Laplace as linear operator and Laplace of derivatives

Well, now's as good a time as any to go over some interesting and very useful properties of the Laplace transform. And the first is to show that it is a linear operator. And what does that mean? Well, let's say I wanted to take the Laplace transform of the sum of the-- we call it the weighted sum of two functions. So say some constant, c1, times my first function, f of t, plus some constant, c2, times my second function, g of t. Well, by the definition of the Laplace transform, this would be equal to the improper integral from 0 to infinity of e to the minus st, times whatever our function that we're taking the Laplace transform of, so times c1, f of t, plus c2, g of t-- I think you know where this is going-- all of that dt. And then that is equal to the integral from 0 to infinity. Let's just distribute the e the minus st. That is equal to what? That is equal to c1e to the minus st, f of t, plus c2e to the minus st, g of t, and all of that times dt. And just by the definition of how the properties of integrals work, we know that we can split this up into two integrals, right? If the integral of the sum of two functions is equal to the sum of their integrals. And these are just constant. So this is going to be equal to c1 times the integral from 0 to infinity of e to the minus st, times f of t, d of t, plus c2 times the integral from 0 to infinity of e to the minus st, g of t, dt. And this was just a very long-winded way of saying, what is this? This is the Laplace transform of f of t. This is the Laplace transform of g of t. So this is equal to c1 times the Laplace transform of f of t, plus c2 times-- this is the Laplace transform-- the Laplace transform of g of t. And so, we have just shown that the Laplace transform is a linear operator, right? The Laplace transform of this is equal to this. So essentially, you can kind of break up the sum and take out the constants, and just take the Laplace transform. That's something useful to know, and you might have guessed that was the case anyway. But now you know for sure. Now we'll do something which I consider even more interesting. And this is actually going to be a big clue as to why Laplace transforms are extremely useful for solving differential equations. So let's say I want to find the Laplace transform of f prime of t. So I have some f of t, I take its derivative, and then I want the Laplace transform of that. Let's see if we can find a relationship between the Laplace transform of the derivative of a function, and the Laplace transform of the function. So we're going to use some integration by parts here. Let me just say what this is, first of all. This is equal to the integral from 0 to infinity of e to the minus st, times f prime of t, dt. And to solve this, we're going to use integration by parts. Let me write it in the corner, just so you remember what it is. So I think I memorized it, because I recorded that last video not too long ago. I'm just going to write this shorthand. The integral of u-- well, let's say uv prime, because that will match what we have up here better-- is equal to both functions without the derivitives, uv minus the integral of the opposite. So the opposite is u prime v. So here, the substitution is pretty clear, right? Because we want to end up with f of x, right? So let's make v prime is f prime, and let's make u e to the minus st. So let's do that. u is going to be e to the minus st, and v is going to equal what? v is going to equal f prime of t. And then u prime would be minus se to the minus st. And then, v prime-- oh, sorry, this is v prime, right? v prime is f prime of t, so v is just going to be equal to f of t. I hope I didn't say that wrong the first time. But you see what I'm saying. This is u, that's u, and this is v prime. And if this is v prime, then if you were to take the antiderivative of both sides, then v is equal to f of t. So let's apply integration by parts. So this Laplace transform, which is this, is equal to uv, which is equal to e to the minus st, times v, f of t, minus the integral-- and, of course, we're going to have to evaluate this from 0 to infinity. I'll keep the improper integral with us the whole time. I won't switch back and forth between the definite and indefinite integral. So minus this part. So the integral from 0 to infinity of u prime. u prime is minus se to the minus st times v-- v is f of t-- dt. Now, let's see. We have a minus and a minus, let's make both of these pluses. This s is just a constant, so we can bring it out. So that is equal to e to the minus st, f of t, evaluated from 0 to infinity, or as we approach infinity, plus s times the integral from 0 to infinity of e to the minus st, f of t, dt. And here, we see, what is this? This is the Laplace transform of f of t, right? Let's evaluate this part. So when we evaluated in infinity, as we approach infinity, e to the minus infinity approaches 0. f of infinity-- now this is an interesting question. f of infinity-- I don't know. That could be large, that could be small, that approaches some value, right? This approach 0, so we're not sure. If this increases faster than this approaches 0, then this will diverge. I won't go into the mathematics of whether this converges or diverges, but let's just say, in very rough terms, that this will converge to 0 if f of t grows slower than e to the minus st shrinks. And maybe later on we'll do some more rigorous definitions of under what conditions will this expression actually converge. But let's assume that f of t grows slower than e to the st, or it diverges slower than this converges, is another way to view it. Or this grows slower than this shrinks. So if this grows slower than this shrinks, then this whole expression will approach 0. And then you want to subtract this whole expression evaluated at 0. So e to the 0 is 1 times f of 0-- so that's just f of 0-- plus s times-- we said, this is the Laplace transform of f of t, that's our definition-- so the Laplace transform of f of t. And now we have an interesting property. What was the left-hand side of everything we were doing? The Laplace transform of f prime of t. So let me just write all over again. And I'll switch colors. The Laplace transform of f prime of t is equal to s times the Laplace transform of f of t minus f of 0. And now, let's just extend this further. What is the Laplace transform-- and this is a really useful thing to know-- what is the Laplace transform of f prime prime of t? Well, we can do a little pattern matching here, right? That's going to be s times the Laplace transform of its antiderivative, times the Laplace transform of f prime of t, right? This goes to this, that's an antiderivative. This goes to this, that's one antiderivative. Minus f prime of 0, right? But then what's the Laplace transform of this? This is going to be equal to s times the Laplace transform of f prime of t, but what's that? That's this, right? That's s times the Laplace transform of f of t, minus f of 0, right? I just substituted this with this. Minus f prime of 0. And we get the Laplace transform of the second derivative is equal to s squared times the Laplace transform of our function, f of t, minus s times f of 0, minus f prime of 0. And I think you're starting to see a pattern here. This is the Laplace transform of f prime prime of t. And I think you're starting to see why the Laplace transform is useful. It turns derivatives into multiplications by f. And actually, as you'll see later, it turns integration to divisions by s. And you can take arbitrary derivatives and just keep multiplying by s. And you see this pattern. And I'm running out of time. But I'll leave it up to you to figure out what the Laplace transform of the third derivative of f is. See you in the next video.