I am very happy to see the vidoe First-order Substitution Methods: Bernouilli and Homogeneous ODE's from you, hopefully the others also are happy for You
after i watched this video First-order Substitution Methods: Bernouilli and Homogeneous ODE's, my insight is very open because the video is very good to give information
he is alright. i can honestly say that his lecture is identical to the one receive at my university. just because he teaches at MIT doesnt mean he is greatest educator of all time. he does his job slightly above average, that is all.
@Styx47ag a small campus in the california state university system called sonoma. it is far from fancy but i also wouldnt care to guess how much cheaper it is than MIT. i would say that this professor is better than my calc 1 professor but i would say that he is either identical or very close to my last two math professors.
Arthur Mattuck's typical lectures are truly thrilling roller-coaster rides inside the hidden potential humor of energetically blazing math, and so I'm simply addicted to them. :)
He is a really good teacher. I am glad that the videos were uploaded to youtube as the opencourseware site said something was wrong with the url. This is the first time I'm doing a diff 1 course and so I am trying to get a headstart on the work. Thanks MIT!!!
Impressive. I live in Ecuador and I study in the best university of my country (ESPOL), and it is know that it has the same academic level as MIT, but the way this professor get's my attention is out of this world. Don't get me wrong, I am proud of my school, but I guess I will watch this lectures before attending to class, just to be a bit ahead.
Sounds like a very good plan. These MIT vids are quite entertaining and interesting. I am pretty much lost when listening, though I think I can grasp a bit of it. Good luck in Ecuador.
This comment has received too many negative votesshow
Now why on earth would one call an equation of the form \dfrac{dy}{dx} = F( \dfrac{y}{x} ) a homogeneous eqn. Take y' = y/x + 1 or in other words xy'-y=x . NOT A HOMOGENEOUS eqn...Abusive terminology
I now found out that by the method of partial fractions, you can write 1/(T^4+1) as the sum of four terms of the form a1/(T+q1) + a2/(t+q2)+..., where q is a fourth root of -1,
that also is samefor dx=dt/1-t^4 wwrite the denominators as product of 1+sqrt (T) and 1-sqrt(T)
applying partial fractions method,then similarly do for 1-t^2 as product of 1+t and 1-t,also dt/1+T^4 can be solved by dividing numerator and denominator both by t^2
then writing denominator (t+1/t)^2-2,then use substitution z=t+1/t,so it simplifies to -dz/z^2-2,hope now u can solve it
En el denominador sumale 2T^2 y restale 2T^2, con esto tienes en el denominador (T^4+2T^2+1)-(2T^2) = [(T^2+1)^2-(SQRT(2)*T)^2], esto te va ayudar muchísimo para formar, las fracciones parciales, y comenzar a integrar.
in00bee, Thanks for your answer, my question should be written better,
1) Restatement of first question, at the end he talks about theta as a function of arctan y/x, isn't this alpha?? But he talks as if he is descibing the curve of the "boat". I do not understand what angle "Theta" he is talking about.
Also, Question 2) Just prior to the Theta calc, he uses the law of logarithims, and gets arctan z= ln(1 + z^2)^1/2 + ln x + c, shouldn't the power be a negative 1/2 for the first term on the right??
Because most professors don't like their job teaching. I am pretty sure that MIT prof's love their job, or well they are damn smart at the very least.
I am very happy to see the vidoe First-order Substitution Methods: Bernouilli and Homogeneous ODE's from you, hopefully the others also are happy for You
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bundawartini 2 weeks ago
after i watched this video First-order Substitution Methods: Bernouilli and Homogeneous ODE's, my insight is very open because the video is very good to give information
anakmudajaman 2 weeks ago 2
geil bin ne sau
ShilaGinafy49 1 month ago
Mind blowing lecture. Whaaah.. but this is so excellent.
agapitoflores001 2 months ago
he is alright. i can honestly say that his lecture is identical to the one receive at my university. just because he teaches at MIT doesnt mean he is greatest educator of all time. he does his job slightly above average, that is all.
vajitarian 3 months ago
@vajitarian what university do you go to? This lecturer blows mine out of the water.
