Uploaded by TheIntegralCALC on 05.12.2011

Transcript:

Today we’re going to be talking about how to solve an initial value problem using Laplace

transforms. And in this particular problem, we’ve been given this second degree function,

2y” or the second derivative of y plus 3y’ or the first derivative of y minus 2y is equal

to t e to the negative 2t. We’ve also been given two initial conditions, y of zero equals

zero and y’ of zero equal to negative 2. Then I’ve gone ahead and written down five

formulas that we’re going to need and I’ve gotten this from a table of Laplace transforms.

We’re going to use the first three here on the left f”t, f’t, and this t to the

n e to the at. We’re going to use this first three to plug in to our original function,

transform it and then we’re going to use the second 2 over here on the right when we

perform the inverse Laplace transform to solve for the function Ft. so pretty self-explanatory

here, at least in the beginning. We’re going to take this formula here, f”t and we’re

going to be plugging in the right side of that formula in place of y”. Then we’re

going to take what’s in the right side of this F’t formula and we’re going to plug

it in for y’. And then we’re also going to make a substitution for y but we don’[t

need a formula for it. Then finally, we’re going to plug in what’s on the right side

of this third formula here for what we have over here on the right side of our original

function. So let’s go ahead and make this substitutions

here. We’ll start on the left-hand side so we’re going to grab here first the 2.

Now, remember, we’re plugging in this right side of this equation here for y”. So we’ll

say s squared. And since our original function is in terms of ys instead of fs, remember

that F of s denotes the Laplace transform. But since our function is in terms of y instead

of F, we’re going to go ahead and use y in place of F. So we’ll say Y of s instead

of F of s. So s squared times Y of s minus s times y of zero, and then lower case y because

we switched to lowercase f minus y’ of zero. Now notice here, we have y of zero and y’

of zero. We’re going to be able to use our initial conditions here in a second to plug

in these values, zero and negative 2 for those things but we’ll get to that in a second.

So now we’ve got plus 3 from my original function times what we’ve got here in our

second formula. So s times Y of s minus y of zero. And then finally, minus 2 times,

and when we plug in here for y, we’re just going to do Y of s.

Alright, so we’ve made all our substitution from the left-hand side. Now on the right-hand

side, we’re going to use this formula that we’ve boxed in here and we’re going to

change this format into the format here on the right-hand side. So notice that in the

numerator here on the right, we’ve got n factorial (n!). So we’re going to be looking

at n here. but in our case, that’s equal to 1. We have t to the first power so we’re

going to have 1! In our numerator. And then in our denominator, we’re going to have

s, remember the s variable always stays, minus a. and you can see here, a is the constant.

That is the coefficient on this t variable on the exponent. So in our case, a is negative

2. So since we have s minus a in our formula, we’ll have s minus negative 2 which is just

going to give us s plus 2. And then that’s all raised to the power n plus 1. We already

know that n is equal to 1 so we’ll have 1 plus 1 which is 2 and so that’s going

to be squared. 1! remember is just 1 so let me just go ahead and just change that to 1

right now. Okay so now we’ve made all our substitutions

and we won’t need these three formulas on the left anymore, what we’re going to do

at this point is use our initial conditions y of zero equals zero and y’ of zero equals

negative 2. And we’re going to plug in for these things here. we can make those. And

then we’re going to solve for Y of s and that will give us our Laplace transform. So

we’ll also kind of distribute these coefficients here, the 2, the 3, and the -2.

So let’s call this first, 2s squared times Y of s minus 2s times y of zero we know to

be zero. We have that there so we plug in a zero minus y’ of zero which we know to

be -2. So we have minus a negative 2 which is going to give us plus 2 but remember we’re

multiplying by the 2 out in front so it’s actually going to be plus 4. Moving on to

our second term, we’ll say plus 3s Y of s minus 3 times y of zero which we know to

be equal to zero. then we have minus 2 times Y of s and we’ll go ahead and leave the

right side as it is, 1 over the quantity s plus 2 squared.

