Hi Chris, I'd like to thank you so much for the time and effort you put into making these videos available to the general public. Unfortunately I can't attend all my math lectures at uni because of other commitments, but with the help of your videos I went from getting pass in maths, to distinction last semester ! Hopefully with the help of your videos, I can maybe get a D (or even HD) :P

you, sir, are my idol....whenever i solve any math question, i pretend i am you and i am explaining it to everyone....lol....man u are a legend....

Pure genius - mathematics at its finest and purest. Thanks

@DrChrisTisdell Having just attempted to compute a convolution the "straight" way (t's and taus) and via Laplace transforms with partial fraction decomposition (and reassuringly obtained the same answer both ways), I think I'll take the Laplace transform with partial fractions any day!!

@DrChrisTisdell thank you verry much.Dr.Chris

@jsm666 In this course we do not discuss the convolution theorem, which is a pity because it can simplify a lot of calculations. :-)

@sabokunogaraa I wouldn't recommend that "trick" for absolutely all problems, but it can work for simple cases.

Great comment! I am sure that students like you will do an even better job of explaining it to others!

Thanks! Glad you are enjoying the videos. Don't forget about my free ebook to accompany this video - the link is on my channel page. Good luck with your mathematics.

Well done!! Hope the ebook is having a positive influence also.

Your ebook and youtube videos have been extremely helpful, thanks so much and keep up the good work. With thanks.... from a dumbass who may be finally seeing the light. Take care and have a good day!

Great suggestion, Ismail. The convolution theorem is definitely on my "to do" list. :-)

hi chris, can you please do a video on convolution theorem. thanks in advance

At 5:46, if you rewrite (e^-s)/[s*(s^2+1)] as [(e^-s)/s]*[1/(s^2+1)], you get the product of the laplace transform of u(t-1) and the laplace transform of cos t. The product of Laplace transforms of two functions equals the convolution of those functions, doesn't it? Or is there something I'm missing?

can we do that trick with every problem or only for limited ones

Hi - I have ignored the e^(-s) in the numerator when I did the calculations around 07:24. Later in the video I put the e^(-s) back where it should be. The reason is because I felt it simplified the algebra.

thanks for the video DrChrisTisdell. I find the laplace logic pretty interesting. However, I am amazed that we're still learning such stuff, developed over 200 years ago. Millions of mathematicians ,scientist and engineers have graduated since Laplace. So no one has developed a replacement for calculus in estimating "rate change". IMHO these techniques are primitive.

Hi, whats the reason of transforming a ODE ? This one can be solved very easily without doing it

Unbelievable acing the course thanks to you, the lecture notes over complicate everything with fancy wording and poor steps - but this is pure ge