 Hi, I'm Zor. Welcome to Unizor Education. Today, I would like to continue overview of ordinary differential equations. This lecture is part of the relatively comprehensive course of advanced mathematics for teenagers presented on Unizor.com. I do suggest you to view this lecture from the website because it provides notes for every lecture, very detailed notes. You can use it as a textbook and also, there are certain exams and the site is completely free. There are no advertisements, so just use it. Okay, now the previous lecture was dedicated basically to introduction to what ordinary differential equations are. We were considering them ordinary because we would like to distinguish them from functions of many arguments, more than one argument. So ordinary is for the functions of one argument. It's differential equations obviously because the derivative of that function is present in the equation. And finally, I would like to concentrate, at least for now, for this particular lecture on differential equations of the first order, which means that in my equation, there are x, y and derivative of y present, where y is unknown function of x. And my equation might be actually looking something like this. This is the general form of ordinary differential equation of the first order. And I'm not going to solve it because it's a very general form and in the very general form we cannot really do anything. It's like with algebraic equations. You can certainly solve the equations of the first and the second order. The third is kind of a problem. Starting from the equations of the fourth order, it's really very difficult. And if you have certain functions put into this equation, like trigonometric function, exponential, logarithm, etc., then forget about general kind of a solution. So there are solutions in certain particular cases. And today I would like actually to concentrate on three major cases with ordinary differential equations of the first order, which can be solved. Now, in algebra, when we were talking about equations, we were talking about polynomial equations or trigonometric equations or logarithmic equations. So we were differentiating by what kind of functions are involved in the equation. In this particular case, I would like to differentiate equations not by what kind of functions participate in it, but how they can be solved. So I'm differentiating them by methodology of solving them. Now, in the introductory previous lecture actually, we were considering a couple of cases when we were separating x from y in the same equation, and basically relatively easily obtained a solution. So this is actually the first and probably the easiest type of differential equations, which can be solved and we know how it can be solved. So let me just give you one relatively general explanation. So if this particular equation can be represented as function of y times dy equals function of x times gx where y is dy by dx, obviously. So if we can represent this function in this format, then we can talk about separation of arguments. So y is on the left, x is on the right, or y is versus, it doesn't matter. But now we can apply indefinite integrals to both of them. So these are two infinitesimals, right? They are infinitesimals, and they are equal in the sense that the difference is an infinitesimal of a higher order than we can integrate them, and we will get the equality. So just one particular example, y plus x, y equals to zero. We can separate them. x, y goes to the right, and we will divide by y, so we will get y divided by y equals minus x, right? Now since derivative of y is this, I can write it as dy divided by y equals to dx goes here. Now I can integrate both of them, and I will get logarithm y. Well, actually it's absolute value of y, but let's just concentrate on positive x and y equals to minus one half x square, right? The derivative of x square is 2x, so 2 will cancel and minus stays. And then I need any constant on the left and on the right, or I can combine them together to be all in the right, from which y is equal to e to the power minus x square plus c, or sometimes it's expressed as e to the power of c, which is another constant c1 times e to the power of x square divided by 2. That's the same thing. Since c is any constant, well this becomes actually positive, but again we are not dealing with delicate nuances of these equations. My point right now is to get to a solution somehow, and then we can obviously spend some time to analyze when the solution exists, when it doesn't exist. That's the whole theory, which I would like actually to stay away from. All right, so this is the method of separation of function and argument, y from x. So this general equation of the first order, if it can be transformed into this case without integrals, then we can apply integrals, and that means that we have integrated or solved the equation, and this is the solution. Our purpose is to find the function which satisfies the equation, right? Well obviously I should actually check this answer, but let's assume that I did the checking, and you should. I might actually skip it just for time saving. All right, fine. So that's my first relatively simple type of equations. We call them separable, and we know how to solve them. It's all predicated obviously on the ability, on our ability to separate x from y into different sides of the equation, and then we can integrate it. Well, not always it's possible, right? So we are considering a couple of more cases when we know how to approach the problem, and I guarantee that there are many, many differential equations which do not fall into any one of these three categories which I'm going to describe today, and then, well, either you are applying certain numerical methodology or think about maybe something special and interesting, and maybe you can guess the solution or something like this. All right, so the next is so-called homogeneous equations. So the first one was separable, the second is homogeneous. Now, homogeneous is the following, and I do want to explain it on the example. So let's say you have equations. Well, it's not obvious how to separate these, so let's just assume that it's impossible. Maybe it is possible, but I don't know how. But I do know it is very interesting peculiarity in this particular equation. If I will change x to something like lambda times x, where lambda is just any number, and I will change y into lambda y, my equation would not change, right? Because this will be lambda x divided by lambda y, and lambda would cancel out. This would be lambda square x square divided by lambda square y square. And again, lambda square can be cancelled out. So my equation will remain absolutely the same. So this is a peculiar quality of this particular equation, and many other equations. For instance, you can have something like y plus x divided by y equals to e to the power minus x divided by y, something like this. It's also x divided by y. Or maybe this cannot be this way. It can be x, but then you can have y here and y here, which means you can actually have exactly the same thing. But it's not obvious that it's explicitly depend on x divided by y, but it's still the same thing. If I will replace y with lambda y and x with lambda x, these lambdas will cancel out, and these lambdas also cancel out. So it's exactly the same type of equation. So we are assuming that this is possible. So if this is possible, so after substitution from x lambda x and from y lambda x, my equation will not change, like in this particular case. Then there is a special methodology to solve these equations. And this special methodology is the following. Since x divided by y plays such an important role, let's just