Interestingly, when solving for a similar differential equation, y''-e^y=0 (the + is replaced with a -), I got the general solution y=ln(c1 csch²(sqrt(c1)(x+c2)/2)/2), but it also gives you ln(2/(x+c2)²) if you take the limit of y as c1 approaches 0. If you plug in c2=0, you get ln(2/x²), which is also very similar to your special solution. I wonder if the same method can be used to get your special solution from the general solution in the video.
There's a technique to solve these. Notice that it's a second order equation where x does not appear. When this happens you set p = dy/dx, and recast it as a first order equation on dp/dy. In this case: d/dx (dy/dx) = dp/dx = dp/dy * p So you have p*dp/dy = -exp(y) p^2 = A - 2exp(y) dy/dx = sqrt(A-2exp(y)) dx = dy/sqrt(A-2exp(y)) Now set y = ln((u^2-A)/2) dy = 2udu/(u^2-A) When you substitute you get the integral of 2du/(u^2-A). Finish it, put y back in, and you're done.
Dumb question but can anyone explain why substitution (let U = e^y) doesn't work? Then y=ln(u) and y'' = -1/U^2. I know I'm wrong because then y becomes a constant ln(1), and plugging it back in doesn't fulfil the 0 condition. So this is wrong, I just don't understand where the argument is invalid. Is it because we must assume y=/=0 for y'' to exist in this solution?
@@kebrongurara1612 the issue is that, unlike an integral, u is not a variable, but a function of x. So, if you do y(x) = ln u(x), you need to apply the chain rule when evaluating derivatives. So, y' = u'/u and y'' = (u''u - (u')^2)/u^2
You can simplify it a lot further, as 1-tanh^2(u) = sech^2(u) ln((c/2)*sech^2(u)) = ln(c/2) - 2*ln(cosh(u)) u = sqrt(c/4)*(x+R), +/- doesn't matter, as function is even in u, let k = sqrt(c/4), and r = k*R f(x) = ln(2) - 2*ln( cosh(k*x+r) / k )
The fact this is improvised is seriously impressive, I wouldn't know where to start or what to do, but regardless was interesting to watch you tackle it
You should try to solve y^(n)=Ae^(By), there is an elementary function solution (though not necessarily a general solution) for almost every A,B complex, n natural.
I think at about 20:39 the -+ should just switch the side and not become a +- right? Because when you want to cancel - you multiply by -1 and when you want to "cancel" + you'd multiply by +1, so when canceling -+ you should multiply by -+1 I think. Nevertheless great video, I'm rt so incredibly happy to see the advent calender coming back! I've been loving it since the beginning and am so happy to watch another one once again! Edit: Though it doesn't change the result so in the end it doesn't matter
I've found the same answer slightly faster by considering the reduced equation in a form y = ln(A - (y')^2) - ln(2) as a d'Alembert equation, where A > 0. Therefore, it may be considered as y = ln(A^2 - (y')^2), where A 0. This form simplifies the integration process drastically.
The equation originates with Emden, in his calculation of density of stars. The parameterized equation is common in engineering, representing mass diffusion coupled with heat generation (as in a catalyst particle), with y representing temperature or concentration of material. Boundary conditions are specified as Neuman, Dirichlet, or in most cases mixed. The parameters are 2 positive real constants in front of the exponent and and in the exponent. The solution(s) to the equation can be analyzed by their representation as fixed points in a Banach space, where it can be shown, that for certain values of the parameters the equation has multiple solutions when the sign before the exponent is +, and any solution is unique when the sign before the exponent is negative. There are as many as 3 solutions to the equation when the sign before the exponent is +, and there is no analytical functional relation between the solutions. Your analysis has identified exactly one of the solutions for parameter values 1 before the exponent and 1 before the y in the exponent for a boundary condition of the example form y(0) + y(1) = fixed. For any other parameter values and boundary conditions there is no closed form of solution, even implicit.
After getting (y')^2 +2exp(y)=C^2 (C = const) introduce the parameter p=y', dy=pdx. So 1) y=Ln((C^2 -p^2)/2). From 1) -> dy=-2pdp/(C^2-p^2)=pdx -> 2) x=(Ln((p+C)/(p-C)))/C + B (B = const). 1) and 2) give a good parametric solution.
Can anyone explain why substitution (let U = e^y) doesn't work? Then y=ln(u) and y'' = -1/U^2. I know I'm wrong because then y becomes a constant ln(1), and plugging it back in doesn't fulfil the 0 condition. So this is wrong, I just don't understand where the argument is invalid. Is it because we must assume y=/=0 for y'' to exist in this solution?
