That's so with all Clairaut DEs. The singular solution is called the envelope of the linear solutions, because it's tangent to all those solutions. There is indeed great beauty in that.
@@nolanrata7537 saying it is of the form u’/2sqrt(u) is doing a substitution. Maybe you mean that you can recognize that it is of this form and do the integration without writing it down? But doing a substitution in your head is still doing a substitution.
@@megalomorphNot really, here it's easy to see that sqrt(1-x^2) is a primitive of -x/sqrt(1-x^2), without any change of variable needed. A similar example would be integrating 2x*exp(x^2), you can see that a primitive is simply exp(x^2) without needing to do any substitution.
@@nolanrata7537 Maybe this is a definitional thing, where we disagree on exactly what u-substitution means. For me, u-substitution is what happens when you use ethe FTC on the integral of the chain rule. It is the act of noting when you have a composition of functions, the outside functions you know the anti-derivative of, and an additional multiplicative factor of the derivative of the inside function. There are formal processes you can go through to aid this when the integral is particularly hairy, or that let you get to an intermediate point when you don't immediately know the anti-derivative of the outside function, but that is an implementation detail that is not strictly necessary. The fact that you can do some substitutions by inspection once you have experience does not make it not substitution. The u-substitution is simply the process by which you recognized that your thing was the derivative of a composition, regardless of what you write down.
I was able to guess a solution based on the fact that the non-sqrt terms look like xdy-ydx which is the polar angle differential. You then get -r^2dtheta = ds. Since circles are rdtheta = ds, this works for a circle of radius 1 (or negative 1, but that’s geometrically the same object)
I'm not used to comment on UA-cam, but I think I have something to say with this one. Without any previous knowledge of non linear ODEs, my first approach was trying to define dy/dx = tan(u), with u some function of x to avoid the sqrt. After substituting in the original equation, we obtain a nice definition of y in terms of u and x. Then, I derivated the whole thing with respect to x, obtaining a new definition of dy/dx. After making equal both definitions of dy/dx, I eventually dealt with the same two scenarios (in my case, du/dx = 0 implied a soution of the form y = xtan(C) + sec(C), equivalent to the one you showed). I found really surprising how you deal with it without struggling with trig as much as I did. Thanks for your effort and nice explanation!
An alternative approach after differentiating is to use the first equation to solve for sqrt(1+y’^2), then substitute into the second equation. This yields a slight different separable equation to solve.
Hey all! Bilingual Fr/En here, the best way to explain it would be: - the word "clay" - a guttural r, coded ʁ in phonetics (I think it doesn't exist in English) - the same "o" sound than in "row", just without the "w" diphthong Ta-daa! klɛʁo in IPA Reader
Something about the solving procedure of Clairaut's equation boggles me; Are we allowed to differentiate y' ? Who said that y should be twice differentiable? I think one should prove this.
@@maths_505 All hail the algo. /s (I look forward to a time when you no longer need it; your mostly useless video titles still bother me, but I understand why you do it. I'm just glad you don't have some "shocked face" graphic on the title card as well.) As an aside, this is one of your most elegant surprises at the end. I imagine someone with deeper imagination than I would have seen it coming sooner, but the Unit Circle falling out at the end is quite beautiful.
It becomes even more beautiful when you realise that the linear solutions are actually ALL tangents to the unit circle
That's so with all Clairaut DEs. The singular solution is called the envelope of the linear solutions, because it's tangent to all those solutions. There is indeed great beauty in that.
For the u-sub near the end, you don’t have to do a trig sub because you have the x in the numerator. Doing u=1-x^2 is simpler.
Can't let Weierstrauss have all the trig substitutions.
You don't need any substitution, the integral is already of the form u'/2sqrt(u) and integrates directly to sqrt(1-x^2) + C
@@nolanrata7537 saying it is of the form u’/2sqrt(u) is doing a substitution. Maybe you mean that you can recognize that it is of this form and do the integration without writing it down? But doing a substitution in your head is still doing a substitution.
@@megalomorphNot really, here it's easy to see that sqrt(1-x^2) is a primitive of -x/sqrt(1-x^2), without any change of variable needed. A similar example would be integrating 2x*exp(x^2), you can see that a primitive is simply exp(x^2) without needing to do any substitution.
