Well hello everyone and welcome to our third lecture on Laplace transforms.
Today we're going to look at taking Laplace transforms of derivatives and using this to solve ordinary differential equations or initial value problem.
So reminding yourselves of the diagram we've seen in previous lectures, we can see that over here on the left hand side we've got the time domain
and we'll have a calculus problem like a differential equation which we would like to solve directly but it's either too complicated or maybe even impossible
using the methods that we have for solving differential equations in the time domain.
So when that situation arises we can take Laplace transforms and convert our calculus time domain problem to an algebra problem in the S domain.
We can solve the algebra problem and we then obtain our transformed solution in the S domain.
Since our original problem was a time domain problem we have to convert back from the S domain to the time domain and we do that by taking inverse Laplace transforms
and we saw how to do that in the last lecture, in our second lecture on Laplace transforms.
The part up at the top here going from the time domain to the S domain we found out how to do that in our first lecture.
In today's class we're going to look at this part here solving the algebra problem to get our transformed solution which as I've said will then invert to get back to the time domain.
So that's the idea for this lecture. We'll get our time domain problem, a differential equation, we'll transform it to the S domain, solve the resulting algebra problem
and then we'll apply inverse Laplace transforms to get back to the time domain with our solution.
Okay so Laplace transforms are derivatives. Well again we're going to denote the Laplace transform of a function of time f of t, little f of t by capital F of s.
And suppose we want to take the Laplace transform of a derivative using the definition, we'll define f dashed t as being the derivative of f.
So the Laplace transform of f dashed t using the definition will be the integral from zero to infinity of e to the minus st of the function whose Laplace transform we want and in this case it's f dashed t.
Now to perform this integral we need to use integration by parts and that will involve us integrating the f dash term and differentiating the exponential.
So I'll just go through this very quickly. This is non examinable but I'll just show you how this comes about.
So we're going to use integration by parts on this integral here. So we'll differentiate the exponential and we'll integrate the f dashed term.
So remind yourselves of the formula for integration by parts. We think of our integral as the product of a function u which we want to differentiate and a function v dashed which we want to integrate.
And then we'll obtain our formula here. It'll be u times v, v you get by integrating v dashed minus the integral of u dashed which is the derivative of u times v dt.
So that's the formula for integration by parts. So as I said we're going to differentiate the exponential and integrate the f dashed. So let's do that.
So we'll go to u equal e to the minus st, differentiating the exponential we get minus s e to the minus st and we integrate the f dashed term.
So we integrate f dashed t with respect to t and if we remember we can think of differentiation integration as the inversions of each other.
So if you integrate a derivative it will strip away the derivative and give us back the function in this case v.
So we'll have v, if v dashed is equal to f dashed of t then when we integrate we'll get v is equal to f of t.
So then we simply have to apply the formula. So here's your integral and the formula says we have the product of u, v and of course that's between the limits.
So u, v, u is e to the minus st times f of t. So that's u, v inside the square brackets here and of course our limits we're going from 0 to infinity.
Now what do we have? We have minus the integral of u dashed v. So it will be minus minus s e to the minus st times f of t.
So we'll have the minus minus come together to give us a plus. The s here can be taken outside the integral because it's an integration with respect to t
and that will leave us e to the minus st times f of t under the integral sign.
So we can now evaluate this. We can simplify what we've got here. So let's focus on the square bracket attempt first of all.
We're integrating from 0 to infinity. So at the upper limit what's going to happen here?
Well we're going to have e to the minus st. Now t will be some massive value. It will be some, you know, infinity is a massive, massive value.
It's not defined. So what you get here is you get exponential decay. So this whole term at the upper limit will die away.
So in the limit it will be equal to 0 and then we sub in the lower limit you get e to the 0 times f of 0.
Now move on to this part here. So we're going to have s times this integral. Now if you look closely at this integral, this is nothing more than the derivative of the Laplace transform of a function f of t and that's given by f of s.
So this is just the Laplace transform of f of t which we call f of s. And then I can rearrange this. I know the d to the 0 is equal to 1.
So what I end up with is s times f of s minus f of 0. So that is the Laplace transform of the derivative f dashed of t.
As I said, this part is not examined, you will not be expected to reproduce this.
So this is just repeating what I said on the previous slide. The Laplace transform of f dashed of t is equal to s times capital f of s minus little f of 0.
So what we've actually been able to do is by taking the Laplace transform of a derivative, we've been able to express it in terms of the Laplace transform of the undifferentiated function.
Remember f of s is the Laplace transform of little f of t. So basically what this has done is that it's stripped away the derivative.
And that's what I'm saying here. In effect the Laplace transform has converted the operation of differentiation into the simpler operation of multiplication by s.
Okay, minus what turns out to be the initial condition. But we'll discuss this shortly.
We can extend this to the second derivative for example and we would get this expression here.
So again you can see that taking the Laplace transform has stripped away the derivative. We've got no appearance of the derivative here.
So you've got s squared times f of s capital f of s which is the Laplace transform of t minus s times f of 0 which is one of the initial conditions minus f dashed of 0 which again is one of the initial conditions and these are supplied.
This will all become clearer shortly.
So when we're solving the order differential equations we tend to use this notation here. So the Laplace transform of x of t is denoted by x bar of s.
The Laplace transform of x dot of t where we saw it was given by that expression there.
We're replacing this as s times x bar minus x little x of 0. So that's what we've got here.
Okay, and then the second derivative is simply defined here and these are actually number 23, 24 and 25 in our tables.
Okay, so moving on, so what do we got? Well we can see that this formula, this can actually be generalized to higher derivatives.
So here is the first and second derivatives. We've already seen them and they're given in the tables.
We can go right up to the nth derivative. Now just be careful. This notation here x with a superscript n in brackets does not mean x to the power n.
It means the nth derivative of x, that's what it means. And the formula here is generalized for the Laplace transform of the nth derivative down in this slide here.
So again you can see, if you could compare this expression here with the expression for the second derivative.
So you have s squared because the second derivative. Here you could s to the n because it's the nth derivative.