So now, we’ll simplify and solve for Y of s. This term here, we’re going to get rid

of this. That’s multiplied by zero so that’s going to go away. This is 3 times zero so

that’s going to go away. And if we now collect like terms, we’ll have 2s squared y of s plus 3s y of s minus 2 times y of s and

then we have plus 4 equal to 1 over the quantity s plus 2 squared. Let’s first actually move

this 4 from the right-hand side. We’ll subtract 4 from both sides, so we’ll get a -4 over

there. Now we go ahead and factor a Y of s because we have it in each term so we can

easily do that. And now we’re multiplying by 2s squared plus 3s minus 2. Let’s go

ahead and get a common denominator over here on the right hand side. So to do that, we’ll

multiply the 4 by s plus 2 squared over s plus 2 squared. and when we combine fractions,

we’ll just get 1 minus 4 times the quantity s plus 2 squared all over quantity s plus

2 squared. Let’s actually start multiplying off the top so we’ll have 4 times the quantity

s squared plus 4s plus 4 all over the quantity s plus 2 squared. And now, we’re going to

divide both sides by this quantity here. 2s squared plus 3s minus 2 in order to solve

for y of s. Dividing the fraction on the right by this quantity here on the left is the same

as multiplying by 1 over that quantity. So we’re going to have actually 1 minus 4s

squared minus 16s minus 16 all over the quantity s plus 2 squared. and bringing that quantity

over right from the left-hand side we’re going to have this multiplied by 1 over 2s

squared plus 3s minus 2. And now, since they are multiplied, we can just combine fractions

and let’s go ahead and consolidate here so we’ve got negative 4s squared minus 16s

and then 1 minus 16 gives us a -15 all over the quantity s plus 2 squared. And we can

factor this 2s squared plus 3s minus 2. If we do that, we’ll get 2s times s and we’ll

say plus 2 minus 1 gives us a 2s squared plus 4s. So we’ll say times 2s minus 1 times

s plus 2. And if you can see here, we’ve got s plus 2 squared and an s plus 2 so let’s

actually go ahead and change this to s plus 2 cubed and we can get rid of this s plus

2 to the first power here. So now, in order to go ahead and solve this

initial value problem, we need to perform a partial fractions decomposition on this

y of s function or the function on which we’ve already performed our Laplace transform. So

we’ve gone on from our original function, we’ve done a Laplace transform and here

we stand. Now we need to use a partial fractions decomposition to solve this. So in order to

do the partial fractions decomposition, remember that we’ll break apart our denominator here.

and because we have a repeated linear factor here in the quantity s plus 2 cubed, remember

that we’ll have to include all factors of lower degree. This 2s minus 1 is also a linear

factor but it’s just a distinct linear factor. So that means that our partial fractions decomposition

is going to look like this. Let’s do the 2s minus 1 first. So we’ll do 2s minus 1.

Then we’ll have plus b, and remember we just need these variables a, b, c, d as opposed

to ax plus b or something like that because we’re dealing with linear factors instead

of quadratic factors. So b over s plus 2 plus c over the quantity s plus 2 squared plus

d over the quantity s plus 2 cubed. And remember that all three of these factors here for b,

c and d are included because we have this repeated linear factor of the quantity s plus

2 cubed and because it’s cubed, we have to include s plus 2 squared and s plus 2 to

the first power in our partial fractions decomposition. So because this is a Laplace transforms problem,

I won’t take the time to do all of the steps of this partial fractions decomposition but

let’s just say that when we solve it for our coefficients a, b, c, and d, we’ll get

a equals -192/125. We’ll find that b is equal to 96/125. c is equal to -2/25 but for

the sake of putting all of these over 12 because it’s going to make it easier on us, we’ll

get -10/125. And then here for d, we’ll get -1/5 which we’ll call -25/125. And you’ll

see in a second why it’s easier to have them all over 125. So now, instead of this

crazy fraction here for y of s, we’ll go ahead and plug in instead our partial fractions

decomposition. And we’re going to pull out a 1/125 outside of our decomposition. And

now we’ll just go ahead and plug in the numerators of all of these fractions, -192,

96, etc. So we’ll get -192 over 2s minus 1 plus 96 over s plus 2. c is a -10 so we’ll

get -10 over s plus 2 squared and then d was a negative 25 so minus 25 over the quantity

s plus 2 cubed. Now, this is where our fourth and fifth formulas

come into play. We have this formula here for 1 over s minus a equal to e to the at.

