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## Differential equations

### Course: Differential equations>Unit 3

Lesson 1: Laplace transform

# Laplace transform 2

Laplace transform of e^at. Created by Sal Khan.

## Want to join the conversation?

• How do you compare complex and real number? I mean he writes s>a...
• For those who are wondering this happens at .

Well firstly, 'a' can also be complex, and secondly, in maths you cannot compare complex numbers, either with real numbers or with other complex numbers.

So what he did is a little more in depth that what he wrote down, and I'm sure that most people don't realise this, as he hasn't defined 's' as a complex number yet.

Let's say that 's = a + bi', and 'a = c + di', then that part on the inside of the integral becomes 'exp((a-s)t) = exp((c + di - a - bi)t)'. Grouping the real and imaginary terms we get 'exp((c-a)t+i(d-b)t)'. We can then split the exponent as such: 'exp((c-a)t)exp(i(d-b)t)'.

Looking at this, as t tends to infinity, we know that the complex exponential will just go round in circles (from Euler's identity, 'exp(i*theta)'). But if we look at the real exponential, we can see that it will either go to infinity or to zero, depending on the value of '(c-a)'. So something that goes around in circles times zero will be zero, and something that goes around in circles times something that tends to infinity will also tend to infinity.

So to make a slight correction (which is only necessary if you want to be strictly mathematical), by writing 's>a', he actually means 'Re(s)>Re(a)', which is entirely possible to compare.

I hope this explanation is clear.
• At , why is e^(-st) * e^(at) combine to e^(a-s)t instead of e^(a-s) ?? Wouldn't the -t and t combine to cancel??
• For multiplying terms which have the same base, (x^a) * (x^b) = x^(a + b). So,
e^(-st) * e^(at) = e^(-st + at). Take out the common factor, "t" from (-st + at) =>
t(a-s), which gives us e^(a-s)t
• What about the case when a = s? We get 0*t, when t tends to infinity, do we consider it an indeterminate form in this case?
• in doing these transformations it is assumed s>a by definition otherwise the function diverges.
(1 vote)
• What is the difference between ordinary differential equations and partial differential equations?
• Ordinary differential equations involve quantities that vary with respect to a single variable.
Partial differential equations involve quantities that vary with respect to more than one variable. For example the quantity of interest could be thermal energy across a metal surface and the variables are time and space.
Note: When I say quantity I am reffering to the output of a function
Thumbs Up!
• How would you go about the Laplace transform of e^(t+2) ?
(1 vote)
• Notice that e^(t+2) could be written as (e^2)*e^t, since e^2 is a constant, you can use the linearity of the Laplace transform, and find (e^2)*L(e^t).
This would equal
(e^2)*1/(s-1).
• Is 's' here something like the 'plus C' in indefinite integrals? I mean it can be nearly any constant, isn't is?
• Not exactly, it was my understanding that the 's' that appears in laplace transforms is actually another variable in a different dimension or with different units of measure (i.e. t -> time domain, while s -> frequency domain).
• What if s=a? would this mean the limit/Laplace Transform DNE because you would have -1/0?
• yes, correct. The Transform still does not work.
(1 vote)
• why does he change the expression to be a limit of a as before?
• It is often a formality to explicitly show why the next steps happen. It is like a justification. And sometimes, with complex systems, it is necessary to show a limit, else the function may behave unexpectedly! :O However, often times we can see what the limit does without putting it down and we don't bother writing it at all! :)
(1 vote)
• At I understand what he is doing, but why does he evaluate to infinity? In the first video, he solves for the limit at a constant. Why is this no longer required?