Then we'll have s to the power 1 times x of 0, we'll have s to the n minus 1 times x of 0, just 1 power less than this one.
And we would keep going all the way down till we get this expression here given to us by the initial condition.
That comes with a differential equation. Okay, so what we're going to do now is we're going to look at what happens as is the case in many situations in control engineering.
For example, when the initial conditions are all 0. So x of 0, 0 here, x of 0, 0 here, x dot of 0, 0 and all these initial conditions given down here.
They're all equal to 0. So when that happens, all that survives are the first terms.
x bar, sx bar, s squared x bar, s to the power n x bar. Because the initial conditions are equal to 0, these terms are all 0.
And all of these terms vanish just as here, these two terms vanish and this term vanishes.
So what you get is here on this slide here. So what's happening here?
So in the time domain, we're differentiating. x has the derivative of x dot, x dot has the derivative of x double dot.
And you could keep on with repeated differentiation all the way down to the nth derivative here.
Now if we look at what's going on here with the initial conditions all set to 0, if the Laplace transform of x is x bar and you differentiate to get x dot,
the Laplace transform is s times x bar. Then for the second derivative, the Laplace transform is s squared times x bar.
What you can see is differentiation in the time domain, every time you differentiate in the time domain, that corresponds to a multiplication by s in the Laplace domain, in the s domain.
So t domain differentiation corresponds to s domain multiplication by s. And you can see that that's actually what's happening.
So all the way down when you calculate the nth derivative in the time domain, it corresponds to multiplication of x bar by s to the power n in the Laplace domain.
So that's what I say in this first line here, differentiation in the time domain corresponds to multiplication by s in the Laplace or the s domain as it's also called.
Now if we go the opposite way, let's start at this one for example.
We've got x double dot here. Now we know that what the integration will strip away at derivatives. So if we integrate this x double dot here, we get x dot.
And the equivalent operation in the Laplace domain, well you get s squared and that becomes s to the power one.
So in other words, you're just dividing by s in the Laplace domain. Or if you prefer to think of it, division by s is the same as multiplication by one over s.
So every time you integrate going from here to here in the time domain, the equivalent operation in the s domain is multiplication by one over s or if you prefer division by s.
So you can see that that works here as well. Again if you integrate x dot, you get x of t and in the s domain you would divide by s or multiply by one over s and you would just get x bar.
And that summarized down here. Integration in the time domain corresponds to division by s or multiplication by one over s in the Laplace domain.
So these are the equivalent effects of differentiation and integration in the time domain become multiplication and division by s respectively.
So our aim here of course is to use Laplace transforms to solve ordinary differential equations. And we've seen how to take the Laplace transforms of the Laplace transform of the derivative of a function.
And we've seen how that can be expressed in by the Laplace transform of the differentiated function. Because when you took Laplace transforms, we saw that if you here in the time domain you're differentiating x dot where the Laplace transform does not involve the derivative at all.
So we're going to break down the process of solving differential equation into some distinct steps. So first of all we'll take a differential equation, we'll transform both sides of the equation, we'll then apply the initial values, solve our algebra problem in the s domain.
Then we have perhaps we sometimes have to manipulate that algebraic solution so that it's in a form that we can use the tables to calculate the inverse Laplace transform and get back to the t domain.
Okay so I'll just say here one area of course where this differs from solving differential equations in the time domain is that we apply the initial conditions very early on.
After we've transformed the differential equation at step one, step two is to substitute the initial values. If you remember when we're solving ODE's in the time domain, the very last step that we did was to apply the initial conditions to identify the unknown constants that appeared in the general solution.
So with Laplace transforms you apply the initial conditions almost at the very start of the solution process. Okay so let's do an example, this is usually the best way to see these things.
So here we've got a first order differential equation dx by dt plus 2x is equal to three. Now and we've got a corresponding initial condition, x of zero is equal to zero.
Now we could solve this equation using the method we saw in our last chapter on differential equations. We could use the integrating factor method in fact to solve this equation and you know we would get a solution x of t would be equal to some function of t and you could plot your solution.
If I just get the pen you could plot your solution and see how x will vary with time times the independent variable and x of t is the dependent variable and you would get some curve that will show you how x varies with time.
Instead we're going to use differential, we're going to use Laplace transforms to solve the differential equation. So first step that we do is here's our differential equation with the corresponding initial condition and I'm going to use the short hand that x dot is equal to dx by dt.
We've seen that before and we've got our diagram here for applying Laplace transforms. We're going to have time to be in problem with differential equation. Here it is along with the initial condition.
We're going to take Laplace transforms off the ODE to get an algebra problem. We place the calculus problem with an algebra problem. So let's take Laplace transforms off the differential equation.
So we've got to do it term by term. So we've got Laplace transforms of x dot plus two times the Laplace transforms of x. I can take the two outside because of the linearity property. I don't need to but it looks a bit neater if I do.
And on the right hand side we've got to take the Laplace transforms of three. Now I can go to my tables to get the Laplace transforms of x dot. That's just simply s times x bar minus x of zero. x of zero of course is our initial condition.
So that's number 24 gives us the Laplace transforms of x dot plus two times the Laplace transforms of x. Well that's just number 23. So x of t has the Laplace transforms of x bar of s.
So we just write x bar as short. We know that these functions are time domain functions and we know that when we take the Laplace these are estimate functions. So I tend to miss out the brackets s term. Again it's just less cluttered really. It's easier to see what's going on.
But just bear in mind these are all estimate functions and these are all time domain functions. So taking the Laplace transform of x we get x bar. So we've got two x bar and then we've got to take the Laplace transform of three on the right hand side. So it's just a constant. So it's number one in the table that we use.
And we get the Laplace transform of three to be equal to three over s.
Okay so the next step involves us applying the initial conditions. Well we know that x of zero is equal to zero. So I just replace this with zero. Everything else stays exactly the same.
So this step two involves applying the initial condition very straightforward. And at zero in this case so I can simplify this expression to obtain this new expression at the foot of the slide.
