This Differential Equation is Nuts
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- Опубліковано 28 гру 2023
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Today we solve y'=x+y in the most elegant way possible, I hope you'll like it! Enjoy! :3
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y'-y=y(1'-1) would have made it so much easier
3:
he said he wanted to do it this way
@@flavioxy I'm not doing a substitution. i'm doing a "incorrect" isolation.
www.youtube.com/@user_math2023
1 prime 🫨
You have a very special talent for presenting mathematics. You stopped making math videos for a while. I really missed them.
Subtract y on both sides, multiply everything by exp(-x). On the left side, apply the reverse product rule and you now just have to integrate both sides to get the general form of y(x).
Yeah, any first order linear differential equation can be solved in this manner. e^(-x) is called an 'integrating factor'. In general, for y'(x)+a(x)*y(x)=b(x), the integrating factor is e^A(x), where A(x) is an antiderivative of a(x).
as far as i remember, isn't this how integerating factor method thingy worked?
That is exactly what I did as well
I did it by substitution.
u = x + y
==> du/dx = 1 + dy/dx
dy/dx = du/dx - 1
dy/dx = x + y = u
du/dx - 1 = u
du/u + 1 = dx
Integrating on both sides
ln |u + 1| = x + C_0
u + 1 = e^x*e^C_0
Since e^C_0 is just a constant,
u + 1 = C_1e^x
x + y + 1 = C_1e^x
y = C_1e^x - x - 1
Loved the observation that the third derivative is equal to the second derivative and working backwards from there. Great video as always Papa Flammy!
thx
How I learned this in my undergrad dif eq class is to first see that the complementary solution is y=ce^x. Then, we make a particular “guess” as to what the other part of the solution will be. In this case, I’ll let y=Ax+B. Then, I’ll evaluate y’ and plug it back into the original equation, to get A=x+Ax+B. Then, we see that x+Ax must equal 0, so A=-1. Then, we see that A=B=-1. We now plug in our constants to the guess and add the complementary solution, to get that y=ce^x-x-1
I used the Laplace Transform and determined that the constant C in the final equation is 1+y(0), as well as that R (kappa) is -1. When checking again with the original differential equation, it works out really well because the coefficient of x is -1. Love your videos!
When I realized, it was just the exponential function when you wrote y'' = y', I was so shocked. Very cool.
I did it the ol' brute force way: y' = y in some way usually means an exponential, so the only thing missing was a -x to remove the x that was present.
It probably doesn't work often, but it worked this time.
You’re basically using the method of undetermined coefficients, which can be used to solve many linear differential equations
Insane video,PapaFlammy! I just want to take a moment to appreciate the fun/light-hearted/educational content you make. As a pure mathematics lover, I find less content on UA-cam which is both amusing and provides fun bits about mathematics. You are a person I resonate with a lot.Looking forward for more great content! You are truly a gem here on UA-cam.
Happy to see the Crab-Boy.
So beautiful and so much substance in this small equation. Great presentation 👏🏼
thx! =)
y'=y+x, y'+1=y+x+1. Substitute u=y+x+1 and you get u'=u, so u=Ce^x meaning y+x+1=Ce^x. Finally you get y=Ce^x-x-1
Happy new year!
13 minutes when you couldve used the fastest DE solving algorithm
"educated" guessing
The problem with using an algorithm solver is like putting something into a computer simulation to see what happens. It is not giving us something outside the box, something enlightening.
@@Mysoi123 i'm referring to its larger meaning, the method used in the video is a sequence of calculations and inferences leading to an answer after all
Knowing the solution is the proven method to solving differential equations
A guy I knew in college said he had a teacher who called it the GWIBNI principle. "Gee, wouldn't it be nice if..." this guess was a solution to the inhomogeneous equation? Hey, look, it is! Now we just add the general solution to the homogeneous equation, and voilà!
My solution before watching the video is multiplying by exp(-x), so the equation is exact, as N_x=M_y=exp(-x)
My intermediate diff eq prof loved power series so i gave that method a go! unless i screwed something up i got the same answer. great vid!
My Calc 1 teacher gave us this differential equation when we were learning separation of variables… the way she solved it was dy/dx=x+y, dy=xdx+ydx, y=1/2x^2+xy+C…
After that I decided to go and find as many ways to solve it correctly that I could, which led me to accidentally discover undetermined coefficients by myself! :D
That solution is incorrect
y = x²/2 + xy + C
y(1-x) = x²/2 + C
y = (x²+C)/(2-x)
In the integral of ydx you cannot treat y as a constant, because y is a function with regards to x
To be clear, I am complaining that the way she solved it was incorrect
@@zapking8209 ahhh okay! Cheers :)
Lol nice but there’s also the pretty standard method of integrating factors for linear first order ode’s
Very cool! Love that you found y"=y³. I get a charge that runs up my spine when I discover some regression that helps me solve an equation. The big emotional rewards in math really show themselves in solving differential equations. (For me that was 2nd year calculus... We would do a semester of derivatives, a semester of integration, then 3rd you do differential equations.) I always wished that differential equations could be presented as a more general abstraction. I felt like the class taught me how to solve 20+ different equations and the rest of the theory was trying to muscle whatever equation given into the form of one of the 20+ equations that you can solve...
