Differential Equations of Motion
ฝัง
- เผยแพร่เมื่อ 14 ก.ย. 2010
- Differential Equations of Motion
Instructor: Gilbert Strang
ocw.mit.edu/highlights-of-calc...
License: Creative Commons BY-NC-SA
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This professor is terrific. I have a degree in mathematics but have learned a lot more re applications and Dr. Strang has no problem showing us where the math is useful. Bravo sir!
Every time i watch this video, my understanding deepens. Thank you :)
22:25 "Can I remember that dumb formula".
Hahaha... I love hearing that from a math teacher!
thank you!
I am a fan of prof. Strang's lectures but also got confused at 27'th minute, searched a little bit and I think this might be a help:
We have two complex basic solutions of the second order equation:
y_1 = e^{-3t}e{it} = e^{-3t}(cost+isint)
y_2 = e^{-3t}e{-it} = e^{-3t}(cost-isint)
and want to 'go back to real'....
any combination (also with complex constants!) of y_1 and y_2 is also a solution.
So take two such combinations of y_1 and y_2 that will result in a cancellation of imaginary:
y_r1=(y_1+y_2)/2=e^{-3t}cost
y_r2=(y_1-y_2)/2i=e^{-3t}sint
only when C is equal to D
Very useful supplement.
Further reading: tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx
e^(-3t)[C.(cos 3t + i.sin 3t) + D(cos 3t - i.sin 3t)] = e^(-3t) [(C+D)cos 3t + (C - D).i.sin 3t)] ->
y= e^(-3t)[ A.cos 3t + B.sin 3t] for A=C+D, B= i.(C-D)
What do you think?
@@erenozturk5796 Dude that was very helpful. Thanks.
Yaaaayyyy, Strang teaching differential equations!! How lucky we, the students, are!
fantastic, I haven't had such a great instruction in my whole life.
thanks for you generosity.
God, I love Gilbert Strang... such an amazing teacher
This is a classical lecture on introductory differential equations. DR. Strang continues to strengthen my knowledge of mathematical tools and concepts.
Beautiful explanation. Thanks Dr. Strang.
No, the formula is right. Normally, for a polynomial ax^2+bx+c=0, the solutions are x = [ -b +/- sqrt (b^2-4ac) ] /2a. In this case, we have ml^2 + 2r l + k =0. So the solutions are
[ -2r +/- sqrt (4r^2 - 4mk) ] /2m, which simplifies to [ -r +/- sqrt( r^2-mk) ] / m, as in the lecture.
Very nice explanation ... Thanks a lot sir for your time.
At 27:28, the solution should be y(t) = Ae^(-3t)cost + i Be^(-3t) sint. The sint would not be cancelled out as C and D are not necessarily equal.
+shakesbeer00 yes i also think he missed out i.
Prof Strang is a world renowned mathematician with contributions to finite elements analysis and wavelets for instance.Prof Strang always reminded me of my former oxford educated mathematics professor for his accent and the ability to think on his feet as if he is doing it for the first time .Prof Strang spent some time at oxford as Rodes scholar if i my memory serve correct?
fantastic, I haven't had such a great instruction in my hole life.
thanks for you generosity.
Maybe because you were living in a hole? Sorry, just kidding, saw your other comment with the correction
It's Amazing lecture ,, by this PARTICULAR LEGEND mathematician,,,,,
For the transform in 27'29", we may derive it in this way: y1 = A * [( exp((-3+i)t) + exp((-3-i)t) ) / 2] = A * exp(-3t) * cos(t); y2 = B * [( exp((-3+i)t) - exp((-3-i)t) ) / (2i)] = B * exp(-3t) * sin(t); y(t) = y1 + y2 = A * exp(-3t) * cos(t) + B * exp(-3t) * sin(t)
+Ting Xu i didn't get how you considered the two equations y1 and y2.
I got that the values of lambda = -3 +/- i
so y(t) will be y(t) = C*exp((-3+i)t + D*exp((-3-i)t now if you put the value of exp(it) and exp(-it) then you will get an "i" term at the last and non pure real terms.
