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 Well hello everyone and welcome to our third lecture on Laplace transforms.

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 Today we're going to look at taking Laplace transforms of derivatives and using this to solve ordinary differential equations or initial value problem.

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 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

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 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

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 using the methods that we have for solving differential equations in the time domain.

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 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.

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 We can solve the algebra problem and we then obtain our transformed solution in the S domain.

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 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

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 and we saw how to do that in the last lecture, in our second lecture on Laplace transforms.

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 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.

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 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.

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 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

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 and then we'll apply inverse Laplace transforms to get back to the time domain with our solution.

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 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.

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 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.

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 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.

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 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.

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 So I'll just go through this very quickly. This is non examinable but I'll just show you how this comes about.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 So if you integrate a derivative it will strip away the derivative and give us back the function in this case v.

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 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.

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 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.

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 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.

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 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.

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 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

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 and that will leave us e to the minus st times f of t under the integral sign.

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 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.

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 We're integrating from 0 to infinity. So at the upper limit what's going to happen here?

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 As I said, this part is not examined, you will not be expected to reproduce this.

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 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.

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 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.

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 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.

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 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.

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 Okay, minus what turns out to be the initial condition. But we'll discuss this shortly.

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 We can extend this to the second derivative for example and we would get this expression here.

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 So again you can see that taking the Laplace transform has stripped away the derivative. We've got no appearance of the derivative here.

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 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.

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 This will all become clearer shortly.

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 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.

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 The Laplace transform of x dot of t where we saw it was given by that expression there.

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 We're replacing this as s times x bar minus x little x of 0. So that's what we've got here.

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 Okay, and then the second derivative is simply defined here and these are actually number 23, 24 and 25 in our tables.

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 Okay, so moving on, so what do we got? Well we can see that this formula, this can actually be generalized to higher derivatives.

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 So here is the first and second derivatives. We've already seen them and they're given in the tables.

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 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.

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 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.

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 So again you can see, if you could compare this expression here with the expression for the second derivative.

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 So you have s squared because the second derivative. Here you could s to the n because it's the nth derivative.

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 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.

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 And we would keep going all the way down till we get this expression here given to us by the initial condition.

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 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.

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 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.

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 They're all equal to 0. So when that happens, all that survives are the first terms.

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 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.

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 And all of these terms vanish just as here, these two terms vanish and this term vanishes.

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 So what you get is here on this slide here. So what's happening here?

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 So in the time domain, we're differentiating. x has the derivative of x dot, x dot has the derivative of x double dot.

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 And you could keep on with repeated differentiation all the way down to the nth derivative here.

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 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,

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 the Laplace transform is s times x bar. Then for the second derivative, the Laplace transform is s squared times x bar.

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 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.

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 So t domain differentiation corresponds to s domain multiplication by s. And you can see that that's actually what's happening.

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 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.

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 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.

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 Now if we go the opposite way, let's start at this one for example.

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 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.

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 And the equivalent operation in the Laplace domain, well you get s squared and that becomes s to the power one.

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 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.

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 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.

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 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.

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 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.

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 So these are the equivalent effects of differentiation and integration in the time domain become multiplication and division by s respectively.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 And we get the Laplace transform of three to be equal to three over s.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 So the inverse or plus time of transform of x bar which is of course a function of s is just x of t.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 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.

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 System whose differential equation was solved but also called the long term behavior.

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 Okay, so here's our next problem.

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 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.

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 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.

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 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.

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 We're going to do our steps that we saw in the previous example to solve this differential equation.

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 Step one take the plus transforms of the full differential equation of both sides of the differential equation.

141
00:31:09.000 --> 00:31:16.000
 So look at the Laplace transform of x dot plus the Laplace transform of x is equal to the Laplace transform of minus t.

142
00:31:16.000 --> 00:31:27.000
 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.

143
00:31:28.000 --> 00:31:30.000
 That's just the plus transforming the right hand side.

144
00:31:32.000 --> 00:31:36.000
 Okay, I don't have to take it out but but it's again easier to work with.

145
00:31:36.000 --> 00:31:42.000
 So we'll go to your table and the Laplace transform of x dot is given by number 24.

146
00:31:43.000 --> 00:31:59.000
 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.

147
00:32:00.000 --> 00:32:01.000
 So that's what we get.

148
00:32:02.000 --> 00:32:18.000
 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.

149
00:32:19.000 --> 00:32:24.000
 Calculus problem in the time domain becomes an algebra problem in the S domain.

150
00:32:25.000 --> 00:32:29.000
 So the next step is to apply the initial condition.

151
00:32:30.000 --> 00:32:40.000
 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.

152
00:32:41.000 --> 00:32:52.000
 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.

153
00:32:52.000 --> 00:32:54.000
 So that's what we've got.

154
00:32:55.000 --> 00:33:01.000
 Okay so the next thing we want to do is we want to solve our algebra problem.

155
00:33:03.000 --> 00:33:10.000
 In other words we want to isolate x bar we want to get x bar is equal to some function of s.

156
00:33:11.000 --> 00:33:22.000
 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.

157
00:33:23.000 --> 00:33:27.000
 So I'll have s here to give me s times x bar.

158
00:33:28.000 --> 00:33:33.000
 Now in this case even though we don't write there's a one written here so that's where that one comes from.

159
00:33:33.000 --> 00:33:42.000
 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.

160
00:33:43.000 --> 00:33:46.000
 And on the right hand side we've got minus one over s squared.

161
00:33:47.000 --> 00:34:01.000
 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.

162
00:34:01.000 --> 00:34:13.000
 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.

163
00:34:14.000 --> 00:34:18.000
 And that's equal to minus one over s squared into s plus one.

164
00:34:19.000 --> 00:34:30.000
 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.

165
00:34:31.000 --> 00:34:45.000
 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.

166
00:34:46.000 --> 00:34:58.000
 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.

167
00:34:58.000 --> 00:35:08.000
 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.

168
00:35:09.000 --> 00:35:23.000
 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.

169
00:35:23.000 --> 00:35:36.000
 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.

170
00:35:37.000 --> 00:35:48.000
 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.

171
00:35:49.000 --> 00:36:10.000
 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.

172
00:36:10.000 --> 00:36:29.000
 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.

173
00:36:29.000 --> 00:36:49.000
 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.

174
00:36:49.000 --> 00:36:54.000
 So in other words minus one is equal to b or b is equal to minus one.

175
00:36:55.000 --> 00:37:11.000
 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.

176
00:37:11.000 --> 00:37:26.000
 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.

177
00:37:27.000 --> 00:37:44.000
 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.

178
00:37:44.000 --> 00:38:11.000
 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.

179
00:38:12.000 --> 00:38:16.000
 Plus to be plus see.

180
00:38:17.000 --> 00:38:31.000
 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.

181
00:38:32.000 --> 00:38:35.000
 I know no a b and c and I substitute them.

182
00:38:36.000 --> 00:38:48.000
 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.

