The Riccati Equation Lesson 

I know it is pretty randomly asking but do anybody know a good site to watch newly released movies online?

@Luca Jensen Flixportal :)

@Ruben Solomon Thanks, I went there and it seems like they got a lot of movies there :D Appreciate it !!

@Luca Jensen Glad I could help xD

Maria Golemi @ hi maria if u need some help in ricati then u can contect me

So do you understand why the form y = y1 + 1/v is the general solution of the Riccati eq? I have a feeling that the proof in the vid does not state that the form y = y1 + 1/v is the general solution. Please, correct me if I misunderstand something.

@@phandinhthanh2295 I have the same question. Did you find out? I know it was 2 years ago hehe. In book my they use different symbols, and a little different method but if I understood right, it is because the general solution to the Riccati equation is formed by adding a particular solution, here y_1, to the general solution to the Bernoulli equation. Where the Bernoulli equation, is the one you get by setting c(x)=0. So they add the particular solution, here y_1, to the general solution for the "adhering" Bernoulli equation, which I here assume is 1/v?? Hope it makes a little sense at least.

www.wolframalpha.com/input/?i=differentiation+erf%28x%29

I think you might be right. D'oh! I hate when that happens.

Jeremy Klassen either way.. please don't exclude this video. The main idea is pretty easy to understand by your explanation.

Yeah... that is a bit confusing. The integral in the denominator gives us the area under the normal curve, and when we let the upper bound of the integral go to infinity, the area is the square root of pi over 2. Now, the thing is, the upper bound in that integral isn't infinity, it is x. So the value of the integral won't exactly be the square root of pi over 2. The erf(x) is an error function that is intended to adjust the answer based on the value of x in the upper limit. Basically it is saying that the value of the integral is going to be the square root of pi over 2 multiplied by some scale value that depends on the value of x. This integral cannot be solved analytically, so the exact nature of that error function is not clear, but we can use computers to give us the value to whatever precision we need.

It has been a while since I have worked with differential equations, but maybe I can help. What is the specific question you are working on?

Thanks for your comment. First I want to say that it has been a long time since I spent a considerable amount of time working with differential equations and so I am not overly confident. If I remember this correctly then the substitution of y=y1+1/v was used to turn the original equation into a linear equation. So I believe you are correct that it isn't the general solution. Again, I am not terribly confident in my memory of how to solve these questions.

@@JeremyKlassenThePiMan Despite that, all the websites I read and the books I have stil claim the form y = y1 + u to be the general solution of the Riccati eq. Perhaps there is some underlying theory here we don't know. Maybe it's proved in more advanced courses. Thanks anyway.

Sorry AHK not AKH .ok .

Sorry it has taken a bit to respond. It's an interesting thing to have a factor of y^2 pulled out. Usually a Riccati equation is quadratic in the dependent variable, and it sounds like yours might not be simply quadratic. It has been awhile since I have worked with this though. Could you send me the equation?

Jeremy Klassen y'+x+y=1+xy^2

Okay. Thanks for sending that. I think I can help. Try letting y_1=1 and make the substitution y=y_1+1/v or in reality you will substitute y=1+1/v. When you simplify the result you should get a linear first order equation that you can solve with an integrating factor. Let me know if that isn't clear.

That is a good question, and it has been asked and answered before with, I think better responses than I have given. If you scroll down a bit in the feed you will find the comments with responses by myself and others. I think they will help with this. Take care.

I can understand why that might be confusing. Once it is established that A is a function of x sometimes we will adopt a lazy way of writing it and not include the (x) part. I don't think it is great notation, but it is not uncommon.

@@JeremyKlassenThePiMan oh ok thanks! That’s about as easy of an explanation as I could hope for.

That is a good question. In a lot of academic scenarios it is given, but often these are solutions that are... "obvious" for lack of a better word. Sometimes there are solutions that clearly present themselves but there are others that are a bit deeper that we need to look for. Please not that solution might not always be easy to see, but it should be easy to verify.

In many cases it is somewhat clear from the question, but honestly, based on an earlier comment, I might not be the best example of this.

@@JeremyKlassenThePiMan OK thank you... But what I understand about particular solution is that we just have to put some value in the give differential equation. And if we get 0 on both sides then it'll be your y1. If I'm not wrong?

@@riya2456 that is my understanding

That is an excellent question. Typically, when you are solving these in a course setting one of two things happens. Either the particular solution is provided, or it is one that is pretty obvious - as in, it should jump out at you as a solution. Usually it takes a bit of practice to gain some experience before that second situation can be expected to happen.

@@JeremyKlassenThePiMan yes! I've been doing research for final project (to be graduated from my bachelor) using differential fractional riccati equation, it turns out the equation is dy/dt=t+y-y^2 and I find it hard to solve manually this type of equation when the particular solution isn't available. I try to find it but still I can't. Could you please help me? I'm really thankful if you did..

I just want you to know I'm working on this. I have taught high school for quite a while and so I am out of practice with DE's. I'm trying to find some time to have a good look at your problem. Were there any boundary conditions provided?

@@JeremyKlassenThePiMan it's very kind of you.. thank you for give me your time to think about it. The value of t is between 0-1 (0

So it turns out this one is a bit more involved. I didn't want to leave you waiting for a long time so I forwarded this to some people I know would have a better, quicker idea of how to approach it. Here is what has come my way. So this is a more difficult one because, although the coefficients are pretty simple, it doesn't reduce to simple functions. In fact, it looks like it might be a big solution. I didn't get a full solution, but this should get you going. First we have to change the form. Try substituting v=-y and you will get v'=-t+v+v^2, so at least the signs are nice. Then let v=u'/u and with a little bit of manipulation you will get u"-u'-tu=0. This appears to be where this gets a bit harder because according to the professor I was in contact with it looks like you are going to get a series solution. She said the the first few terms of u(t) were u(t)=t+1/2t^2+1/6t^3+1/16t^4+1/80t^5+.... Then to get your solution you let y=-v=u'/u which is a ratio of two infinite series... not cool. This is about all I can do right now. I hope it helps, although I suspect I have missed the window where this is useful information.

I'm not sure what you mean.

Assume it as only y1 you will get the solutions

Hi. I am working on this, but I spend most of my time with the high school curriculum so I have to brush the dust off. After playing with it for a bit I found a particular solution y=x^2. So right now I am using the substitution y=x^2+1/v to reduce his to a linear first order DE. Now I am working to solve that, but the integral is taking a bit longer than I wanted and I have to look up a couple of steps that I haven't used in a long time. I wasn't sure if the steps I just mentioned might be of help to you.

Here is my attempt, but I had some difficulty at one point and I am not super confident with this material anymore. I just hope it gives you the help you need. I'm sorry if it isn't helpful. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-wb3WJZWtslI.html

Hang in there bud

Same situation in department of mathematics. I'm still trying to pass but mental health is not good.. i hope you get through it