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### Course: Differential equations>Unit 3

Lesson 1: Laplace transform

# Laplace transform 1

Introduction to the Laplace Transform. Created by Sal Khan.

## Want to join the conversation?

• 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
• 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.
• Would you do a video about the proof of the Laplace Transform definition ?
• Huh? How can you prove a definition?
• at "" Sal mentions anti derivitive. what is that?
• It's another name for the integral (without any bounds).
• 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.
• why does this transform actually work in helping to solve differential equations?
• It's a property of Laplace transform that solves differential equations without using integration,called"Laplace transform of derivatives".
Laplace transform of derivatives: {f'(t)}= S* L{f(t)}-f(0).
This property converts derivatives into just function of f(S),that can be seen from eq. above.
Next inverse laplace transform converts again function F(S) into f(t).
If my ans. looks confusing .Just observe am example of solving D.E. using laplace,i hope droughts will disappear.
• and what is actually the difference between laplace transform and laplace operator ?
• The Laplace operator involves partial derivatives.
The Laplace transform involves an improper integral to transform to a function of a different variable.
• If I am asked to use the Laplace transform on f(x)=x, would that be the same as using f(t)=t?
• A laplace transform on a polynomial with degree n is n!/s^(n+1).

Given that you have f(x) = x, your Laplace transform should be:
L(s) = 1/s^2
(1 vote)
• Hi,
May you please do the Derivation of the Fourier Transform Pairs?
I tried to search online, but I can NEVER find any.