MU PRIME!!!! You just answered the question that I had when I first took diff eq over a decade ago!!!!! Wow!!! This is an amazing and exciting video. I enjoyed every second of it!!

@johnzhu94

2 жыл бұрын

Came here from your video!

@nikhilnagaria2672

2 жыл бұрын

@@johnzhu94 same

I'm here because you were mentioned by bprp. Man, it was so worth it.

O man! I feel like I re-learn Laplace transform today! Thanks for explaining step by step the purpose behind! Keep going in this direction!

I love teaching math by walking through a sort of discovery/"play-around" process. This is the first I've seen for the Laplace transform - absolutely wonderful!

There's one thing and one thing only I have to say: you my friend are quite literally THE GOAT. Your approach to breaking down the concepts of how and why we use the Laplace transform the way we do is REALLY helpful. Thank you!

Maaan, can’t thank you enough. It’s not often seen that the concept is actually explained in a way that you clearly see what was the reasoning behind it and not some stroke of some genius. That’s what’s maths all about. Once again thanks - you should be proud of this vid.

I am so grateful for your guidance. I never understood this concept before, but now it all makes sense. Thank you for being an amazing teacher 😍

I wish I had this information while I was doing my engineering. I asked many math professors what’s going on. They could not explain anything beyond “it’s defined that way”. Great video. Videos like this will inspire younger generation to learn mathematics!

At 11:55 you gloss over the fact that you rewrite the function into a very useful form. That requires insight because in many of its other forms it is very hard to get back to what needed to be integrated to get you there. At 15:56 = mind blown! It is amazing what happens when someone actually explains WHY things work the way they work. I really had an AHA! moment during the first half of the video! Thank you, thank you, thank you!

Thanks! This is very useful for my differential equations course. By the way, now I know how to solve homogeneous and non homogeneous differential linear equations thanks to you and my teachers. 👍

This was a very interesting intuition for why we use Laplace Transforms, since I've no idea how it worked beyond writing it off as "sorcery" when I was only taught the table of Transforms and Inverses. Seeing the step at 8:00 has me curious now. Most elementary functions we know of such as the trigonometric or polynomial functions are dwarfed by the exponential function's growth rate, which allows for the upper limit to go to infinity and result in a value of 0. What if the function y(t) grows faster than e^(st)? Does this mean there are some functions that can't be solved using Laplace transforms?

@MuPrimeMath

4 жыл бұрын

Yes, you're right! Some functions don't have a Laplace transform, so we can't use this method to solve differential equations that include them. Two examples of functions that don't have Laplace transforms are f(t) = e^(t^2) and f(t) = 1/t (vertical asymptote as t approaches 0) For these functions, we will need a different method.

@angelmendez-rivera351

4 жыл бұрын

Other examples of such functions are Γ(t) and t^t.

@angelmendez-rivera351

4 жыл бұрын

Laplace transforms are not the only transform that work for solving differential equations. They may be the simplest of their kind, but in reality, any invertible transformation pair of the form Integral{t -> (a, b), k(s, t)y(t)} such that k(s, t) is differentiable, and such that the image of k(s, t)y(t)|(a, b) - Integral{t -> (a, b), Partial[t, k(s, t)]·y(t) is in the codomain of the transformation in consideration, will work as well. Laplace transforms are only considered special because Partial[t, k(s, t)] = -s·k(s, t), since k(s, t) = e^(-st), and because k evaluated at the boundary points is independent of s. This makes it extremely useful for applications, but in theory, there are infinitely many such transformations that can help solve the equations, all working for different initial conditions and for different classes of functions y(t). So, when dealing super-exponential growth functions, such as those mentioned above, we can use some other transform among the infinitely many. Alternatively, we may choose to not use transforms at all and instead find the solutions via a constructive proof, which is in reality how most equations are solved.

@MarkMcDaniel

4 жыл бұрын

@@angelmendez-rivera351 -- You might be able to solve t^t if you convert it to e^(t•ln t).

this is one of the most intuitive explanations of the laplace transform. the fact that you are the same age as me and explaining this better than my prof is super humbling but this series is the only thing getting me through my winter diffEq course.

YOU ARE AMAZING!!!!!! I NEVER THOUGHT THERE WOULD BE SUCH A BEAUTIFUL, SIMPLE MOTIVATION FOR LAPLACE TRANSFORMS, BUT HERE YOU EXPLAINED IT SO ELOQUENTLY!!!!!!

