i'll fly over your pronounciation of Riccati and I'll grant you with an ad honorem italian citizenship

@maths_505

16 күн бұрын

I'm supporting them in the Euros along with the Netherlands and Portugal

@banjo2402

16 күн бұрын

​@@maths_505NETHERLANDS MENTIONED🧡🦁🦁🧡🧡, WTF IS A WORLD CUP?!!🇳🇱🌷🌷🌷🇳🇱🇳🇱

Your pronunciation of "Riccati" is lovely, I would say "cool". Greetings from Italy. Ciao!

Now that u mention it, I really wanna see an Italian impression from u🤣

found it quite weird having an ad for financial aid to israel via the AJU on this video, knowing your view on the subject. kind of sent a chill down my spine

What a coincidence, I was just watching some old videos of Flammy and I came across one of his videos on Riccati equations...

Hi, How quick this demonstration was! Personally I don't know by heart the formula for integrating a differential equation with an integrating factor. "terribly sorry about that" : 1:55 , 3:34 , "ok, cool" : 3:19 , 3:54 .

Nice video, next I would like to see you solving an integro-differential ecuation

Good play. Thank you.

I've looked over my work a few times. I'm trying to understand why I'm getting a slightly different answer. I'm getting that y = x^2 (1 - 2/(ce^(x^2)+1)). I took the original equation and multiplied by 1/x^2. This gave me x - (y^2/x^3) = d/dx(y/x^2). Substituting u=y/x^2 gave me x(1-u^2) = du/dx. Separating and integrating both xdx and du/(1-u^2) gave me (1/2)x^2+k = (1/2)ln((1+u)/(1-u)) for some constant k. Letting ln|c| = 2k gives me ce^(x^2) = (1+u)/(1-u). Solving for u gives me u = 1 - 2/(ce^(x^2)+1). Substituting back in terms of y gave me my answer.

Nice evaluation.

I don't have much to add to the video beside these 3 points: THe 1st thing is the importance of including the domain and codomain of the solution because, formally, a function is a triple (A,B,G) where A is the domain, B is the codomain, And G is the graph (rule of assignment). THe domain and codomain are a part of the function itself and thus , not including them means basically that one have given an incomplete answer. If 2 functions have different domains or codomains but defined by the same expression, then they are different functions. As an example, the functions f: R-->R , x |-->f(x) = x^2 and g: R+ --> R , x |--->g(x) = x^2 are different functions. Even in their graphic presentations are not the same!! one is a parabola and the other is half a parabola. The 2nd thing is to specify the range of values of integration constants because if you don't then you haven't specified all solutions. It could be the case that some values for the integration constants may correspond to false solutions. The last thing is to be careful while solving for y . One may do an unjustified step like dividing by y from both sides which means you are assuming that y is nonzero everywhere when in reality it could be zero somewhere. A mistake happened during the solving procedure which was assuming that y is equal to yp + 1/u , it could be the case that y is equal tp yp + 0 , On Wikipedia, the proposed substitution is y + z where z is the solution of some Bernoulli equation of degree 2. It is true that one solves this by making the substitution 1/z = u but this means one is asuming that z is nonzero everywhere when z = 0 is indeed a solution to Bernoulli equation. As for the equation in the video, here is the full solution if one requires the domain to be D\{0}. (we agree that the codomain is always R ) y: D\{0}-->R ; x|--> y(x) = x^2 or y(x) = x^2[1+2/(c1exp(x^2)--1)] if x>0 and y(x) = x^2 or y(x) = x^2[1+2/(c1exp(x^2)--1)] if x

1:49 y1 can also be -x^2

@maths_505

15 күн бұрын

Yes indeed. Doesn't affect the solution though.

Wait, I don't quite see how one can recover the particular solution y = x^2 from the final solution. Am I missing something here?

@maths_505

15 күн бұрын

You can solve the final equation for the desired value of C

At 2:16 Why is the general solution y=x^2 +1/u. ?

what drawing app do you use to do math?