We’re going to be using that for the first two terms here, the first two fractions in

our partial fractions decomposition. And then we have the second formula here for n factorial

over the quantity s minus a raised to the n plus 1. We’re going to be using that formula

for these two. And basically, we’re going to be performing the inverse Laplace transform.

That’s the final step that we need to do in order to solve the function y of t and

therefore have solve our initial value problem. So in order to do that, we need to make each

of these fractions into the same format as the formulas we were given.

So let’s focus on the first two with the formulas here for 1 over s minus a. As you

can see, essentially if we factored out this -192 and then the denominator here, if we

factor out a 2, then that would be multiplied by 1 over, the 2 is factored out in the denominator

so we’re actually going to get s minus 1/2. Now, this 1 over s minus 1/2 is in the same

form as in our formula 1 over s minus a so we can easily convert it to e to the at. So

that’s what we’re going to do with the 96 over s plus 2 as well. We’ll factor out

the 96 and what we’ll be left with is 1 over, and we could leave it as plus 2 but

let’s go ahead and say s minus a -2 which is the same thing. So we have it in exactly

the same format as 1 over s minus a. so those are our first two terms. And we’ll perform

the inverse Laplace transforms in a second. The second two terms here, we need to get

into the same format as our second formula and again these formulas are from our table

of Laplace transforms. So in order to do that, keep in mind that we have this s minus a so

we’ll do that. But then the exponent here, n plus, we need to figure out what n is in

this n plus 1 exponent and then make sure that our numerator matches that for this n

factorial. So we’re going to pull out a 10 and we’re going to say minus 10 times.

We’ll leave the numerator for now. We can see we have s plus 2 here so we’re going

to say s minus a negative 2 to get them into the same form. And then notice that our exponent

here is squared, which is 1 plus 1. So if we have 1 plus 1, then we know that n is equal

to 1. So we’ll go ahead and say that we have 1 factorial in the numerator which is

just 1, and that’s fine. So we have 1 over the quantity s minus negative 2 squared. then

for our fourth term, we’ll do similar thing. We’ll pull out the negative 25 then we’ve

got this multiplied by, in the denominator we have again s minus a -2 then the exponent

here, we have a 3, which essentially is 2 plus 1. If we’re looking at the exponent

here, n plus 1. If our exponent is equal to 3, then n has to be equal to 2. So we’ll

leave this as a 3 but in the numerator say 2 factorial over 2 factorial because we know

from the n plus 1 exponent in the denominator that n is equal to 2. So we need to make the

numerator n factorial. And the only way we can do that is by multiplying by 2 factorial

over 2 factorial. And now, we can pull out a 2 factorial. We’ll pull out the bottom

one here. So we’ll just be left with 2 factorial over this denominator.

So now that we’ve converted everything into the same format as our formulas, we can go

ahead and perform the last step of our problem which is the inverse Laplace transform. So

let’s go ahead and do that. What we’ll get is y of t which is going to be equal to

1/125. That’ll stay out in front. The -192/2 is going to be -96. And our formula tells

us that this part here that we’ve set up, 1 over s minus 1/2, in the same format as

this formula here, 1 over s minus a, tells us that a is equal to 1/2. So we’re going

to convert that 1 over s minus 1/2 to e to the 1/2t. so we’ll get e to the 1/2t. and

similarly we’re going to have plus 96 times e to the -2t. then in our second formula,

we’re going to have minus 10 and notice that we’re converting it into something

that looks like this. so minus 10 times t so we have t to the first power which is just

t and then multiplied by e to the at. Well, we know that a is -2 so we have e to the -2t.

And then for our last fraction, we’ll have minus 25/2 and again here multiplied by t

raised to the n power. Well, we know that n is equal to 2 because we have 2 factorial

in the numerator and because our exponent here in the denominator is equal to 3 and

we know that n plus 1 has to equal 3 which means n has to equal 2. So we have t squared

and then e to the at where we know a to be -2. So we have a to the -2t. And that’s

it. That’s our final answer. So again, we used our formulas here from the

table of Laplace transforms to perform the Laplace transform on our original function,

we plugged in our initial conditions, then we had to use partial fractions decomposition

to get our function to the point where we could perform an inverse Laplace transform.