183
00:38:48.000 --> 00:39:07.000
 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.

184
00:39:08.000 --> 00:39:35.000
 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.

185
00:39:35.000 --> 00:39:58.000
 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.

186
00:39:59.000 --> 00:40:18.000
 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.

187
00:40:18.000 --> 00:40:47.000
 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.

188
00:40:48.000 --> 00:41:10.000
 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.

189
00:41:10.000 --> 00:41:23.000
 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.

190
00:41:24.000 --> 00:41:34.000
 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.

191
00:41:35.000 --> 00:41:40.000
 Is one over s minus three because if you think of it.

192
00:41:42.000 --> 00:41:59.000
 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.

193
00:42:00.000 --> 00:42:03.000
 Now I'll apply the initial condition.

194
00:42:03.000 --> 00:42:14.000
 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.

195
00:42:15.000 --> 00:42:22.000
 I then move on to step three which requires me to solve for x bar.

196
00:42:22.000 --> 00:42:26.000
 I want to get a solution to my algebra problem.

197
00:42:26.000 --> 00:42:29.000
 So this is just repeated from the previous slide.

198
00:42:29.000 --> 00:42:38.000
 I want to isolate x bar because I want a solution that will state x bar is equal to some function of s.

199
00:42:38.000 --> 00:42:54.000
 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.

200
00:42:54.000 --> 00:42:57.000
 Now the minus two.

201
00:42:57.000 --> 00:43:02.000
 I'll move across the other side or if you prefer I add two to both sides.

202
00:43:02.000 --> 00:43:05.000
 So I've done two steps in one here.

203
00:43:05.000 --> 00:43:13.000
 I've taken the x bar out as a common factor and I've added two to both sides.

204
00:43:13.000 --> 00:43:15.000
 So that's what I obtain.

205
00:43:15.000 --> 00:43:28.000
 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.

206
00:43:28.000 --> 00:43:33.000
 So x bar is equal to one over s minus three squared.

207
00:43:33.000 --> 00:43:44.000
 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.

208
00:43:44.000 --> 00:43:47.000
 Okay so let's have a look.

209
00:43:47.000 --> 00:43:51.000
 So in this case we're just going to use the tables directly.

210
00:43:51.000 --> 00:43:58.000
 So I want to take the inverse Laplace transform of both sides.

211
00:43:58.000 --> 00:44:07.000
 I don't need to do any manipulation I don't think because these terms are actually pretty much in the tabulated form.

212
00:44:07.000 --> 00:44:11.000
 Well okay this one's been multiplied by two but yeah we can handle that.

213
00:44:11.000 --> 00:44:15.000
 So we could take the two outside if we wanted to.

214
00:44:15.000 --> 00:44:19.000
 So the inverse Laplace transform of x bar.

215
00:44:19.000 --> 00:44:25.000
 Well we know x bar is a function of s and its inverse Laplace transform is x of t.

216
00:44:25.000 --> 00:44:30.000
 The inverse Laplace transform of one over s minus three of squared.

217
00:44:30.000 --> 00:44:36.000
 Well that's going to be number six with alpha equal to minus three.

218
00:44:36.000 --> 00:44:45.000
 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.

219
00:44:45.000 --> 00:44:47.000
 So that's where that comes from.

220
00:44:47.000 --> 00:44:57.000
 My second term again this times number five with alpha equal to minus three.

221
00:44:57.000 --> 00:45:08.000
 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.

222
00:45:08.000 --> 00:45:18.000
 And that is the solution of my differential equation of the initial value problem that I was given.

223
00:45:18.000 --> 00:45:26.000
 Okay so let's just have a look at one of the Mobius questions.

224
00:45:26.000 --> 00:45:32.000
 So I've taken this question from Mobius so I just expand this so we can see.

225
00:45:32.000 --> 00:45:34.000
 Here's my differential equation.

226
00:45:34.000 --> 00:45:45.000
 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.

227
00:45:45.000 --> 00:45:51.000
 We have find the solution of the differential equation that's really what we're trying to do here.

228
00:45:51.000 --> 00:45:55.000
 So this is all set up in Mobius.

229
00:45:55.000 --> 00:45:58.000
 Assume x of t is the solution of the problem.

230
00:45:58.000 --> 00:46:08.000
 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.

231
00:46:08.000 --> 00:46:14.000
 Okay so we want to find an expression for x bar.

232
00:46:14.000 --> 00:46:17.000
 So let's do that.

233
00:46:17.000 --> 00:46:19.000
 So this is just one step here.

234
00:46:19.000 --> 00:46:24.000
 This is step one in Mobius.

235
00:46:24.000 --> 00:46:27.000
 But with us there's a bit more work.

236
00:46:27.000 --> 00:46:30.000
 We've got a couple of steps involved here.

237
00:46:30.000 --> 00:46:34.000
 So here's my differential equation written again with the corresponding initial condition.

238
00:46:34.000 --> 00:46:37.000
 There it is.

239
00:46:37.000 --> 00:46:41.000
 Now what I'm going to do here, there's my differential equation.

240
00:46:41.000 --> 00:46:44.000
 Over here it is again.

241
00:46:44.000 --> 00:46:53.000
 Just to get things a little easier for me, I can multiply both sides of this differential equation by minus one.

242
00:46:53.000 --> 00:46:56.000
 I don't like these too many minuses.

243
00:46:56.000 --> 00:46:59.000
 So I'm going to multiply both sides by minus one.

244
00:46:59.000 --> 00:47:04.000
 If I do that, I get 2x dot plus x is equal to minus two.

245
00:47:04.000 --> 00:47:06.000
 Just be careful.

246
00:47:06.000 --> 00:47:08.000
 This is a minus two.

247
00:47:08.000 --> 00:47:13.000
 If I multiply the left hand side by minus one, I must multiply the right hand side by minus one.

248
00:47:13.000 --> 00:47:16.000
 And of course the initial condition will just stay the same.

249
00:47:16.000 --> 00:47:19.000
 x of zero is equal to zero.

250
00:47:19.000 --> 00:47:25.000
 I now want to take Laplace transforms of this differential equation.

251
00:47:25.000 --> 00:47:30.000
 This differential equation is completely equivalent to this one.

252
00:47:30.000 --> 00:47:33.000
 So I take Laplace transforms of it.

253
00:47:33.000 --> 00:47:37.000
 So two times the Laplace transform of x dot will be two times.

254
00:47:37.000 --> 00:47:40.000
 And that will be number 24.

255
00:47:40.000 --> 00:47:44.000
 Two times s times x bar minus x of zero.

256
00:47:44.000 --> 00:47:46.000
 The Laplace transform of x is x bar.

257
00:47:46.000 --> 00:47:50.000
 And I've got minus two times the Laplace transform of one.

258
00:47:50.000 --> 00:47:53.000
 Just rewriting this right hand side.

259
00:47:53.000 --> 00:47:55.000
 And that's just minus two over s.

260
00:47:55.000 --> 00:48:00.000
 So the Laplace transform of a constant is just a constant over s.

