 So, what can you do with the Laplace transform? Now, before we make extensive use of the Laplace transform, it will be useful to know what its properties are. And the important question is, given functions f, g, and constants a and b, what is the Laplace transform of the linear combination af plus bg? Definitions are the whole of mathematics. All else is commentary. We have the Laplace transform of af plus bg is going to be, but except for the fact that this is an improper integral, this is just an integral, and so all of our properties of definite integrals apply. We can distribute this e to power minus sx. The integral of a sum is the sum of the integrals. The integral of a constant times a function is the constant times the integral. Definitions are the whole of mathematics. We recognize these as the Laplace transforms of f and g. And if we put all these together, this means that our Laplace transform is a linear operator. We can take any linear combination af plus bg and evaluate it as the linear combination of the Laplace transforms of the individual functions. We like linear operators because they make our life so much easier. For example, let's find the Laplace transform of y, where y prime plus y equals e to power x, and we know that y of zero is equal to zero. So we'll apply the Laplace transform to both sides. The linearity of the Laplace transform allows us to simplify the left-hand side. The Laplace transform of a sum is the sum of the Laplace transforms. The reduction relationship allows us to rewrite the Laplace transform of an nth derivative as the Laplace transform of an n minus first derivative plus some other stuff. So if the Laplace transform of y prime, well, that's just s times the Laplace transform of y minus y of zero. And l of y is still there. Since we assumed y of zero was equal to zero, this becomes... and we'll clean up the algebra by combining the like terms. Now, because the right-hand side is the Laplace transform of a specific function, we can actually evaluate it. The Laplace transform of e to power x is going to be... and we'll evaluate that... getting one over s minus one. And now, since we want to find the Laplace transform of y, we'll solve for it, getting... So for the differential equation y prime plus y equals e to power x with y of zero equal to zero, we found the Laplace transform of y is one over s minus one times s plus one. And this raises the important question. Suppose we could find a function f where the Laplace transform of f was one over s minus one times s plus one. Could we then conclude that y is f of x? And the answer is... in general, yes. This suggests an approach to solving nth-order differential equations. First, apply the Laplace transform. Collapse the Laplace transforms to the Laplace transform of y itself, and then try to find f with the same Laplace transform. Now, in order for this to be useful, we need a library of Laplace transforms. Well, since the Laplace transform is just defined as a particular improper integral, then with some effort, we can find the Laplace transform of many functions. And we can find more. So let's see how our approach might work. Let's try to solve the differential equation y prime plus y equals zero. So, applying the Laplace transform to both sides, by linearity, we know that the Laplace transform of the sum is the sum of the Laplace transforms. By the reduction relationship, we know the Laplace transform of y prime is s times the Laplace transform of y minus y of zero. And the Laplace transform of y that we started with is still there. We'll do a little algebra. Now, on the right hand side, we'll have to find the Laplace transform of zero. By definition, this is going to be, which is zero. And so now this gives us an equation in the Laplace transform. So we'll solve our equation for the Laplace transform of y. Now, this factor one over s plus one, that came from the Laplace transform of some function. So we'll check our library. And checking our library, we see that the Laplace transform of e to the power ax is one over s minus a. And so comparing this Laplace transform with this Laplace transform, we conclude that the Laplace transform of e to the power minus x is one over s plus one. And so this is the Laplace transform of e to the minus x. And everything else stays the same. Linearity allows us to move the constant into the Laplace transform. And since the Laplace transform of y is equal to the Laplace transform of y of zero e to the power minus x, then y must be y of zero e to the power minus x. And there's the solution to our differential equation.