Yes that would be correct, what you're noticing is that the hyperbolic trig functions are related to the regular trig functions when you make the arguments of the functions imaginary. For example, cos(ix) is the same as cosh(x), and sin(ix) is -i*sinh(x). Tanh(x) is the same as -i tan(ix). Someone can check my signs, but that's the idea If you want to prove those formulae, start with Euler's formula and take the real and imaginary parts by adding and subtracting z to/from z*. Anyway, with that in mind, yeah if you make the thing in the denominator a negative square (i.e. i times what it was as a square root) you turn the tanh to a tan!
How the fuck does one improvise all this shit and their answer perfectly matches without a single mistake!? My adhd ass is gonna spend 2 hrs trying to figure out the step I fucked up
Hey flammy! You have a lot of single variable calculus, could you upload some fun multivariable / vector calculus stuff possibly? /happy advent from Sweden
@@PapaFlammy69 consider it bought. I don’t normally do stuff like this, but it’s actually a pretty funny clock, and I really like your channel. Cheers!
Unless you’re just cheating and saying you’re over complex numbers, you didn’t solve for the case where c is non positive, which leads to two more solutions (for c=0 and c
Would be easy if you were an engineer: could just set e^y=1 for all real numbers y, then it's trivial If you were a physicist you could use e^y=1+y, then it would be a nice harmonic approxillator
📝 Summary of Key Points: The speaker announces the start of their Advent calendar series, where they will be posting a video every day from December 1st to December 24th. They also mention a discount code for their merchandise and encourage viewers to check it out and support the channel. The main focus of the video is on solving a differential equation. The speaker explains their thought process and approach to solving the equation, relating it to concepts from classical mechanics and conservation of energy. They manipulate the equation, separate variables, and integrate both sides to find the solution. They go through several substitutions and calculations to arrive at the final solution for the differential equation. The speaker mentions their interest in physics and hints at a future physics video coming to their channel. They conclude the video by thanking viewers for watching and encouraging them to subscribe and check out their other channels for more content. 💡 Additional Insights and Observations: 💬 [Quotable Moments]: "Welcome to my Advent calendar series! I'll be posting a video every day from December 1st to December 24th." 📊 [Data and Statistics]: No relevant statistics or data were presented in the video. 🌐 [References and Sources]: No crucial references or sources were shared in the video. 📣 Concluding Remarks: The video introduces the speaker's Advent calendar series and encourages viewers to support the channel. The main focus of the video is on solving a differential equation, with the speaker explaining their approach and calculations. They also mention their interest in physics and hint at future physics content. Overall, the video serves as an introduction to the Advent calendar series and showcases the speaker's expertise in solving differential equations. Generated using Talkbud (Browser Extension)
y'' = -e^y ln(y'') = ln(-e^y) = ln(-1)+y ln(y'') = ln(e^(ipi(1+2k))+y ln(y'') = ipi(1+2k)+y... ...and now its real messy. I bet you thought this was going somewhere, didn't you :D
y'=u(y) y'' =u'(y)y' u'(y)u(y) = -exp(y) 2u(y)u'(y) = -2exp(y) u(y)^2 = -2exp(y)+C u(y) = \pm\sqrt(C-2exp(y)) Here wh have two cases y' = \sqrt(C-2exp(y)) dy/sqrt(C-2exp(y)) = dt \int{\frac{1}{sqrt(C-2exp(y)}dy} z^2 = C - 2exp(y) 2zdz = -2exp(y)dy zdz = -exp(y)dy -exp(y) = 1/2(z^2-C) zdz = 1/2(z^2-C)dy 2zdz = (z^2-C)dy dy = 2z/(z^2-C)dz \int{\frac{2}{z^2-C_{1}}}dz Here we have three cases depending on sign of C C < 0 = 2/sqrt(-C)arctan(z/sqrt(-C)) C = 0 =-2/z C > 0 Here good idea is to use partial fraction decomposition It is quite long , not difficult
I very much appreciate your videos, just brilliant mathematics. The heavy German accent is not a big deal (I’m German myself). But please, please, do me a favour and try to pronounce the “i’s” and the “y’s” as /ai/ where needed. A “y” and the word “why” is pronounced correctly /wai/ and not /va/. It means corporal pain to me to hear it like 100 times in a row in your videos.😢
*_Hope you enjoyed, stay tuned for more
that discount code totally made my day! :D
:D
Interestingly, ln(-2/x²) seems to be a special solution. Formally it works. However it isn't defined on ℝ->ℝ.