Thanks for explaining this in a way that means something! Numerical analysis was the salvation of my engineering career. Using numerical methods I found I could solve anything! I was using it as a crutch, but desperate times called for desperate measures. Your video is beginning to make an honest engineer out of me. ;)

Excellent presentation of the Laplace transform! Many thanks 👍

A careful, beautiful, and clear dissection. Loved it

Thanks for this video! Back in engineering school we were taught the Laplace transform but had no idea why it worked. It seemed like magic. Now I understand. :-)

Very well explained. I like the way you introduce notions and techniques by explaining the intuition behind them. You're definitely on the right track. Keep it up.

This is the first time I am putting a comment for a KZread Video (or may be second) ... you killed it man ! This is one of the greatest videos ever made on this topic !! Same with likes too. I put a like after a very very very long time on KZread !!

gawd damn best video on laplace transforms anywhere

Thank you so much for this! This is so clearly explained that even someone, who has just begun really learning calculus can follow through and understand the concepts presented. Well done!

This explanation is so good I literally cried

Another derivation that's sorely need in texts and lectures. It would void lots of head scratching.

Wish my lecturer would of taught me that way instead 'rote' .I learnt how solve Intergral/diff problem without true understanding , didn't even realise the role of Intergration by parts! Top U-tube.Well done👍

Absolutely marvelous. Huge thanks for this

Ok, I watched it when it came out, and I was amazed and I re watched it just now. I think that's it. This one might be one of the best Laplace transform videos on youtube

That's a really well explained video! I've used the Laplace transform method, but I don't think I've ever really thought about how I'd derive it if I hadn't seen it before.

@stevematson4808

Жыл бұрын

So it seems that the Laplace transform can be used without completely understanding it? Would that be plug and chug method?

Great explanation! you just saved my entire degree with this intuition. Thanks you

Excellent exposition!

Awesome explanation! Thanks!

Applause, applause, applause. Amazing. I never knew why in gods name we cared about this thing and why it worked, but now I get it. I GET IT. OH MY GOD. THANK YOU.

This is GOLD here

This video is a gem

Recently I fall in love with your videos. Because I like how deep you think about problems And this video ... One of the best (maybe the best) videos I see about Laplace, till now. Great ... Great Thank you so much ♥️

this video deserves more views.

Wow very nice work, it was all clear and well explained.

Thanks dude. This was awesome

You deserved it, here, take my LIKE. Keep up the good work.

When I took engineering mathematics in my first year of University I just continued Laplace without knowing what it is, how did it come to be, why do we need it and all. But as a engineering major, the question always bothered me, why laplace, why do we use it? But you answered this in very very clear and precise manner. Loved and enjoyed every part of the video.

Nice one man great video.

great explanation. thank you sir.

Beautiful explanation

I am just amazed. You are the best :D

Great derivation! Always wondered about defining the laplace transform’s limits in terms of positive and negative infinity (instead of zero); I think Fourier transform has limits of negative and positive infinity

Something that μ' didn't mention: At 11:56 we get the two new fractions with partial fraction expansion. The partial fractions expansion is used a lot when we get complex expressions of s at the right hand side of the equation. E.g. L{y(t)} = (4s+s^2)/(s^2+2s+2)

@carultch

7 ай бұрын

With your example, we can start by adding zero in a fancy way to the top, to form a term we can cancel: (s^2 + 2*s + 2 - 2*s - 2 + 4*s)/(s^2 + 2*s + 2) Regroup: (s^2 + 2*s + 2)/(s^2 + 2*s + 2) + (2*s - 2)(s^2 + 2*s + 2) Cancel the first term: 1 + (2*s - 2)/(s^2 + 2*s + 2) Complete the square on the second term: s^2 + 2*s + 2 = (s + 1)^2 + 1 Regroup its numerator so s+1 appears in it: 2*s - 2 = 2*(s + 1) - 4 Thus: 1 + (2*(s+1) - 4)/((s + 1)^2 + 1) Since L-1{F(s + a)} = e^(-a)*L{F(s)}, this means we have an exponential envelope enclosing sine and cosine, both with a frequency of 1, and a decay constant of 1. Coefficient on cosine is 2, and coef on sine is -4. Thus our result is: delta(t) + e^(-t)*[2*cos(t) - 4*sin(t)]

I was very confused with Laplace transforms when I took that class. (not only before you were born, but maybe your parents, too) There was no mention of this background. When I went to my professor, all he did was show me how to do them, rather than the theory behind them. I just couldn't understand transforms, so it only confused me more. It seemed some kind of cheating. I just has an update -- math sometimes takes a long time before the flash -- thanks to this explanation, it hit me what the connection is between "piecewise continuous functions" (which drove me nuts) and this explanation. It avoids the gaps by avoiding the derivatives, which are undefined among the pieces. Now I understand it better. Congratulations to our distinguished young man, and future PhD. Again, I wish you had been my professor.