261
00:48:00.000 --> 00:48:03.000
 And in this case the constant is minus two.

262
00:48:03.000 --> 00:48:07.000
 It's not two, it's minus two.

263
00:48:07.000 --> 00:48:10.000
 And that's why we're minus two over s.

264
00:48:10.000 --> 00:48:13.000
 The next step of course is apply the initial condition.

265
00:48:13.000 --> 00:48:15.000
 Well x of zero is zero.

266
00:48:15.000 --> 00:48:18.000
 So I replace x of zero by zero.

267
00:48:18.000 --> 00:48:28.000
 I can simplify this expression then to get two sx bar plus x bar is equal to minus two over s.

268
00:48:28.000 --> 00:48:32.000
 Now I'm going to do step three here as well.

269
00:48:32.000 --> 00:48:35.000
 I'm going to solve for x bar.

270
00:48:35.000 --> 00:48:40.000
 So here I've got x bar appeared in both terms on the left hand side.

271
00:48:40.000 --> 00:48:42.000
 So I can take it out as a common factor.

272
00:48:42.000 --> 00:48:48.000
 I get two s plus one bracketed times x bar is equal to minus two over s.

273
00:48:48.000 --> 00:48:51.000
 So if I divide both sides by two s plus one,

274
00:48:51.000 --> 00:48:57.000
 I'll obtain my solution to the algebra problem x bar.

275
00:48:57.000 --> 00:49:02.000
 And x bar is equal to two over s times two s plus one.

276
00:49:02.000 --> 00:49:05.000
 So that's the solution to my algebra problem.

277
00:49:05.000 --> 00:49:11.000
 This is the expression I now want to invert to get my time domain solution x of t.

278
00:49:12.000 --> 00:49:20.000
 So the x bar of s inverts to give me x of t, which is the solution that I want.

279
00:49:20.000 --> 00:49:29.000
 So what Mobius does next is, as you saw, step one was all of this.

280
00:49:29.000 --> 00:49:33.000
 They wanted us to find an expression for x bar.

281
00:49:33.000 --> 00:49:35.000
 That was step one.

282
00:49:35.000 --> 00:49:37.000
 So that involved a little bit of work.

283
00:49:37.000 --> 00:49:43.000
 In this case, I multiplied the differential equation through by minus one to make it easier to work with.

284
00:49:43.000 --> 00:49:49.000
 I then took Laplace-Ransom's of the differential equation, both sides of the differential equation.

285
00:49:49.000 --> 00:49:53.000
 That's what I did at step one in this part here.

286
00:49:53.000 --> 00:49:58.000
 And then I applied the initial conditions x of zero is zero, simplified the things a bit,

287
00:49:58.000 --> 00:50:01.000
 and then solved for x bar.

288
00:50:01.000 --> 00:50:05.000
 And I've got my expression for x bar in the blue box here.

289
00:50:05.000 --> 00:50:08.000
 So there it is, there it is.

290
00:50:08.000 --> 00:50:16.000
 And even if you get it wrong, Mobius will give you the correct expression for x bar.

291
00:50:16.000 --> 00:50:23.000
 So that you're not working with the wrong expression in the rest of the question.

292
00:50:23.000 --> 00:50:33.000
 So we want to decompose this term that we got at step one to get its partial fraction representation.

293
00:50:33.000 --> 00:50:38.000
 There's two distinct linear terms here, the s and the 2s plus one.

294
00:50:38.000 --> 00:50:40.000
 So I need a fraction for each of these.

295
00:50:40.000 --> 00:50:44.000
 Multiply both sides by the denominator of the left hand side.

296
00:50:44.000 --> 00:50:49.000
 Well, multiply this by its own denominator, the cancel to leave me minus two.

297
00:50:49.000 --> 00:50:55.000
 Multiply this by this denominator, then the s is cancelled to give me a into 2s plus one.

298
00:50:55.000 --> 00:51:04.000
 And if I multiply this term here by this denominator, well, the 2s plus one's cancel leave me b times s.

299
00:51:04.000 --> 00:51:06.000
 So that's what I get here.

300
00:51:06.000 --> 00:51:13.000
 And then as always, I now try and set s equal to some value that will simplify this equation for me.

301
00:51:13.000 --> 00:51:20.000
 And I see that if I set s equal to zero, this table go and leave me just an equation involving a,

302
00:51:20.000 --> 00:51:23.000
 which I can hopefully solve quite easily.

303
00:51:23.000 --> 00:51:31.000
 So if s is equal to zero, well, I'll have minus two is equal to a times zero plus one or just a.

304
00:51:31.000 --> 00:51:36.000
 So minus two is equal to a plus zero because s is zero with that function.

305
00:51:36.000 --> 00:51:39.000
 So in other words, a is equal to minus two.

306
00:51:39.000 --> 00:51:47.000
 I now set s equal to minus a half because that will make this bracket equal to zero.

307
00:51:47.000 --> 00:51:50.000
 I want 2s plus one equal to zero.

308
00:51:50.000 --> 00:51:55.000
 So in other words, 2s equal to minus one or s equal to minus a half.

309
00:51:55.000 --> 00:52:03.000
 That will make this term vanish and leave me an expression that will enable me to solve for b.

310
00:52:03.000 --> 00:52:10.000
 So I'll have minus two is equal to zero plus b times minus a half.

311
00:52:10.000 --> 00:52:13.000
 In other words, minus two equals minus a half b.

312
00:52:13.000 --> 00:52:15.000
 So b must be equal to four.

313
00:52:15.000 --> 00:52:18.000
 I've found a, I've found b.

314
00:52:18.000 --> 00:52:24.000
 I can substitute them in the right hand side here and that gives me my partial fraction representation.

315
00:52:24.000 --> 00:52:39.000
 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,

316
00:52:39.000 --> 00:52:42.000
 the original ODE.

317
00:52:42.000 --> 00:52:45.000
 So let's see now.

318
00:52:45.000 --> 00:52:52.000
 So step three in Mobius is to obtain the time domain solution.

319
00:52:52.000 --> 00:52:57.000
 In other words, calculate the inverse Laplace transform here of x bar.

320
00:52:57.000 --> 00:52:59.000
 So that's what I'm going to do over here.

321
00:52:59.000 --> 00:53:04.000
 So I just take my solution from the previous slide, my partial fraction representation.

322
00:53:04.000 --> 00:53:08.000
 Here it is and invert both sides.

323
00:53:08.000 --> 00:53:13.000
 So x bar was equal to minus two over s plus four over two s plus one.

324
00:53:13.000 --> 00:53:18.000
 So I'm taking the inverse Laplace transform of each of these three terms.

325
00:53:18.000 --> 00:53:21.000
 So the inverse Laplace transform of x bar of s.

326
00:53:21.000 --> 00:53:22.000
 Well, that's pretty easy.

327
00:53:22.000 --> 00:53:27.000
 It's just x of t Laplace transform, inverse Laplace transform minus two over s.