@Commonstories-lz1mk

16 күн бұрын

Note of samsung

@geoffhuang4314

16 күн бұрын

@@Commonstories-lz1mk thanks

Is there a easy way of finding particular solutions to a differential equation? Also nice video

@maths_505

16 күн бұрын

Bro what is easier than guessing 😂😂😂

@kappascopezz5122

16 күн бұрын

For this problem, it should work to try the approach y=ax^n, which would only need some simple algebra to directly tell you that y=x² is a solution. Though it of course still needs some intuition to pick the right approach in the first place

@hrperformance

16 күн бұрын

There might be! I don't know of any such route myself and clearly this guy doesn't either but that doesn't mean there isn't one to be found. Also an easy method for you might be hard for someone else and vice versa. Don't feel bad or be put off if you find something hard that someone else easy. He did a load of baby steps starting with a guess (near the beginning) to find a particular solution, and then using the "integration factor method" to find the general solution. If your not familiar with it, it can seem very random 😅. But it's a pretty standard method. I got taught it year 1 of undergrad physics and never used it since (I just finished year 3). Don't worry if this seems too much now. You will get there! Even guessing can be tough when you don't know where to start sometimes. Especially when you first start looking at differential equations 👍🏼

@kappascopezz5122

16 күн бұрын

@@hrperformance The point of their question was that Maths 505 didn't do any "baby steps" to find a particular solution, but instead, that his guess seems to be too specific to just guess easily. But if you make a more general guess, like y=x^n, then x³-y' = y(y-2)/x becomes x³-n x^(n-1) = x^(n-1) (x^n - 2) which you can multiply out to become x³-n x^(n-1) = x^(2n-1) - 2 x^(n-1). Comparing component-wise, you get 3=2n-1 and n=2, and there you got your y=x² that Maths 505 guessed directly.

@hrperformance

16 күн бұрын

@@kappascopezz5122 cool. I actually think we both misinterpreted the question, reading it again 😅, but the person asking it can clarify if they want to. I like your route to getting the initial specific solution. But I wouldn't describe that as easier personally. But that's just my personal opinion. Also I define rearranging an equation step by step as baby steps...you know, simple arithmetic operations. But again that's just me, it's not a technical term as far as I'm aware. All the best

Why did you choose y in this way (min 2:10) ? By the way: Thanks a lot ❤

@maths_505

15 күн бұрын

What do you mean?

when we integrate like ( cos(lnx) )we can subtitue cos(lnx) by the real part of x^i then integrate, what's the proof of that?

@maths_505

16 күн бұрын

Dude didn't you ask the exact same question on a different video and someone already answered it? You can look for a proof on Google if you want but I think the comment answered your query.

@jejnsndn

16 күн бұрын

​@@maths_505 No no, he didn't understand my qustion

@jejnsndn

16 күн бұрын

​@@maths_505 I mean that how we can get the ( Re ) out of the integral

@maths_505

16 күн бұрын

@@jejnsndn a "casual" sort of proof would be to write out the integral as a Riemann sum and work from there. Give that a try and you'll definitely figure it out.

@jejnsndn

16 күн бұрын

@@maths_505 I didn't study these, sorry, originally I'm still young

I feel slightly guilty for saying Scottish Vampire. 🤦🏼‍♂️🏴󠁧󠁢󠁳󠁣󠁴󠁿🦇 Please consider it friendly roasting. :) 💜🔥

@maths_505

15 күн бұрын

No problem bro😂😂

x^3-dy/dx = y/x(y-2) x^3-y/x(y-2) = dy/dx dy/dx = -y^2/x+2y/x+x^3 -(d/dx(x^2) = -((x^2)^2/x) + 2x^2/x+x^3) dy/dx - d/dx(x^2) = -1/x(y^2-x^4)+2/x(y-x^2) d/dx (y-x^2) = -1/x(y-x^2)(y+x^2)+2/x(y-x^2) d/dx (y-x^2) = -1/x(y-x^2)(y-x^2+2x^2)+2/x(y-x^2) d/dx (y-x^2) = -1/x(y-x^2)(y-x^2)+2x^2(y-x^2)+2/x(y-x^2) d/dx (y-x^2) = -1/x(y-x^2)^2+(2/x+2x^2)(y-x^2) Let w = y-x^2 So we have dw/dx - (2/x+2x^2)w = -1/xw^2 And this is Bernoulli equation easy to solve In this equation it was easy to guess particular solution If particular solution is difficult to guess we can try to reduce Riccati equation to its canonical form or reduce it to linear second order (but coefficients might not be constant)

No, not the Irish accent 😭

early

If I'm gonna be honest, after I post this comment, there will be one more comment on this video

Isaticefied.withyour.pronounciation..nokidding.italian..!!!!