Okay so I've taken my calculus problem. I've taken the Laplace transforms and I've got an algebra problem now, And this is the algebra problem down here that I want to solve. And I want to solve it for x bar.
I want to get an expression for x bar because when I invert x bar of s I'll get x of t, the solution that I want. So I need to solve this expression for x bar. This is my algebra problem. This is the algebra problem we're going to solve to get the transformed solution.
So if I solve this I'll take this over to the next slide. That's just repeated what I had on the foot of the previous slide. So I can take the x bar out as a common factor because it appears in both terms on the left hand side.
So I get s plus two times x bar is equal to three over s. I can simplify this to get an expression for x bar. So I get x bar is equal to three over s into s plus two. So that is my transformed solution. That is the solution to my algebra problem.
This is the expression now. A function of s, x bar of s is equal to three over s into s plus two. This is what I need to invert to get my x of t expression, my solution in the time domain to the original differential equation.
Now in this particular case I could go to my tables and I could use number seven directly to calculate the inverse Laplace transform. But as you'll see with the Mobius exercises they tend to ask you to use partial fractions to obtain an expression for x bar that we can use any sets of tables for two.
So what do we have here? Well we've got x bar is equal to three over s into s plus two. Now let's focus on the denominator here. The denominator here contains two linear terms s and s plus two. So I'll need a partial fraction for each of them.
So I'm going to my first term would be a over s plus the second term b over s plus two. So just remember linear terms, that's power one terms in the denominator means we have constants on the top line.
Because with partial fractions the expression in the numerator has degree one less than the denominator. So this is if you like s to the power one terms, linear terms, so we need an s to the zero in the top line in other words a constant.
And we saw this in our last class anyway when we were evaluating, when we were obtaining partial fraction representation for expression. So if you remember what we did was we multiplied both sides by the denominator of the left hand side.
So multiplying this expression by its own denominator, well that'll cancel. So basically if I just write this down just in this case just to explain it just a little so we've got s into s plus two.
So I would take this quantity here, multiply that first fraction by it so that cancels with itself and it just leaves me three. I move this over and you can see here the s is cancelled it give me a times s plus two.
And then finally I multiply the second term on the right hand side by s into s plus two. The s plus two is cancelled giving me b times s. So now we basically have to derive a and b.
We saw how to do this, we try and set s to a value that will eliminate one of the variables a or b. So for example if I set s equal to zero I'll get three is equal to two a plus zero so that vanishes so three is equal to two a so a is equal to three over two.
I can then set s equal to minus two and this term will vanish so I'll get three is equal to minus two b or in other words b is equal to minus three over two.
I can then substitute these values for a and b into my right hand side up here and that will give me the partial fraction representation of three over s into s plus two. So I'll just do that so I'll get three over two times one over s minus three over two times one over s plus two and that equals x bar.
So I should now be able to move on to step five and calculate the inverse or plus transform. So there's the expression I got in the previous slide by using the partial fraction representation and I'm down here now.
I want to invert I want to go from the s domain to the t domain and thereby obtain the solution of my differential equation or initial value problem because I had a differential equation and an initial condition.
So all I'm doing here is the plus inverse the plus of each of the three terms that appear here. I'm taking the constants outside that's constant that appeared in the right hand side.
They can get taken out because the linearity property again. I don't have to but it just makes things I think a little easier.
So the inverse or plus time of transform of x bar which is of course a function of s is just x of t.
Here I'll have three over two times one because the inverse or plus transform of one over s is just one so I get a three over two here and I'll have minus three over two times e to the minus two t and that's because this is of the form one over s plus alpha.
That's number five in the table which was seen quite a number of times now and the inverse or plus of that is e to the minus alpha t in our case alpha is equal to two just match the s plus alpha and s plus two so alpha is equal to two so get e to the minus two t and that's where that comes from.
So that is my time domain solution to my differential equation and that's exactly the same solution you would have obtained if you'd use the integrating factor method which we saw a few lectures ago.
Okay so again this is optional but how does this solution behave over time what does it look like well along the horizontal axis we've got t we've got time and on the vertical axis we've got x of t the solution to a differential equation so let's just have a look how what happens here so first of all what happens when t is equal to zero.
Well my solution will be three over two minus three over two e to the zero e to the zero is one so my solution my function x of t when t is equal to zero will have the value zero so it will start at the origin here.
Now as t starts to increase what we have here is a negative exponential and this term will decay you get exponential decay this term will go to zero over time and go fairly quickly so what happens is as t it starts when t is zero it's got the value three over two we just saw that.
This part here will make less and less of a contribution as time increases as t increases because this term is dying away so what happens is over time this solution because this will become effectively zero or very close to zero this solution will actually tend towards three over two and you can see that that's exactly what's happened here.
Three over two is one point five so after a fairly short time this exponential term this negative exponential has decayed away pretty much to zero it's a little never get to zero because the exponential function is never zero but it will get very close to it and to all intents and purposes our solution will take the value three over two or one point five so that's what's shown there this is called a steady state response of our exponential.
System whose differential equation was solved but also called the long term behavior.
Okay, so here's our next problem.
And as you can see again we've been asked we've been given differential equation and we've been asked to use the plus transform to solve this differential equation.
Of course again we have an initial condition supply the zero initial condition in other words when t is zero x is zero just as in the last example that we looked at.
Okay, so we've got a differential equation again with short hand here x dot plus x is equal to minus t and x of zero is equal to zero x dot of course is just dx by dt.
We're going to do our steps that we saw in the previous example to solve this differential equation.
Step one take the plus transforms of the full differential equation of both sides of the differential equation.
So look at the Laplace transform of x dot plus the Laplace transform of x is equal to the Laplace transform of minus t.
That's like minus one times t so I could take the minus one outside we don't bother writing the ones with just it's just minus Laplace transform of t.
That's just the plus transforming the right hand side.
Okay, I don't have to take it out but but it's again easier to work with.
So we'll go to your table and the Laplace transform of x dot is given by number 24.
