a general solution and we are interested in a particular solution, we have to define this constant c using something else outside of this ordinary differential equation. Well, for instance, we can actually say, okay, from certain knowledge which we know, we know that this function y of x, if x is equal to 1, then the function would be equal to, let's say, 1 as well. So this is an initial condition. So not only we know the differential equation which basically defines the relationship between x and y, we also know that at certain point function gets certain value. For example, you can have the differential equation like, for instance, the second law of Newton and it actually defines the second derivative. Now, if you will integrate it, if you will really solve it to find the function x as a function of t, you will not be able to do it unless you know where exactly the movement started, at what particular point on the coordinate axis it started, and at what initial speed it started. Because unless you know the initial speed and initial position, your law of movement, the Newton's second law, does not really give you exactly the coordinate at certain time in the future. You really have to know these two additional parameters, because if you, for instance, shift your beginning point from 0, let's say, to 10, then obviously all other coordinates will be shifted by 10. So you have to know the initial position. And simultaneously, since the second law of Newton involves the second derivative, so you have to know the first derivative initial value. So initial values are important if you want to have an exact particular solution to your equation. Now, in this case, this is an initial condition which might be satisfied. Well, let's just find the c from this, all right? So what will be 1 equals to c times e to the power of minus 1. e to the power of minus 1 is actually 1 over e, right? So c is equal to e. And my function, and this is a particular solution which satisfies these initial conditions, is e to the power e to the 1 over x, or you can add e to the power 1 minus 1x, right? So this is a particular solution which satisfies not only this, but also this. So now we know the difference between general solution and particular solution which satisfies certain initial parameters. And we need only, if we have only one constant, we obviously need only one parameter. If we had the second derivative involved, we would need two. But that's a little bit further for another lecture. Now, let me talk about the second. So the second, I had tangent x times y derivative equals to y squared. And our solution was y is equal to minus 1 over logarithm of c times sine of x, okay? Now, what kind of initial condition might be? Well, just for example, if x is equal to pi over 2, my function should be equal to 1. Okay, using this, let's try to determine c. So the function is equal to 1, so 1 is equal to minus 1 over logarithm of sine of pi over 2 is 1, so it's c. So from this, logarithm c is equal to minus 1, right? From this c is equal to 1 over i, all right? Because logarithm of 1 over i is minus 1. So I can put it here, and now I have a particular solution which satisfies our condition, which is y is equal to minus 1 over logarithm of sine of x divided by e. Now if x is equal to pi over 2, this is 1, 1 over e is 1 over e, logarithm of 1 over e is minus 1, minus 1, minus 1. So we will get 1 here. So that's a particular solution which satisfies our initial condition. So what's important in this particular case is we had an ordinary differential equation. Ordinary means from one argument. In this case we're using x and the function y of x. The first derivative is participating in the equation, and the equation is separable. So in both our examples we were using separation y to the left, x to the right, and then we integrated them. And then whenever we got the general solution, we had some constant involved, which is kind of any number. And to determine this any number, we need additional initial conditions on the function which we are looking for. And that actually is defined like for instance if x is equal to something then y is supposed to be equal to something. That's one of the form of initial condition. And we were able to determine c in that particular case. I suggested to read the notes for this particular lecture on Unisor.com. And then we will continue with other things. We will describe what kind of types of differential equations you might actually meet. There are not too many actually and how to solve them. Other than that, that's it. Thank you very much and good luck.