And once we’ve performed our inverse Laplace transform, we get a function y of t which

gives us our final answer. So that’s it. I hope this video helped you

guys and I will see you in the next one.

transforms. And in this particular problem, we’ve been given this second degree function,

2y” or the second derivative of y plus 3y’ or the first derivative of y minus 2y is equal

to t e to the negative 2t. We’ve also been given two initial conditions, y of zero equals

zero and y’ of zero equal to negative 2. Then I’ve gone ahead and written down five

formulas that we’re going to need and I’ve gotten this from a table of Laplace transforms.

We’re going to use the first three here on the left f”t, f’t, and this t to the

n e to the at. We’re going to use this first three to plug in to our original function,

transform it and then we’re going to use the second 2 over here on the right when we

perform the inverse Laplace transform to solve for the function Ft. so pretty self-explanatory

here, at least in the beginning. We’re going to take this formula here, f”t and we’re

going to be plugging in the right side of that formula in place of y”. Then we’re

going to take what’s in the right side of this F’t formula and we’re going to plug

it in for y’. And then we’re also going to make a substitution for y but we don’[t

need a formula for it. Then finally, we’re going to plug in what’s on the right side

of this third formula here for what we have over here on the right side of our original

function. So let’s go ahead and make this substitutions

here. We’ll start on the left-hand side so we’re going to grab here first the 2.

Now, remember, we’re plugging in this right side of this equation here for y”. So we’ll

say s squared. And since our original function is in terms of ys instead of fs, remember

that F of s denotes the Laplace transform. But since our function is in terms of y instead

of F, we’re going to go ahead and use y in place of F. So we’ll say Y of s instead

of F of s. So s squared times Y of s minus s times y of zero, and then lower case y because

we switched to lowercase f minus y’ of zero. Now notice here, we have y of zero and y’

of zero. We’re going to be able to use our initial conditions here in a second to plug

in these values, zero and negative 2 for those things but we’ll get to that in a second.

So now we’ve got plus 3 from my original function times what we’ve got here in our

second formula. So s times Y of s minus y of zero. And then finally, minus 2 times,

and when we plug in here for y, we’re just going to do Y of s.

Alright, so we’ve made all our substitution from the left-hand side. Now on the right-hand

side, we’re going to use this formula that we’ve boxed in here and we’re going to

change this format into the format here on the right-hand side. So notice that in the

numerator here on the right, we’ve got n factorial (n!). So we’re going to be looking

at n here. but in our case, that’s equal to 1. We have t to the first power so we’re

going to have 1! In our numerator. And then in our denominator, we’re going to have

s, remember the s variable always stays, minus a. and you can see here, a is the constant.

That is the coefficient on this t variable on the exponent. So in our case, a is negative

2. So since we have s minus a in our formula, we’ll have s minus negative 2 which is just

going to give us s plus 2. And then that’s all raised to the power n plus 1. We already

know that n is equal to 1 so we’ll have 1 plus 1 which is 2 and so that’s going

to be squared. 1! remember is just 1 so let me just go ahead and just change that to 1

right now. Okay so now we’ve made all our substitutions

and we won’t need these three formulas on the left anymore, what we’re going to do

at this point is use our initial conditions y of zero equals zero and y’ of zero equals

negative 2. And we’re going to plug in for these things here. we can make those. And

then we’re going to solve for Y of s and that will give us our Laplace transform. So

we’ll also kind of distribute these coefficients here, the 2, the 3, and the -2.

So let’s call this first, 2s squared times Y of s minus 2s times y of zero we know to

be zero. We have that there so we plug in a zero minus y’ of zero which we know to

be -2. So we have minus a negative 2 which is going to give us plus 2 but remember we’re

multiplying by the 2 out in front so it’s actually going to be plus 4. Moving on to

our second term, we’ll say plus 3s Y of s minus 3 times y of zero which we know to

be equal to zero. then we have minus 2 times Y of s and we’ll go ahead and leave the

right side as it is, 1 over the quantity s plus 2 squared.