It's just s times x bar minus x of zero where x of zero is the initial condition Laplace transform of x is x bar and the Laplace transform of t is number two in the table and it's one over s squared but don't forget the minus that's out here.
So that's what we get.
So we've taken the Laplace transform of our differential equation and we've converted our time domain problem differential equation we've converted it to an S domain algebra problem.
Calculus problem in the time domain becomes an algebra problem in the S domain.
So the next step is to apply the initial condition.
Well I told us that when t is zero x is zero so x of zero is equal to zero so we can apply that directly here and replace x of zero by zero and that's what I've done.
And again just like in the last example we can simplify this to get s x bar plus x bar is equal to minus one over s squared.
So that's what we've got.
Okay so the next thing we want to do is we want to solve our algebra problem.
In other words we want to isolate x bar we want to get x bar is equal to some function of s.
So I carried this over from the previous slide so I can take x bar out as a common factor it appears in both terms on the right hand side.
So I'll have s here to give me s times x bar.
Now in this case even though we don't write there's a one written here so that's where that one comes from.
So I've expanded this bracket of term I would get s x bar plus one times x bar and that would be the left hand side here.
And on the right hand side we've got minus one over s squared.
Now to solve the problem of course I divide through by the coefficient of x bar so I divide by both sides by s plus one and I end up with this solution to my algebra problem.
x bar of s remember x bar is a function of s although we don't bother including the brackets here we know x bar is a function of s so we just write x bar like so.
And that's equal to minus one over s squared into s plus one.
Again I'm going to use partial fractions but if we look carefully at the denominator this is not the same type of partial fraction as in the previous example.
We had two distinct linear terms in this case we could have repeated linear term in the denominator and we saw how to handle these last time so we need three fractions.
So my function is minus one over s squared times s plus one there it is I need three fractions I'll have a over s plus b over s squared plus c over s plus one.
So you can look back at the last lecture if you need a refresher on the partial fractions when you get a repeated linear term in the denominator.
Again I multiply both sides by the denominator of the left hand side so let me just write that out and I'll just show that so that should be s squared of course.
So if I take this expression here and I multiply each term so multiply both multiply the term on the left hand side first of all well by its own denominator they'll cancel to give me minus one.
Move this over to the first term and as you can see what happens here is that s cancels with one of the s's here so you get a into s times s plus one.
Then we're to the next term what happens the s squared cancel with x squared gives me b times s plus one there and finally multiplying the last term on the right hand side by this denominator of the left hand side the s plus one's cancel and I get c times s squared.
So that is where this line comes from one equation three unknowns but we can manipulate this expression to simplify it by setting the values that will reduce the number of unknown constants.
So for example if I set s equal to zero then this term vanishes because s will be zero this term also vanishes because s is zero and that just leaves me this term here with s equal to zero and b times one would be equal to minus one.
So in other words minus one is equal to b or b is equal to minus one.
I then look at the other any other values I can substitute to eliminate terms well yes if I set s equal to minus one that bracket would be zero so this will all vanish that bracket would be zero with s equal to minus one so that will vanish.
I'll have c times minus one squared minus one all squared is one so I'll have c here and that's equal to minus one so I've no found c so the only one I've got to find now is a.
Now can I substitute any other values for s that will force terms to disappear well no I've used zero I've used minus one so there's nothing else that can really substitute that will make these any of these terms vanish.
But what I do in this case as we saw last time the last lecture is I choose a value of s a sensible value for s that hasn't been used before and well I've used zero I've used minus one I'm just going to say s equal to one that's what I wouldn't take s equal to one and I'll substitute it in here into the right hand side here so if I do that I get minus one is equal to two a.
Plus to be plus see.
So there's my equation to a plus to be plus easy to minus one I know what B and C are so I can substitute them in and it's very straightforward I can find that a is equal to one.
I know no a b and c and I substitute them.
In this expression here and that gives me the partial fraction representation of minus one over s squared times s plus one so there's my partial fraction representation.
And this is a fairly straightforward expression to obtain the inverse Laplace transform of so that is my solution my my manipulated solution in the estimate manipulated so that I can use the Laplace tables.
So if I do that I now take the inverse Laplace transform of each of these terms I'll just do them one by one as I've written here that could be number one the inverse Laplace transform one over s is just one the inverse Laplace transform of one over s squared is just T so I'll have one minus T and the inverse Laplace transform one over s plus one well it's we know this one by now it's one over s plus alpha.
Alpha is equal to one so the inverse Laplace transform of one over s plus one is just e to the minus one T or e to the minus T combining all of these these three individual inverse Laplace transform combine them together gives me the solution x of t is equal to one minus T e to the minus T so that is the solution.
To my original differential equation given here or if you prefer given here so my independent variable is T my dependent variable is x and I could if I wanted produce a plot of x against T to see how x varies over time.
Okay so that's a couple of examples done let's move on now and look at another example this example is going to be slightly different perhaps slightly more complicated because the initial condition is not zero x of zero is equal to two so let's just go and have a look at this again you could solve this.
Problem using the integrating factor method but we're going to stick with differential with Laplace transform so here's my differential equation step one take the plus transforms off the ODE so I just do that term by term I can take the constant three outside if I like because of the linearity property so.
That's me setting up the Laplace transform so I then use the tables number 24 gives the Laplace transform of x dot number 23 gives me the Laplace transform of x is x bar.
That's there number 23 and number five which we're quite familiar with now tells me that the Laplace transform of e to the three T.
Is one over s minus three because if you think of it.
Three T you match it with a minus alpha T you would have minus minus three T so alpha must be minus three and that tells me that the Laplace transform of e to the three T is one over s minus three just as I've got here.
Now I'll apply the initial condition.
In this case. unlike the previous two examples. x of zero is not equal to zero in this case x of zero is equal to two so I replace x zero with two.
I then move on to step three which requires me to solve for x bar.
I want to get a solution to my algebra problem.
So this is just repeated from the previous slide.
I want to isolate x bar because I want a solution that will state x bar is equal to some function of s.
