Laplace Transforms and Integration by Parts with Three Functions




Uploaded by TheIntegralCALC on 12.03.2012

Transcript:
Today, we’re going to talk about how to use the definition of the Laplace transform
to find the Laplace transform of t cosh 3t. This is also going to be a great example of
how to use integration by parts when you have three functions inside your integral. Before
we get started let me say that I would not recommend watching this example if you’re
just wanting to learn how to use the definitions of Laplace transforms. I would recommend watching
this video if you’re interested in using the definition to find the Laplace transform
of an integral that ends up having three functions inside of it or if you’re just interested
in integration by parts with three functions. So remember that what we’re trying to do
here is that we’ve been given the function t cosh 3t and we’ve been asked to find the
Laplace transform using the definition. Remember that the definition asks us to take the integral
from 0 to infinity of e^-st times our original function. So we’ve got our original function
which we’ll call f(t) as 3 cosh 3t and we have to plug that into our integral. The problem
we have and the problem a lot of people run into is that once you do that, you have three
functions. You have e^-st as your first function. You have t as a second function. And now you
have cosh 3t as a third function and you’re not quite sure how to take the integral. As
it turns out, you’re going to use integration by parts but it’ll get a little complicated
and a little bit involved. Keep in mind that we’re going to use integration
by parts many times in order to get to our final Laplace transform. If you’re just
using integration by parts in general to find the integral of three functions, basically
in a nutshell what you’re going to do is put two of the functions together. For example
for our first step here, we’ll call u = t and dv is going to be equal to the other two
functions, e^-st (cosh 3t). And what you’ll do is kind of group things together like this
and then you’ll sort it out later as you do multiply integration by parts steps. Like
I said, for our first step, we’ll set u = t and then we’ll find du to be dt, the
derivative. dv we set equal to the other two functions. And now in order to find v, we
ran into something a little bit difficult because we have to take the integral of e^-st
cosh 3t in order to find v. So that’s going to take a little bit of time but I want to
show you guys how to do that. If I want to find the integral and we’re
looking at the integral here of this dv inorder to find v, we’re going to find the integral
of e^-st cosh 3t dt. In order to do that, we need to use integration by parts again
because we have two functions e^-st and cosh 3t. We’ll set u = cosh 3t and in these kinds
of problems, we’ll always set u equal to the trigonometric identity here. and then
we’ll set dv = e^-st dt. So to find the integral of dv to get v, that’s not too
bad. Remember that s is a constant. So we’ll just bring that –s out here into the denominator
of our coefficient so -1/s e^-st will be v. Define du, we’ll just take the derivative
of u and that will be hyperbolic sine of 3t but then we’ll have to multiply the 3 out
in front. So we have 3 times hyperbolic sine of 3t dt. Now that we have that, we can go
ahead and plug those in to our integral. Remember the integration by parts formula
is u dv = uv minus the integral of v du. So we have -1/s e^-st cosh 3t minus the integral
of v du. Because we have v = -1/s e^-st, we’ll take this negative sign and combine it with
this negative sign and we’ll turn this into a positive. We’ll also bring that 1/s out
in front of our integral because it’s a constant as is a constant so we can pull that
out in front. So plus 1/s times the integral of e^-st times hyperbolic sine of 3t. We’ll
bring the 3 out in front here so now we’ve got 3/s and then hyperbolic sine of 3t dt.
So we’re in good shape except that our integral over here needs integration by parts again.
