 One very powerful method of solving nth order differential equations is based around what's called the Laplace transform. We define the following. Let f of x be a function with suitable properties. The Laplace transform of f, which we'll write as l of f, is given by the improper integral from 0 to infinity of e to power minus sx f of x dx, provided that the limit as x goes to infinity of the integrand is equal to 0. So let's find the Laplace transform of x itself. So from our definition, our Laplace transform of x is going to be the improper integral from 0 to infinity e to minus sx times x dx. Now we have to evaluate this integral, and so we'll use integration by parts. And the useful thing to remember here is that if I integrate or differentiate e to power minus sx, it doesn't change very much, but if I differentiate x, it will go away. So we'll let u equals x, let dv equal e to power minus sx dx, and we'll differentiate u and anti-differentiate dv. And so our anti-derivative is going to be u times v minus the integral of v du. And this is still an improper integral, so we'll include those limits of integration. So remember this is an improper integral, so we should at least dot all of our t's and cross all of our i's. We can't really let this upper bound be infinity. We have to rewrite it as a limit as t goes to infinity. We'll evaluate at the upper bound. We'll evaluate at the lower bound. We'll do a little algebra. And now we have a problem. If we're going to let t go to infinity, the value of this expression is going to depend on s and on t. But remember, definitions are the whole of mathematics, all else is commentary. If we retrieve our definition of the Laplace transform, we see that we made a very important assumption. The limit as x goes to infinity of our integrand has to be zero. And that means that this limit must also be zero. And so the only thing left is the second integral. Well, we can do that, too. So we'll find the antiderivative. Again, we're dealing with the improper integral from zero to infinity, so we should rewrite that using our limit formulation. We'll evaluate. And again, we have this limit as t goes to infinity of this expression. And the real problem here is finding the limit as t goes to infinity of e to power minus s t. But again, remember, we made the assumption that the limit as t goes to infinity of t e to power minus s t must be zero. And since this is true, it's necessary that the limit as t goes to infinity of e to power minus s t must also be zero. No, no, no, no. Don't trust b. This is something you should be able to prove. In any case, this value is going to go to zero, leaving one over s squared. Now, this example illustrates a useful property. So remember, our Laplace transform is based on the improper integral of e to power minus s x times f of x, but we have to require that the limit as x goes to infinity of our integrand is equal to zero. And the question you might ask yourself is self, how often does that happen? And if the universe were a kind and gentle place, this would happen all the time. Unfortunately, the universe is not a kind and gentle place. And yet, it's still true that for most functions, this limit will be zero. So every now and then, the universe makes things easy for us. Don't count on it. So suppose our nth derivative of f is such a function, then we can find the Laplace transform. So again, we'll use integration by parts, but unlike in the previous example, where we knew that if we differentiated f enough times, it would disappear. In this case, we want to anti-differentiate the nth derivative, because if we anti-differentiate it enough times, we'll get the function. So we'll let u equals e to power minus s x. dv will be the nth derivative, dx, and so we can find v by anti-differentiation and du by differentiation. And so our integral becomes uv minus the integral of vdu. And we'll clean that algebra up a little bit. There's no need to carry around a minus minus s. So now let's evaluate. The improper integral can be evaluated by taking the limit as t goes to infinity. Remember, the important assumption is that the integrand is going to go to zero as t goes to infinity. Meanwhile, this part doesn't depend on t at all, so it's going to remain. And so I just get minus the n minus first derivative at zero. But wait, you say, this improper integral still exists, so we haven't really evaluated it. Well, since s is a constant, we can at least remove that constant factor. Definitions are the whole of mathematics. All else is commentary. This improper integral here is the Laplace transform of f n minus one. And so we can rewrite our expression. And this gives us a very useful result, provided that it exists. The Laplace transform of an nth derivative is s times the Laplace transform of the n minus first derivative minus the n minus first derivative at zero. Now this is an amazingly useful property, because this allows us to express the Laplace transform of any nth derivative in terms of the initial values of the lower order derivatives and the Laplace transform of the function. How does that work, you ask? Well, let's take a look. Let's find the Laplace transform of the third derivative of a function. So our theorem says that the Laplace transform of an nth derivative is s times the Laplace transform of the n minus first derivative minus the n minus first derivative at zero. So the Laplace transform of that third derivative is s times the Laplace transform of the second derivative minus the second derivative at zero. But wait, we can go further. We have this Laplace transform of the second derivative. But our theorem says that the Laplace transform of the second derivative is s times the Laplace transform of the first derivative minus the first derivative at zero. But wait, we have the Laplace transform of the first derivative. And our theorem says that the Laplace transform of the first derivative is s times the Laplace transform of the zeroth derivative, the function itself, minus the function at zero. But wait, we have the Laplace transform of the function. Well, actually, we can't do anything else with that. However, we can expand out our products and get... So a useful mantra in mathematics is lather, rinse, repeat, and through this process we have the following general result. Suppose f is a suitable function. In other words, it's a function that we could find the Laplace transform of. Then the Laplace transform of the nth derivative is going to be given by a nice formula that relies only on the Laplace transform of the function itself, along with the values of the nth derivative at zero. We'll see what we can do with this result next.