328
00:53:27.000 --> 00:53:30.000
 What's just minus two.

329
00:53:30.000 --> 00:53:37.000
 And this one, I need to do a little bit of manipulation before I can use the table.

330
00:53:37.000 --> 00:53:45.000
 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.

331
00:53:45.000 --> 00:53:52.000
 But that's okay because I can take two s plus one over here.

332
00:53:52.000 --> 00:54:01.000
 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.

333
00:54:01.000 --> 00:54:08.000
 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.

334
00:54:08.000 --> 00:54:11.000
 So I'll write s plus what?

335
00:54:11.000 --> 00:54:19.000
 Well, I'm going to multiply whatever goes in here by two, then I must divide what we had by two.

336
00:54:19.000 --> 00:54:29.000
 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.

337
00:54:29.000 --> 00:54:39.000
 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.

338
00:54:39.000 --> 00:54:41.000
 So that's what I've got here.

339
00:54:41.000 --> 00:54:46.000
 So that line there, I've just replaced the denominator in here.

340
00:54:46.000 --> 00:54:50.000
 Okay, so I can simplify things a little bit.

341
00:54:50.000 --> 00:54:56.000
 I've got four and two, so the two will cancel, it'll go into the four twice.

342
00:54:56.000 --> 00:54:59.000
 So I will have what will I have?

343
00:54:59.000 --> 00:55:03.000
 I'll have the inverse Laplace transform of two over s plus a half.

344
00:55:03.000 --> 00:55:13.000
 I can take the two outside and that just leaves me number five, one over s plus alpha, where alpha is a half.

345
00:55:13.000 --> 00:55:20.000
 And I can invert that to give me e to the minus a half t because alpha is equal to a half.

346
00:55:20.000 --> 00:55:29.000
 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.

347
00:55:29.000 --> 00:55:38.000
 And that is the solution to the initial value problem to the differential equation that I was given here in Mobius.

348
00:55:38.000 --> 00:55:46.000
 So that is a solution to one of the differential equations in Mobius.

349
00:55:46.000 --> 00:55:52.000
 Same procedure exactly as we've been seeing in the examples I've done up to now.

350
00:55:52.000 --> 00:55:58.000
 Okay, so there we are now and I could plot that if I want to do it.

351
00:55:58.000 --> 00:56:06.000
 Again, this is totally optional, you don't need to really look at this, you're not expected to reproduce this.

352
00:56:06.000 --> 00:56:13.000
 Okay, so we've looked at first order differential equations.

353
00:56:13.000 --> 00:56:26.000
 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.

354
00:56:26.000 --> 00:56:34.000
 So let's have a look at a second order differential equation that's been solved using Laplace transforms.

355
00:56:34.000 --> 00:56:46.000
 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.

356
00:56:46.000 --> 00:56:57.000
 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.

357
00:56:57.000 --> 00:57:14.000
 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.

358
00:57:14.000 --> 00:57:19.000
 But we're going to use Laplace transforms to solve it today.

359
00:57:20.000 --> 00:57:29.000
 So the first step requires us to take Laplace transforms of the differential equation, both sides of the differential equation.

360
00:57:29.000 --> 00:57:32.000
 So let's use the dot notation for our derivative.

361
00:57:32.000 --> 00:57:45.000
 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.

362
00:57:45.000 --> 00:57:50.000
 In each of these cases, in these cases I've just taken a constant outside the Laplace operator.

363
00:57:50.000 --> 00:57:55.000
 I didn't have to, but I just think it makes life a little bit easier.

364
00:57:55.000 --> 00:57:59.000
 So the Laplace transform of x double dot, well that's given by number 25.

365
00:57:59.000 --> 00:58:01.000
 There it is there.

366
00:58:01.000 --> 00:58:10.000
 s squared x bar minus s times x of zero minus x dot of zero.

367
00:58:10.000 --> 00:58:17.000
 So the x of zero and the x dot of zero, they're given to us as the initial conditions.

368
00:58:17.000 --> 00:58:19.000
 So we know what they are.

369
00:58:19.000 --> 00:58:21.000
 They're given to us. They're supplied.

370
00:58:21.000 --> 00:58:23.000
 In this question they're both equal to zero.

371
00:58:23.000 --> 00:58:27.000
 Then I've got two times the Laplace transform of x dot.

372
00:58:27.000 --> 00:58:31.000
 So if I write that, I'll have two times all of this.

373
00:58:31.000 --> 00:58:33.000
 And that's exactly what I've got here.

374
00:58:33.000 --> 00:58:39.000
 Two times s x bar minus x of zero.

375
00:58:39.000 --> 00:58:44.000
 Plus five times the Laplace transform of x gives me five times x bar.

376
00:58:44.000 --> 00:58:48.000
 Of course the Laplace transform of x of t is just x bar of s.

377
00:58:48.000 --> 00:58:51.000
 We don't bother with the brackets s notation.

378
00:58:51.000 --> 00:58:53.000
 We know these are s domain functions.

379
00:58:53.000 --> 00:58:59.000
 And on the right hand side, I need the Laplace transform of eight e to the minus three t.

380
00:58:59.000 --> 00:59:04.000
 Or in other words, eight times the Laplace transform of e to the minus three t.

381
00:59:04.000 --> 00:59:06.000
 I'll go back a couple of slides.

382
00:59:06.000 --> 00:59:10.000
 The Laplace transform of e to the minus three t.

383
00:59:10.000 --> 00:59:12.000
 Well alpha is equal to three.

384
00:59:12.000 --> 00:59:14.000
 It'll be just one over s plus three.

385
00:59:14.000 --> 00:59:17.000
 So that is exactly what I've got here.

386
00:59:17.000 --> 00:59:20.000
 And don't forget to multiply by eight.

387
00:59:20.000 --> 00:59:26.000
 So I've obtained the Laplace transform of my differential equation.

388
00:59:26.000 --> 00:59:30.000
 The next thing to do now is apply the initial conditions.

389
00:59:30.000 --> 00:59:36.000
 And that'll be quite easy because the initial conditions are both zero.

390
00:59:36.000 --> 00:59:40.000
 So just copying that expression from the previous slide, there it is.

391
00:59:40.000 --> 00:59:43.000
 And there are my initial conditions in the green box.

392
00:59:43.000 --> 00:59:45.000
 I just apply them.

393
00:59:45.000 --> 00:59:48.000
 So s times x of zero is just s times zero.

394
00:59:48.000 --> 00:59:52.000
 So that's a zero minus x dot of zero.

395
00:59:52.000 --> 00:59:58.000
 Well that's zero plus two times s x bar minus x of zero.

396
00:59:58.000 --> 01:00:00.000
 Well x of zero we know is zero.

397
01:00:00.000 --> 01:00:02.000
 So that's where that zero comes from.

398
01:00:02.000 --> 01:00:05.000
 And the other terms just stay as they are.

