Lec 3 | MIT 18.03 Differential Equations, Spring 2006
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- Опубліковано 15 січ 2008
- Solving First-order Linear ODE's; Steady-state and Transient Solutions.
View the complete course: ocw.mit.edu/18-03S06
License: Creative Commons BY-NC-SA
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i just started studying it during lockdown . At first,i was unsatisfied with the quality but after getting to know the content i was really impressed its really a great help for students who dont get quality education .These type of things create unbiased education for every citizen across the world .
A big thanks to MIT !
All the professors and staff sharing knowledge for free are beautiful souls. Blessed are all.
If you thought this lecture was excellent, Mattuck is even more impressive in person.
He rides his bike every morning to MIT from Brookline ~10 miles, even in the rain/snow
His explanation for finding the integrating factor was simple and elegant - very nice.
Love this professor. Very lucid explanations --- he makes sure you know what you're supposed to be looking for. He also obviously has fun with what he's doing!
Professor Arthur Mattuck, thank you for explaining and Solving First-order Linear ODE's; Steady-state and Transient Solutions. This part of differential equations is fairly easy to understand.
This lecture alone was more helpful than wasting my time on reading many hours of Calculus without understanding one thing.
the teacher is so clear with explaning. everything sounds so simple and reasonable.
Making this free on youtube... Is a gift to humanity.
they know how to teach at MIT
i could finaly understand the theory behind this method just by the way he ponders his calculations. i love this teacher's lectures
RIP Prof. Mattuck. Thank you for these lectures.
Your explanation about the integrating factor is very good. Thank you.
Now this is what I call lecture. My professor didn't even say a thing about all these basic stuffs. Awasome, and thanks for posting.
Beautiful conclusion to the lesson.
Models (heat diffusion, salt concentration), and Integrating factor determination and solution of first-order linear differential equations..
heey MIT how did you find this amazing teacher this is no ordinary teacher man i would love to go to all his lectures !
This is an amazingly good course.
-sinx is the derivative of (1+cos x) And thus integrating -sinx/(1+ cos x) .dx we get ln(1+cos x) whi9ch is then exponentiated to e^ln(1+ cos x)
I would love if you add video lecture on more theoretical math too, like 18.034 and 18.100c.
Thank you for what you have provided though. You people are a big help. I habe completed 4 courses from ocw, and I dont know how many more I will.
Excellent lecture....I learned a lot .
The diff.eq. he attempted at 44:36 is also separable. He could have solved by seperation of variable. It turns to logarithmic and exponential form later.
He just makes you love math! Impressive!
@agentorion Professor was right, derivative of ln(1+cos(x)) is "-sin(x)/(1+cos(x))". The negative sign was not forgotten actually.
thanks for posting these
The convection is totally independent of the conduction, in this example. Even if the cup was a perfect insulator the coffee would transfer heat to the surrounding air ( considering there is no lid on ) because of the density gradient present. Although, the heat transfer mode is considered to be convection there is actually conduction within a small boundary layer.
Lucid explanation, didn't know when it completed!!..
One thing to add as well. As amazing as is sounds, this phenomena can produce turbulence. The density gradient might be high enough to consider the fluid to behave in the turbulent region. Of course there is a lot more to this subject but I thought it would catch your attention if you were new to it.
many many many many many many many thaaanks for mit
This guy is so solid.
this man is awesome!!! he knows too many! =D i'm learning diff eq. thanks to him!!!
his penmanship is impecable!
brilliant teacher, imo
awesome lecture...
Desde Venezuela. Muchas gracias.
Excelente
He is a good lecturer!
your voice is awesome !
Great lectures!!
great video! right around 28:00 it got very exciting!!
Convection can occur while fluids aren't moving or are quasi static. This is caused by the density gradient between both fluids.
If you think of it, a cup of coffee will normalize to room temperature if let standing in a mug and in a room where no noticeable movement of air air present. In this case there is both convection and conduction, but I hope you get what I'm saying. You could find more info in any text dedicated to convective heat transfer.
Good Job Professor
It's interesting comparing this lecture to what my own are like with the University of Toronto's Engineering Science. It seems this professor explains things a lot more than my own, but on the other hand it's more of a challenge to have to discover these things independently.
omg, so much better than my prof! thank you :)
Great video
The chalk is so smooth!
Wonder if it's Hagoromo. Don't even know if it was a thing at the time.
great clear explanations (just bad video quality). if someone could re-digitally fix the video, it would be awesome.