Nice video! Being a stickler for details, I'd only mention that we need to specify y(0) and the result at 11:04 is invalid when s < 3.

Amazing!

Amazing video

Thanks for the clarification what if the y is raise to a power like y²

fourrier => decomposition of a function as a sum of sinusoids, laplace => decomposition of a function as a sum of decaying sinusoids, wavelets => decomposition of a function as a sum of sinusoids which increase then decrease. fun fact : the 3 guys that made these transformations are french.

Nice starting POV!

could you do videos on the different types of Fourier transform PLEASE?!!?

When I verified your solution I came. Very good explanation

wow ! GOLD !

Fantastic

Aradığım bilgi, teşekkürler.

Nice !

Amazing

Very intuitive explanation of the laplace transform. The pinnacle would be to make the crazy inverse formula with the line integral along a vertical line in the complex plane whose real part is greater than all singularities of F(s). I haven't find a video on that topic yet and would be super hyped if you could be the first one to do that. Of course only if that's possible and you actually want to :D

@MuPrimeMath

4 жыл бұрын

I've never taken a complex analysis class so I haven't learned those kinds of formulas! I can link you to some videos about it, but because I don't know the method myself I don't know if they will help or not: kzread.info/dash/bejne/o5-YsKuNiNLTqNY.html kzread.info/dash/bejne/d6Nt1KaDj7u_o5M.html

@Rundas69420

4 жыл бұрын

@@MuPrimeMath Thanks man, I'll definitely check them out. Then I can see for myself whether they are helpful or not. I'm questioning myself, why I didn't find them on my own though xD

For anyone who's confused how 11:56 came, use partial fraction method.

Thanks a lot, you are my god.

Integration by part is the method which you integrate something twice in one integral and that's the trick

Thank you for this explanation! Question: How worried do I have to be about restricting the interval of s? I notice for example that for L{exp(3t)}, s must be greater than 3, otherwise the integral diverges. Can I just assume that s will be such that the integral converges, or should we explicitly restrict s to assure convergence?

@MuPrimeMath

4 жыл бұрын

To be rigorous, it would be correct to restrict the interval of s. In that case, we would say that the Laplace transform of exp(3t) is defined for s>3. For the purpose of differential equations, it's not a big deal because the Laplace transform is still unique even if we restrict the values that s can be!

Simply magnificent. I have been searching the idea behind Laplace Transform and found this video. Very well explained, thank you for the video. By the way, if you can make a video about Fourier Transform it would be very nice.

You've got what i'm looking for, the reason why there is laplace transform.

In differential equations are we always stuck with not being able to determine a constant such as (in this case) y(0) similar to how integration results in an unknown constant to add? In other words the original differential equation here holds true for all functions y of the form ae^(-t) + e^(3t) whatever "a" happens to be, but are there always one or more constants like this that crop up in the result?

Great explanation about Laplace transform! Many thanks for making this video. I just have one question at 10:52 when you evaluate the e^-(s-3)t at infinity. Does it only become 0 when s-3 is positive or s > 3? Thanks!

@MuPrimeMath

4 жыл бұрын

Yes, that is correct! Most Laplace transforms have some kind of restriction like that on the domain of s.

I had three questions. How do you know in advance that y is defined in t=0 ? Before knowing the result, one should assume it might be an asymptote, no? At 10:55 you say "e to minus infinity" but what if s At 11:57 should it not be 4/(s+1)(s-3) instead of 1/(s-3) ? Further reading of comments gave me the answer to the two last ones.

You're awesome

Can you explain why you changed s to s+1 at 9:45 ? Doesn’t that change the formula for integration by parts and make it not valid anymore?