399
01:00:05.000 --> 01:00:08.000
 Simplifying this, well that's very straightforward.

400
01:00:08.000 --> 01:00:10.000
 I obtained this expression here.

401
01:00:10.000 --> 01:00:20.000
 s squared x bar plus two s x bar plus five x bar is equal to eight over s plus three.

402
01:00:20.000 --> 01:00:26.000
 Okay, so step three, I need to get the solution to my algebra problem.

403
01:00:26.000 --> 01:00:31.000
 In other words, I need to get x bar equal to some function of s.

404
01:00:31.000 --> 01:00:36.000
 And to do that, I need to isolate x bar.

405
01:00:36.000 --> 01:00:41.000
 And if I look at the left hand side, there's three terms and x bar appear in each of them.

406
01:00:41.000 --> 01:00:46.000
 So what I do is I take x bar out as a common factor.

407
01:00:46.000 --> 01:00:48.000
 And that's what I've done here.

408
01:00:48.000 --> 01:00:55.000
 Down in this next line here on the left hand side, I've got s squared plus two s plus five.

409
01:00:55.000 --> 01:00:57.000
 And that's all times x bar.

410
01:00:57.000 --> 01:01:01.000
 The right hand side stays the same.

411
01:01:01.000 --> 01:01:08.000
 I know to get my x bar solution, I just divide both sides by this quadratic term here.

412
01:01:08.000 --> 01:01:10.000
 So that's what I get.

413
01:01:10.000 --> 01:01:15.000
 I get eight over s plus three times this quadratic expression.

414
01:01:15.000 --> 01:01:19.000
 And there is my solution for x bar.

415
01:01:19.000 --> 01:01:21.000
 So that's x bar of s.

416
01:01:21.000 --> 01:01:25.000
 So that is my solution to the algebra problem.

417
01:01:25.000 --> 01:01:30.000
 And this is the thing that I now have to invert.

418
01:01:30.000 --> 01:01:38.000
 But as I've mentioned before, and we've seen that quite often these expressions are not in tabulated form.

419
01:01:38.000 --> 01:01:45.000
 We need to do a little bit of algebraic manipulation to get these into a form that's given in the tables.

420
01:01:45.000 --> 01:01:52.000
 Certainly if we look in our tables, we won't find an expression that looks like this.

421
01:01:52.000 --> 01:02:00.000
 But we know that we can use partial fractions to get the partial fraction representation of this expression here.

422
01:02:00.000 --> 01:02:04.000
 And hopefully that will allow us to use our tables.

423
01:02:04.000 --> 01:02:07.000
 So let's have a look at this.

424
01:02:07.000 --> 01:02:08.000
 So what do we go?

425
01:02:08.000 --> 01:02:10.000
 We've got one linear term in the denominator.

426
01:02:10.000 --> 01:02:15.000
 We're going to focus on the denominator to get our partial fraction set up.

427
01:02:15.000 --> 01:02:17.000
 So we've got one linear term.

428
01:02:17.000 --> 01:02:22.000
 So I'm going to have a constant over s plus three, a over s plus three.

429
01:02:22.000 --> 01:02:25.000
 Now this second term here, this is a quadratic.

430
01:02:25.000 --> 01:02:35.000
 And we would generally ask ourselves, can this be factorized into s plus or minus some constant times s plus or minus some constant.

431
01:02:35.000 --> 01:02:37.000
 And if we can, we'd have three linear terms.

432
01:02:37.000 --> 01:02:41.000
 I would have a over, the first one plus b over the second plus c over the third.

433
01:02:41.000 --> 01:02:45.000
 But in this case, this does not factorize easily.

434
01:02:45.000 --> 01:02:47.000
 You know, five is a prime number.

435
01:02:47.000 --> 01:02:54.000
 And the only way you can form five working with integers is doing one times five.

436
01:02:54.000 --> 01:02:58.000
 And there is no way that one five will add together to give two.

437
01:02:58.000 --> 01:03:05.000
 So there's no way you can factorize this into s plus an integer times s plus some other integer.

438
01:03:05.000 --> 01:03:07.000
 It's just not going to work.

439
01:03:07.000 --> 01:03:09.000
 So what do we do?

440
01:03:09.000 --> 01:03:12.000
 Well, this is what we call an irreducible polynomial.

441
01:03:12.000 --> 01:03:16.000
 We saw an example of this in the previous lecture.

442
01:03:16.000 --> 01:03:19.000
 So it's a quadratic in the denominator.

443
01:03:19.000 --> 01:03:24.000
 So the numerator must always have degree one less when we're using partial fractions.

444
01:03:24.000 --> 01:03:33.000
 So if this is a quadratic, the numerator will have the most general form of a linear term, and that's bs plus c.

445
01:03:33.000 --> 01:03:36.000
 So we need to find a, b and c.

446
01:03:36.000 --> 01:03:39.000
 That's what our challenge is now.

447
01:03:39.000 --> 01:03:43.000
 So again, we'll do what we always do.

448
01:03:43.000 --> 01:03:51.000
 We get rid of these fractions and we do that by multiplying both sides by the denominator of the left hand side.

449
01:03:51.000 --> 01:03:59.000
 So first of all, if you multiply the left hand sides by zone denominator, well, they'll cancel and just need you eight.

450
01:03:59.000 --> 01:04:05.000
 Moving to the right hand side and multiplying by the denominator, what happens?

451
01:04:05.000 --> 01:04:16.000
 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.

452
01:04:16.000 --> 01:04:18.000
 That term there.

453
01:04:18.000 --> 01:04:24.000
 If I then multiply my second right hand term by this denominator here, what will happen?

454
01:04:24.000 --> 01:04:32.000
 Well, this term will cancel with that term and just give me bs plus c times s plus three.

455
01:04:32.000 --> 01:04:35.000
 And that's what I've got here.

456
01:04:35.000 --> 01:04:38.000
 Okay.

457
01:04:38.000 --> 01:04:47.000
 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.

458
01:04:47.000 --> 01:04:49.000
 Let's go about it as we usually do.

459
01:04:49.000 --> 01:04:58.000
 Let's see, can we set s equal to some value that will eliminate terms?

460
01:04:58.000 --> 01:05:01.000
 Well, yes, we can.

461
01:05:01.000 --> 01:05:07.000
 We can set s equal to minus three and then this bracket minus three plus three would be zero.

462
01:05:07.000 --> 01:05:09.000
 So all of this will vanish.

463
01:05:09.000 --> 01:05:14.000
 So if I set s equal to minus three, what do I get?

464
01:05:14.000 --> 01:05:23.000
 Well, if I have s equal to minus three, that would be minus three squared, which is nine.

465
01:05:23.000 --> 01:05:25.000
 s that'd be minus six.

466
01:05:25.000 --> 01:05:28.000
 Nine minus six is three plus five is eight.

467
01:05:28.000 --> 01:05:34.000
 So I'll get eight is equal to eight a or in other words, a is equal to one.

468
01:05:34.000 --> 01:05:36.000
 So that's fine.