@@nolanrata7537 Maybe this is a definitional thing, where we disagree on exactly what u-substitution means. For me, u-substitution is what happens when you use ethe FTC on the integral of the chain rule. It is the act of noting when you have a composition of functions, the outside functions you know the anti-derivative of, and an additional multiplicative factor of the derivative of the inside function. There are formal processes you can go through to aid this when the integral is particularly hairy, or that let you get to an intermediate point when you don't immediately know the anti-derivative of the outside function, but that is an implementation detail that is not strictly necessary.
The fact that you can do some substitutions by inspection once you have experience does not make it not substitution. The u-substitution is simply the process by which you recognized that your thing was the derivative of a composition, regardless of what you write down.
Really beautiful to come that far to end on an unit circle🎉🎉🎉
Interesting timing, I just finished a quick project on the Clairaut equation in my ODE class.
I was able to guess a solution based on the fact that the non-sqrt terms look like xdy-ydx which is the polar angle differential. You then get -r^2dtheta = ds. Since circles are rdtheta = ds, this works for a circle of radius 1 (or negative 1, but that’s geometrically the same object)
I'm not used to comment on UA-cam, but I think I have something to say with this one. Without any previous knowledge of non linear ODEs, my first approach was trying to define dy/dx = tan(u), with u some function of x to avoid the sqrt. After substituting in the original equation, we obtain a nice definition of y in terms of u and x. Then, I derivated the whole thing with respect to x, obtaining a new definition of dy/dx.
After making equal both definitions of dy/dx, I eventually dealt with the same two scenarios (in my case, du/dx = 0 implied a soution of the form y = xtan(C) + sec(C), equivalent to the one you showed).
I found really surprising how you deal with it without struggling with trig as much as I did.
Thanks for your effort and nice explanation!
Your approach was very creative so points to you my friend.
Interesting DE. Thank you for your featured effort.
An alternative approach after differentiating is to use the first equation to solve for sqrt(1+y’^2), then substitute into the second equation. This yields a slight different separable equation to solve.
Hi,
"terribly sorry about that" : 3:23 ,
"ok, cool" : 7:48 .
I found it rather fun to transform the problem into polar coordinates and getting the same answers there.
We are half way to ur medal in yt❤
if he's french the pronounciation would be kinda like you say laurent in laurent series (same kind of o at the end, and same accent)
Agree Clah - ROH
Hey all! Bilingual Fr/En here, the best way to explain it would be:
- the word "clay"
- a guttural r, coded ʁ in phonetics (I think it doesn't exist in English)
- the same "o" sound than in "row", just without the "w" diphthong
Ta-daa!
klɛʁo in IPA Reader
@@Calcprofmore like claire - o
@@imTyp0_ok
4:30 I would do a u substitution instead
Clairaut: 18th century French mathematician. Clair rhymes with English bear, aut with o.
So kear-o?
@@incrxdible9896 klear-o. You missed the l.
Something about the solving procedure of Clairaut's equation boggles me; Are we allowed to differentiate y' ? Who said that y should be twice differentiable? I think one should prove this.
“Where the general form is y = xy’ + f(y’) where f is a continuously differentiable function and *Clairaut is dead* ”
I'm worried about branching issues when you square both sides. Is that not a problem here?
Doesn't look like it especially because we checked by plugging in the information into the equation.
How to find this type of questions, what book prefer you?
Bro I just make up problems for myself 😭
@@maths_505 ok ok😂
Ok, How to find your answer 100% correct?
@@MRGamesStreamer just f**king around with it till it makes sense, investigating various scenarios and working backwards always helps.
Is there method behind the black/white background on the title cards?
The algorithm likes these better
@@maths_505 All hail the algo. /s (I look forward to a time when you no longer need it; your mostly useless video titles still bother me, but I understand why you do it. I'm just glad you don't have some "shocked face" graphic on the title card as well.)
As an aside, this is one of your most elegant surprises at the end. I imagine someone with deeper imagination than I would have seen it coming sooner, but the Unit Circle falling out at the end is quite beautiful.
Be honest, did you start at the unit circle and work backwards toward the differential equation? 👀
im curious now
Well I was with the unit circle alone at home the other day and things .....kinda escalated.
She might not love me anymore, but I'll always love math.
VERY NICE. Pronunciation will be: kleh-rOH the t is silent!!
Clairoh is the way you pronounce his French name.
Alexis Clairaut was a Frenchman. So his name is pronounced CLAIR-OH
lovely but difficult i found.
Clairaut is dead?! I'm sorry, I'm going to need a minute here...
This channel is my daily dose of entertainment