But I do love me a Laplace transform and don't get me started on convolutions!🎉❤
Papa flammy ended the year with a banger
Papa flammy, I have an interesting integral question for you. I do not know the answer or if it is possible.
Look at the two graphs: 1/(1+x)^2 and 1/sqrt(1-x). Geometrically, it is easy to see that that the area under 1/sqrt(1-x) from 0 to 1 is equal to the area under 1/(1+x)^2 from 0 to infinity, with a difference of 1 between them from the area of the unit square. That is, the Integral from 0 to 1 of 1/sqrt(1-x) = 1 + Integral from 0 to infinity of 1/(1+x)^2.
I noticed a nice way of bringing 1 into the integral is by turning it into the Integral from 0 to infinity of (2/pi)(1/(1+x^2)). So now we can write Integral from 0 to 1 of 1/sqrt(1-x) = Integral from 0 to infinity of 1/(1+x)^2 + (2/pi)(1/(1+x^2)).
My question for you is what substitution(s) are required for turning the 0 to infinity integral into the 0 to 1 integral if it is possible at all. I think it should be possible since they have the same area but I'm not sure how. Thank you for your time!
Truly crazy
Cool video Professor Flammy. Is there any chance you could make a video discussing some interesting concepts (i.e. L-Functions, etc) from the langlands program?
woah that differential equation is 🥜🥜🥜🥜
The Annihilator Method! It's very cool. As long as you have a Differential Operator that 'Annihilates' the non-homogenous part of the ODE, you can recover a homogenous ODE, solve it, and work backwards to solve the original, precisely as in the video.
The Differential Operator in this example would of course be the second derivative operator. If you'd used a more redundant operator, like the third derivative, or some more complicated operator that still reduced 'x' to 0, you can still recover a solution, but with a more redundant/tedious system of equations at the end.
If you use transcendental-order differential operators, like e^D, Γ(D), sin(D) and cos(D), 1/(D+s) for parameter s, etc., then you can annihilate more functions, but you also make the resulting DE more complicated. Fun!
Dude, I am getting uncomfortable with all that kissing and fondling 😂 Happy new year Jens et al.
:D
y'=x+y
*with* loss of generality, assume y' differentiable
y''=1+y'
dy'/dx=1+y'
dy'/(1+y')=dx
ln(1+y')=x+c, c in C
y'=ce^x-1, c in C (new constant)
y=ce^x-x+d
original equation: d=-1
y=ce^x-x-1
consider the original equation.
now, without loss of generality, let
y=ce^x-x-1+z
where z is differentiable once in x.
y can always be expressed this way because by choosing z correctly, we can cancel out the other terms.
our equation now becomes
ce^x-1+z'=x+ce^x-x-1+z
z'=z
z=ke^x
y=ce^x-x-1 remains general because the constants just combine.
using the fact that y'' = y''', we can integrate backwards, fix some constants and arrive at y = a * exp(x) - x - 1
this -differential equation- approach is nuts FTFY
Interesting idea. My chain of reasoning would have been that I can solve the equation y' = y + c by making the substitution Y = y + c to turn it into Y' = Y. This won't work here because x is not a constant, but I can turn any polynomial into a constant by differentiating enough times. Ergo, I'd differentiate once to turn the x into a constant: y'' = 1 + y' and now I can solve via the substitution Y = y' + 1.
wow, it was the first math video that I actually followed a 100% with no rewinding :D very cool you dont go too fast.
Congrats on coming up with it yourself!
Prime Newtons did the exact same problem 3 days ago.
@@programmingpi314 thats true, although he did it in a different way, so what's your point?
Nice!
Flammy wearing _the boy_ and holding it in intro is too cute!!1! I'd like to commission a fanart
Define u=x+y => u’-1=u => u=ce^x-1 => y+x= ce^x-1 => y=ce^x-x-1. This time we integrate both sides of u’dx/(u+1) = dx without introducing a second derivative. Half a dozen of one, six of the other.
Technically an nth order ODE should have n unknown constants so it wasnt surprising that kappa was unique
Can you do a video on the equation y''-xy = 0 and the derivation of the Airy functions from it?