introduce a new function, and I will convert this equation in terms of x and y, into an equation in terms of x and z. And that would probably be simpler. I mean, I hope that this will be simpler, because this ratio seems to be important in this particular case. It's kind of a universal, right? Well, let me see if it helps. So what can I say about y is equal to x times z? y derivative is equal to derivative of the first times derivative of the first times the second, which is z, because derivative of x is one, right? Plus x derivative of z. Okay, so we can substitute it here, and we will have z plus x z prime, z derivative, plus x divided by y is one over z, x square divided by y square is one over z square. So this is my new equation. Is it better than the original one? Yes, and here is y, because we can always separate z times x equals to minus z minus one over z minus one over z square. Now, z derivative is, as we know, dz by dx. So we can put dz here, and dx and x will stay here, and we will flip to get reverse. I will use minus here dz z plus one over z plus one over z square. So we have separated. So you see, that particular equation, even if it's not separable by itself, it will become separable after this type of substitution. So as long as you know that the equation is homogeneous, which means if you multiply x by lambda and y by some number lambda, lambda will cancel out. It will be exactly the same equation. Then you can use this technique. So technique is, instead of resolving by y and x, in this particular case, you transform the whole thing in terms of z and x, solve it for z. And as soon as you get the solution for z, because this is supposed to be integrated, and I'm not going to do it, but obviously it is already as simple as it can be. Whenever you are reducing your differential equation to just a bunch of integrals, that's it. I mean, that's the maximum you can do. If you can do it, fine. If you can't do it, well, too bad, you can just leave it as is, but it's still better than the original equation. And then after you find z, you calculate y by multiplying by x. That's it. So that's the approach which you're using. So again, my purpose is to characterize a differential equation with certain quality. In this particular case, it's homogeneity, and then suggest a methodology how to solve this particular equation. In this case, it's this particular technique. And now I have the third type of equation where I can actually suggest the methodology of solving. So let's assume our equation is not separable, and it's not homogeneous, and you can actually try. First, you can try for separability. Then you can try for homogeneity. And then there is a third. It's not like everything else. It's just one more type of equation which I can recommend you how to solve it. And there are a ton of others which I cannot. All right. So this one is called linear non-homogeneous. But basically, the general format of this is this, where fg and h some functions. Well, the first thing which can be actually done to simplify it, we can divide it by f at x, and we will get something like y plus g divided by f, I will put u of x. It's a known function, right? fg and h are known functions. It's part of the equation. So the ratio is also known. And h divided by f, I put v of x. So I will consider this particular equation. So how to solve? Now, why is it called linear? Well, because it's linear by y and y derivative. You see? It's not y derivative square or y square here. Nothing like this. It's just a linear form of y and derivative of y. Okay. Now, here is the methodology which I can suggest which will bring you the results. Well, it will bring you to the moment when you can just write some integral which represents the solution. If you can take it in explicitly, like a function, it's good. If not, it's not so bad as well. It's still better than the equation. Okay. So what's the approach? Approach is the following. I will look for a solution in terms of product of two functions. Now, what does it give me? Well, obviously I can write this p times q, p, q, right? Where p and q sum functions. So I will transform this equation into a couple of equations where p and q will be participating. And I will solve them for p and q correspondingly. And then by combining them as a product, I will get y. So that's my approach. Now, why is it working in this particular case? Well, here is why. Let's just see what happens here. So our equation is transformed into the following. Derivative of y, which is this. So it's p times q prime plus p prime times q plus u times y and y is, so it's this u times p times q. Now, all of them are functions of x. I just don't want to write the function. Plus v equals to zero. This is our new equation. Let me just transform it just a little bit. So I will do this. I will put p, this and this. I will combine together. And I will have q plus u, q plus whatever is remaining. Okay. Now, what I'm going to do, I'm going to find a function q from this expression by equating it to zero. So that's my plan. First, I will find q from this equation, which is very simple. It's a separable equation, right? You can have dq by dx plus u, q equal to zero. u, q goes to the right. So it would be minus minus u, q. q goes here. dx goes there. And I have dq divided by q equals to minus u as a non-function of x times dx. So I can integrate both of them. It's a separable equation. So that's my first thing. That would, if I can find such a solution, it would bring this identically to zero. So this whole thing would be identically to zero if I will find solution to this equation. Now, as soon as I found this, so q now is a known function. I will do the next step. So since q is a known function, I will find a solution to this, which is also a separable equation, right? q is known. Now, we have found it. v is known as well. So it's basically like p is equal to minus v over q. This is dp by dx. So dx goes here. And we have a separable equation for p. And that's how I find the function p. Now, known q and known p, I get my y from here. So this is approach. Is it always working? Well, not necessarily. Maybe this is such an equation which never actually has a solution or something like this. But this is an approach. And you can definitely try to do that. So three most likely occurring cases where you really can solve something. Maybe they're happening in real life. Maybe they're happening on exams, whatever it is. But you probably would be given equations to solve. 90% of the cases would be one of these three. And knowing the methodology how to solve each one of these three types, and let me just repeat, first was a separable equation. The second was homogeneous. When we were replacing instead of y, we were using z times x. And the third one was linear where we are using the product of two functions to solve the equation for y. So these are three methodologies, three types of equations which I think is very helpful to know. Now, next lectures, I will try to concentrate on each of those guys and have some examples. Basically, next three lectures will be dedicated to these three types of ordinary differential equations. I will just try to solve as much as possible. The time allows basically, right? So the next lectures will be separately for separable for homogeneous and for linear differential equations. So far, that's it for today. Thank you very much. I do recommend you to read notes for this particular lecture on Unison.com. That's part of the calculus, ordinary differential equations. Then they have overview. This is part of the overview. So look at the website. It's very helpful, as I was saying, it's like a textbook. That's it. Thank you very much and good luck.