Inorder to do that, we’ll set u again equal to our trigonometric identity so hyperbolic
sine of 3t. We’ll take the derivative to get du and we’ll get 3 cosh 3t and then
dv is going to be equal to the other part of our integral so e^-st dt. And then the
integral v, we’ll get -1/s e^-st. Now that we’ve got that parts, we’ll take this
whole equation and we’ll say that the integral of e^-st cosh 3t dt = -1/s e^-st cosh 3t +
3/s. Now, we’re going to take this information here and plug it in for our integral so we’ve got -1/s e^-st
times hyperbolic sine of 3t minus the integral of v du. So we’ll have plus 3/s in front
again. That will come out in front and we’ll be left with e^-st cosh 3t dt. Let’s do
some multiplication. Let’s distribute this 3/s right here and when we combine these two
over here, we’ll have -3/(s^2) so let’s go ahead and write that down, -3/(s^2). When
we distribute here, we’ll end up with 9/(s^2). Now, we’ll end up with -3/(s^2) here and
we’ll end up with plus 9/(s^2). What’s great about this now is that if you
can see, our original integral was equal to e^-st cosh 3t dt. And you can see that our
new integral over here, e^-st cosh 3t dt are equal to one another. So, we can subtract
this whole part here from the right-hand side and bring it over to the left. So what we’ll
have is the integral of e^-st cosh 3t dt – 9/(s^2) times the integral of e^-st cosh 3t dt = -1/s
e^-st cosh 3t – 3/(s^2) e^-st hyperbolic sine of 3t. Now, on the left-hand side, since
these two are equal to one another, we can factor them out and what we have is 1 – 9/(s^2).
Instead of writing the integral, let me just say integral dot, dot, dot. So we have that
equal to the right-hand side. What we can do is in order to solve for the integral,
we’ll just end up dividing both sides by this 1 – 9/s^2 to bring it over to the right.
So what we’ll have is the integral of e^-st cosh 3t dt is equal to, we can take this whole
thing and divide by 1 – 9/s^2. Let’s do some work on that. Let’s combine the two
terms in our denominator so let’s call this, instead of 1, we’ll call it (s^2)/(s^2).
That’ll just give us s^2 – 9/s^2 when we combine those two fractions so we’ll
get s^2 – 9/s^2 and now we can multiply by the inverse. We’ll get – 1/s e^-st
cosh 3t and we’ll multiply here by [s^2/(s^2 – 9)] – 3/s^2 e^-st
hyperbolic sine of 3t [s^2/(s^2 – 9)]. Now, we can start canceling a little bit. You can
see that we’ve got s^2 in our numerator and s in the denominator, so this s will go
away and the square will go away in the numerator. In our second term, we’ve got s^2 in the
denominator and s^2 in the numerator so those will both go away. This will just turn into
a negative sign here and now we can really just combine. Let’s go ahead and make this
–s/(s^2 – 9), we’ll combine those two fractions and then we’ll have minus 3/s^2
– 9 and then this part will go away because we combine those fractions. That’s pretty
much as simple as we’re going to get it. We know now that the integral here was basically
what we had to do in order to find v up here. So v = (-s e^-st cosh 3t – 3e^-st hyperbolic
sine of 3t)/(s^2 - 9). Now we can go back to our original integration by
parts. Now that we have u and du, v and dv, we can perform our original integration by
parts and we’ll be much closer to solving our original integral. We’re going to say
that the integral from zero to infinity of e^-st t cosh 3t dt is equal to uv so we have
u and v. Really, we’re just going to multiply this huge fraction we just found by t so we’ll
get –s and let’s go ahead and put that t in there, e^-st cosh 3t – 3t e^-st hyperbolic
sine of 3t all over s^2 – 9. Because with the definition, we’re evaluating from the
range 0 to infinity, we can evaluate this on the range zero to infinity. We’re only
halfway through with our integration by parts formula so we have to subtract and still do
the vdu here. So minus the integral of v du. v is going to be equal to the huge fraction
again that we just found and du is just dt. What we’ll do is see when we have the two
negative in front of these two terms. A negative here and a negative here and we have a negative
in front of our integral. We’ll just go ahead and make that a positive and then the
negatives in our fraction will go away. So we have (se^-st cosh 3t + 3e^-st hyperbolic
sine of 3t)/(s^2 - 9) and then we’ve got dt in here. And we’re going to be evaluating
this on the range zero to infinity. Okay. Now that we have that integral, let’s
go ahead and evaluate this first fraction on the range zero to infinity. Remember that
we’re going to be plugging in for t. s is a constant. It’s not a variable so we’ll
be leaving s in there, plugging in for t. Essentially, you can plug in infinity fort
first. When we plug in, notice that we’ll get e^-infinity. e^-infinity = 0. So that
whole first term in the numerator is going to go to zero then again here when you plug
in infinity for t, that e^-infinity will go to zero so you’ll have zero again and then
s^2 – 9. And then we subtract and plug in our lower limit of integration which is zero.
you can see because we’re plugging in zero here for t, that the first term again will
go to zero. When we plug in again for t in our second term, we’ll get 0/(s^2 - 9).
