 Hi, I'm Zor. Welcome to Inezor Education. Today we will talk about ordinary differential equations of a specific type. It's called homogenous. I have introduced the concept of homogenous differential equations in the introductory lecture, and today we will just solve a couple of problems. I have three different equations which I will try to solve, and basically I think it would be a good demonstration of what is actually homogenous equations are and how to solve them. This lecture is part of the course of advanced mathematics presented on Unizor.com. If you are watching this lecture from YouTube directly or from some other website, I do suggest you to switch to Unizor.com because every lecture has detailed notes, so you can just read it as a textbook, and certain topics have exams, so you can test yourself. The site is completely free and there are no advertisements, so basically it's easier to use it than anything else. Alright, so first of all, let me just repeat what exactly homogenous ordinary differential equations are. Well, first of all, ordinary differential equations of the first order, that's something like this. So we have some kind of function which involves argument function and its first derivative, so we are talking about first order differential equations. Now, the homogenous differential equations have the following property. If instead of x, you will put lambda times x, where lambda is just any real number, not equal to zero, and instead of y, you substitute the same lambda times y, do not touch the first derivative. You will get exactly the same function, well, simply speaking, lambda would cancel each other. Well, an example, for instance, like y derivative is equal to y divided by x. Well, if you will multiply lambda by x and by lambda, lambda will cancel out and it will be exactly the same function. So the function of this type is y divided by x minus y equals to zero. And if you will substitute lambda y and lambda x, this function is exactly the same as this function, because lambda is canceling out. Another example, 3y plus 2x equals x times y derivative. If you substitute lambda x instead of x and instead of x and lambda y instead of y, lambda lambda lambda will cancel each other, so you can just factor it out and it will be the same. So these are examples of homogeneous equations. Now, homogeneous equations can be solved using a specific technique, which I am going to talk right now. And again, I did mention before, this is just a repetition. So if you have some kind of an equation of this type and this equation has this property of being homogeneous, then the suggested methodology is, let's substitute instead of function y, function z times x. So y is a function of x and z is a function of x. Now, by substituting this and substituting this, now this is a product. So it's the derivative of the first one times the second plus derivative of the second, which is one times the first. So that would be my derivative. So if I will substitute this and this in case my equation is homogeneous, that would allow to separate z from x and solve using the method of separation. And then, as soon as you find z, then you can find y from this, right? So here are my three examples, which I would like to use and to demonstrate this particular property. This particular technique of how to solve differential equations which are homogeneous. My first is x times y derivative equals x times sin of y over x plus y. So first, we have to check that this is a homogeneous differential equation. So if I will substitute instead of x lambda x and instead of x lambda x and instead of y lambda y, you see that we can very easily cancel lambda out. So this is a homogeneous equation. Okay, since it's a homogeneous equation, by the way, I didn't really put it here, lambda and lambda, it will also cancel out. Forgot to do it. So lambda here, here, here and here and here. So all lambda's cancel out, so it's a homogeneous equation. Now, how to solve it? Okay, this is a suggested methodology. Now if I will substitute this to this, what will happen? To make my life easier, let me just divide this by x. So I will have my original equation can be rewritten as this. And it's easier because y divided by z is x. Now let me just mention here, I divided by x. Well, obviously this function is not defined as x is equal to zero, right? So when we are trying to solve differential equations, we are not as, I would say, rigorous with dividing by something which contains an unknown x, etc. Because we are not looking for the roots of the equation in algebraic sense. We are talking about the function and if this function is not defined at x is equal to zero, I can just divide by x is equal to zero because this function is also not defined if x is equal to zero. So that's why we are not really paying attention to these nuances and in the future probably I will skip all these considerations. Now this is easier because y divided by x is z, right? So what do I have is, so instead of y derivative I have to substitute this equals to sin of z plus z, right? Now z cancels out and what do I have? I have z derivative which is dz by dx times x is equal to sin of z. Looks very simple, right? So I can separate dx and x goes to the right, sin z goes to the left and my final equation in terms of z and x is dz divided by sin z equals dx divided by x. So what's my next step in this case? How can I define, how can I find z in this particular case? Well, obviously I integrated to get rid of the differentials, right? And hopefully I will have some kind of a function on the left of z and some kind of a function of the x and from their equality I can or cannot find z doesn't really matter but at least I will not have differentials, right? Okay, so now on the right it's easy, right? On the right it's basically logarithm of absolute value of x. Sometimes I skip absolute value and just say that, okay, let's just forget about negative x. Let's consider the function which is defined only for positive x and then I don't have to have this little complication like absolute value. Well, sometimes not, doesn't really matter. Now, this is the difficult part. How can I find this particular integral? Okay, yes, that's not