A riccati differential equation

Here's an example on how to solve a class of differential equations called ricatti equations.

My complex analysis lectures:

If you like the videos and would like to support the channel:

You can follow me on Instagram for write ups that come in handy for my videos and DM me in case you need math help:

My LinkedIn:

Advanced MathWear:

[silviatotaro]

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

[kingzenoiii]

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

[Tosi]

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

[aryaghahremani]

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

[OptiInfo]

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

[CM_France]

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 .

[onegreengoat]

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.

[ericthegreat]

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

[YouTube_username_not_found]

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<0 where c1 <= 0 and c2 <= 0 .

It is important to exclude the positive values of c1 and c2 because including them would require the solution to be undefined on the values that sets the denominator to equal zero ( cexp(x^2)--1 ) meaning that the domain is no longer equal to D\{0}.

[azmath]

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

[RandomBurfness]

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?

[michaelneal]

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

[holyshit]

Have you tried to find solution of general Riccati equation without particular solution
and without reducing to second order linear
Why not reducing to second order linear ?
Because we wont be able to use Riccati to solve second order linear

[wuyqrbt]

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

[jejnsndn]

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?

[Vendine]

what drawing app do you use to do math?

[orionspur]

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

[holyshit]

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) =
d/dx (y-x^2) =
d/dx (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)

[threepointone]

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