469
01:05:36.000 --> 01:05:44.000
 Can I set s equal to some sensible value that will make something else disappear?

470
01:05:44.000 --> 01:05:54.000
 Well, we've set s equal to zero here and s equal to, sorry, minus three here and that made this term vanish.

471
01:05:54.000 --> 01:06:04.000
 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.

472
01:06:04.000 --> 01:06:09.000
 So we know that we can substitute in here for s to make it vanish directly.

473
01:06:09.000 --> 01:06:14.000
 So what we have to do is we have to choose a sensible value for s that we haven't used before.

474
01:06:14.000 --> 01:06:17.000
 Well, we've used minus three, so let's choose a sensible value.

475
01:06:17.000 --> 01:06:22.000
 Why not just choose s equal to zero and substitute into the right-hand side?

476
01:06:22.000 --> 01:06:24.000
 So if I do that, what do I get?

477
01:06:24.000 --> 01:06:27.000
 That'll be zero, zero plus five.

478
01:06:27.000 --> 01:06:30.000
 I'll have eight is equal to five a.

479
01:06:30.000 --> 01:06:33.000
 Well, with s equal to zero, that will vanish.

480
01:06:33.000 --> 01:06:35.000
 That will vanish.

481
01:06:35.000 --> 01:06:37.000
 That'll just be three c.

482
01:06:37.000 --> 01:06:39.000
 So a is equal to five plus three c.

483
01:06:39.000 --> 01:06:41.000
 I know what a is.

484
01:06:41.000 --> 01:06:50.000
 a is equal to, sorry, I don't even need to know what, yes, yes, of course, a is equal to one.

485
01:06:50.000 --> 01:06:54.000
 So that will be a is equal to five plus three c.

486
01:06:54.000 --> 01:07:00.000
 Or in other words, take five from both sides, three is equal to three c, c is equal to one.

487
01:07:00.000 --> 01:07:01.000
 So that's fine.

488
01:07:01.000 --> 01:07:04.000
 I've got a and c, just leaves me b to find now.

489
01:07:04.000 --> 01:07:10.000
 And again, I must choose a value that I haven't used before, a sensible value, of course,

490
01:07:10.000 --> 01:07:16.000
 and s equal one is probably the most next most sensible value after zero,

491
01:07:16.000 --> 01:07:18.000
 which we've already used, of course.

492
01:07:18.000 --> 01:07:21.000
 So s equal to one goes in here.

493
01:07:21.000 --> 01:07:22.000
 So what happens?

494
01:07:22.000 --> 01:07:31.000
 Well, I'll have eight is equal to, that'll be one plus two plus five.

495
01:07:31.000 --> 01:07:33.000
 So one plus two is three plus five is eight.

496
01:07:33.000 --> 01:07:35.000
 So eight is equal to eight a.

497
01:07:35.000 --> 01:07:39.000
 So that's that b is equal to one.

498
01:07:39.000 --> 01:07:45.000
 So I'll have b plus c times four.

499
01:07:45.000 --> 01:07:47.000
 So it'll be four b plus four c.

500
01:07:47.000 --> 01:07:53.000
 So there is my expression eight is equal to eight a plus four b plus four c.

501
01:07:53.000 --> 01:07:57.000
 I know that a and c are equal to one.

502
01:07:57.000 --> 01:07:59.000
 I know that a and c are equal to one.

503
01:07:59.000 --> 01:08:05.000
 So that when I simplify everything here, it's fairly straightforward.

504
01:08:05.000 --> 01:08:09.000
 It'll give me that b is equal to minus one.

505
01:08:09.000 --> 01:08:11.000
 So I know a, b and c.

506
01:08:11.000 --> 01:08:16.000
 I can substitute them into the right hand side to obtain the partial fraction representation.

507
01:08:16.000 --> 01:08:20.000
 Of this expression, which was equal to x bar.

508
01:08:20.000 --> 01:08:22.000
 So this is what I've got now.

509
01:08:22.000 --> 01:08:31.000
 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.

510
01:08:31.000 --> 01:08:36.000
 If I look at this, I know I can handle this.

511
01:08:36.000 --> 01:08:37.000
 This is easy.

512
01:08:37.000 --> 01:08:38.000
 This is just number five.

513
01:08:38.000 --> 01:08:40.000
 That's one over s plus alpha.

514
01:08:40.000 --> 01:08:42.000
 That's very easy to work with.

515
01:08:42.000 --> 01:08:44.000
 But what about this one?

516
01:08:44.000 --> 01:08:47.000
 This isn't quite so easy.

517
01:08:47.000 --> 01:08:51.000
 I don't have a denominator in my tables that looks like this.

518
01:08:51.000 --> 01:08:55.000
 I can't factorize it into two product or two linear terms.

519
01:08:55.000 --> 01:09:02.000
 We know that at least with integers because it's not easy to factorize.

520
01:09:02.000 --> 01:09:04.000
 So what do I do?

521
01:09:04.000 --> 01:09:08.000
 Well, what I'm going to do is I'm going to complete the square in the denominator.

522
01:09:08.000 --> 01:09:15.000
 Remember, we saw this in the last lecture as well when we're struggling to factorize.

523
01:09:15.000 --> 01:09:18.000
 We can always complete the square.

524
01:09:18.000 --> 01:09:22.000
 So here's my expression s squared plus two s plus five.

525
01:09:22.000 --> 01:09:25.000
 So I'll just get the tablet.

526
01:09:25.000 --> 01:09:27.000
 So let's just remind ourselves.

527
01:09:27.000 --> 01:09:33.000
 Part complete the square s squared plus two s plus five.

528
01:09:33.000 --> 01:09:36.000
 What we do is we have a brackets here.

529
01:09:36.000 --> 01:09:39.000
 We have an s here.

530
01:09:39.000 --> 01:09:41.000
 This is squared.

531
01:09:41.000 --> 01:09:43.000
 We'll minus something.

532
01:09:43.000 --> 01:09:45.000
 And don't forget the plus five.

533
01:09:45.000 --> 01:09:47.000
 Right. Let's fill this in.

534
01:09:47.000 --> 01:09:48.000
 We've got a plus here.

535
01:09:48.000 --> 01:09:50.000
 So we have a plus in here.

536
01:09:50.000 --> 01:09:54.000
 We take half the square efficient, just one and write it in there.

537
01:09:54.000 --> 01:10:00.000
 Then we subtract whatever we wrote in here, we subtract it squared.

538
01:10:00.000 --> 01:10:05.000
 And when we simplify this, we'll get s plus one all squared minus one plus one.

539
01:10:05.000 --> 01:10:09.000
 So that's what I've got here.

540
01:10:09.000 --> 01:10:13.000
 And that simplifies to s plus one all squared plus four.

541
01:10:13.000 --> 01:10:17.000
 Again, like I said last time, remember to be careful here.

542
01:10:17.000 --> 01:10:20.000
 This is subtract one squared.

