Today, we’re going to be talking about how to do a Laplace transform when we have access
to a table of Laplace transforms.
There are two ways that you’ll perform a Laplace transform.
The first is using the definition of a Laplace transform which is where you take the integral
from zero to infinity of your function multiplied by e to the negative st.
That’s using the definition.
Often times though, you’ll have access to a table of Laplace transforms and the table
will show you common functions and their corresponding Laplace transforms.
So for example, in this particular problem, we’ve been given this function g of t and
have identified three functions in our Laplace transform table that are formulas that will
help us out here.
Keep in mind that if we’re taking the Laplace transform of g of t, we can take the Laplace
transform of each of these three terms separately and then put them together.
So in this case, we have e to the 3t and we’re going to be using the Laplace transform equation
here for a function e to the at where a is a constant.
In our case, a is equal to 3.
Then we have a second term in our original function, cos 6t.
And when we look at our Laplace transform table, and any Laplace transform table should
have this so if you’re using a table, you should see a Laplace transform equation or
formula for cos at where a is a constant, in our case it’s equal to 6.
Then we’ve got our last term here in our original function, e to the 3t cos 6t.
And any Laplace transform table should have this e to the at cos pt, where a and p are
constants.
So if you’re using a table, this becomes really easy.
If you want to learn how to use the definition, please look at my video about how to perform
Laplace transforms with the definition of a Laplace transform.
But in this case, it’s really just plugging in our constants a and b.
So the Laplace transform here of g of t, we’re going to call it F of s which is our notation
for our Laplace transform.
Remember that when we’re doing this transformation, we’re moving from an equation or a function
that’s defined in terms of t or whatever our variable is to a different equation that’s
in terms of s.
So to perform the Laplace transform here, we’re going to look at our first term, e
to the 3t.
We can see that a is equal to 3 so our Laplace transform according to the formula is going
to be 1/s-3.
Keep in mind that the s is always going to stay.
You’re only going to be plugging in for a and b.
So that’s our first term.
Then we have this plus here so we go ahead and add to that cosine of 6t.
Using our formula here we can see that a is equal to 6 so our formula is going to give
us s over s squared plus 6 squared.
So we’ll go ahead and put that in there.
And then, we have a minus sign so we go ahead and subtract.
We can see here from our third term e to the 3t cos 6t and our formula that a is going
to be equal to 3 and b is going to be equal to 6 in our original function.
So if we plug that into our formula, we’ll have s minus 3 divided by the quantity s minus
3 squared plus 6 squared.
And that’s it.
We just plug in our constants and now all we need to do is simplify.
You don’t have to and you probably shouldn’t combine these fractions into one so we’ll
go ahead and leave them separately.
We’ll say 1 over s minus 3 plus s over s squared plus 36 minus s minus 3 divided by
the quantity s minus 3 squared plus 36.
So we’ll just multiply those out.
The 6 squared we’ll evaluate that, but other than that, we’ll leave the functions as
it is.
So this function here F of s is our Laplace transform.
It’s our final answer.
I hope that video helped you, guys and I will see you in the next one.