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Differential equations

Unit 3: Lesson 1

Laplace transform

Laplace transform 1

Introduction to the Laplace Transform. Created by Sal Khan.

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• At of the video Sal begins integration. He starts with -1/s times e to the -st but it gets hairy for me because what happened to adding 1 to the exponent??
• It involves integration by substitution, wherein:
Let -st=u
=> du = -s.dt
Thus int e^-st = int (-1/s) e^u du
= -1/s e^u
Substituting back
int e^-st = -1/s.e^-st
• My doubt is with the equation at , why we multiply the f(t) with e^-st why not with e^+st?
• You can also ask why do we do the integral from 0 to inf instead of 0 to -inf, or -inf to 0, or something completely different. There are, in fact, other different transformations (like the Fourier transformation), and each of these transformations is useful for something different.

Why is exactly the integral from 0 to inf of e^-st * f(t) so useful that it has a proper name and that we want to study is something that has not yet been explained (it's explained later, in this video http://www.khanacademy.org/video/laplace-transform-to-solve-an-equation?playlist=Differential+Equations). Hint: it's very useful to solve differential equations.
• This is a classic case of u-substitution.
∫ e^(-st) dt. u=-st. du/dt=-s. dt=-du/s
∫ e^(u) * -du/s = -1/s * ∫ e^u du = -1/s * e^u = -1/s * e^(-st) + c
With enough practice, you will be able to do u-sub problems like this in your head.
• Why does e show up so much I can't seem to get away from it its used everywhere in math.
• e is very common in nature. If you look up the definition of e it arrises with any form of natural growth (think of a graph of compound interest). That's financial growth or even the growth (increase) of randomness around a norm (i.e. Gausian/normal distribution). It's everywhere so you see it in lots of places.
• Would you do a video about the proof of the Laplace Transform definition ?
• Huh? How can you prove a definition?
• sir my question is that in laplace transform 1 ,we use definite integral can we not use indefinite integral?
• the definition of the laplace transform is: the integral from 0 to infiniti of (e^(-st))*f(t)dt
this is just a definition, the laplace transform is a specific operation you can perform on a function, and removing the limits would give you a different operation that may or may not be useful for solving differential equations
• at "" Sal mentions anti derivitive. what is that?
• It's another name for the integral (without any bounds).
• Why can you assume that s > 0 at ?
• S is representing a frequency as in a "per second" or 1/second which is why you get 1/s for the Laplace of 1. From a physics perspective there is no such thing as a negative frequency because in reality it is the same thing as the frequency, saying something happens -3 times a second does not make a lot of sense. You can also tell that s cannot equal zero from the 1/s.
• What if s < 0?

Then this integral would diverge.
• Yes it will diverge.

Remember that a laplace transform is essentially telling you how close the function is to e^(st). If the integral diverges that just means the function gets farther and farther away from e^(st) as s gets closer to 0 (in this case).

Really what it means is you need to pick an s much larger than 0.