Hello Super Papa! All the best to you and your family in 2024.
oey thats a nice crab
question for try: for what alpha the integral of 1/(1+x^a*sin(x)^2) in convergent? Too curious
Power series solution works out.
When you use wolfram alpha so much that you write log() instead of ln()
has nothing to do with WA
In pure math, log() without a specified base means complex log, while ln() means the real number log. Pure math always specifies otherwise, to use logs of other bases, either with subscripts or the change of base rule. For positive real values of x, where we only care about the principal value of complex log, log(x) and ln(x) are interchangeable.
This drives me nuts, when I copy a result from Wolfram Alpha to another program, and that other program assumes log() means log base ten. If it were my choice, I'd call it lnc(z) to specify the complex logarithm.
maybe try power séries?
You could have also used linear algebra in terms of differential operators and annihilators to do it in less than half the time while still being formal, quite overkill for a 1st degree equation but works really well.
This has nothing to do with efficiency, I think you missed the point pf the video. Noone's in a rush... lol
I can tell you get no bitches
good video
Solved it in my head it's a classic isolate out the homogenous part then guess
the final answer shouldnt have a degree of freedom in the constant term. You can plug in the final answer and see it only works when kappa equals -1
6:04 makes me wonder if he said "relationshit" on purpose or not.
After seeing this video, I realised that I can use this in my exams and confuse my teachers.😂
I blame you for my budding math degree... the terrible terrible analysis you've caused me (on a serious note, you make awesome videos and legitimately are one of the reasons I'm pursuing higher math, thank you!)
If this solution is elegant, surely the fact that the function y is supposed differentiable 2 times has to do with it here. Though, I wonder what we get if we don't assume anything more than the original equation itself (as well as y being differentiable one time of course), does someone know ??
It's fine, you can just take that result and prove that it's general.
Just set y=f(x)+z where z is differentiable once and f(x) is the "cheated" solution. this is without loss of generality because we can still get anything by choosing z to cancel out f(x).
If we substitute back into the equation we started with, you see that everything cancels out except for z'=z. that gives you z=ke^x and you can combine the constants with the ce^x term.
For other equations, the z might not already be part of the f(x) term. In these cases, solving for z gives you the extra terms you need to add for the general solution which were lost when assuming extra differentiability. The equation for z will generally be much simpler than the original equation.
The function is smooth (C infinite) since, from the equation itself, we know that its derivative is the sum of the identity (differentiable) and itself (differentiable), so y' is differentiable. By induction we can prove the existence of any order of derivative.
We assume that y' exists in some domain; Clearly, if this is not true, there is no differential equation at all.
@@octaviocarpinetti4326 wow that's even better. I'll have to keep that trick in mind!
Hmm.. interesting video to watch, but that method does seem like a long way around!
If you say u=x+y then dy/dx = (du/dx -1); substituting that back in you can immediately separate the ODE, integrate and get y=ce^x -x -1 right away
being quicc wasn't the point of the video ..
@@PapaFlammy69 Fair enough! As I said, it was indeed an interesting video to watch 😁
I've never done differential equations and I have a question about the solution
The first derivative of y should be y+x but isn't the derivative y - 1 here?
I like math, but I'm not skilled at all, so I think I'm just getting wrong something trivial.
Edit: I needed sleep, I got it now
nice vdo
omg bro ppl in the comments keep saying "this is a much faster solution bla bla" but they missed the main point said in intro
always the same on the internet lol
@@PapaFlammy69 hah true lol also wish you a happy new year :) (in advance)
If y' = x + y:
Derive both sides
y'' = 1 + y'
Derive both sides
y''' = y''
So y'' = c*e^x for some real number c since it equals its own derivative
y'' = 1 + y'
y' = y'' - 1 = c*e^x - 1
y' = x + y
y = y' - x = c*e^x - x - 1
Boom, done. :D
why the derivety of the solution dont give you the original y`=x+y
break to mid january?!?!? i'm back in school on tuesday 😭
lol
Didn’t Mathority just make a vid on this
I did:
y' = x + y
Dy = x + y
Dy - y = x
(D - 1)y = x
y = (D - 1)^-1 * x
y = (-1 - D^2 - D^2 - ...)x
y = -x - 1
then just add ce^x because it will stay constant
y = ce^x - x - 1
:)
well, u can easily guess the solution of y=-x -1 and then add the c*exp(x) for generality
This is the way we are taught to solve non-homogeneous linear differential equations in high schools in Russia:
y’ - y = x
=================
General homogeneous:
y’ - y = 0
y’ = y
dy/y = dx
ln y = x + const
y = exp(x + const) = Ce^x
=================
Particular non-homogeneous:
y’ - y = x
y = Ax + B (since y’ - y is a linear polynomial)
y’ = A
A - Ax - B = x
(A + 1)x + (B - A) = 0
A = -1, B = -1
y’ = -x - 1
=================
General non-homogeneous:
y = Ce^x - x - 1
It can also be solved as a homogeneous of first order or a linear of first order
@@wooeidikd9412 I suppose that one of those is the way I presented above
Another is through the characteristic equation
@@adammizaushev solving it with characteristic like you did is probably the easiest. I would have solved it as a homogeneous differential equation and replaced my x with vy.
