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- 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.

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