So I spot here that x bar appears in two of the terms so I can take the x bar out as a common factor so that will give me s minus three times x bar.
Now the minus two.
I'll move across the other side or if you prefer I add two to both sides.
So I've done two steps in one here.
I've taken the x bar out as a common factor and I've added two to both sides.
So that's what I obtain.
And then it's a fairly straightforward matter of dividing through by s minus three to obtain my solution to the algebra problem to get x bar, x bar of s.
So x bar is equal to one over s minus three squared.
Remember I'm dividing through by s minus three so this would become s minus three squared and I'll have plus two over s minus three.
Okay so let's have a look.
So in this case we're just going to use the tables directly.
So I want to take the inverse Laplace transform of both sides.
I don't need to do any manipulation I don't think because these terms are actually pretty much in the tabulated form.
Well okay this one's been multiplied by two but yeah we can handle that.
So we could take the two outside if we wanted to.
So the inverse Laplace transform of x bar.
Well we know x bar is a function of s and its inverse Laplace transform is x of t.
The inverse Laplace transform of one over s minus three of squared.
Well that's going to be number six with alpha equal to minus three.
So if I do that I find I'll have t e to the minus minus three t or in other words t e to the three t.
So that's where that comes from.
My second term again this times number five with alpha equal to minus three.
I could take the two outside and I would have two e to the minus minus three t or in other words two e to the three t.
And that is the solution of my differential equation of the initial value problem that I was given.
Okay so let's just have a look at one of the Mobius questions.
So I've taken this question from Mobius so I just expand this so we can see.
Here's my differential equation.
First order differential equation that says given the initial value problem minus two d x by dt minus x equals two with x of zero equals zero.
We have find the solution of the differential equation that's really what we're trying to do here.
So this is all set up in Mobius.
Assume x of t is the solution of the problem.
With the plus transform, the plus transform of x of t is x bar of s and we are going to use the plus transform to solve this equation.
Okay so we want to find an expression for x bar.
So let's do that.
So this is just one step here.
This is step one in Mobius.
But with us there's a bit more work.
We've got a couple of steps involved here.
So here's my differential equation written again with the corresponding initial condition.
There it is.
Now what I'm going to do here, there's my differential equation.
Over here it is again.
Just to get things a little easier for me, I can multiply both sides of this differential equation by minus one.
I don't like these too many minuses.
So I'm going to multiply both sides by minus one.
If I do that, I get 2x dot plus x is equal to minus two.
Just be careful.
This is a minus two.
If I multiply the left hand side by minus one, I must multiply the right hand side by minus one.
And of course the initial condition will just stay the same.
x of zero is equal to zero.
I now want to take Laplace transforms of this differential equation.
This differential equation is completely equivalent to this one.
So I take Laplace transforms of it.
So two times the Laplace transform of x dot will be two times.
And that will be number 24.
Two times s times x bar minus x of zero.
The Laplace transform of x is x bar.
And I've got minus two times the Laplace transform of one.
Just rewriting this right hand side.
And that's just minus two over s.
So the Laplace transform of a constant is just a constant over s.
And in this case the constant is minus two.
It's not two, it's minus two.
And that's why we're minus two over s.
The next step of course is apply the initial condition.
Well x of zero is zero.
So I replace x of zero by zero.
I can simplify this expression then to get two sx bar plus x bar is equal to minus two over s.
Now I'm going to do step three here as well.
I'm going to solve for x bar.
So here I've got x bar appeared in both terms on the left hand side.
So I can take it out as a common factor.
I get two s plus one bracketed times x bar is equal to minus two over s.
So if I divide both sides by two s plus one,
I'll obtain my solution to the algebra problem x bar.
And x bar is equal to two over s times two s plus one.
So that's the solution to my algebra problem.
This is the expression I now want to invert to get my time domain solution x of t.
So the x bar of s inverts to give me x of t, which is the solution that I want.
So what Mobius does next is, as you saw, step one was all of this.
They wanted us to find an expression for x bar.
That was step one.
So that involved a little bit of work.
In this case, I multiplied the differential equation through by minus one to make it easier to work with.
I then took Laplace-Ransom's of the differential equation, both sides of the differential equation.
That's what I did at step one in this part here.
And then I applied the initial conditions x of zero is zero, simplified the things a bit,
and then solved for x bar.
And I've got my expression for x bar in the blue box here.
So there it is, there it is.
And even if you get it wrong, Mobius will give you the correct expression for x bar.
So that you're not working with the wrong expression in the rest of the question.
So we want to decompose this term that we got at step one to get its partial fraction representation.
There's two distinct linear terms here, the s and the 2s plus one.
So I need a fraction for each of these.
Multiply both sides by the denominator of the left hand side.
Well, multiply this by its own denominator, the cancel to leave me minus two.
Multiply this by this denominator, then the s is cancelled to give me a into 2s plus one.
And if I multiply this term here by this denominator, well, the 2s plus one's cancel leave me b times s.
So that's what I get here.
And then as always, I now try and set s equal to some value that will simplify this equation for me.
And I see that if I set s equal to zero, this table go and leave me just an equation involving a,
which I can hopefully solve quite easily.
So if s is equal to zero, well, I'll have minus two is equal to a times zero plus one or just a.
So minus two is equal to a plus zero because s is zero with that function.
So in other words, a is equal to minus two.
I now set s equal to minus a half because that will make this bracket equal to zero.
I want 2s plus one equal to zero.
So in other words, 2s equal to minus one or s equal to minus a half.
That will make this term vanish and leave me an expression that will enable me to solve for b.
So I'll have minus two is equal to zero plus b times minus a half.
In other words, minus two equals minus a half b.
So b must be equal to four.
I've found a, I've found b.
I can substitute them in the right hand side here and that gives me my partial fraction representation.
And thereby an expression for x bar that I can invert fairly easily to get my time domain solution x of t to the original initial value problem,
the original ODE.
So let's see now.
So step three in Mobius is to obtain the time domain solution.
In other words, calculate the inverse Laplace transform here of x bar.