So now, we’ll simplify and solve for Y of s. This term here, we’re going to get rid

of this. That’s multiplied by zero so that’s going to go away. This is 3 times zero so

that’s going to go away. And if we now collect like terms, we’ll have 2s squared y of s plus 3s y of s minus 2 times y of s and

then we have plus 4 equal to 1 over the quantity s plus 2 squared. Let’s first actually move

this 4 from the right-hand side. We’ll subtract 4 from both sides, so we’ll get a -4 over

there. Now we go ahead and factor a Y of s because we have it in each term so we can

easily do that. And now we’re multiplying by 2s squared plus 3s minus 2. Let’s go

ahead and get a common denominator over here on the right hand side. So to do that, we’ll

multiply the 4 by s plus 2 squared over s plus 2 squared. and when we combine fractions,

we’ll just get 1 minus 4 times the quantity s plus 2 squared all over quantity s plus

2 squared. Let’s actually start multiplying off the top so we’ll have 4 times the quantity

s squared plus 4s plus 4 all over the quantity s plus 2 squared. And now, we’re going to

divide both sides by this quantity here. 2s squared plus 3s minus 2 in order to solve

for y of s. Dividing the fraction on the right by this quantity here on the left is the same

as multiplying by 1 over that quantity. So we’re going to have actually 1 minus 4s

squared minus 16s minus 16 all over the quantity s plus 2 squared. and bringing that quantity

over right from the left-hand side we’re going to have this multiplied by 1 over 2s

squared plus 3s minus 2. And now, since they are multiplied, we can just combine fractions

and let’s go ahead and consolidate here so we’ve got negative 4s squared minus 16s

and then 1 minus 16 gives us a -15 all over the quantity s plus 2 squared. And we can

factor this 2s squared plus 3s minus 2. If we do that, we’ll get 2s times s and we’ll

say plus 2 minus 1 gives us a 2s squared plus 4s. So we’ll say times 2s minus 1 times

s plus 2. And if you can see here, we’ve got s plus 2 squared and an s plus 2 so let’s

actually go ahead and change this to s plus 2 cubed and we can get rid of this s plus

2 to the first power here. So now, in order to go ahead and solve this

initial value problem, we need to perform a partial fractions decomposition on this

y of s function or the function on which we’ve already performed our Laplace transform. So

we’ve gone on from our original function, we’ve done a Laplace transform and here

we stand. Now we need to use a partial fractions decomposition to solve this. So in order to

do the partial fractions decomposition, remember that we’ll break apart our denominator here.

and because we have a repeated linear factor here in the quantity s plus 2 cubed, remember

that we’ll have to include all factors of lower degree. This 2s minus 1 is also a linear

factor but it’s just a distinct linear factor. So that means that our partial fractions decomposition

is going to look like this. Let’s do the 2s minus 1 first. So we’ll do 2s minus 1.

Then we’ll have plus b, and remember we just need these variables a, b, c, d as opposed

to ax plus b or something like that because we’re dealing with linear factors instead

of quadratic factors. So b over s plus 2 plus c over the quantity s plus 2 squared plus

d over the quantity s plus 2 cubed. And remember that all three of these factors here for b,

c and d are included because we have this repeated linear factor of the quantity s plus

2 cubed and because it’s cubed, we have to include s plus 2 squared and s plus 2 to

the first power in our partial fractions decomposition. So because this is a Laplace transforms problem,

I won’t take the time to do all of the steps of this partial fractions decomposition but

let’s just say that when we solve it for our coefficients a, b, c, and d, we’ll get

a equals -192/125. We’ll find that b is equal to 96/125. c is equal to -2/25 but for

the sake of putting all of these over 12 because it’s going to make it easier on us, we’ll

get -10/125. And then here for d, we’ll get -1/5 which we’ll call -25/125. And you’ll

see in a second why it’s easier to have them all over 125. So now, instead of this

crazy fraction here for y of s, we’ll go ahead and plug in instead our partial fractions

decomposition. And we’re going to pull out a 1/125 outside of our decomposition. And

now we’ll just go ahead and plug in the numerators of all of these fractions, -192,

96, etc. So we’ll get -192 over 2s minus 1 plus 96 over s plus 2. c is a -10 so we’ll

get -10 over s plus 2 squared and then d was a negative 25 so minus 25 over the quantity

s plus 2 cubed. Now, this is where our fourth and fifth formulas

come into play. We have this formula here for 1 over s minus a equal to e to the at.