Styx47ag 3 months ago
@Styx47ag a small campus in the california state university system called sonoma. it is far from fancy but i also wouldnt care to guess how much cheaper it is than MIT. i would say that this professor is better than my calc 1 professor but i would say that he is either identical or very close to my last two math professors.
vajitarian 3 months ago
This has been flagged as spam show
I love him <3
kaychaos5 5 months ago
Comment removed
kaychaos5 5 months ago
great lecture as always, damn this is getting difficult lol
Liaomiao 8 months ago 2
I love watching maths nerds snicker when they teach it. Another great lecture.
CostaDelBarto 8 months ago 2
It is the best explanation...superb
stan021 9 months ago
It seems like every MIT teacher is amazing!
alexwhb122 9 months ago 2
HE RULES
QED
x1x2x3ct 9 months ago 3
Arthur Mattuck's typical lectures are truly thrilling roller-coaster rides inside the hidden potential humor of energetically blazing math, and so I'm simply addicted to them. :)
AA10Megaviv 10 months ago 3
Only a genius could mention a drug boat in a pure math class and got away with it!
HGoodwyn 10 months ago
really maths GURU
trgopan 10 months ago
great lecture!!! if you don't understand this, you probably need to watch the previous lectures or go back to high school.
naturallymagic88 1 year ago
damn..that was intense x.x
yyourfacee 1 year ago
I really wish all the schools would use the same book so that they teach all the same components! =/ His lectures are great though...
lovericecake 1 year ago
He is a really good teacher. I am glad that the videos were uploaded to youtube as the opencourseware site said something was wrong with the url. This is the first time I'm doing a diff 1 course and so I am trying to get a headstart on the work. Thanks MIT!!!
paultech1988 1 year ago 2
Gracias maestro!
ducefalo21 1 year ago
This comment has received too many negative votes show
it's really poor video not able to understand anything
anuragk833 1 year ago
Three seconds or three grahams. haha.
ivionday 1 year ago
Good lecture
odinheim 1 year ago
What lecture does he explain "exact equations"?
tennisIS4pussys 1 year ago
@tennisIS4pussys I need this, too. And I bet he does a damn good job.
brco2003 11 months ago
Its great they still use chalk.
PukkPukk 2 years ago 31
Drug boat? This guy is ridiculous..!
robinhoode 2 years ago
Impressive. I live in Ecuador and I study in the best university of my country (ESPOL), and it is know that it has the same academic level as MIT, but the way this professor get's my attention is out of this world. Don't get me wrong, I am proud of my school, but I guess I will watch this lectures before attending to class, just to be a bit ahead.
darkrathamantis 2 years ago 5
Sounds like a very good plan. These MIT vids are quite entertaining and interesting. I am pretty much lost when listening, though I think I can grasp a bit of it. Good luck in Ecuador.
PukkPukk 2 years ago 2
@PukkPukk Thanks :D
darkrathamantis 1 year ago
This comment has received too many negative votes show
Now why on earth would one call an equation of the form \dfrac{dy}{dx} = F( \dfrac{y}{x} ) a homogeneous eqn. Take y' = y/x + 1 or in other words xy'-y=x . NOT A HOMOGENEOUS eqn...Abusive terminology
MATH462PDE 2 years ago
where is the lecture for exact equations?
Hivernya 2 years ago
This comment has received too many negative votes show
THE LORD says that this man works for 23 !!!!
BRAINONBOARD 2 years ago
How do you solve the problem dT/dx=1/(1-T^4)? I am very curious about this one, and i think he didn't solve it here. Thanks.
jsmith754 2 years ago
you multiply both sides by (1-T^4)dx, so the equation becomes(1-T^4)dT=dx then you integrate boht sides (left to dT and dx)
that results in T-T^5/5= x + C but to further reduce that to a explicit function of x is pretty much impossible
sikory 2 years ago
I guess I was writing it the wrong way. I meant to say dx=dT/(1+T^4). Any idea if it can be integrated over T?