543
01:10:20.000 --> 01:10:22.000
 So it's minus one.

544
01:10:22.000 --> 01:10:25.000
 It's not minus one times minus one.

545
01:10:25.000 --> 01:10:29.000
 You do not write that as a plus one and get a plus six here.

546
01:10:29.000 --> 01:10:34.000
 It's minus one plus five when you simplify it.

547
01:10:34.000 --> 01:10:40.000
 So we've got s plus one all squared plus four.

548
01:10:40.000 --> 01:10:44.000
 Or we can write it as s plus one all squared plus two squared.

549
01:10:44.000 --> 01:10:48.000
 And then we would...

550
01:10:48.000 --> 01:10:51.000
 We've got this expression here.

551
01:10:51.000 --> 01:10:54.000
 Now, I can break this up.

552
01:10:54.000 --> 01:10:57.000
 I can look at the table first of all at the denominator.

553
01:10:57.000 --> 01:10:59.000
 I've got a focus on the denominator.

554
01:10:59.000 --> 01:11:03.000
 And I would look at my table and I would see that there's three canals.

555
01:11:03.000 --> 01:11:07.000
 And I would see that there's three candidates that have of this form.

556
01:11:07.000 --> 01:11:11.000
 s plus a number all squared plus some number squared.

557
01:11:11.000 --> 01:11:13.000
 That's exactly what I've got here.

558
01:11:13.000 --> 01:11:16.000
 s plus alpha all squared plus omega squared.

559
01:11:16.000 --> 01:11:20.000
 The denominator in 14, 15 and 16, they're all the same.

560
01:11:20.000 --> 01:11:24.000
 And I've got to identify which one of these...

561
01:11:24.000 --> 01:11:28.000
 Or more than one of these, which ones I can apply here.

562
01:11:28.000 --> 01:11:31.000
 So I would look at the top line.

563
01:11:31.000 --> 01:11:34.000
 This one's only got a constant. So I can't use that one.

564
01:11:34.000 --> 01:11:40.000
 What about this one down here?

565
01:11:40.000 --> 01:11:43.000
 This one here.

566
01:11:43.000 --> 01:11:46.000
 Well, it's not quite of the form that I want.

567
01:11:46.000 --> 01:11:53.000
 I've got s plus alpha, but in here I've got minus s plus a number, plus alpha, if you like.

568
01:11:53.000 --> 01:11:58.000
 It's minus s I've got here, not s.

569
01:11:58.000 --> 01:12:02.000
 So what can I do?

570
01:12:02.000 --> 01:12:05.000
 Well, this one won't work either because that's just got an s.

571
01:12:05.000 --> 01:12:10.000
 But what I can do is, of course, I can break this into two fractions, add it together.

572
01:12:10.000 --> 01:12:15.000
 I can minus s over all of this plus one over all of this.

573
01:12:15.000 --> 01:12:18.000
 And that's exactly what I've done here.

574
01:12:18.000 --> 01:12:20.000
 That's exactly what I've done here.

575
01:12:20.000 --> 01:12:24.000
 And this is certainly looking a bit more promising.

576
01:12:25.000 --> 01:12:31.000
 This one I've said I can use number five directly on.

577
01:12:31.000 --> 01:12:33.000
 What about this one?

578
01:12:33.000 --> 01:12:36.000
 Well, actually this is number sixteen.

579
01:12:36.000 --> 01:12:40.000
 S over s plus alpha all squared plus over omega squared.

580
01:12:40.000 --> 01:12:42.000
 That is number sixteen.

581
01:12:42.000 --> 01:12:44.000
 So I can use number sixteen on this one.

582
01:12:44.000 --> 01:12:48.000
 What about my third term here?

583
01:12:48.000 --> 01:12:51.000
 Now, that's a constant in the top line.

584
01:12:51.000 --> 01:12:55.000
 Well, I can't use that because that's got the plus variable s.

585
01:12:55.000 --> 01:12:57.000
 I can't use this one either.

586
01:12:57.000 --> 01:12:58.000
 That's got s plus alpha.

587
01:12:58.000 --> 01:13:00.000
 That's got the plus variable as well.

588
01:13:00.000 --> 01:13:03.000
 But number fourteen only has a constant in the top line.

589
01:13:03.000 --> 01:13:05.000
 So that is certainly promising.

590
01:13:05.000 --> 01:13:09.000
 I can certainly look at number fourteen.

591
01:13:09.000 --> 01:13:11.000
 But what have I got to check?

592
01:13:11.000 --> 01:13:19.000
 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.

593
01:13:19.000 --> 01:13:20.000
 And is it?

594
01:13:20.000 --> 01:13:21.000
 Well, no, it's not.

595
01:13:21.000 --> 01:13:24.000
 The denominators are two here.

596
01:13:24.000 --> 01:13:27.000
 Omega squared with omega c equal to two.

597
01:13:27.000 --> 01:13:31.000
 So that means I should have a two in the top line.

598
01:13:31.000 --> 01:13:33.000
 These things have to match.

599
01:13:33.000 --> 01:13:36.000
 So if I'm going to use number fourteen, I must have these things matching.

600
01:13:36.000 --> 01:13:39.000
 But we saw how to do this last time.

601
01:13:39.000 --> 01:13:41.000
 We can easily do this.

602
01:13:41.000 --> 01:13:46.000
 We can write down exactly what the table would force me to have.

603
01:13:46.000 --> 01:13:50.000
 And that would be two over s plus one all squared plus two squared.

604
01:13:50.000 --> 01:13:56.000
 And if I have that, then my omega here in the numerator matches the omega down below.

605
01:13:56.000 --> 01:14:01.000
 But I've changed what I started out with because it was just a one ahead in the numerator.

606
01:14:01.000 --> 01:14:03.000
 But I can easily fix this.

607
01:14:03.000 --> 01:14:07.000
 I can multiply by this fraction here.

608
01:14:07.000 --> 01:14:14.000
 And if you remember the fraction in the top line will contain what the table has forced me to use.

609
01:14:14.000 --> 01:14:16.000
 So sorry, big apartment.

610
01:14:16.000 --> 01:14:23.000
 The fraction in the top line in the numerator will contain the thing that I want, which is a one.

611
01:14:23.000 --> 01:14:24.000
 There it is.

612
01:14:24.000 --> 01:14:30.000
 And in the bottom line, I write what the table forced me to introduce, which is a two.

613
01:14:30.000 --> 01:14:33.000
 So that's where that half comes from.

614
01:14:33.000 --> 01:14:40.000
 And if you see that two, cancel with that two, and just leave me one numerator, which is exactly what I started out with.

615
01:14:40.000 --> 01:14:47.000
 So I narrowed the field down to number 14 for this expression, this term here.

616
01:14:47.000 --> 01:14:56.000
 But I realized that my numerator constant value did not match the constant down here, the omega value.

617
01:14:56.000 --> 01:15:01.000
 I then wrote down exactly what the table forced me to have, which was a two.

