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
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 .
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.
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
He skipped one step. In fact you first substitute y by y = y1+y2 (y1 is the particular solution and y2 the complementary one). That is, you substitute by y = x^2 + y2. Once you substitute the result is a Bernoulli differential equation in terms of y2, which then again we can solve by substituting y2 = 1/u. All in all you can directly substitute by y = x^2 + 1/u in the Riccati equation.
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
He skipped one step. In fact you first substitute y by y = y1+y2 (y1 is the particular solution and y2 the complementary one). That is, you substitute by y = x^2 + y2. Once you substitute the result is a Bernoulli differential equation in terms of y2, which then again we can solve by substituting y2 = 1/u. All in all you can directly substitute by y = x^2 + 1/u in the Riccati equation.
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 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.
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
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 👍🏼
@@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.
@@kappasphere 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
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)
i'll fly over your pronounciation of Riccati and I'll grant you with an ad honorem italian citizenship
I'm supporting them in the Euros along with the Netherlands and Portugal
@@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
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 .
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.
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
What a coincidence, I was just watching some old videos of Flammy and I came across one of his videos on Riccati equations...
Nice video, next I would like to see you solving an integro-differential ecuation
Nice evaluation.
is there a trick to find the particular solution quickly then using a lot of combinations?
Good play. Thank you.
1:49 y1 can also be -x^2
Yes indeed. Doesn't affect the solution though.
At 2:16 Why is the general solution y=x^2 +1/u. ?
He skipped one step. In fact you first substitute y by y = y1+y2 (y1 is the particular solution and y2 the complementary one). That is, you substitute by y = x^2 + y2. Once you substitute the result is a Bernoulli differential equation in terms of y2, which then again we can solve by substituting y2 = 1/u.
All in all you can directly substitute by y = x^2 + 1/u in the Riccati equation.
Great!
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?
You can solve the final equation for the desired value of C
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
Why did you choose y in this way (min 2:10) ?
By the way:
Thanks a lot ❤
What do you mean?
He skipped one step. In fact you first substitute y by y = y1+y2 (y1 is the particular solution and y2 the complementary one). That is, you substitute by y = x^2 + y2. Once you substitute the result is a Bernoulli differential equation in terms of y2, which then again we can solve by substituting y2 = 1/u.
All in all you can directly substitute by y = x^2 + 1/u in the Riccati equation.
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?
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.
@@maths_505
No no, he didn't understand my qustion
@@maths_505 I mean that how we can get the ( Re ) out of the integral
@@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.
@@maths_505
I didn't study these, sorry, originally I'm still young
what drawing app do you use to do math?
Note of samsung
@@Commonstories-lz1mk thanks
Is there a easy way of finding particular solutions to a differential equation? Also nice video
Bro what is easier than guessing 😂😂😂
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
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 👍🏼
@@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.
@@kappasphere 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
I feel slightly guilty for saying Scottish Vampire. 🤦🏼♂️🏴🦇 Please consider it friendly roasting. :) 💜🔥
No problem bro😂😂
No, not the Irish accent 😭
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)
If I'm gonna be honest, after I post this comment, there will be one more comment on this video
early
Isaticefied.withyour.pronounciation..nokidding.italian..!!!!