Laplace Transforms and Differential Equations 

Sure, same feeling :)

Repeating is the mother of learning. And recognition is why you got this excitement.

I'm always exhausted from work, and my work isn't exactly related to his lectures, but it's so entertaining and interesting to learn from Steve! Great lecture!

such a nerd

Why are u sure it's a drop of olive oil? :P

No Professor Steve is left-handed, I guess he shoots the video reversed then flips it to normal in editing

I wonder about it too, how to make such a video?? I think it would be fun to use this kind of video in my class, anyone can explain? Or maybe give me the link to a youtube video about how to create this kind of video.

Sir, please recommend some PIV data repository for jets, plumes

Lots of undergrad physics curricula do cover this. But my experience is that it often really sinks in the second time students see it in grad school.

@@Eigensteve I believe I only first saw Laplace transforms when I took control theory at the University of Pennsylvania (which was a grad level class). Then it helped me in my math classes as an alternative way to solve differential equations. Never made any sense to me why the Laplace transforms weren't just taught earlier in math or physics curricula as opposed to learning them in a graduate level engineering class. They're super useful.

@@MrSandman213 For what it's worth, I learned Laplace transforms as a sophomore undergraduate in Physics.

They are not necessarily in the same direction, but in the special case that the mass is moving away from the equilibrium position, they are. If the position is x=2 and the velocity is x' = 1, then the restoring force is acting against the fact that it at a positive x-position, and the drag force acts against the fact that it has a positive x-velocity. The two forces are therefore both to the left, at this particular instant in time. So if x and x' both have the same sign, then the restoring force and drag force will have the same sign.

Here's how it would work with a forcing function. Given the free response equation of motion of: x" + 5*x' + 4*x = 0 x(0) = 2 x'(0) = -5 Suppose the force function (right hand side) were 10*cos(2*t), with the same initial conditions. x" + 5*x' + 4*x = 10*cos(2*t) Find the Laplace transform of each term: (s^2 + 5*s + 4)*X - 2*s + 5 - 5*2 = 10*s/(s^2 + 4) Shuffle initial conditions to the right: (s^2 + 5*s + 4)*X = 10*s/(s^2 + 4) + 2*s + 5 Rework the RHS, so we have one single fraction: 10*s/(s^2 + 4) + 2*s + 5 = (10*s + (2*s + 5)*(s^2 + 4))/(s^2 + 4) Expand numerator: (2*s^3 + 5*s^2 + 18*s + 20)/(s^2 + 4) Isolate X: X = (2*s^3 + 5*s^2 + 18*s + 20)/((s^2 + 4)*(s^2 + 5*s + 4)) Factor the quadratic: s^2 + 5*s + 4 = (s + 1)*(s + 4) Thus: X = (2*s^3 + 5*s^2 + 18*s + 20)/((s^2 + 4)*(s + 1)*(s + 4)) Set up partial fractions: X = A/(s + 1) + B/(s + 4) + (C*s + D)/(s^2 + 4) Heaviside coverup finds A and B: at s = -1, A = (2*(-1)^3 + 5*(-1)^2 + 18*-1 + 20)/(((-1)^2 + 4)*covered*(-1 + 4)) A = 1/3 at s = -4, B =(2*(-4)^3 + 5*(-4)^2 + 18*-4 + 20)/(((-4)^2 + 4)*(-4 + 1)*covered) B = 5/3 Construct result thus far: (2*s^3 + 5*s^2 + 18*s + 20)/((s^2 + 4)*(s + 1)*(s + 4)) = 1/3/(s + 1) + 5/3/(s + 4) + (C*s + D)/(s^2 + 4) Let s = 0, to solve for D: 20/((4)*(1)*(4)) = 1/3/(1) + 5/3/(4) + D/4 5/4 = 1/3 + 5/3/4 + D/4 5 = 4/3 + 5/3 + D D = 2 Let s =1 to solve for C: (2 + 5 + 18 + 20)/((1 + 4)*(1 + 1)*(1 + 4)) = 1/3/(1 + 1) + 5/3/(1 + 4) + (C + 2)/(1 + 4) 9/10 = 1/6 + 1/3 + (C + 2)/5 C = 0 Laplace result: X = 1/3/(s + 1) + 5/3/(s + 4) + 2/(s^2 + 4) Invert the transform to find solution for x(t): x(t) = 1/3*e^(-t) + 5/3*e^(-4*t) + sin(2*t)

You are right, but c1=c2=1 because of initial cond-ns

@@Troll-cl1fh thanks

He's not writing mirrored, it's just flipped

It's complicated. The s doesn't necessarily stand for anything in particular that starts with the letter s. You might think of it as standing for "state", I don't know if this is the actual term it stands for. What s represents, is complex frequency. The real part of a complex frequency is an exponential decay, while the imaginary part usually comes as a complex conjugate pair, that corresponds to oscillatory frequency of the sine and cosine terms. What a Laplace transform is, is a spectrum of sine waves, cosine waves, and exponential decays, that add up to the function in question.

Thanks

I am an experienced engineer and my daughter is now going to attend one of CA leading Engineering school.. i was looking for "good stuff " for reference for her and myself, stumbled on your channel and lectures... Watched one .. then two ... So i completed the F.transform serie ... And found myself feeling sorry for not having seen such good lectures when i was a student .... Then I laughed as these set of lectures really show that in person lecture can not be so immersive. This medium of glass projection , the fact that you dont need to turn your back to the class , while writing, sharing slides, computer sims, also means that you dont need to limit yourself to pre defined set if slides. ++ You are very clear and able to take a topic that usually perceived as technical tool and breath life into it . Awsome!!