@keijigo @analemma2345 Actually he meant to say "du over dx (all of that) over u" (the derivative of u with respect to x divided by u itself). That is called the differential quotient (u'/u or du/xdx depending on the notation you're using) and equals (lnu)'. So then we have (lnu(x))'=p(x), we integrate on both sides with respect to x and use some simple algebra to get u(x). Hope you understand my explanation despite my crappy english.
@analemma2345 @analemma2345 Actually he meant to say "du over dx (all of that) over u" (the derivative of u with respect to x divided by u itself). That is called the differential quotient (u'/u or du/xdx depending on the notation you're using) and equals (lnu)'. So then we have (lnu(x))'=p(x), we integrate on both sides with respect to x and use some simple algebra to get u(x). Hope you understand my explanation despite my crappy english.
31:05 Integrating Factor Method
@MrAlb0t Yeah, I saw that too. I guess we should just focus on the differential equations part of this and not the Calculus involved. But I think the problem as a whole is unsolvable because I can't do the final integration.
I like the 'sine of x' instead of 'sign of x' in the captioned version 06:52
Fantastic :P Thanks
WARNING: The subtitles around 7:00 state that the sin(p) should be carefully scrutinized when deciding which way to represent this equation (and ultimately solve it). The subtitles should have stated the SIGN (e.g. +/-) of p. He is referring to the algebraic manipulation of the equation and dropping the (-) sign from the p(x)y portion of the representation, which is perfectly OK to do as long as one keeps them straight.
It is my understanding that it is only considered convection when the fluid is in motion. In this case, the fluid is static, which would indeed make this conduction.
Am I wrong?
In fact, Newton's law of cooling governs heat transfer by conviction not by conduction as stated in the lecture. The law governing heat transfer by conduction is Fourier laws and it says the the rate of change of temperature is propostional to the temperature gradient.
Also, steady state means, as far as I know, that the temperatures becomes constant i.e. fixed not as mentioned approaching infinity!
It applies both ways-check out the original version from the book Principia by Newton or come to Cambridge where you will find one in the Library.
Fawzy Hegab Steady-state can mean that the function approaches a sinusoid. There's no definite way of saying it though, because you're just looking at which terms of the function dominate as t goes to infinity. Terms that approach zero are easier to eliminate.
Dave Yen Thanx for clarifying that
Fawzy Hegab NP. I had a lot of issues with that in circuits because they always say (solve steady state) , but I thought that was an unimportant aspect. but if it wasn't steady state, then circuit would require you to assume the solutions of voltage, current are not sinusoidal and thus would require you to use diff eqns to get the solution, whereas the steady state assumption in that case allows you to work only a constant that's a multiple of a sinusoid with a particular frequency (or superposition of these ). Additionally, it wasn't until vibration mechanics (dynamics) that steady state solution looked important, because then we would just assume it was a sinusoid and get a constant out of that simplifying the hell out of the diff eqns soln (just using phasors).
I WILL be back to watch this very video in around 3 years from now, seeing as im only in calc 1 :)
did you do it?
Did you did it?
@@TheSharkyBoyCostyn hell no lmao!
he left calculus after getting to know it
Surya Singh :))))
@@suryasingh9526 Actually, about to complete a CS degree with a Math minor. I took Diff EQ a couple years ago. ;)
I Really Like The Video From Your Solving First-order Linear ODE's; Steady-state and Transient Solutions
TV 2 FILM No again please, I saw you at 18.06!!
At 40:20 he should not get the same eqn, he forgot a minus sign when computing the integral Pdx, so if you followed that same mistake, you'll you cannot check with the product rule, I guess even that counts as his education of why he uses that step to his students.
At 28:09 he seems to imply that u'/u = du/u, so u'=du. But since u is a function of x, isn't u' equal to du/dx? Can somebody please explain this to me?
MIT needs to pass out some lozenges to those students...
@algorithMIT what does the avg home work look like after/for one of this lectures out there?
How can we find the differential eq of order 3 with nonconstant coefficients? e.g. y''' + P(x).y'' + G(x).y' + L(x).y = F(x).
The integral of 1/x is ln(|x|).
I like this video lecture
very end: core concept of steady state
I don´t think you can get anyway better explained than this ...
they have lecture notes on the MIT site for a lot of courses if that is helpful
Se Deus quiser, eu vou estudar aí.