@@wooeidikd9412 I forgot about the “x = vy” substitution, thanks.
Turns out we’ve already mentioned 3 different ways of solution
Watched a video on the same equation recently, solved via substitution (with other suggestions in the comments): ua-cam.com/video/F7ANuRyjzeg/v-deo.htmlsi=kRQX6mxRdcH2Selj
(Edit) I need to ask: coincidence or no?
coincidence
very cool solution :D also mostly understandable to stopid people like me who can only do basic calc 1
um dosen't differenting y'=x+y with respsect to x give y"x'=1+y'x'
edit: realised that's differenting with respect to y later
it is y"x'=1+y'x', but x' = 1 so its the same
These brilliant ads are getting to like 30% of the video length
wah
I'm suddenly seeing this equation everywhere these days
really?
@@PapaFlammy69 yes, it's the 3rd or 4th time youtube recommends me someone solving this particular equation this week. It also seems that each youtuber used a different method.
oh xD
Here's how I did it: y = ∫(x + y) = x^2/2 + C + ∫y, so that (1 - ∫)(y) = x^2/2 + C, and we get y = (1 - ∫)^(-1)(x^2/2 + C) = sum from n = 0 to ∞ of ∫^n(x^2/2 + C) = sum from n = 0 to ∞ of x^(n+2)/(n+2)! + Cx^n/n! = (C + 1)e^x - x - 1 = Ke^x - x - 1.
2038😮
Let x + y = u
y' = u' - 1
Therefore the original equation becomes u' - 1 = u
so (du)/(1+u) = dx
Integrate both sides and you'll get this answer.
Kappa????
Are you fromm 1970
ye
@@PapaFlammy69 ap sih ye ngejek kah kamu
What’s that? 😅😂🤭🤪👐🏫✌️
y(x) = c e^x - x - 1
let me give an alternative solution:
y' = x + y (*)
=> y'' = 1 + y' = 1 + x + y
=> y''' = 1 + y' = 1 + x + y
=> y'' = y'''
=> y'' = a*exp(x)
=> y' = a*exp(x) + C
=> y = a*exp(x) + Cx + D
=> a*exp(x) + C = y' = x + y = x + a*exp(x) + Cx + D = a*exp(x) + (C + 1)x + D
=> C = D; C+1 = 0
=> C = D = -1
=> y = a*exp(x) - x - 1
this also shows that your kappa has to be -1.
x+y=t
1+y' = t'
y'=x+y
t' -1 = t
dt/(t+1)=dx
t+1=c e^x
y= c e^x -1-x
Wow! You solve it like a REAL man, instead of just relying on pre-derived integrating factors or ansätze.
Next, can you solve y' = x + y^2?
I don't see why you need to solve a standard first-order linear differential equation with a quasi polynomial on the right side in such a complex way. Neither the solution, nor the answer of this particular equation seem to show anything essential to students who are unable to grasp the general method yet.
1 + x = 70
My pronouns are now -1/12
bro, just...
y'=x+y
y'-y=x
y'e^-x - ye^-x = xe^-x
(ye^-x)'= xe^-x
You integrate and boom finished.
noone cares
@@PapaFlammy69 who cares about a solution that long for nothing, seriously, going to the 3rd ordrer derivative for a thing so simple
@FreeGroup22 have you even watched the first minute of the video? holy moly
@@PapaFlammy69 Who looks at "brilliant, blah blah blah, lets get into this"
Wrong method still interesting one... U should just l t x+y to be k .. i hopw it works
And i saw ur previous vid... By my name u have guessed it m Indian .. although I know it was AJOKE but plz dont pass Ramanujan please 😢... Even as a joke it hurts... And dont laugh at Brahmagupta too...
IK IT WAS A JOKEEEEE BUT STILL....
Those people have contributed so much to maths ....
And plz check out about sir Madhava too🙂 ...from 14th century
bruh these kinds of comments are why he has a prejudice against Indians lmao
u=y+x. du/dx = dy/dx + 1.
u'-1=u
du/dx=u+1
du/(u+1)=dx
u+1=Ce^x
Now, y=Ce^x - x - 1