So that's what I'm going to do over here.
So I just take my solution from the previous slide, my partial fraction representation.
Here it is and invert both sides.
So x bar was equal to minus two over s plus four over two s plus one.
So I'm taking the inverse Laplace transform of each of these three terms.
So the inverse Laplace transform of x bar of s.
Well, that's pretty easy.
It's just x of t Laplace transform, inverse Laplace transform minus two over s.
What's just minus two.
And this one, I need to do a little bit of manipulation before I can use the table.
As you can see, looking at the denominator here, number five, it's of the form s plus r for the coefficient of s is one.
But that's okay because I can take two s plus one over here.
I can make the coefficient of s equal to one, what I need for the table by taking the two out as a common factor.
Now, I can't just write a one here because if I did, I would have two times one, which is two, which is not what I started with.
So I'll write s plus what?
Well, I'm going to multiply whatever goes in here by two, then I must divide what we had by two.
So that's why I've got one over two, half in here because just checking it and I'm taking two s plus two times a half, which is one.
Everything stays exactly as it was, but I've now got this in a form that I can use number five in the tables to obtain the inverse Laplace transform.
So that's what I've got here.
So that line there, I've just replaced the denominator in here.
Okay, so I can simplify things a little bit.
I've got four and two, so the two will cancel, it'll go into the four twice.
So I will have what will I have?
I'll have the inverse Laplace transform of two over s plus a half.
I can take the two outside and that just leaves me number five, one over s plus alpha, where alpha is a half.
And I can invert that to give me e to the minus a half t because alpha is equal to a half.
Put all of this together, I then find that x of t is equal to minus two plus two e to the minus a half t.
And that is the solution to the initial value problem to the differential equation that I was given here in Mobius.
So that is a solution to one of the differential equations in Mobius.
Same procedure exactly as we've been seeing in the examples I've done up to now.
Okay, so there we are now and I could plot that if I want to do it.
Again, this is totally optional, you don't need to really look at this, you're not expected to reproduce this.
Okay, so we've looked at first order differential equations.
Well, as mentioned earlier, we can apply this process to differential equations of any order, but we don't really look at equations of order higher than two.
So let's have a look at a second order differential equation that's been solved using Laplace transforms.
So here's my differential equation here, so if I wrote it in the dot notation I would have x double dot plus two x dot plus five x is equal to eight e to the minus three t.
Second order differential equations, so we need two initial conditions and we've got x of zero equals zero and x dot of zero is equal to zero.
So this is an initial value problem, second order initial value problem and we could certainly solve this using the method of undetermined coefficients that we saw in the previous section on differential equations.
But we're going to use Laplace transforms to solve it today.
So the first step requires us to take Laplace transforms of the differential equation, both sides of the differential equation.
So let's use the dot notation for our derivative.
So we've got the Laplace transform of x double dot plus two times the Laplace transform of x dot plus five times the Laplace transform of x is equal to eight times the Laplace transform of e to the minus three t.
In each of these cases, in these cases I've just taken a constant outside the Laplace operator.
I didn't have to, but I just think it makes life a little bit easier.
So the Laplace transform of x double dot, well that's given by number 25.
There it is there.
s squared x bar minus s times x of zero minus x dot of zero.
So the x of zero and the x dot of zero, they're given to us as the initial conditions.
So we know what they are.
They're given to us. They're supplied.
In this question they're both equal to zero.
Then I've got two times the Laplace transform of x dot.
So if I write that, I'll have two times all of this.
And that's exactly what I've got here.
Two times s x bar minus x of zero.
Plus five times the Laplace transform of x gives me five times x bar.
Of course the Laplace transform of x of t is just x bar of s.
We don't bother with the brackets s notation.
We know these are s domain functions.
And on the right hand side, I need the Laplace transform of eight e to the minus three t.
Or in other words, eight times the Laplace transform of e to the minus three t.
I'll go back a couple of slides.
The Laplace transform of e to the minus three t.
Well alpha is equal to three.
It'll be just one over s plus three.
So that is exactly what I've got here.
And don't forget to multiply by eight.
So I've obtained the Laplace transform of my differential equation.
The next thing to do now is apply the initial conditions.
And that'll be quite easy because the initial conditions are both zero.
So just copying that expression from the previous slide, there it is.
And there are my initial conditions in the green box.
I just apply them.
So s times x of zero is just s times zero.
So that's a zero minus x dot of zero.
Well that's zero plus two times s x bar minus x of zero.
Well x of zero we know is zero.
So that's where that zero comes from.
And the other terms just stay as they are.
Simplifying this, well that's very straightforward.
I obtained this expression here.
s squared x bar plus two s x bar plus five x bar is equal to eight over s plus three.
Okay, so step three, I need to get the solution to my algebra problem.
In other words, I need to get x bar equal to some function of s.
And to do that, I need to isolate x bar.
And if I look at the left hand side, there's three terms and x bar appear in each of them.
So what I do is I take x bar out as a common factor.
And that's what I've done here.
Down in this next line here on the left hand side, I've got s squared plus two s plus five.
And that's all times x bar.
The right hand side stays the same.
I know to get my x bar solution, I just divide both sides by this quadratic term here.
So that's what I get.
I get eight over s plus three times this quadratic expression.
And there is my solution for x bar.
So that's x bar of s.
So that is my solution to the algebra problem.
And this is the thing that I now have to invert.
But as I've mentioned before, and we've seen that quite often these expressions are not in tabulated form.
We need to do a little bit of algebraic manipulation to get these into a form that's given in the tables.
Certainly if we look in our tables, we won't find an expression that looks like this.
But we know that we can use partial fractions to get the partial fraction representation of this expression here.
And hopefully that will allow us to use our tables.
So let's have a look at this.
So what do we go?
We've got one linear term in the denominator.
We're going to focus on the denominator to get our partial fraction set up.
So we've got one linear term.
So I'm going to have a constant over s plus three, a over s plus three.
Now this second term here, this is a quadratic.
