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.

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.