We’re going to be using that for the first two terms here, the first two fractions in

our partial fractions decomposition. And then we have the second formula here for n factorial

over the quantity s minus a raised to the n plus 1. We’re going to be using that formula

for these two. And basically, we’re going to be performing the inverse Laplace transform.

That’s the final step that we need to do in order to solve the function y of t and

therefore have solve our initial value problem. So in order to do that, we need to make each

of these fractions into the same format as the formulas we were given.

So let’s focus on the first two with the formulas here for 1 over s minus a. As you

can see, essentially if we factored out this -192 and then the denominator here, if we

factor out a 2, then that would be multiplied by 1 over, the 2 is factored out in the denominator

so we’re actually going to get s minus 1/2. Now, this 1 over s minus 1/2 is in the same

form as in our formula 1 over s minus a so we can easily convert it to e to the at. So

that’s what we’re going to do with the 96 over s plus 2 as well. We’ll factor out

the 96 and what we’ll be left with is 1 over, and we could leave it as plus 2 but

let’s go ahead and say s minus a -2 which is the same thing. So we have it in exactly

the same format as 1 over s minus a. so those are our first two terms. And we’ll perform

the inverse Laplace transforms in a second. The second two terms here, we need to get

into the same format as our second formula and again these formulas are from our table

of Laplace transforms. So in order to do that, keep in mind that we have this s minus a so

we’ll do that. But then the exponent here, n plus, we need to figure out what n is in

this n plus 1 exponent and then make sure that our numerator matches that for this n

factorial. So we’re going to pull out a 10 and we’re going to say minus 10 times.

We’ll leave the numerator for now. We can see we have s plus 2 here so we’re going

to say s minus a negative 2 to get them into the same form. And then notice that our exponent

here is squared, which is 1 plus 1. So if we have 1 plus 1, then we know that n is equal

to 1. So we’ll go ahead and say that we have 1 factorial in the numerator which is

just 1, and that’s fine. So we have 1 over the quantity s minus negative 2 squared. then

for our fourth term, we’ll do similar thing. We’ll pull out the negative 25 then we’ve

got this multiplied by, in the denominator we have again s minus a -2 then the exponent

here, we have a 3, which essentially is 2 plus 1. If we’re looking at the exponent

here, n plus 1. If our exponent is equal to 3, then n has to be equal to 2. So we’ll

leave this as a 3 but in the numerator say 2 factorial over 2 factorial because we know

from the n plus 1 exponent in the denominator that n is equal to 2. So we need to make the

numerator n factorial. And the only way we can do that is by multiplying by 2 factorial

over 2 factorial. And now, we can pull out a 2 factorial. We’ll pull out the bottom

one here. So we’ll just be left with 2 factorial over this denominator.

So now that we’ve converted everything into the same format as our formulas, we can go

ahead and perform the last step of our problem which is the inverse Laplace transform. So

let’s go ahead and do that. What we’ll get is y of t which is going to be equal to

1/125. That’ll stay out in front. The -192/2 is going to be -96. And our formula tells

us that this part here that we’ve set up, 1 over s minus 1/2, in the same format as

this formula here, 1 over s minus a, tells us that a is equal to 1/2. So we’re going

to convert that 1 over s minus 1/2 to e to the 1/2t. so we’ll get e to the 1/2t. and

similarly we’re going to have plus 96 times e to the -2t. then in our second formula,

we’re going to have minus 10 and notice that we’re converting it into something

that looks like this. so minus 10 times t so we have t to the first power which is just

t and then multiplied by e to the at. Well, we know that a is -2 so we have e to the -2t.

And then for our last fraction, we’ll have minus 25/2 and again here multiplied by t

raised to the n power. Well, we know that n is equal to 2 because we have 2 factorial

in the numerator and because our exponent here in the denominator is equal to 3 and

we know that n plus 1 has to equal 3 which means n has to equal 2. So we have t squared

and then e to the at where we know a to be -2. So we have a to the -2t. And that’s

it. That’s our final answer. So again, we used our formulas here from the

table of Laplace transforms to perform the Laplace transform on our original function,

we plugged in our initial conditions, then we had to use partial fractions decomposition

to get our function to the point where we could perform an inverse Laplace transform.

And once we’ve performed our inverse Laplace transform, we get a function y of t which

gives us our final answer. So that’s it. I hope this video helped you

guys and I will see you in the next one.