And we would generally ask ourselves, can this be factorized into s plus or minus some constant times s plus or minus some constant.
And if we can, we'd have three linear terms.
I would have a over, the first one plus b over the second plus c over the third.
But in this case, this does not factorize easily.
You know, five is a prime number.
And the only way you can form five working with integers is doing one times five.
And there is no way that one five will add together to give two.
So there's no way you can factorize this into s plus an integer times s plus some other integer.
It's just not going to work.
So what do we do?
Well, this is what we call an irreducible polynomial.
We saw an example of this in the previous lecture.
So it's a quadratic in the denominator.
So the numerator must always have degree one less when we're using partial fractions.
So if this is a quadratic, the numerator will have the most general form of a linear term, and that's bs plus c.
So we need to find a, b and c.
That's what our challenge is now.
So again, we'll do what we always do.
We get rid of these fractions and we do that by multiplying both sides by the denominator of the left hand side.
So first of all, if you multiply the left hand sides by zone denominator, well, they'll cancel and just need you eight.
Moving to the right hand side and multiplying by the denominator, what happens?
Well, this term here, all this term will multiply this fraction and the s plus threes will cancel to give me a into s squared plus two s plus five.
That term there.
If I then multiply my second right hand term by this denominator here, what will happen?
Well, this term will cancel with that term and just give me bs plus c times s plus three.
And that's what I've got here.
Okay.
So this looks a bit more complicated than the ones we've seen earlier on in this class, but let's just go about it.
Let's go about it as we usually do.
Let's see, can we set s equal to some value that will eliminate terms?
Well, yes, we can.
We can set s equal to minus three and then this bracket minus three plus three would be zero.
So all of this will vanish.
So if I set s equal to minus three, what do I get?
Well, if I have s equal to minus three, that would be minus three squared, which is nine.
s that'd be minus six.
Nine minus six is three plus five is eight.
So I'll get eight is equal to eight a or in other words, a is equal to one.
So that's fine.
Can I set s equal to some sensible value that will make something else disappear?
Well, we've set s equal to zero here and s equal to, sorry, minus three here and that made this term vanish.
We would really like this term to vanish, but we know there's no integer value that will make this vanish because, you know, we know we can't factorize it.
So we know that we can substitute in here for s to make it vanish directly.
So what we have to do is we have to choose a sensible value for s that we haven't used before.
Well, we've used minus three, so let's choose a sensible value.
Why not just choose s equal to zero and substitute into the right-hand side?
So if I do that, what do I get?
That'll be zero, zero plus five.
I'll have eight is equal to five a.
Well, with s equal to zero, that will vanish.
That will vanish.
That'll just be three c.
So a is equal to five plus three c.
I know what a is.
a is equal to, sorry, I don't even need to know what, yes, yes, of course, a is equal to one.
So that will be a is equal to five plus three c.
Or in other words, take five from both sides, three is equal to three c, c is equal to one.
So that's fine.
I've got a and c, just leaves me b to find now.
And again, I must choose a value that I haven't used before, a sensible value, of course,
and s equal one is probably the most next most sensible value after zero,
which we've already used, of course.
So s equal to one goes in here.
So what happens?
Well, I'll have eight is equal to, that'll be one plus two plus five.
So one plus two is three plus five is eight.
So eight is equal to eight a.
So that's that b is equal to one.
So I'll have b plus c times four.
So it'll be four b plus four c.
So there is my expression eight is equal to eight a plus four b plus four c.
I know that a and c are equal to one.
So that when I simplify everything here, it's fairly straightforward.
It'll give me that b is equal to minus one.
So I know a, b and c.
I can substitute them into the right hand side to obtain the partial fraction representation.
Of this expression, which was equal to x bar.
So this is what I've got now.
This is what I've got, but I'm not quite finished with the manipulation process yet because this isn't quite in a tabulated form.
If I look at this, I know I can handle this.
This is easy.
This is just number five.
That's one over s plus alpha.
That's very easy to work with.
But what about this one?
This isn't quite so easy.
I don't have a denominator in my tables that looks like this.
I can't factorize it into two product or two linear terms.
We know that at least with integers because it's not easy to factorize.
So what do I do?
Well, what I'm going to do is I'm going to complete the square in the denominator.
Remember, we saw this in the last lecture as well when we're struggling to factorize.
We can always complete the square.
So here's my expression s squared plus two s plus five.
So I'll just get the tablet.
So let's just remind ourselves.
Part complete the square s squared plus two s plus five.
What we do is we have a brackets here.
We have an s here.
This is squared.
We'll minus something.
And don't forget the plus five.
Right. Let's fill this in.
We've got a plus here.
So we have a plus in here.
We take half the square efficient, just one and write it in there.
Then we subtract whatever we wrote in here, we subtract it squared.
And when we simplify this, we'll get s plus one all squared minus one plus one.
So that's what I've got here.
And that simplifies to s plus one all squared plus four.
Again, like I said last time, remember to be careful here.
This is subtract one squared.
So it's minus one.
It's not minus one times minus one.
You do not write that as a plus one and get a plus six here.
It's minus one plus five when you simplify it.
So we've got s plus one all squared plus four.
Or we can write it as s plus one all squared plus two squared.
And then we would...
We've got this expression here.
Now, I can break this up.
I can look at the table first of all at the denominator.
I've got a focus on the denominator.
And I would look at my table and I would see that there's three canals.
And I would see that there's three candidates that have of this form.
s plus a number all squared plus some number squared.
That's exactly what I've got here.
s plus alpha all squared plus omega squared.
The denominator in 14, 15 and 16, they're all the same.
And I've got to identify which one of these...
Or more than one of these, which ones I can apply here.
So I would look at the top line.
This one's only got a constant. So I can't use that one.
What about this one down here?
This one here.
Well, it's not quite of the form that I want.
I've got s plus alpha, but in here I've got minus s plus a number, plus alpha, if you like.
It's minus s I've got here, not s.
So what can I do?
Well, this one won't work either because that's just got an s.
But what I can do is, of course, I can break this into two fractions, add it together.
