In the next few lectures, I'm going to introduce a new technique for solving differential equations called the Laplace transform technique. It's mainly useful in differential equation solving when you have in an inhomogeneous term that has a discontinuity. But the technique itself is also kind of a useful idea. So, we'll talk about that in this video. How by using the Laplace transform you can convert a differential equation into another space where the equation is easier to solve. That idea is applicable for other types of transforms that we won't cover in this course, but you might see as an engineer such as Fourier transform. So, what is the definition of the Laplace transform? So, we consider the Laplace transform of some function of time, f of t, will become another function, which is a function of a different variable s. So, we talk about the Laplace transform is transforming a function in t space to another function in s space, and what is the definition of that? The Laplace transform is an integral transform. So, the definition is that, this is the integral from 0 to infinity of e to the minus st times f of t dt. So, you have to do an integral over this function f of t to get a function of s. The s here enters in this exponential function. The Laplace transform has some nice properties. One is that it's a linear transformation. So, it's linear. What that means is that, if you take the Laplace transform of a constant times f1 of t plus another constant times f2 of t, when you substitute this into the integral, you can use the fact that the integral of a sum of two functions is a sum of the two integrals. So, the integrals separate, and you can also pull constants out of integrals. So, this is equal to c1 times the Laplace transform of f1 of t plus c2 times the Laplace transform of f2 of t. That's the property of linearity of a transform. There is one more property of the Laplace transform that is necessary for us to be able to use it in our solution. You can take the Laplace transform of f of t to get another function, capital F of s. But the transform is invertible, so you can also take the inverse Laplace transform of f of s to get back the function f of t. The Laplace transform is done by an integration. The inverse Laplace transform, we will do using a table. How does such a table get constructed? We can look at an example to see how that's possible. So, we can take say the Laplace transform of some function that may occur when solving a differential equation, such as the Laplace transform of the exponential function, because our exponential function is our important ansatz for the homogeneous differential equation with constant coefficients. So, what is the Laplace transform of e to the at? While we use the definition, so that means f of t equals e to the at. So, this is the integral from 0 to infinity. Then we have an e to the, we can have e to the negative, and then we have s minus at dt. We can do this integral. This is just the integral of an exponential function. This is equal to negative 1 over s minus a times e to the minus s minus at. As t goes from zero to infinity. The upper limit we can do, provided that s is larger than a. So, we're going to have to assume s is larger than a, then the upper limit will go to zero. So, we assume here that s is greater than a. The upper limit goes to zero and then the lower limit we subtract the value at t equals zero. So then, we'll just get 1 over s minus a. So, that's just done by a straightforward integration. The Laplace transform of e to the at is just 1 over s minus a. So, you do this for a whole bunch of functions that are relevant for solving a differential equation, and you construct a table. So, let me show you now what the table looks like. You can see the table in the lecture notes that you've downloaded. if you look at the table, you'll see that the result that I just derived is in line three, and the left-hand side is the f of t, e to the at and the right-hand side is 1 over s minus a. In line four, is the Laplace transform that you would use to construct the transform of a polynomial. You get t to the transform of t to the n. In line five, it's a polynomial times an exponential function. Six and seven are the sine and cosine functions. Eight and nine are the exponential functions times the sine and cosine functions. Ten and 11 are the, is t times sine and t times cosine, which is relevant for problems with resonance. We'll discuss lines 12 and 16 later. So, we assume now that we have the table of Laplace transforms and we can use that in solution. So now finally, in this video, I want to take a bigger picture of what we're doing. What does it mean to solve a differential equation by the method of Laplace transform? Let me draw a grid to try and explain this better. We have a two-by-two grid. So, the left side, I'm going to call this t-space and the right side I'm going to call s-space. So, you move from t-space to s-space by Laplace transform, and you move from s-space to t-space by the inverse of the Laplace transform. The first row here, I'm going to call the equation, and the second row of this table, I'm going to call the solution. So, what is the problem? The problem is that we're given a differential equation, so this is the ODE here. That's where we start. We want to end up in the box below, which is the solution of the ODE. So, we have an equation for x of t say, and we want to find a solution for x of t. So, we can use the direct method. We can go from the ODE to the solution, so this one will be our x of t. That's what we've been doing since we started this course. We have a differential equation in time. For the second-order one we use an ansatz method, and then we using the principle of superposition, we construct a general solution, and then we put in the initial conditions and find the solution. What we're going to do now in the Laplace transform technique, is we're going to take this differential equation and we're going to take the Laplace transform of it. So, we're going to take Laplace transform here. Then what do we get in s-space? What we will get is an algebraic equation. In fact, there will be a linear equation that's easily solved for our Laplace transform function, which is our capital X. So, starting with x of t, we're going to end up in the equation for capital X of s. It will be an algebraic equation for capital X, and we can easily solve it. The solution then will be here. This is an equation in this box here. It will be the x equation, right? Then in this box here, will be the capital X solution. So, we're going to go from the equation for little x, to the equation for big X, and then we're going to solve for big X and that will be our solution. Then the last step then will be going to find the solution for x of t and that will require the inverse Laplace transform. So, we're going to see then two different methods for solving differential equation. The first one we've seen already, which is the direct method. We use ansatz and principles of superposition, and we calculate a solution. The method that we're gonna do in the next few lectures is an indirect method. We're going to Laplace transform the differential equation to get an algebraic equation for x of s, and then we're going to solve that very easy, and then we're gonna inverse Laplace transform back. So, this one is easy. This one is easy. So, the hard ones will be the inverse Laplace transform. This one might be hard, and as we know that the solution by the direct method can sometimes also be hard. Let me summarize. We're going to be looking at a new method for solving differential equation called the Laplace transform technique. In this video, I defined the Laplace transform, it's a transform of a function of t into a function of s by means of an integral. The Laplace transform is linear and it's also invertible. Because of the property that the Laplace transform is invertible, we're going to be able to use a table to take the inverse of the functions of s to get functions of t. I'm Jeff Jasnoff. Thanks for watching and I'll see you in the next video.