Hi, everyone. Welcome back to integralcalc.com. Today, we’re going to be talking about inverse
Laplace transforms and in this particular problem, we’ve been given the function f
of s equals s plus 3 divided by s squared plus 4s plus 13. This function here has already
been put through the Laplace transform process. What we’ve been asked to do in this problem
is reverse that process and find the original function that we must have put through the
Laplace transform process to get here to f of s.
So the way we’re going to do that is using these two formulas here for e to the at cosine
of bt and e to the at sine of bt. I’ve got these formulas from an inverse Laplace transform
table. You’ll use them fairly commonly in inverse Laplace transforms just because you’ll
often times have this e to the at and sine of cosine will commonly occur with all these
kinds of problems. You may not know right off the bat that these are the formulas you’ll
need to use. But what you should recognize is that in the numerator here, you have s
plus 3 which you can also write as s minus -3 which is kind of similar for what we’ve
got here, s minus a. a is a constant. b is a constant in both of these formulas. And
then in the denominator, you have s squared plus 4s plus 13 which is a polynomial. And
if you see over here on this formula, we’ve got two squareds. We’ve got the quantity
s minus a squared and we have b squared. We know that with this denominator over here,
if we use the method of completing the square, we might be able to pull out a squared term
from it. So this formula here for e to the at times cosine of bt could be promising.
But we may not know that originally. So we’re just going to try to complete the
square on the numerator and see how this goes. So remember that with completing the square,
when you’ve got a polynomial like this, s squared plus 4s plus 13, you’re going
to be looking at this coefficient here on the s to the first power term. Remember that
you have to get the sign. If that was minus 4s, you would need to get the negative sign
too. But you’ve got this positive 4, you’re going to take that number and you’re going
to divide it by 2. And remember this is the method for completing the square. So you divide
it by 2 and of course you get 2. Whatever you get for your answer, you’re going to
square it and of course 2 squared gives you 4. So 4 is what you’re going to add to this
polynomial in order to complete the square. So what we’re going to have now is f of
s equals s plus 3 and then in the denominator, we’ll pair this together so the 4 that we’re
going to add is going to complete the square with s squared and 4s. So we’ll have s squared
plus 4s plus 4 and then we’ve got plus 13 outside. Because we added 4 to the denominator,
we also need to subtract 4 because if we just add 4, that fundamentally changes the function.
So now that we have plus 4 and minus 4, it’s like we haven’t even changed the function
at all. We’re just manipulating it in a way that will make it more convenient to work
with. So now, we’ve got s plus 3 in the numerator and then in the denominator, we
can call this s plus 2 squared. If we factor that s squared plus 4s plus 4, we would get
s plus 2 times s plus 2 which gives us s plus 2 squared. Then here on the right-hand side,
this plus 13 minus 4 gives us a plus 9. So we’re starting to see this look more
and more like our formula here for e to the at times cosine of bt. You can see though
this s minus a squared, we need that to be negative sign. So instead of s plus 2, we’re
going to make it s minus negative 2 squared which also doesn’t change fundamentally
the function because we have a minus a negative 2 which is the same as s plus 2. But at least
we’ve got that negative 2 now. And then notice here we need a b squared. Well we have
plus 9 so we know that we can change that into 3 squared right? So now our denominator
looks very similar to the denominator of our formula and we just need to change the numerator
because this s minus a here needs to match this s minus a in the denominator. So in order
to do that, we need that to be s minus -2 in the numerator. The way we’re going to
do it is we’re going to say s minus a -2. What we essentially have done here is we’ve
added 2 because this is the same as s plus 2. So since we’ve added 2, we’re also
going to subtract 2 and of course we have to leave in that plus 3. So when we do that,
we’ll have this s minus negative 2 and now we got here minus 2 and plus 3 so that’s
going to turn into plus 1. So we’ll get f of s equals s minus -2 and then we get plus
1 all over s minus negative 2 squared plus 3 squared.
So now our function looks exactly the same as our formula here for e to the at times
cosine of bt. The only thing that we need to deal with here is that what I’ve boxed
here in orange matches our formula exactly except for this plus 1. We need to get that
out of there because that is not included here in the formula. We don’t have like
another plus b or plus c or some other constant to account for that plus 1. So we got to get
it out. So what we end up doing is splitting our fraction. So we’ll end up with s minus
a -2 all over s minus -2 squared plus 3 squared. And that plus 1, we’re going to break out
into a separate fraction so we’ll have plus 1 over the same denominator.
So now this is where our second formula comes into play, this e to the at times sine of
bt. Notice that we have the same s minus a squared which we’ve already taken care of.
We have the b squared which we’ve already taken care of. The denominator is the same.
The only thing that’s different is that we have this b term here in our numerator
and obviously this b variable here has to match this b variable. They’re the same.
So we know that in our denominator here, we’ve got b equal to 3 which means we need to get
our numerator equal to 3 and the way we’re going to do that is by multiplying this by
3/3. So when we do that, we’ll leave the first fraction the same, s minus -2 divided
by s minus -2 squared plus 3 squared. But the second term here, we want the 3 here to
stay in the fraction and this 1/3 here, we’re going to move that out in front. So we’ll
do plus 1/3 times 3. This 3 right here is what’s staying in the numerator. So 3 divided
by s to the -2 squared plus 3 squared. So now, this part here is in the exact same form
as our formula here. And our first fraction is in the exact same form as this formula
for e to the at times cosine of bt. So now that we have these things in the same
format, we can do the inverse Laplace transform to find the function f of t. We just need
to identify here a and b. So we can see from our first fraction that a is equal to -2.
And we know that b is equal to 3. In our second fraction, don’t forget about the 1/3 here,
we’ll get to that in a second; but we have same thing a equals negative 2 which we get
from this part right here and we have b equals 3. So now we can go ahead and use our formulas.
So for our teal fraction here, we use the formula e to the at times cosine of bt. So
our a is -2 so we’ll get -2t times cosine. We know that b is equal to 3 so times cosine
of 3t. Don’t forget about the 1/3 here. We’re going to make sure we have that so
we’ll do 1/3 times, we’ll use our green formula, e to the at times sine of bt. Again
a is -2 so we’ll get e to the -2t times sine of bt. We know b is equal to 3 so we
get sine of 3t. And that’s it. That’s our final answer.
This is the answer we get when we do the inverse Laplace transform on our original function,
F of s. This is f of t, our original function, that if we now performed a Laplace transform
on f of t, we will wind back up at F of s. So that’s it. I hope this video helped you
guys and I will see you in the next one. Bye!