easy and now let me just think about it. Sine under the integral is always good if it's also, if under integral you also have a cosine which can be brought into the differential, right? Because the cosine is the derivative of sine and that's how I can get rid of the sine. Well, I can actually introduce cosine here by doing the following. Integral dz. Now sine of z I will represent as sine of z over 2 times cosine of z over 2 because this is a double angle, right? This is a sine of z over 2 plus z over 2, right? So that would be sine of z. So that's the same thing. Now I have the cosine. However, I have it in the denominator. I would like to have it in the numerator because then if I had it here, I can combine cosine z and dz into differential of sine of z, right? Or z over 2 doesn't matter. I mean, what matter is it's in the wrong place. So what should I do? Well, what I can do is I can actually multiply cosine z over 2 and multiply the bottom cosine over z, right? Now I have this cosine which I can combine with dz to get something like differential of z. Now this 2, I can combine to z and I will have dz over 2, right? This dz and this dz and this 2, it's a differential of z over 2 times cosine of z over 2. So this is differential of sine of z over 2 and now I have sine of z over 2 and cosine square which I can also express as 1 minus sine square z over 2. So now I have everything expressed as sine. It's d of sine z over 2, right? Divided by sine of z over 2 times 1 minus sine z over 2. So what do I do in this case? So you see everything is sine square. So everything is expressed in terms of sine. So what people do in this particular case? Obviously we have to just substitute my sine of z over 2 as t and my integral is integral of dt divided by t times 1 minus t squared or dt dt, t 1 minus t 1 plus t, right? 1 minus t squared is 1 minus t times 1 plus t. So I can actually take this integral and this is not very difficult. All I have to do is whenever I have some kind of a polynomial which is represented as a product of first degree polynomials t1 minus t1 plus t it's always good to represent it as sum of this thing. So this can always be represented in this thing where a, b and c are some kind of coefficients. How can I find these a, b, c? Well, this is easy. Well, if I will take this common denominator which is this one I will have a times 1 minus t times 1 plus t. So it's a times 1 minus t squared plus b times t times 1 plus t plus c, c and 1 minus t. And this should be equal to 1. Now this is what? a minus at squared plus bt plus bt squared plus ct plus minus minus ct squared equals to 1. So what does it mean? It means that my coefficients are t squared. What do I have at t squared? Minus a plus b coefficient at t would be bt, so it's b plus c. Oh, by the way, I forgot ct squared, sorry, minus c. Now t would be b plus c and the free is a. Now, and this should be identical for any t, which means this should be equal to 0, this should be equal to 0, otherwise my dependency on t would be substantial and a should be equal to 1. If a is equal to 1, so I will get minus 1 plus b minus c is equal to 0 and b plus c is equal to 0. So I have a system of two linear equations with two variables, which can be easily solved and that's how I will get my a, b and c. I mean a I have already got. So b minus c is equal to 1, b plus c is equal to 0, so b is equal to minus c. Going to here minus c minus c is equal to 1, c is equal to minus 1 half and b is equal to 1 half and a is equal to 1. Okay, so I've got the representation. I have my integral of a. It's this d t, b is this and c is minus 1 half t. So each one of them is easily taken. Now t, by the way, is sine of z over 2. Remember this, right? So each one is, this is logarithm t. This is minus logarithm t, logarithm 1 minus t and this is again minus logarithm 1 plus t. I can use absolute value or whatever. Alright, so basically where t is this. So basically I have expressed this left part as the function of z and the right part is logarithm x. Unfortunately, I cannot really resolve this, at least not easily, for z as a function of x in this particular case because this is logarithm and under the logarithm I have sine of z divided by 2 and this is all this is equal to this integral which is logarithm of x. It's not easy, alright? So in many cases solution is basically retained in this not exactly clear format as y is equal to y of x as a function, like in this particular case. I'm actually indirectly defining my y as some kind of a function which satisfies certain equation f of x, y equals to 0. For instance, if I cannot resolve it for y. And this is unfortunately an example of such kind of a thing. Yes, I can probably use this as an exponent. So I will have a product of t times 1 minus t times 1 plus t is equal to x, but still there will be lots of signs in between signs of this function. So it's not easy to resolve it for z and consequently for y. So in this case we just retain it as is. I think my other two examples are a little bit more resolvable if you wish, they do have some final product. But this, it looks like a simple thing, right? And yet the result is unfortunately kind of very difficult to resolve for function y to get an exact explicit expression for y as a function of x. It happens. But as an example, I think it's just good. Next. So what's my next? Next looks a little bit more complicated, but result will be a little bit more resolute. y minus x, y divided by x in the power of x is equal to e to the power of y. Okay, so first of all we have to check if this is the heterogeneous equation. Well, within these brackets everything is obviously heterogeneous because I can always represent it as y divided by x minus y. And if I will substitute y lambda instead of y and lambda x instead of x, these will cancel out. But how about these guys? Well, why don't we raise both sides into the power of 1 over x? 1 over x. What happens? Whenever you are a to the power of b to the power of c is equal to a to the power of bc. Remember this, right? So if this, this is a to the power of x and then to the power of 1 over x, I will have a to the power of x times 1 over x, which is 1, right? So I will have just this. So on the left