So are you saying that the initial conductive transfer of heat to the air (in the coffee example) would cause a density gradient in the air and thus cause convection? Or are you saying that they both occur simultaneously (the convection is not a result of the conduction) simply due to the fact that the coffee is denser than air? If the former, I understand what you're saying. If the latter, I have something interesting to learn :-)
I wish they were able to improve the video quality. Is there anyway MIT can get this up to 480?
nice sir
@Satchindra meeeee :DD
search 18.01 MIT on youtube... that's pretty much all the background you need (at least for this lecture)
Please process the videos with a CNN that puts it in HD
Did he forget the y at 39.30 in equ. (1+cosx)y'-sinx"y"=2x?
Thanks for answers.
@38:41 what hapend there?????? i think he got that wrong and what will happend if its negative??? todo cambiaria por que tendras -(1+cos(x)) de facor integrante, - ( 1 + cos(x)) y' + sin(x) y = - 2 x
+Javier Parapar - ( 1 + cos(x)) y' + sin(x) y = - 2 x multiply both sides by (-1) and you get the same thing, am I wrong?
+Cesar Cardoso If I am not wrong minus should be in front of ln(1+cosx) so when you you elevate "e" to that you should get 1/(1+cosx)
I don't know if this is still relevant to you but the reason for there not being a "-" in front of ln(1+cosx) is that you get ( by use of the chain rule and the fact that (cosx)' = -sinx )
( ln(1+cosx) )' = (1+cosx)^-1 * (-sinx) = p(x)
where p(x) is the factor in front of y in the standard form introduced in this lecture.
Damn good teacher. Makes my university look like crap.
About 7 minutes in, he starts talking about the errors that occur with the 'change of sine of p' (according to the subtitles). Surely it should be the 'sign' of p and not 'sine'? Just thought I would mention it seeing as its a little confusing otherwise.
he has real nice handwriting
@jimmylovesyouall from 38:15 by the way.
What if you can't integrate p
@yaymynameispete The MIT transcripts have many errors, try not to pay attention to them if you can.
1+cosx not multiplied in right hand side
Do I have to be an actual MIT student to get help with calculus?
good silence in the lecture ..we have a rock party every lecture
I´m not a mechanical engineer and I would call that a conduction model.
I'm surprised no one picked up on his mistake
+reardelt What mistake are you talking about? Conduction (convection)? If so, I don't think that is a mistake. The temperature of the metal box inside increases via conduction. The way the heat arrives at the walls of that metal box (via convection) is wholly irrelevant.
tokamak no, they’re talking about 39:40 when he forgot to multiply sin(x) by y.
It's First Order Unear DE, apparently.
why does he says if you want to drop the course at 00:57?
I will fell honor just for sitting in one of those chairs
17:45 Semipermeable membrane. Sounds like the dialysis bag lab for AP Biology.
heys people... i have a doubt.. at the end, when he solves the second differential equation, and he's finding the integration factor, didn't he forgot the minus? i mean, he has got : - integr(sen(x)/(1+cos(x))) ... so the answer it would be (u)^-1 ... it wouldn't match with the first equation ... am i right?
Nah mate, I'm not trying for Cambridge hahaha. But I did have FP1, FP2 and FP3 in Further Maths besides the normal general maths.
26:50
At minute 41 he made a mistake and forgot to write that the (-sin(x)) term was multiplied by y, sad confusing some students
At least I got c=c over one plus 1.
38:27, where does the minus sign go when he does e^ln(1+cosx) :S
lovely lecture tho imo :P
If you're talking about the absolute value you get after integration, you can neglect it because 1+ cos x if always greater than or equal to 0 because cos x has a minimum value of -1
fuck I just replied to a 7 year old comment
I think he was talking about the minus sign in front of the integral.
d/dx (1 + cos x) = - sin x, so - integral[sin x / (1 + cos x) dx] = integral[-sin x / (1 + cos x) dx] = log(1 + cos x).
wait, someone answered him 4 years later
Diffusion ...isnt it 2nd order usually???
Different cases, some case could use first order
It was supposed to be "sign of p", not "sine of p"
But in lecture 30 of MIT 18.02 Multivariable Calculus course (ua-cam.com/video/seO7-TwXH_I/v-deo.html) we learn in the first 3 minutes a complete different diffusion equation:
∂u/∂t = k ▽²u
where u is the spatial concentration.
4K UHD
didn't he miss to multiply 2x by (1+cosx) in the example # 2 ....???
This man emphasized multiplying BOTH sides of the equation and neglected to do that because he was too focused on his other mistake. Somehow nobody in the class nor the comments caught that except you. I would just like to say I still love this professor and I could be dumb and missed something but this was extremely confusing.
He is just tryin to say that even if you drop from this course you would have learnt 2 important things about 1st order differential.