So you can really see that that whole thing is going to cancel out so this really just
goes away. And then we’re just left with the integral.
And the integral we can split into two pieces. We’ll have the integral from zero to infinity.
We’re going to split it here at the plus sign. We’ll have (se^-st cosh 3t)/(s^2 - 9)
dt and then we’ll say plus and we’ll just separate our integral like this. We’ve got
(3e^-st hyperbolic sine of 3t)/(s^2 - 9) dt. Now, this looks more complicated than it actually
is. Remember that s is a constant. What we can do is bring some things out in front here.
First of all, s^2 – 9 is going to be a constant because if s is a constant, imagine of it’s
2, you’ve got 2^2 = 4, 4 – 9 = -5. That whole denominator turns into a constant. So
we can pull that out in front as well as the s in the numerator. We have s/(s^2 - 9) times
the integral of zero to infinity of e^-st cosh 3t dt and then here again in the second
integral, we can pull out 3/(s^2 – 9) so times the integral of zero to infinity of
e^-st hyperbolic sine of 3t. the good news is that we already know that the integral
of e^-st cosh 3t dt is the exact same thing as dv and we already know that the integral
is v. So we can go ahead and plug that in. We’ll have s/(s^2 - 9) [(-se^st cosh 3t
– 3e^-st hyperbolic sine of 3t)/(s^2 - 9). And we’re going to be evaluating that on
the range zero to infinity. Then for our other integral, we could do the same thing that
we did when we found the integral of dv to get v. It’s the exact same thing we did
before except instead of setting u equal to cosh 3t, we’ll set it equal to hyperbolic
sine. So we’d find that du would be equal to 3 cosh 3t and you’d say that dv was equal
to e^-st dt and find that v = -1/s (e^-st). You would plug that in and you would do integration
by parts and you would find that fraction. What you’d find that the integral is equal
to the fraction (–e^-st s hyperbolic sine of 3t – 3e^-st cosh 3t)/(s^2 - 9) and of
course we’ll be evaluating that on the range zero to infinity as well.
We’re done integrating. We don’t have to do any more integration and we’ll just
be evaluating our definite integrals. Now this will be fairly quick. Remember that when
we’re evaluating our definite integrals, we’re plugging in for t. So let’s take
our first one here. We’ll plug in for infinity first. When we plug in for infinity here for
this t, we’ll get e^-infinity which is going to go to zero so that whole first term will
be zero. The second one is the same thing, we’re going to plug in here and we’ll
get e^-infinity which is going to give us zero and then when we plug in zero, we’ll
get (–se^0 cosh 0 – 3e^0 hyperbolic sine 0)/(s^2 - 9). And then over here, we’ve
now evaluated our definite integrals so we just have to simplify that. For our second
fraction, we’ll get 3/(s^2 - 9) times when we plug in infinity for t, we’ll get e^-infinity
for this term which is going to give us zero so that whole first term is zero minus 0 over
s^2- 9 minus whatever we get when we plug in zero. So we get –e^0 s hyperbolic sine
of 0 – 3e^0 cosh 0 all over s^2 – 9. So s/(s^2 - 9), this term is going to go away.
The first term is going to be 1. These two negative signs are going to turn into positive
signs so those will go away. So we’ll just have 1 here for e^0. Cosh 0 is also 1 so that’s
going to go away. We’ll have s + 1. Hyperbolic sine of 0 is 0 so we’re going to have zero
for that whole term. We’ll be left with s^2 – 9 + 3/(s^2 - 9). This term’s going
to go away. Hyperbolic sine of zero is going to be zero so that whole term will go away.
These two will turn into a positive so we’re just left with 3/(s^2 – 9) and we can cross
that out. So when we simplify (s^2)/[(s^2 - 9)^2] + 9/[(s^2 - 9)^2], since we have common
denominators we can combine our numerators and we get (s^2 + 9)/[(s^2 - 9)^2]. And that’s
going to be equal to F(s) which is going to denote our function in terms of s. And this
is our Laplace transform of the function t cosh 3t using the definition of the Laplace
transform. So that’s it. I hope that video helped you
guys and I will see you in the next one.