This is how I wish differential equations was taught when I was in undergrad, and methods 1 and 3 are the approach I'm taking for a video I want to make on rotorcraft flapping. Around time 31:42, you start trying to solve c1 and c2 in terms of the initial values x0 and v0. My response to this is, why bother? You already established that c1 and or c2 must be complex valued in order to make the final equation real. They are arbitrary constants, and i is a constant as well. So, when I demonstrate this, I just say "let a=c1+c2, let b=(c1-c2)i" and then establish a=x0 and b=v0. No need for complex analysis, and the result matches the intuitive guess (assuming you were able to guess the sine component) and the Taylor series result, while still being completely general.
Keep uploading the content Professor. I am taking Engineering courses and this has been very useful for even providing motivation how all things are fundamentally related .
Linear operator! Superposition! Great! 27:50 So, I guess we're assuming some linear algebra already. Not the first time in this series that I've seen messages from the future. It's not that free stuff is useless, but anyone watching these is sinking a half-hour at a time into following the presentation. It seems more that these videos are intended as refresher, or as quick review of formal lecture material given in a physical classroom by Steve himself. Anyone else is going to have to forego any pretense of rigor or fill in a lot of blanks with a lot of legwork. This series is not a substitute for a complete course.
Around 17:28 I think you left the minus sign out on the board which I have put in single quotes (although you do say it): => C3 = '-'1/3! * V0 - (from an example I calculated it also seems like there should not be a minus here: => C5 = '-'1/5! * V0)
in time 30:00 if λ is complex like a±ib then the answer should be x=e^at (x_0 cos(bt) + v_0 sin(bt) ) and here a=0 and b=1 then x=x_0 cost+v_0 sin(t)
The Laplace Transform is basically guessing a mixture of exponential solutions (3rd way) and converting analytic problems into algebraic problems. In this case, the guess is also that it is a discrete mixture.
There is also the characteristic equation. That being said, I suspect that he is going to hit in that with the eigenvalue determinant or if he covers Laplace Transforms.
Strange. In Method II the answer is x(t) = cos(t)x_zero + sin(t)v_zero. But here we are adding a length to a velocity!? The answer x(t) is a length. Perhaps the second term should have been multiplied by 't'? Am I missing something here?
the simplification not to write (k/m) which has dimension 1/second^2 causes the "problem" with units. I had same thought as you to add length and velocity which is not possible due to different units. But considered the (k/m) factor and the units are then correct as in corresponding terms of the taylor serie (1/second^2) (1/second^4) ... appears forcing each term to have dimension of length. i.g: - 1/(3!)*v0*t^3*(1/second^2) (1/second^2) stems from k/m and so on.
hello. Can someone please explain how x0 got divided it by 2 at 32:35? this makes no sense to me… method III after the fire alarm in particular is very confusing to me
I'm having a bit of trouble using the second method for any k/m instead of just 1. Would the Xo and Vo terms be multiplied by k/m as well? I tried to find such solutions online but had no luck.
In the general solution: x = x0*cos(w*t) + "v0*sin(w*t)" v0 isn't really the coefficient on the sine term, unless w=1 (which would be the case if k/m were 1, since w=sqrt(k/m)). I put this term in quotes, because it really needs to be something else. We want dx/dt to equal v at t=0. Let A and B be defined such that the general solution is: x = A*cos(w*t) + B*sin(w*t) Take the derivative: dx/dt = -A*w*sin(w*t) + B*w*cos(w*t) Evaluate at t = 0: x(0) = A dx/dt at t=0 = B*w So as you can see, v0 = B*w. Thus, when we solve for the coefficient B, this really is equal to v0/w. Thus, the general solution is: x = x0*cos(w*t) + v0/w * sin(w*t)
So how do you get the notes you're writing to appear correctly? That is, writing right to left when I'm pretty sure you would have to be writing backwards to achieve that since you're facing the viewer. Enjoying the talks, btw!
And there are also 2 other ways to establish the differential equation of the oscillator. 1. From the conservation of the energy : d/dt(1/2.m.V**2 + 1/2.K.X**2) = 0 then mV*d/dt(V)+KX*d/dt(X)=0 and then ... 2. From the momentum : d/dt(mV)=kX then ... Many ways to establish the equation, many ways to solve the equation. Why are there so many ways to deal with the oscillator ? Why are there so many points of view to deal with the oscillator ? I really don't know what it means ...
The Taylor series is too tedious for most humans. But if it's general, isn't it something that can be handled by computers? Computer solvers? (And I mean "symbolic" solvers, not "data fitting".) Or is the "suspend variables" method actually used?
Thanks for not cutting fire alarm!
I'm loving this whole series, thanks a lot professor for putting forward a whole course!
This is how I wish differential equations was taught when I was in undergrad, and methods 1 and 3 are the approach I'm taking for a video I want to make on rotorcraft flapping.
Around time 31:42, you start trying to solve c1 and c2 in terms of the initial values x0 and v0. My response to this is, why bother? You already established that c1 and or c2 must be complex valued in order to make the final equation real. They are arbitrary constants, and i is a constant as well. So, when I demonstrate this, I just say "let a=c1+c2, let b=(c1-c2)i" and then establish a=x0 and b=v0. No need for complex analysis, and the result matches the intuitive guess (assuming you were able to guess the sine component) and the Taylor series result, while still being completely general.
Keep uploading the content Professor. I am taking Engineering courses and this has been very useful for even providing motivation how all things are fundamentally related .
Have you graduated yet?
A perfect topic by an excellent teacher... thank you.
Linear operator! Superposition! Great! 27:50 So, I guess we're assuming some linear algebra already. Not the first time in this series that I've seen messages from the future.