If k = sqrt(c/4), and r = k*R, as in my other comment, I think you get that solution as k -> +/-i * 0, and r = +/-i *(pi/2)
Complex numbers go brrr
Interestingly, when solving for a similar differential equation, y''-e^y=0 (the + is replaced with a -), I got the general solution y=ln(c1 csch²(sqrt(c1)(x+c2)/2)/2), but it also gives you ln(2/(x+c2)²) if you take the limit of y as c1 approaches 0. If you plug in c2=0, you get ln(2/x²), which is also very similar to your special solution. I wonder if the same method can be used to get your special solution from the general solution in the video.
There's a technique to solve these. Notice that it's a second order equation where x does not appear. When this happens you set p = dy/dx, and recast it as a first order equation on dp/dy.
In this case:
d/dx (dy/dx) = dp/dx = dp/dy * p
So you have p*dp/dy = -exp(y)
p^2 = A - 2exp(y)
dy/dx = sqrt(A-2exp(y))
dx = dy/sqrt(A-2exp(y))
Now set y = ln((u^2-A)/2)
dy = 2udu/(u^2-A)
When you substitute you get the integral of 2du/(u^2-A). Finish it, put y back in, and you're done.
Exactly what I did.
You get the same dy = 2udu/(u^2 - A) with this so you don't need to worry about that.
😂😂😂 This would be easy Only if someone paid more attention to his ODE textbooks, and also search the net. Obviously, in this video...
Dumb question but can anyone explain why substitution (let U = e^y) doesn't work? Then y=ln(u) and y'' = -1/U^2. I know I'm wrong because then y becomes a constant ln(1), and plugging it back in doesn't fulfil the 0 condition. So this is wrong, I just don't understand where the argument is invalid. Is it because we must assume y=/=0 for y'' to exist in this solution?
@@kebrongurara1612 the issue is that, unlike an integral, u is not a variable, but a function of x.
So, if you do y(x) = ln u(x), you need to apply the chain rule when evaluating derivatives.
So, y' = u'/u and y'' = (u''u - (u')^2)/u^2
I feel like in a math anime this guy would be the non-evil antagonist
Only the most rigorous mathematicians divide by dx
That was so impressive and fun to watch! Thank you Papa Flammy
AYY ADVENT CALENDAR IS BACK.
You can simplify it a lot further, as 1-tanh^2(u) = sech^2(u)
ln((c/2)*sech^2(u)) = ln(c/2) - 2*ln(cosh(u))
u = sqrt(c/4)*(x+R), +/- doesn't matter, as function is even in u, let k = sqrt(c/4), and r = k*R
f(x) = ln(2) - 2*ln( cosh(k*x+r) / k )
The fact this is improvised is seriously impressive, I wouldn't know where to start or what to do, but regardless was interesting to watch you tackle it
I think that you can remove the +- from inside the tanh() because it is odd and the next step squares so you can drop it all together!
Papa flammy, your extreme conditioning of the mathematical karate training is unparalleled. I kneel before you oh great master flammy
😂😂🤣
“i thought this integral would fuck me in the arse right here”
- Flammable Maths 2023
You should try to solve y^(n)=Ae^(By), there is an elementary function solution (though not necessarily a general solution) for almost every A,B complex, n natural.
The first problem on the “simple” part of my real analysis exam 👍👍👍
I think at about 20:39 the -+ should just switch the side and not become a +- right? Because when you want to cancel - you multiply by -1 and when you want to "cancel" + you'd multiply by +1, so when canceling -+ you should multiply by -+1 I think. Nevertheless great video, I'm rt so incredibly happy to see the advent calender coming back! I've been loving it since the beginning and am so happy to watch another one once again!
Edit: Though it doesn't change the result so in the end it doesn't matter
exactly what I noticed
that one popped up in my research
it's a wonderful recap exercise for calculus things
My first advent calendar I can’t wait!!!!
14:10 is the best moment in this video! :D hahaha
I've found the same answer slightly faster by considering the reduced equation in a form y = ln(A - (y')^2) - ln(2) as a d'Alembert equation, where A > 0.
Therefore, it may be considered as y = ln(A^2 - (y')^2), where A 0. This form simplifies the integration process drastically.
@21:29 "now we can square this bitch on both sides" 😂😂
Since (y')^2 >= 0 and exp(y) > 0 we must have C > 0 so we can as well call it C^2 to avoid using √C later.
so fancy, using manim for the end credits haha. great video as always papa
14:39 can't we just use the integral 1/(a^2-x^2) dx result here?
