A small mistake at 5:17 , the constants should be c1, c2, and c3. Thanks so much for all of your playlists! They're really helpful for my self-studying!
Thanks for doing these videos. I am a self-taught roboticist. You have become my go-to channel for math. Whenever I come across something in my reading I don't understand, I go to your playlists and look for relevant topics. It is pretty common for me to watch a video a couple of times over a few days before I become comfortable relating the material to my work. If I am still struggling, I look for an OCW lecture. With the good foundation your material provides, even the stuff from MIT is reasonably understandable.
@@DrTrefor Quick follow-up to my earlier comment. I like your use of graphics. I occasionally print screenshots. Then I use the screenshot as mnemonic devices to recall an aspect of a particular topic. I put the screenshots in a pile near where I eat breakfast. Every morning as I eat breakfast I go through the pile and try to answer two questions. Why did I take this particular screenshot and why is it important to what I am trying to do?
Viewing these vids after some exposure to ODEs has been great. It's really nice to already have some context as it makes the inform stick a bit better the second (or third) time round.
If you were an instructor of my diff class, even if the class was at 8.am all of your coffes'd be from me. Thank you so much man i really appreciate your work and efforts.
Very Nice and crisp explanation of how the basic idea of solving homogenous homogeous ODEs (with constant coeff) of orders>2 is SIMPLY an extension of finding the solution of ODE's of 2nd order. I recall in college we learnt about the 'D' (differential) operator in the similar to the "r's" in "characteristic" equation. I would request you to make a separate video on this 'D' operator (attributed to the genius American engineer/mathematician Oliver Heaviside), and its inverse '1/D' ( which is the integral operator) and use the Laplace Trasform and makes solving homogenous ODE's of any order look so simple- almost like school algebra and is a standard tool used to solve complex circuits in electrical engineering. Looking forward to more gems from you Prof Bazett
Professor Bazett, thank you for an excellent video/lecture on Higher Order Constant Coefficient Differential Equations. DR. Bazett, when you factored r^3 +r equal to zero, you have r(r+i)(r -1)equal to zero) instead of r(r + i)(r - i) equal to zero. There is also two c2 in your final solution. Please correct these small errors in the video.
All sorts of things in fluid dynamics in particular. Stopping at acceleration is fairly arbitrary, we can also ask questions like how does acceleration change in time?
at the end, when finding the c4 term, instead of multiplying the c3 term by t, could you multiply the c1 term by t to get c4*t as your final term instead?
Good question, but the answer is no. Multiplying by unnecessary factors of t, will produce "solutions" that don't work. As an example, consider: y" + 6*y' + 8*y = e^(-2*t) The correct solution is: y(t) = A*e^(-4 t) + B*e^(-2*t) + 1/2*t*e^(-2*t) Since our given RHS matches one of the homogeneous solutions, we have to multiply by t to create a linearly independent term to account for the overlap. What if we tried C*t^2*e^(-2*t) as one of the solutions? d/dt C*t^2*e^(-2*t) = C*(2*t - 2*t^2)*e^(-2*t) d^2/dt^2 C*t^2*e^(-2*t) = 2*C*(2*t^2 - 4*t + 1)*e^(-2*t) See if this works, as a solution to this diffEQ: 2*C*(2*t^2 - 4*t + 1)*e^(-2*t) + 6*C*[(2*t - 2*t^2)*e^(-2*t)] + 8*C*t^2*e^(-2*t) =?= e^(-2*t) Cancel the common factor: 2*C*(2*t^2 - 4*t + 1) + 6*C*[(2*t - 2*t^2)] + 8*C*t^2 =?= 1 Expand, and combine like terms: 4*C*t^2 - 8*C*t + 2 - 12*C*t^2 + 12*C*t + 8*C*t^2 =?= 1 2*C + 4*C*t =?= 1 We have a linear expression of t on the left, and just a constant on the right. No matter what we make the unknown C equal to, we can't make a multiple of t^2*e^(-2*t) be a solution to the original DiffEQ. Doing this by assuming a solution of C*t*e^(-2*t), we can find a value of C that works. Combining it's derivatives per the LHS, we get: 2*C*e^(-2*t), which is proportional to the given RHS. We have no issue letting C=1/2, to find the particular solution, by having guessed C*t*e^(-2*t).