It's not that free stuff is useless, but anyone watching these is sinking a half-hour at a time into following the presentation. It seems more that these videos are intended as refresher, or as quick review of formal lecture material given in a physical classroom by Steve himself. Anyone else is going to have to forego any pretense of rigor or fill in a lot of blanks with a lot of legwork. This series is not a substitute for a complete course.
Around 17:28 I think you left the minus sign out on the board which I have put in single quotes (although you do say it): => C3 = '-'1/3! * V0 - (from an example I calculated it also seems like there should not be a minus here: => C5 = '-'1/5! * V0)
Just saw that you spotted and corrected these later on in the video :)
Great video! Would be great videos of Lagrangian's dynamics
Every time I need to counter differential equation in my Engineering journey I come here to refresh the stuff.
Seems like magic. Thank you very much!
in time 30:00 if λ is complex like a±ib then the answer should be x=e^at (x_0 cos(bt) + v_0 sin(bt) ) and here a=0 and b=1 then x=x_0 cost+v_0 sin(t)
What about Laplace Transform? I think we could use it as well.
I came to write this. :)
The Laplace Transform is basically guessing a mixture of exponential solutions (3rd way) and converting analytic problems into algebraic problems. In this case, the guess is also that it is a discrete mixture.
I learned so much in less than 40 min.! Thanx! ❤ 😂
Thanque very much for this beautiful lecture.
jajaja, a fire alarm short break
28:00 ✨ok that was just the fire alarm✨
There is also the characteristic equation. That being said, I suspect that he is going to hit in that with the eigenvalue determinant or if he covers Laplace Transforms.
Strange. In Method II the answer is x(t) = cos(t)x_zero + sin(t)v_zero. But here we are adding a length to a velocity!? The answer x(t) is a length. Perhaps the second term should have been multiplied by 't'? Am I missing something here?
the simplification not to write (k/m) which has dimension 1/second^2 causes the "problem" with units. I had same thought as you to add length and velocity which is not possible due to different units. But considered the (k/m) factor and the units are then correct as in corresponding terms of the taylor serie (1/second^2) (1/second^4) ... appears forcing each term to have dimension of length. i.g: - 1/(3!)*v0*t^3*(1/second^2) (1/second^2) stems from k/m and so on.
@@Oberbremser Thanks for your comment. I need to think about this. Old brain..
great lecture! is x_o cos(t) - v_o sin(t) soln dimensionally incorrect? (taylor series soln)
I was sooooo close to skinning that cat…phew!🧐😉😆
hello. Can someone please explain how x0 got divided it by 2 at 32:35? this makes no sense to me… method III after the fire alarm in particular is very confusing to me
Because c1=c2.
I'm having a bit of trouble using the second method for any k/m instead of just 1.
Would the Xo and Vo terms be multiplied by k/m as well?
I tried to find such solutions online but had no luck.
In the general solution:
x = x0*cos(w*t) + "v0*sin(w*t)"
v0 isn't really the coefficient on the sine term, unless w=1 (which would be the case if k/m were 1, since w=sqrt(k/m)). I put this term in quotes, because it really needs to be something else. We want dx/dt to equal v at t=0.
Let A and B be defined such that the general solution is:
x = A*cos(w*t) + B*sin(w*t)
Take the derivative:
dx/dt = -A*w*sin(w*t) + B*w*cos(w*t)
Evaluate at t = 0:
x(0) = A
dx/dt at t=0 = B*w
So as you can see, v0 = B*w. Thus, when we solve for the coefficient B, this really is equal to v0/w. Thus, the general solution is:
x = x0*cos(w*t) + v0/w * sin(w*t)
Hello @@carultch, thank you for the reply.
I was asking about the second method followed in the video, that is, the Taylor series solution.
Hi, did you find what you were looking for somewhere/figured it out somehow? I'm attempting the same rn.
I jumped at the fire alarm at ~27:30
I was looking for this comment!
how do we visualize second order DE equations like we understand by seeing slope field in 1st order DE ?
So how do you get the notes you're writing to appear correctly? That is, writing right to left when I'm pretty sure you would have to be writing backwards to achieve that since you're facing the viewer. Enjoying the talks, btw!
He probably mirrors the video, so it appears normal. If you saw him in person, your side of his writing would appear backwards.
Are you actually writing left to right from your perspective?
A mathematical magic carpet ride !
thank you so much sir
Please tell me how you use this display board technology?
they write normally on a transparent board, and then flip the video, so we are to read it
And there are also 2 other ways to establish the differential equation of the oscillator.
1. From the conservation of the energy : d/dt(1/2.m.V**2 + 1/2.K.X**2) = 0 then mV*d/dt(V)+KX*d/dt(X)=0 and then ...
2. From the momentum : d/dt(mV)=kX then ...
Many ways to establish the equation, many ways to solve the equation.
Why are there so many ways to deal with the oscillator ?
Why are there so many points of view to deal with the oscillator ?
I really don't know what it means ...
Why? I think Jesse Pinkman summed it up best, "Yeah, science bitch!"
Don’t forget the Lagrangian formulation.
@@navsquid32 L=1/2.m.V**2 - 1/2.k.X**2
dL/dV= mV
dL/dX=-kX
So from eq of Lagrage : d/dt(dL/dV) = dL/dX and ... You're right !
MUITO OBRIGADO
Why F = - kx?
The Taylor series is too tedious for most humans. But if it's general, isn't it something that can be handled by computers? Computer solvers? (And I mean "symbolic" solvers, not "data fitting".) Or is the "suspend variables" method actually used?
bro the fire alarm XD
I thought I had a dead pixel on my monitor, lol.
I got ads after the fire alarm
the harmonic approxillator hehe
anyone got the the homework?