618
01:15:01.000 --> 01:15:10.000
 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.

619
01:15:10.000 --> 01:15:12.000
 And I did that by the top line.

620
01:15:12.000 --> 01:15:22.000
 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.

621
01:15:22.000 --> 01:15:29.000
 So I should now be able to use the tables directly on each of these three terms.

622
01:15:29.000 --> 01:15:39.000
 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.

623
01:15:39.000 --> 01:15:43.000
 And the inverse Laplace is e to the minus alpha t.

624
01:15:43.000 --> 01:15:55.000
 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.

625
01:15:55.000 --> 01:15:58.000
 Because you get omega squared and you get two squared.

626
01:15:58.000 --> 01:16:09.000
 And this one here perfectly matches number 14, where omega is equal to two and alpha again is equal to one.

627
01:16:09.000 --> 01:16:17.000
 So just copying that down again to the next slide, I'm ready to do the inverse Laplace transform.

628
01:16:17.000 --> 01:16:22.000
 So as we know, number five, this will just give me e to the minus three t.

629
01:16:22.000 --> 01:16:24.000
 There it is.

630
01:16:24.000 --> 01:16:29.000
 Now, we've said we're going to use number 16 with alpha equal to one and omega equal to two.

631
01:16:29.000 --> 01:16:42.000
 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.

632
01:16:42.000 --> 01:16:47.000
 So it's minus a half sine two t cos omega equal to two.

633
01:16:47.000 --> 01:16:49.000
 So that's where that comes from.

634
01:16:49.000 --> 01:17:02.000
 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.

635
01:17:02.000 --> 01:17:09.000
 And don't forget, of course, this one, you had the half multiplying the expression.

636
01:17:09.000 --> 01:17:12.000
 So that's what we've got here.

637
01:17:12.000 --> 01:17:15.000
 I can tidy it up a little bit.

638
01:17:15.000 --> 01:17:20.000
 I can open out these brackets and if I do that, I'll obtain these four terms here.

639
01:17:20.000 --> 01:17:26.000
 Actually, the exponential I can't do anything with, so we leave that as it is.

640
01:17:26.000 --> 01:17:33.000
 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.

641
01:17:33.000 --> 01:17:41.000
 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.

642
01:17:41.000 --> 01:17:45.000
 That will just give me one e to the minus t sine two t.

643
01:17:45.000 --> 01:17:52.000
 And that is the full solution to my initial value problem.

644
01:17:52.000 --> 01:17:58.000
 That's second order differential equation with the zero initial conditions that I was given.

645
01:17:58.000 --> 01:18:01.000
 And we've done that using Laplace transforms.

646
01:18:01.000 --> 01:18:10.000
 And as I said, you could, alternatively in this case, you could have used the method of undetermined coefficients to obtain this solution.

647
01:18:11.000 --> 01:18:15.000
 Okay, and again, I've plotted it what it looks like.

648
01:18:15.000 --> 01:18:24.000
 I can either write it like this or I can rearrange it like this.

649
01:18:24.000 --> 01:18:34.000
 This is totally optional, but you could negative exponentials e to the minus t, so they'll correspond to exponential decay.

650
01:18:34.000 --> 01:18:38.000
 They will die away over time and they'll die away pretty quickly actually.

651
01:18:38.000 --> 01:18:43.000
 And the larger the value in front of the t, the quicker the term dies away.

652
01:18:43.000 --> 01:18:45.000
 So this term will die away pretty quickly.

653
01:18:45.000 --> 01:18:52.000
 And you've got then got this exponential, negative exponential, which will show decay, but you're going to cause an assign.

654
01:18:52.000 --> 01:18:55.000
 Cause and assign cause oscillations.

655
01:18:55.000 --> 01:19:05.000
 And you can see that for a short time there are oscillations until this eventually takes over until the negative exponential eventually takes charge.

656
01:19:05.000 --> 01:19:11.000
 And everything decays away really fast, heads towards zero.

657
01:19:11.000 --> 01:19:15.000
 But again, as I said, this is purely optional.

658
01:19:15.000 --> 01:19:23.000
 And but it helps me understand what these solutions look like.

659
01:19:23.000 --> 01:19:32.000
 So just to summarize then, we've got five distinct steps for solving differential equations.

660
01:19:32.000 --> 01:19:37.000
 Sometimes you can combine them, sometimes you don't need particular ones.

661
01:19:37.000 --> 01:19:40.000
 So we transform both sides of the differential equation.

662
01:19:40.000 --> 01:19:46.000
 In other words, we take the Laplace transforms of both sides of the differential equation.

663
01:19:46.000 --> 01:19:53.000
 We substitute in the initial values, we then solve our algebra problem in the S domain.

664
01:19:53.000 --> 01:20:04.000
 In fact, in most cases, we have to perform some sort of manipulation to get our expression for X bar.

665
01:20:04.000 --> 01:20:06.000
 That's the solution to our algebra problem.

666
01:20:06.000 --> 01:20:16.000
 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.

667
01:20:16.000 --> 01:20:18.000
 And that is the final step.

668
01:20:18.000 --> 01:20:25.000
 So but with practice, you'd be able to combine a lot of these steps together.

669
01:20:25.000 --> 01:20:39.000
 So just to summarize, in this lecture, we've seen how we can apply the Laplace transform to convert a time domain calculus problem.

670
01:20:39.000 --> 01:20:45.000
 For example, that's an initial value problem to allow a class domain estimate algebra problem.

671
01:20:45.000 --> 01:20:50.000
 And the algebra problem, as we've said, was solved.

672
01:20:50.000 --> 01:20:56.000
 And then the inverse Laplace transform was applied to obtain the required time domain solution.

673
01:20:56.000 --> 01:21:03.000
 So, you know, we've now seen examples of first and second order initial value problems.

674
01:21:03.000 --> 01:21:11.000
 So with different right hand side functions, and you should be able to solve these with a bit of practice using Laplace transforms.

675
01:21:11.000 --> 01:21:17.000
 Our next lecture will introduce what's called the unit step function.

676
01:21:17.000 --> 01:21:29.000
 And unit step function appears quite often in the right hand term of differential equations that model real world situations.

677
01:21:29.000 --> 01:21:39.000
 And solving the resulting differential equations, which appear in electrical engineering, mechanical engineering, and many other areas,

678
01:21:39.000 --> 01:21:45.000
 would be considerably more difficult without Laplace transforms.

679
01:21:45.000 --> 01:21:55.000
 So Laplace transforms are an ideal way to solve differential equations that involve step functions and other types of special functions.

680
01:21:55.000 --> 01:22:01.000
 So just to finish, you should now be able to attempt questions one to ten.

681
01:22:01.000 --> 01:22:03.000
 That's a bobyist questions one to ten.

682
01:22:03.000 --> 01:22:12.000
 And I've also included here some of the questions that you should be able to attempt from the notes.

683
01:22:12.000 --> 01:22:18.000
 So that's the end of this class. So I'll just stop the recording there.