Need to think in terms of bubbles and symbiotic oils...
And now I will go ahead and watch this clip :D
The equation originates with Emden, in his calculation of density of stars. The parameterized equation is common in engineering, representing mass diffusion coupled with heat generation (as in a catalyst particle), with y representing temperature or concentration of material. Boundary conditions are specified as Neuman, Dirichlet, or in most cases mixed. The parameters are 2 positive real constants in front of the exponent and and in the exponent. The solution(s) to the equation can be analyzed by their representation as fixed points in a Banach space, where it can be shown, that for certain values of the parameters the equation has multiple solutions when the sign before the exponent is +, and any solution is unique when the sign before the exponent is negative. There are as many as 3 solutions to the equation when the sign before the exponent is +, and there is no analytical functional relation between the solutions. Your analysis has identified exactly one of the solutions for parameter values 1 before the exponent and 1 before the y in the exponent for a boundary condition of the example form y(0) + y(1) = fixed. For any other parameter values and boundary conditions there is no closed form of solution, even implicit.
The problem with non-linear differential equations is that you can’t prove your solutions are the only solutions/general solution
After getting (y')^2 +2exp(y)=C^2 (C = const) introduce the parameter p=y', dy=pdx. So 1) y=Ln((C^2 -p^2)/2). From 1) -> dy=-2pdp/(C^2-p^2)=pdx -> 2) x=(Ln((p+C)/(p-C)))/C + B (B = const). 1) and 2) give a good parametric solution.
What would have worked well is if he split c into three different cases, c>0, c=0, and c
Not good enough! Gotta make some graphs!
ADVENT CALENDAR THIS YEAR LETS GOOOO
i love when u use explitives
Thanks
y=e^iy 😂
(Don’t kill me)
That choice of substitution symbols is unbelievably cursed
xD
This looks like the Korteweg de Vries soliton, I didn't know it obeyed this equation too!
Can anyone explain why substitution (let U = e^y) doesn't work? Then y=ln(u) and y'' = -1/U^2. I know I'm wrong because then y becomes a constant ln(1), and plugging it back in doesn't fulfil the 0 condition. So this is wrong, I just don't understand where the argument is invalid. Is it because we must assume y=/=0 for y'' to exist in this solution?
y(x)=ln(u(x)) => y'(x)=u'(x)/u(x)
You are forgetting the chain rule.
Instead of going hyperbolic could you still have left in generic trig form?
most certainly!
Imagine at 20:05 you just say c = some -r and that results into an arctan function. Would that be correct though?
Yes that would be correct, what you're noticing is that the hyperbolic trig functions are related to the regular trig functions when you make the arguments of the functions imaginary. For example, cos(ix) is the same as cosh(x), and sin(ix) is -i*sinh(x). Tanh(x) is the same as -i tan(ix). Someone can check my signs, but that's the idea
If you want to prove those formulae, start with Euler's formula and take the real and imaginary parts by adding and subtracting z to/from z*.
Anyway, with that in mind, yeah if you make the thing in the denominator a negative square (i.e. i times what it was as a square root) you turn the tanh to a tan!
How the fuck does one improvise all this shit and their answer perfectly matches without a single mistake!? My adhd ass is gonna spend 2 hrs trying to figure out the step I fucked up
routine I guess
Wow a mathematician that actually swears, this might be the only solution to that equation
I would just do energy conservation - that allows you to get t(y) then you just need to see if you can invert that
Yeah, that’s what you did - hard part is the integral for time and then inverting that to get x
I'm an engineer, when in doubt, I Taylor series it. Keep the second term, or just use Runge Kutta
Can somebody explain why there is a log instead of ln at the solution?
log is ln
@14:11 I thought this integral would f*** me in the a**
Hey flammy! You have a lot of single variable calculus, could you upload some fun multivariable / vector calculus stuff possibly? /happy advent from Sweden
y = - ln((x^2)/2) satisfies the equation man.
If I buy the clock will it make up for me using adblock?
yup!
@@PapaFlammy69 consider it bought. I don’t normally do stuff like this, but it’s actually a pretty funny clock, and I really like your channel. Cheers!
That +- inside of tanh² is not necessary cuz tanh(x) = -tanh(-x), so tanh²(x) = tanh²(-x).
Welcome back papa
tanh^2 is even so you could remove the plus or minus in the argument of it too, right?