IIUC the equality y(t) = y1 + y2 implies C = D = (A/2 + B/2i). So this seems to (i) require C=D, which y(t) doesn't guarantee, and (ii) hide the imaginary part in the C and D, where Strang claims this is a purely real formula.
@@euccastro he used the principle of superposition regarding homogeneous d.e. that is why the imaginary part is eliminated
Fantastic! Just fantastic!
This video should be preceded by Differential Equations of Growth video.
fantastic Profesor!
he is a very good teacher.
Brilliant!
wow I wish my professors were as smooth as this was!
Excellent lecture 🙏🙏🙏🙏
Best video on the subject ! At 27:26, B is a complex number isn't it ?
God bless you, Gilbert Strang!
@cesfigas Yes, he is an amazing teacher.
Dang. It seems like MIT profs. make it so easy to learn. Any other school, you'd be teaching yourself out of the book.
We normally write ax^2 + bx + c =0, and get the solutions x = (-b +/- sqrt( b^2 - 4ac) ) / 2a. He, the equation is ml^2 + 2r l + k =0. This is a second degree equation in the variable l. In this case, what corresponds to the coefficient b of the equation ax^2 + bx + c = 0 is the coefficient of l, namely, 2r. Now, what is (2r)^2? It is 4r^2. And this is where the 4 comes from. Does it help?
27:18 Professor Strang, how did you cancel out the virtual part? The cancellation should be given that C = D I think.
We have two complex basic solutions of the second order equation:
y_1 = e^{-3t}e{it} = e^{-3t}(cost+isint)
y_2 = e^{-3t}e{-it} = e^{-3t}(cost-isint)
and want to 'go back to real'....
any combination (also with complex constants!) of y_1 and y_2 is also a solution.
So take two such combinations of y_1 and y_2 that will result in a cancellation of imaginary:
y_r1=(y_1+y_2)/2=e^{-3t}cost
y_r2=(y_1-y_2)/2i=e^{-3t}sint
Got it, thanks!
One other thing you might be able to clear up,
I was confused about the same thing as JanisMac314 who asked: at minute 27:24 " I.. have found the coefficient B to have an imaginary in it
(comming from i*sin t )"
Where professor Strang states that the equation is meant to go back into it's real form.
i came to this lecture searching an answer for this question. Did you get that?
thank you very much
where does the 4 come from in the 4r^2
Was he teaching the students , how to solve diffrential equations or was he just revising them of the previously taught forms of D.E. ?
Very nice
Oro puro
2010: a good time to live.
hi, someone knows some good video about dynamic optimitation?
tnx
I got differential equation next semester, but this lecture is still a bit too new for me to fully understand
27:24 "back to real"
really? I have tried it and found the coefficient B to have an imaginary in it
(comming from i*sin t )
can someone explain please?
i came to this lecture searching an answer for this question. Did you get that?
Gilbert strang is like a father to me..... :-)
Seems like a nice guy
I too can't see where the i went in the B part at minute 27. He rushed that part anf its the critical bit!!!
Couldn't we solve the second one with -e^-Wt
30 years after trying to learn it the first time, and realizing it ain't gonna happen. I'll just stick to solids modeling. No supersonic jet designs for me.
The equation is z( k o)=t
All this is trivial basic maths.
Likes God....
Biri tercüme etsin lan :D
There's a fatal error in this lecture: the quadratic formula used misses two numbers:
There's a fatal error in this lecture: the quadratic formula used misses two numbers:
4 to multiply K(that's a constant number) in the square root and 2 to multiply m(same case). If you insert any example, including the ones he mentioned, if not using them, the results will not coincide, therefore the formula in the lecture is wrong!
No, this is accounted for since the coefficient of x is "2r" (which would be used in the standard quadratic formula) but "r" is picked up in his adjusted quadratic formula, ie the x coefficient is halved from 6 to 3, which takes account of the 2 and 4 factors you mention.