@MuPrimeMath

4 жыл бұрын

That comes from the fact that when we take the Laplace transform of the left side of the original equation, we have to include the Laplace transform of y. The +1 is adding the Laplace transform of y because that's the integral that's multiplied on the right!

Nice

eyw reis

At 8:34 when we evaluate the lower bound of the first part, why do we evaluate y at 0 when we are taking our bounds with respect to t? Is it because y is a funtion of t, so we can just evaluate them at the same point in order to get a constant for the y term? Thank you for the video!

@MuPrimeMath

3 жыл бұрын

Yes, that's correct! When we're doing these differential equations, we assume that y is a function of t, so evaluating y at t=0 gives a constant.

@I0MSammy

3 жыл бұрын

@@MuPrimeMath Perfect! Thanks so much for the quick reply! I worked along with pen and paper and now it is much more intuitive for me. Liked it and subcribed!

How do you solve for y(0)

11:57 where did the 4(s+1) go

Hi! Came here from bprp! I was wondering if you could solve DEs that don’t have constant coefficients/non linear.

@MuPrimeMath

3 жыл бұрын

See kzread.info/dash/bejne/X6h6speIpLrZcaQ.html

@bensonkwok951

3 жыл бұрын

@@MuPrimeMath thanks!!

at 1:24 : "easy to solve for y dividing by 2"... that equation is true for all y

Can I ask how do you integrate y prime in the T world how you end up with y + integrate y respect to T

@MuPrimeMath

2 жыл бұрын

At 1:58 I am taking the integral of y' + y, I just didn't write down the integral of y on the first line.

10months after back with a question why is it -y(0)(1)instead of just-y(1), 8:45

Can you please explain what you did at 11:50 ?

@MuPrimeMath

2 жыл бұрын

I solved for the integral. You can think of it like we're treating the integral expression as a variable, then solve it like any other equation using algebra.

How do prove the Laplace transform is unique?

@MuPrimeMath

4 жыл бұрын

Here are two similar proofs: www.ctr.maths.lu.se/media11/MATC12/2013ht2013/uniqueness.pdf web.mit.edu/jorloff/www/18.03-esg/notes/extra/laplaceuniqueness.pdf

Must be over 40 years since I did these . Might need to watch it another couple of times for it to sink in.

Is that tabular method of integration at 5:00 ?? Or is that integration by parts?

@MuPrimeMath

Жыл бұрын

The tabular method is just a way of writing integration by parts.

@te-kowski

Жыл бұрын

@@MuPrimeMath I realized that after commenting, but still a little confused. Doesn't tabular method go diagonal? Why would the (-se^-st) be multipled across to (y) instead of diagonally to the next antiderivative-- the antiderivative of y? So it would look like ((-se^-st)*(integral(y)))

@MuPrimeMath

Жыл бұрын

The purely diagonal version of the tabular method only works when the function we're differentiating is a polynomial, so that its derivatives eventually go to zero. In general, we eventually have to stop and integrate a row. See this video for an explanation: kzread.info/dash/bejne/ZH1hwbWPaMbdo9o.html

@te-kowski

Жыл бұрын

@@MuPrimeMath AHHH THANKS SO MUCH, I forgot that part.

Just WoW

that's exactly the way that things should be taught. i think this is the logical way to explain for humans, because they have a brain work by logical thinking. the other way that teachers explaining in schools and universities, it's more suitable for monkeys and parrots because it depends on just imitating what the teacher is doing, without any logical thinking or explanation.

but why is "s" a complex variable??

It took one hour to get it. Because of me. Started from the bottom.

Why did you add s+1?

@MuPrimeMath

3 жыл бұрын

If you're referring to 9:30, the +1 is from taking the laplace transform of y. The extra y term is from the equation that we started with.

It clicked when u did (s+1)

what about initial conditions

@carultch

5 ай бұрын

Initial conditions are accounted for, in the full answer to the Laplace transform of the derivatives of the dependent variable. For instance: L{y'} = s*Y - y(0) L{y"} = s^2*Y - s*y(0) - y'(0) Where L{y} = Y. In general: L{y^[n]} = s^n * Y - sum of s^(n - k - 1)*y^[k] (0), from k =0 to k=n-1 where k is the indexing variable of the sum, and the exponents in brackets is my notation for the order of differentiation. This means, you descend the power of s, until it gets to zero. Meanwhile, you ascend the order of differentiation for the initial condition. The whole sum is subtracted from s^n * Y, and all initial conditions can be shuffled to the right side as part of the stimulus of the system. You'll find that initial conditions "cut in line", and shift themselves to the highest power of s in the numerator, after you make Y the subject, and simplify. This is due to the initial value theorem.

