Laplace Transform: First Order Equation 

@123 321 Rarely stated by teachers

@123 321 not really. It’s basic knowledge that you sound like you belong on the subreddit /Iamverysmart though :)

That's true in ODEs, but not in PDEs.

It depends on your application, in electrical engineering we use it for signal analysis and also, it can be seen as an extended Fourier series . However this is only true when the double sided Laplace transform is taken into consideration, where s is a complex not real variable. The Laplace transform has many other applications aside for solving DEs. Another example is the property of its convolution equalling to multiplication in the S domain.

@@astroandriodrox2356 In Electrical Engineering the Laplace Transform is used for analysis of systems, while Fourier Transform is used for signal analysis.

Nice picture if you look closely in mine you can see Nietzsche in there

Love you professor!

tienes redes sociales? soy de Perú tamb. tengo los mismos inetereses!

I wonder why, though. Dr. Strang didn't go into the intuition behind Laplace transform at all, which is what you need to have to even understand it in the first place.

@@AnindyaMahajan if you haven't already see "where the laplace transform comes from" parts 1 and 2 by Arthur Mattuck, MIT

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-an5E940fqZQ.html Try this one. Herbert Gross is one of a kind as a distance lecturer.

Ass Möde Thank you!!,

@@Zack-xz1ph Thanks man

I love this why am I just seeing this…

which teacher in India? I'm curious. IITB?

@@ashishkumarsharma1323 His face is flashing across my eyes.. but my memory is failing me terribly. It is a shame, of course, not to remember the names of your great teachers.,.. find me guilty of that ..

Carlos Ivan Saa you learn such stuff in 4th grade?..

Very late but Tibees did a video on this!!

Carlos, take a look at Newton-Raphson iteration. It lets you approximate square and cubic roots - and much, much more!

Because they suck at teaching?

أتفق.

If s

I understand it as an extension to Fourier transform that adds decay over time to insure more functions' transforms converge

I heard they are 2 of major founders of MATLAB

Who IS it. This french please merci d'avance cordialement FrèdéØ ČrèdéÔ CöœL bisous

Omar Al-Ani the limit of n-> infinity of e^(a-s)n is zero when the exponent is negative. a - s < 0 => a < s.

We are putting the limit t=infinity. e^infinity=0.

I think its because e^(-st) is decaying, so it doesn’t grow at all?

In general when doing Laplace transform Re(s) (real part of s) must be large enough to ensure f(t)e^{-st} is decaying to zero as t grows

ocw.mit.edu/resources/res-18-009-learn-differential-equations-up-close-with-gilbert-strang-and-cleve-moler-fall-2015/differential-equations-and-linear-algebra/

Welcome to university

@Tzabek So in other words we define s > a to make it work. It reminds me of the chicanery they indulge in in some aeronautical engineering texts I've read, where they integrate something and arbitrarily define the constant of integration as zero. Clearly it's valid, because the airplanes they designed using the math were historically known to work, but it always seemed a bit dishonest to me.

@@Ensign_Cthulhu It isnt that you define it to work. The transform takes a function of 't' and outputs a function of 's'. The domain of the new function of 's' is all 's' for which the integral exists. This is the case for all transforms. Its kind of like when you take derivatives. The formulas are only valid for x values that the derivative actually exists. You just say derivative of ln(x) is 1/x. But the formula is only valid for x's that you actually have a derivative. In this case, x>0. Even though it is perfectly reasonable to plug -1 into 1/x after the fact, it is nonsensical in terms of the derivative.

get out😂

that's what happened when they are trying to record a video and limit to only 1 (2) board(s)

In order to be able to evaluate the integral, it must converge, implying that the coefficient in front of the variable t must be negative, i.e. s > a.