I can minus s over all of this plus one over all of this.
And that's exactly what I've done here.
That's exactly what I've done here.
And this is certainly looking a bit more promising.
This one I've said I can use number five directly on.
What about this one?
Well, actually this is number sixteen.
S over s plus alpha all squared plus over omega squared.
That is number sixteen.
So I can use number sixteen on this one.
What about my third term here?
Now, that's a constant in the top line.
Well, I can't use that because that's got the plus variable s.
I can't use this one either.
That's got s plus alpha.
That's got the plus variable as well.
But number fourteen only has a constant in the top line.
So that is certainly promising.
I can certainly look at number fourteen.
But what have I got to check?
I've got to check that this omega here in the top line in the numerator is exactly the same as the omega in the denominator.
And is it?
Well, no, it's not.
The denominators are two here.
Omega squared with omega c equal to two.
So that means I should have a two in the top line.
These things have to match.
So if I'm going to use number fourteen, I must have these things matching.
But we saw how to do this last time.
We can easily do this.
We can write down exactly what the table would force me to have.
And that would be two over s plus one all squared plus two squared.
And if I have that, then my omega here in the numerator matches the omega down below.
But I've changed what I started out with because it was just a one ahead in the numerator.
But I can easily fix this.
I can multiply by this fraction here.
And if you remember the fraction in the top line will contain what the table has forced me to use.
So sorry, big apartment.
The fraction in the top line in the numerator will contain the thing that I want, which is a one.
There it is.
And in the bottom line, I write what the table forced me to introduce, which is a two.
So that's where that half comes from.
And if you see that two, cancel with that two, and just leave me one numerator, which is exactly what I started out with.
So I narrowed the field down to number 14 for this expression, this term here.
But I realized that my numerator constant value did not match the constant down here, the omega value.
I then wrote down exactly what the table forced me to have, which was a two.
But that made I had to do a little bit more manipulation so that I wasn't changing the term that I started out with.
And I did that by the top line.
I multiply by the thing that I want, which is a one down below in the denominator and write what the table forced me to introduce, which was a two.
So I should now be able to use the tables directly on each of these three terms.
As we said, number, this one here is number five with alpha equal to three, because number five in our table gives us one over s plus alpha.
And the inverse Laplace is e to the minus alpha t.
Here, this one here matches perfectly with number 16 with alpha equal to one, so s plus one all squared, and omega equal to two.
Because you get omega squared and you get two squared.
And this one here perfectly matches number 14, where omega is equal to two and alpha again is equal to one.
So just copying that down again to the next slide, I'm ready to do the inverse Laplace transform.
So as we know, number five, this will just give me e to the minus three t.
There it is.
Now, we've said we're going to use number 16 with alpha equal to one and omega equal to two.
So number 16 with alpha equal to one gives me e to the minus t cos two t minus alpha equal to one omega equal to two.
So it's minus a half sine two t cos omega equal to two.
So that's where that comes from.
And then using number 14 with omega equal to two and alpha equal to one, I get e to the minus t sine two t.
And don't forget, of course, this one, you had the half multiplying the expression.
So that's what we've got here.
I can tidy it up a little bit.
I can open out these brackets and if I do that, I'll obtain these four terms here.
Actually, the exponential I can't do anything with, so we leave that as it is.
e to the minus t cos two t, well that's the only expression of that type, so that has to be left as it is.
But if you look at this one, I've got a half e to the minus t sine two t plus a half e to the minus t sine two t.
That will just give me one e to the minus t sine two t.
And that is the full solution to my initial value problem.
That's second order differential equation with the zero initial conditions that I was given.
And we've done that using Laplace transforms.
And as I said, you could, alternatively in this case, you could have used the method of undetermined coefficients to obtain this solution.
Okay, and again, I've plotted it what it looks like.
I can either write it like this or I can rearrange it like this.
This is totally optional, but you could negative exponentials e to the minus t, so they'll correspond to exponential decay.
They will die away over time and they'll die away pretty quickly actually.
And the larger the value in front of the t, the quicker the term dies away.
So this term will die away pretty quickly.
And you've got then got this exponential, negative exponential, which will show decay, but you're going to cause an assign.
Cause and assign cause oscillations.
And you can see that for a short time there are oscillations until this eventually takes over until the negative exponential eventually takes charge.
And everything decays away really fast, heads towards zero.
But again, as I said, this is purely optional.
And but it helps me understand what these solutions look like.
So just to summarize then, we've got five distinct steps for solving differential equations.
Sometimes you can combine them, sometimes you don't need particular ones.
So we transform both sides of the differential equation.
In other words, we take the Laplace transforms of both sides of the differential equation.
We substitute in the initial values, we then solve our algebra problem in the S domain.
In fact, in most cases, we have to perform some sort of manipulation to get our expression for X bar.
That's the solution to our algebra problem.
We have to do some manipulation to get this to get that into a form that appears in our tables so we can invert to get back to the time domain.
And that is the final step.
So but with practice, you'd be able to combine a lot of these steps together.
So just to summarize, in this lecture, we've seen how we can apply the Laplace transform to convert a time domain calculus problem.
For example, that's an initial value problem to allow a class domain estimate algebra problem.
And the algebra problem, as we've said, was solved.
And then the inverse Laplace transform was applied to obtain the required time domain solution.
So, you know, we've now seen examples of first and second order initial value problems.
So with different right hand side functions, and you should be able to solve these with a bit of practice using Laplace transforms.
Our next lecture will introduce what's called the unit step function.
And unit step function appears quite often in the right hand term of differential equations that model real world situations.
And solving the resulting differential equations, which appear in electrical engineering, mechanical engineering, and many other areas,
would be considerably more difficult without Laplace transforms.
So Laplace transforms are an ideal way to solve differential equations that involve step functions and other types of special functions.
So just to finish, you should now be able to attempt questions one to ten.
That's a bobyist questions one to ten.
And I've also included here some of the questions that you should be able to attempt from the notes.
So that's the end of this class. So I'll just stop the recording there.