Unless you’re just cheating and saying you’re over complex numbers, you didn’t solve for the case where c is non positive, which leads to two more solutions (for c=0 and c
Buy i really love all your videos just so you know ❤
Different kind of logics used here 😊😊
We are so back
Hard core, man. Hard core.
Is the "del x"-Term the same as d/dx?
yes
Hello papa Flammy please let me scape Latin America. I could live in your attic. I will clean your house and make you breakfast.
attempt 3 of asking where the blackboard is from
Would be easy if you were an engineer: could just set e^y=1 for all real numbers y, then it's trivial
If you were a physicist you could use e^y=1+y, then it would be a nice harmonic approxillator
Wikipedia: Autonomous system (mathematics).
I need 15% off of my math major tuition😂!
There's a slight glare in the video that makes it difficult to read what you have on the board. I suggest you adjust the lighting. Thanks.
Not how I've usually seen people handle y^11, but okay
📝 Summary of Key Points:
The speaker announces the start of their Advent calendar series, where they will be posting a video every day from December 1st to December 24th. They also mention a discount code for their merchandise and encourage viewers to check it out and support the channel.
The main focus of the video is on solving a differential equation. The speaker explains their thought process and approach to solving the equation, relating it to concepts from classical mechanics and conservation of energy. They manipulate the equation, separate variables, and integrate both sides to find the solution. They go through several substitutions and calculations to arrive at the final solution for the differential equation.
The speaker mentions their interest in physics and hints at a future physics video coming to their channel. They conclude the video by thanking viewers for watching and encouraging them to subscribe and check out their other channels for more content.
💡 Additional Insights and Observations:
💬 [Quotable Moments]: "Welcome to my Advent calendar series! I'll be posting a video every day from December 1st to December 24th."
📊 [Data and Statistics]: No relevant statistics or data were presented in the video.
🌐 [References and Sources]: No crucial references or sources were shared in the video.
📣 Concluding Remarks:
The video introduces the speaker's Advent calendar series and encourages viewers to support the channel. The main focus of the video is on solving a differential equation, with the speaker explaining their approach and calculations. They also mention their interest in physics and hint at future physics content. Overall, the video serves as an introduction to the Advent calendar series and showcases the speaker's expertise in solving differential equations.
Generated using Talkbud (Browser Extension)
This got to be AI generated
my man is chat gpt
y = wa
y'' = -e^y
ln(y'') = ln(-e^y) = ln(-1)+y
ln(y'') = ln(e^(ipi(1+2k))+y
ln(y'') = ipi(1+2k)+y...
...and now its real messy.
I bet you thought this was going somewhere, didn't you :D
why not rewrite it as sech^2 at the end
no need, I got an answer, I'm good
please NO CURSING !! ITS DEMEANING !!
yo
nice
can you solve this abdomination e^y'' - e^-y = Y(x)
y'=u(y)
y'' =u'(y)y'
u'(y)u(y) = -exp(y)
2u(y)u'(y) = -2exp(y)
u(y)^2 = -2exp(y)+C
u(y) = \pm\sqrt(C-2exp(y))
Here wh have two cases
y' = \sqrt(C-2exp(y))
dy/sqrt(C-2exp(y)) = dt
\int{\frac{1}{sqrt(C-2exp(y)}dy}
z^2 = C - 2exp(y)
2zdz = -2exp(y)dy
zdz = -exp(y)dy
-exp(y) = 1/2(z^2-C)
zdz = 1/2(z^2-C)dy
2zdz = (z^2-C)dy
dy = 2z/(z^2-C)dz
\int{\frac{2}{z^2-C_{1}}}dz
Here we have three cases depending on sign of C
C < 0
= 2/sqrt(-C)arctan(z/sqrt(-C))
C = 0
=-2/z
C > 0
Here good idea is to use partial fraction decomposition
It is quite long , not difficult
😅😅😅, I'm surprised that you curse like a sailor.
kann es sein das du deutsch bist ``Ansatz`` 2:17
I very much appreciate your videos, just brilliant mathematics. The heavy German accent is not a big deal (I’m German myself). But please, please, do me a favour and try to pronounce the “i’s” and the “y’s” as /ai/ where needed. A “y” and the word “why” is pronounced correctly /wai/ and not /va/. It means corporal pain to me to hear it like 100 times in a row in your videos.😢
what grade is this?😭
Solution become a tough problem...
झिला दिया भाई...😂😂😂
Mind your swearing words. They are not needed.
when he said he thought the integral was gonna fuck him in the ass 😂
Try this one:)
y' = e^(xy)