 Hi, I'm Zor. Welcome to Unisor Education. We have started talking about ordinary differential equations, and I would like to continue basically exemplifying one particular type of these equations, type in terms of method knowledge of solving using the separation method. All I will be doing today is just solve a few problems precisely using this particular method. This lecture is part of the course of advanced mathematics presented on Unisor.com. I suggest you to watch the lecture from this website. You just go to the main menu, calculus, ordinary differential equations and types of equations. That's where you will find this particular lecture on separable differential equations. Before using this website rather than let's say straight YouTube, for instance, or whatever else what other website provides you is that every lecture has very detailed notes and also some topics are accompanied by exams which you can just take. The site is completely free, no advertisement. Alright, so let's talk about one particular type of differential equations, the separable differential equations. Now we are talking about ordinary differential equations which means only the function of one argument is getting involved. Like y is equal to y of x. That's one function of one argument. Now we will be talking only about differential equations of the first order which means only the first derivative is involved. And in addition we will be talking about separable equations. This was actually defined in the previous lecture which was kind of introductory to all differential equations. I explained and exemplified actually what this is and today I will continue talking about these separable equations. So our methodology to solve differential equations can be different, obviously. Different equations require different approach. But there is one particular class of equations which I did talk about in the introductory but I will repeat right now. The equations which eventually can be exemplified as this type. Or if you wish you can change it. Or, and this is the final form we would like our equation to be represented because we can do this. Now this integral is basically integral of one variable y. This is one variable x and presuming we can take these indefinite integrals and we will have something like f of y equals to g of x plus c where this is an indefinite integral of f and this is indefinite integral of lowercase g and from this we, well, if we are lucky we can actually take it, resolve it for y and that's completely the function we need. Well, in terms of these, that would be y is equal to f minus one g of x plus c but this is an inversion of f. So that's the plan. Now in some cases this plan works and I'm going to present four different problems where in transformations this plan will actually be working and to tell you the truth it's probably one of the more often occurring cases of differential equations which you might actually meet in tests, exams, whatever. Not as much in the practical life. I mean there are different differential equations in practice. I mean in real science like physics, chemistry, stock market, etc. and they're not necessarily so easily solvable but this is a particular type which you can just think about and consider this as one of the fundamental types of equations which you can solve. So let me just go to examples and the best thing to learn this thing is practice of this. The more equations you will solve the better you will be equipped for the future. Plus again the purpose of the whole course is maybe not as much to put some knowledge which you remember into your heads but rather to develop your creativity your logic, your ingenuity, etc. and manipulation of the equations is a great tool to develop your creativity. Alright, so let me just go. First example y derivative plus xy plus y minus x equals to 0 where y is the function of x obviously. Now it doesn't look immediately like this type of equation which can be separable but you can actually do some manipulation with it or equal to 1. So what kind of manipulation you can take? Well let's say this y is equal to, now we will factor out y so it's minus yx plus 1 because these are on that side of the equation which would be also on this side. Plus x plus 1 Now what we do, we obviously factor out x plus y and I will have 1 minus y right? If you factor out x plus y 1 minus y. Now I can actually do this separation so there are certain transformations which you might take which will lead you to a transformable separable equation. So now I can say that now this is obviously dy by dx so I will put 1 minus y to the left dx to the right and I will have dy divided by 1 minus y equals 2 dx Now to make it even simpler I will do the following what is differential of 1 minus y well this is a constant so it doesn't really matter so it's differential of minus y which is minus dy. So I can replace dy as with a minus sign because I can take the minus here. Now differential of x plus 1 and differential of x are exactly the same so I can put it this way. Now that's easy because if I will integrate these two things this would be minus goes out, it would be logarithm of 1 minus y. Now in theory integral should be with absolute value of this but let me just ignore this for a while just for simplicity I will use regular so let's assume that I am looking for solutions among positive x and positive 1 minus y. Yes it restricts a little bit the solution but I don't want to get into the details about considering y is greater than 1, y is less than 1 this is not the purpose of this lecture, the purpose is to get to some kind of a solution and show you the way. Now this is technicality this is probably more important where you really have to think about how to convert it into a separable equation but now this is a technicality and this is equal to what? 1 half x plus 1 square plus c right? Now obviously I can put minus on this side. Now plus or minus it doesn't matter because c is just a constant, it's any constant and finally 1 minus y is equal to e to the power so I am basically raising both, I am using both as exponent so e to the power of logarithm is 1 minus e y minus y and e to the power of this I will just retain as x plus 1 square divided by 2 and e to the power of c might be actually in front of e as another constant because it's a multiplication e to the power of this plus this times e to the power of this and since this is any constant I can put any constant here as a multiplier well from which y is equal to 1 minus y is equal to 1 minus c power minus x plus 1 square over 2 by the way plus or minus again it doesn't really matter here because this is the constant which can be positive, negative etc alright now I did kind of shorthanded certain piece of this solution by not actually being very specific about logarithm of the module of absolute value of