@Kurosaka

5 ай бұрын

mannn had I seen this 5 months ago. Thanks alot man cleared up some doubts fr @@carultch

Can someone explain the difference between Fourier and Laplace transforms, I've used the Fourier transform for partial and ordinary differential equations under the assumption that the equation you're solving for decays at infinity (e.g. quantum mechanics), however I've never studied Laplace transforms.

@carultch

7 ай бұрын

The Fourier transform is a special case of the Laplace transform, that focuses on the steady state, in the limit as time approaches infinity, and the real part of s diminishes to zero. i*omega (or j*omega) takes the place of s in the Fourier transform. The Laplace transform scans the original function for a spectrum of hidden exponential functions, while the Fourier transform scans the original function for its hidden spectrum of sine and cosine waves.

@carultch

7 ай бұрын

As an example, consider solving the following DiffEQ with both transforms. Both initial conditions start at zero. y" + 6*y' + 9*y = 50*cos(t) With the Laplace transform: s^2*Y + 6*s*Y + 9*Y = 50*s/(s^2 + 1) Y = 50*s/((s^2 + 1)*(s + 3)^2) Y = -4/(s + 3) - 15/(s + 3)^2 + (4*s + 3)/(s^2 + 1) y = -4*e^(-3*t) - 15*t*e^(-3*t) + 4*cos(t) + 3*sin(t) With the Fourier transform: y" + 6*y' + 9*y = 50*cos(t) (i*w)^2*Y + 6*i*w*Y + 9*Y = 50*pi*[delta(w - 1) + delta(w + 1)] Let C and S be Fourier transforms of cosine & sine. Solve for Y: Y = 50*C/(9 - w^2 + 6*i*w) Y = 50*C/(8 + 6*i), at w=1 Y = (4 - 3*i)*C Y = 4*C + 3*S y = 4*cos(t) + 3*sin(t) As you can see, we get the same steady state solution, which is 4*cos(t) + 3*sin(t). The Fourier method only gives these solutions, while the Laplace transform gives us the exponential decay from the initial conditions, as it settles on this steady state solution.

I really enjoyed this video, because most people try to explain this only in theory, using some big words, then throwing some exaples, so at the end you know how to use Laplace transform, not why it works. Well it maybe that did't happen in your case, but it certainly did in mine. Thanks for providing a more intuitive backround on the topic. :)

This was an interesting view but it doesn't get to the heart of WHY Laplace transforms make solving these differential equations easier. The key idea is that the derivative of a Laplace transform is just S * the transform or S*L(S). So the second derivative is S^2*L(S). So if you take a differential equation and you transform it then you factor our L(S) and everything is just algebra. Finding the poles of the algebraic expression give you the exponential and sinusoidal components of the solution.

@carultch

5 ай бұрын

Essentially, you're replacing the given functions with a hidden spectrum of exponential decays and sine waves, that are represented in the Laplace transform. Since exponential decays and sine waves are easiest to use when solving differential equations, this allows you to convert calculus into algebra.

That’s Kanye’s favorite equation, in fact he’ll tell you it was his idea 😂

Where did this Laplace guy come from???? OUTER SPACE?

@carultch

7 ай бұрын

France

It gives a very nice understanding, however, how the FUCK do you get from 11:53 to 11:57 ? It seems I am unable to process some of the most basic algebraic manipulation ever, so that really grinds my gears, could someone help ?

@MuPrimeMath

4 жыл бұрын

In that case, we're trying to solve for the integral, just like isolating x in a normal algebra problem. First add y(0) on both sides, then divide by (s+1) to isolate the integral. The one thing that I didn't mention (which might be the confusing part) is that I did partial fraction decomposition on 4/[(s+1)(s+3)] at the very end to split it into two parts!

@tryphonunzouave8384

4 жыл бұрын

@@MuPrimeMath Ah yes, that's it, now I understand, thank you for the swift response man :)

at 8:15 No,zero is not a nice number in exam result😂

I came here from blackpenredpen video ... kzread.info/dash/bejne/mKii0q-YgrKWesY.html Both of you guys are awesome.!

Needs to learn to write properly and make sure that his 4 does not look like a Y