I will have just this. On the right I will have e to the power of just multiplication. y divided by x. So now I also have this y divided by x, right? So whenever I am replacing y and x with lambda y and lambda x these lambda will also cancel each other. So it is heterogeneous, homogenous, sorry, homogenous equation. Alright, so let's try to solve it now. Homogeneous equation and I will try to solve it from this latest incarnation because it looks simpler. Now, why does it look simpler? Because y over x would immediately be z, right? So it is z minus. Derivative of y is this. So I will subtract zx minus z and it is equal to e to the power of z. So this cancels out and I have dz by dx equals minus e to the power of z, right? Minus, I put it on the right side. Or let's divide both sides by e to the power of z and multiply by e to the power of minus z and I will have dz e to the power of minus z is equal to minus dx divided by... Oh, sorry, I didn't put... It should be x here, divided by x. And this is easier to integrate explicitly because the integral of this is minus logarithm of absolute value of x. Integral of this is minus e to the power of minus z, all right? The derivative of this is minus and then derivative of e to the power of minus z which is e to the power of minus z times derivative of minus z which is another minus, it will be plus. So I have this particular equation. Am I right? I think I missed a sign somewhere. Minus dz in x. So this is z is x is equal to minus e to the power of z. So dz by dx times x is equal to minus z, okay? dx divided by x e to the power of minus z and I integrate this. Yeah, seems to be fine. Seems to be fine. So we can get rid of this. So minus z is equal to... I have to logarithm both sides. Logarithm of logarithm of absolute value of x. Or z is equal to minus and y is equal to zx, minus x times logarithm of logarithm of x. It's kind of ugly function, but it's an explicit solution. Well, obviously there is a constant somewhere which I kind of ignored. So it's probably here. And if I do this constant, if I will... Yeah, I have to edit here probably as part of the solution. Okay, anyway. What's most important? What's most important is that using this substitution y is equal to zx. I'm converting that ugly equation that I had before into another equation where I can separate function from its argument. Okay? And finally I have the third example and I don't think that it's resolvable explicitly. But nevertheless... Okay, x times y times y derivative is equal to x plus y square. Okay, now what can I do about this? Well, first of all, is it homogeneous? Well, yes, because if I will replace x with lambda x and y with lambda x I will have lambda square here. And here I will also have lambda and lambda square. It will be lambda square outside of the parenthesis. So lambda square and lambda square would cancel each other out. So it is homogeneous. Okay, now, y is equal to zx. So what do I have in this case? And y's derivative is equal to zx plus z. Alright, so I have x times y, which is zx, times y derivative, which is this, equals to x plus y. It's x plus zx. So x can be factor out. So it will be x square, 1 plus z square. Right? So this x square goes out. And what do I have? I have z times z derivative times x, plus z square is equal to... If I will open this, it's 1 plus 2z plus z square. And z square also goes out. So what do I have? I have z divided by 1 plus 2z, if I will go down. dz is equal to dx divided by x, right? x goes to the right, to the denominator. dx used to be here. It goes to numerator. This goes to denominator. And z and dz is a numerator. So, all I have to do now is take these two integrals. Now this one again, it's easy, it's logarithm. Now how about this one? It's the same kind of a problem which I had before, which I can resolve by representing this as a sum of two simpler fractions. It's 1 half 2z, right? Instead of z, I need 2z, plus 1 minus 1, divided by 2z plus 1, right? That's the same thing. Now this is equal to 1 half of this divided by this. It's 1 minus 1 divided by 2z plus 1. And now my integral is very easy to take because it's equal to integral 1 half minus 1 divided by 2z plus 1. And I can actually do it as two different integrals, dz and dz. Each one is easy to take, so this one is dz. It's 1 half z. This one is minus. I need, well, this is what? 4z plus 2, right? 4z plus 2, so I need 1 quarter to get 4z here, plus 2. So I will have minus 1 quarter logarithm of 2z, no, 4z plus 2. And it's equal to this integral which is logarithm of x plus z. All right, so again, this is something which is not easy to resolve for z because you see this is a combination of logarithm and polynomial function, the first degree. But nevertheless, it allows you to represent y and x in one bigger formula. So instead of z, we can always put y divided by x and that would be an expression which combines together x and y, which defines the function y as a function of x, not explicitly but still defines. Okay, basically these are examples I wanted to present to you. A couple of lessons. Well, first of all, differential equations do not necessarily get resolved in explicit function like y is equal to some kind of function of x, not necessarily. Sometimes we have to be satisfied if we have just some kind of functional dependency which is not explicitly represented as a formula. Well, which is still better than original differential equation. So we still have to appreciate the fact that we got rid of the differential equation which came up with just a functional dependency between x and y. And then obviously there are numerical ways of addressing the problem of what exactly my y, if my x is this, because all these equations, which are not explicitly defining y as a function of x, can be numerically defined and numerically resolved. Okay, that's it. Thanks very much. And I do suggest you to look at the Unisor text for this lecture, notes for this lecture. It's a little bit more in more details maybe than whatever I was just presenting to you because I wanted to convey the idea of how to approach homogeneous differential equations. And there are some more details maybe in the text. That's it. Thanks very much and good luck.