something or c should be positive or negative again these are technicality which I don't want to spend too much time on my most important goal was to convert to transform my original equation to separable format and then to separate y from x so I can integrate it and then yes obviously to do it very very rigorously I should have really be more accurate with absolute value and constant c alright next y minus y derivative minus x plus 1 e to the power x plus y equals to 0 well again it doesn't look immediately that this is a separable because this is x plus y but let's just remember that this is the sum in the exponent and this thing is actually e to the power of x times e to the power of y now let's move everything to the right and I will have y is equal to now we know how we can separate it right instead of multiplying here we will multiply here by reverse right which in turn goes to e to the power of minus y dy equals x plus 1 e to the power of x not x plus 1 x dx and now I can integrate both how to integrate this well this is easy that's if I will differentiate e to the power of minus i y it will be e to the power of minus y times minus so I need plus so it will be this that's the integral of this plus c but we will use plus c on this side now this thing how can I integrate this well this is a typical example of how I can use the integration by parts if you remember the integral u dv is equal to uv minus integral of v du right remember this integration by parts so this is u and this is this is v because differential of e to the power of x is equal to e to the power of x times dx because the derivative from e to the power of x is e to the power of x so I can replace this with this so this is u and this is v so I will have x plus 1 1 times e to the power of x minus integral e to the power of x d of x plus 1 well the derivative of x plus 1 is the same as d of x which is equal to x plus y e to the power of x minus what's the definite integral of e to the power of x it's e to the power of x plus c now what is this e to the power of x can be factor factor out I will have x plus 1 minus 1 so it's x times e to the power of x now we obviously transfer this minus to this c is a constant so it doesn't really matter plus or minus from which logarithmic logarithm of this would give me would give me minus y logarithm of this c minus x e to the power of x and obviously this is the final expression for y that's it so what was important here I just recall that e to the power of x plus y is e to the power of x times e to the power of y and that allowed me to separate things now next logarithm of y derivative equal to x plus y well this is actually very close to whatever it was before because right now I can raise a e to these two things so e to the power of logarithm of something would be y derivative and e to the power of x plus y would be e to the power of x times e to the power of y which is separable so I will put e to the power of y on the left I will have e to the power of minus y dy equals e to the power of x dx so now I can very easily integrate it we did this before that's minus e to the power of minus y with e to the power of x plus c so let's change the sign here so minus y is equal to logarithm of c minus e to the power of x and plus y is equal to the minus here so that's the answer that's even simpler than the previous one right and the last I wanted to do something with trigonometry just for illustration and so you don't really forget the trigonometry so my example is sin of y and y derivative equals sin x plus y plus sin x minus y well sounds too complicated right for separation I mean not immediately obviously I have to do something now although this thing is good because everything with y now this thing is a mixture but let's just recall that sin of x plus y is equal to sin of x okay I need another sin of x plus y is equal to sin of x cos sin of y plus cos sin x sin y now sin of x minus y the second one is equal to almost the same but with a minus sign here this is a plus this is a minus now this is the sum of these so I have to sum them up if I will sum them up this thing is going down it cancels out and all I have is 2 sin of x cos sin of y and now obviously I can separate them by bringing cos here so I will have sin of y dy by cos sin of y equals 2 sin of x dx clear separation now what can you do with this now this is easier well let's recall that the cos sin of y if I will take a derivative I'm talking about y only as an independent variable that's minus sin right of y so this thing can be converted into minus d cos sin of y divided by cos sin of y sin times dy is differential of cos sin but I need a minus sin because of that here I have already what I have 2 sin of x dx now how can I integrate this well that seems to be easy now this is what the logarithm of cos sin of x with a minus sin right that's what it is derivative of logarithm is 1 over argument derivative of this well again sin is a derivative of minus cos sin so that would be minus 2 cos sin of x plus c almost done okay so obviously minus on the left and minus on the right cancel each other I raised e to this power and I will have cos sin x is equal to e to the power of 2 cos sin x plus c and now again I will do some not exactly rigorous thing I will just put this is y and this is y and I will just put instead of cos sin I will take arc the reverse function arc cos sin right now if I will take arc cos sin from the left I will get y and on the left on the right I will have arc cos sin of the power of 2 cos sin of x plus c well quite a crazy function but function nevertheless now why did I talk about this arc not being exactly rigorous well because again it's the function which this is not completely equivalent to this that's all I'm saying this is one particular solution but there are some others because I can add periodicity obviously stuff like this don't want to think about this right now so again my approach was right now to transform the original equation in such a way that it will allow separation of x from y x on the right y on the left something like this and then if my derivative is just stand alone multiplier I can represent it as divided by dx put dx to the right and now I have two integrated expressions completely independent of each other y on the left x on the right which I can integrate separately and get something like this and then resolve it by y if possible alright so what I suggest you to do is take a look at the notes to this lecture and do yourself everything whatever four examples I suggested if you get the same solutions great if you don't look at the concrete notes I mean notes contain the solutions as well I'm just trying to say that don't really pay attention to solution try to find this solution yourself and check it with the answer right okay that's it for today thank you very much and good luck