jsmith754 2 years ago
I typed it in Wolfram Mathematica and that gives me a integral that i guess nobody would have found otherwise:
1/(4 Sqrt[2])(-2 ArcTan[1 - Sqrt[2] T] + 2 ArcTan[1 + Sqrt[2] T] - Log[-1 + Sqrt[2] T - T^2] + Log[1 + Sqrt[2] T + T^2])
Log being the natural logarithm, but i don't think this would be something you would guess or get by things like integration by parts
sikory 2 years ago
I now found out that by the method of partial fractions, you can write 1/(T^4+1) as the sum of four terms of the form a1/(T+q1) + a2/(t+q2)+..., where q is a fourth root of -1,
sikory 2 years ago
it's pretty simple
take the term(1-T^4) to left,integrate both sides to get T-T^5=x+arbitrary constant
nirmalagrawal 2 years ago
I meant the less obvious dx=dT/(1-T^4). I think it is using mixed fractions up to some point. Any ideas? Is dT/(1+T^4) solvable by formula?
jsmith754 2 years ago
that also is samefor dx=dt/1-t^4 wwrite the denominators as product of 1+sqrt (T) and 1-sqrt(T)
applying partial fractions method,then similarly do for 1-t^2 as product of 1+t and 1-t,also dt/1+T^4 can be solved by dividing numerator and denominator both by t^2
then writing denominator (t+1/t)^2-2,then use substitution z=t+1/t,so it simplifies to -dz/z^2-2,hope now u can solve it
nirmalagrawal 2 years ago
En el denominador sumale 2T^2 y restale 2T^2, con esto tienes en el denominador (T^4+2T^2+1)-(2T^2) = [(T^2+1)^2-(SQRT(2)*T)^2], esto te va ayudar muchísimo para formar, las fracciones parciales, y comenzar a integrar.
HoNiAn 2 years ago
try tangent inverse (1+ (T^2))
wasssabii 2 years ago
he speaks english!... unlike all my profs.
loafair 3 years ago
why no longer available.Help!
laiwencong 3 years ago
I wonder if this is the first class in the math sequence at MIT.
rpdwpb 3 years ago
At the very end he say's the arctan(y/x) was theta,, does he mean alpha plus 45 degrees is theta?? Seems weird cuz y/x is theta??
PS Do enjoy these MIT lectures.
kmharrower 3 years ago
Theta is the azimuth in polar coordinates. See wikipedia's article on the polar coordinate system
in00bee 3 years ago
in00bee, Thanks for your answer, my question should be written better,
1) Restatement of first question, at the end he talks about theta as a function of arctan y/x, isn't this alpha?? But he talks as if he is descibing the curve of the "boat". I do not understand what angle "Theta" he is talking about.
kmharrower 3 years ago
Also, Question 2) Just prior to the Theta calc, he uses the law of logarithims, and gets arctan z= ln(1 + z^2)^1/2 + ln x + c, shouldn't the power be a negative 1/2 for the first term on the right??
Thanks for your help.
kmharrower 3 years ago
Opps, I just noticed he moves that term from side of the equation to the other, making it a positive term.
kmharrower 3 years ago
yes i think he missed that part...actually is a different y..
ribakuka 3 years ago
This guy is good and he has legible handwriting!
That's it...I'm never showing up to class, I'm just going to watch these videos. I wish I went there :(
xxsmangiexx 3 years ago
MAn, I am totally lost in my dif. eq. course. I wish this man was my instructor!
yosonie 3 years ago 3
Professor Arthur is the best!
pollardrho06 3 years ago 3
HOW WILL I GET ADMISSION IN mit?????
Royalshippie 3 years ago
Why can't my math profs be like this guy? :(
kingslice75 3 years ago 3
Because most professors don't like their job teaching. I am pretty sure that MIT prof's love their job, or well they are damn smart at the very least.
rpdwpb 3 years ago
He is the best teacher!
amhara23 3 years ago 21
how come when he find v prime, he always adds the extra y prime on the end? is the some chain rule going on there or sumin??
stokeajh 3 years ago
yes it is the chain rule
MathMikie 3 years ago
these are going to save my ass for my final..
tehdennis 3 years ago 2
I noticed there were over 20 lectures , I got time today so...
LongShlong125 3 years ago
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THE LORD says that you are stupid.... don't say that you even understand one of hes lectures not even in one year!!!!
BRAINONBOARD 2 years ago
*( -1) sorry, thank you for reminding me....
BRAINONBOARD 2 years ago
congratulations MIT!!
lhyd7hak 3 years ago 3