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##### Laplace Transform solves an equation 2 : Second part of using the Laplace Transform to solve a differential equation.

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- Welcome back.
- We're finally using the Laplace Transform to do
- something useful.
- In the first part of this problem, we just had this
- fairly straightforward differential equation.
- And I know it's a little bit frustrating right now, because
- you're like, this is such an easy one to solve using the
- characteristic equation.
- Why are we doing Laplace Transforms?
- Well I just want to show you that they can solve even these
- problems. But later on there are going to be classes of
- problems that, frankly, our traditional methods aren't as
- good as the Laplace Transform.
- But anyway, how did we solve this?
- We just took the Laplace Transform of both sides of
- this equation.
- We got all of this hairy mess.
- We used the property of the derivative of functions, where
- you take the Laplace Transform, and we ended up,
- after doing a lot of algebra essentially, we got this.
- We got the Laplace Transform of y is equal to this thing.
- We just took the Laplace Transform of both sides and
- manipulated algebraically.
- So now our task in this video is to figure out what y's
- Laplace Transform is this thing?
- And essentially what we're trying to do, is we're trying
- to take the inverse Laplace Transform of both sides of
- this equation.
- So another way to say it, we could say that y-- if we take
- the inverse Laplace Transform of both sides-- we could say
- that y is equal to the inverse Laplace
- Transform of this thing.
- 2s plus 13, over s squared plus 5s plus 6.
- Now we'll eventually actually learn the formal definition of
- the inverse Laplace Transform.
- How do you go from the s domain to the t domain?
- Or how do you go from the frequency
- domain to the time domain?
- We're not going to worry about that right now.
- What we're going to do is we're going to get this into a
- form that we recognize, and say, oh,
- I know those functions.
- That's the Laplace Transform of whatever and whatever.
- And then we'll know what y is.
- So let's try to do that.
- So what we're going to use is something that you probably
- haven't used since Algebra two, which is I think when
- it's taught in, you know, eighth, ninth,
- or 10th grade, depending.
- And you finally see it now in differential equations that it
- actually has some use.
- Let me write it.
- We're going to use partial fraction expansion.
- And I'll do a little primer on that, in case you don't
- remember it.
- So anyway, let's just factor the bottom part right here.
- And you'll see where I'm going with this.
- So if I factor the bottom, I get s plus 2 times s plus 3.
- And what we want to do, is we want to rewrite this fraction
- as the sum of 2-- I guess you could
- call it partial fractions.
- I think that's why it's called partial fraction expansion.
- So we want to write this as a sum of A over s plus 2, plus B
- over s plus 3.
- And if we can do this, then-- and bells might already be
- ringing in your head-- we know that these things that look
- like this are the Laplace Transform of functions that
- we've already solved for.
- And I'll do a little review on that in a second.
- But anyway, how do we figure out A and B?
- Well if we were to actually add A and B, if we were to--
- let's do a little aside right here-- so if we said that A--
- so if we were to give them a common denominator, which is
- this, s plus 2 times s plus 3.
- Then what would A become?
- We'd have to multiply A times s plus 3, right?
- So we'd get As plus 3A.
- This, as I've written it right now, is the same thing as A
- over s plus 2.
- You could cancel out an s plus 3 in the top and the bottom.
- And now we're going to add the B to it.
- So plus-- I'll do that in a different color-- plus-- well,
- if we have this as the denominator, we could multiply
- the numerator and the denominator
- by s plus 2, right?
- To get B times s, plus 2B, and that's going
- to equal this thing.
- And all I did is I added these two fractions.
- Nothing fancier than there.
- That was Algebra two.
- Actually, I think I should do an actual
- video on that as well.
- But that's going to equal this thing.
- 2s plus 13, all of that over s plus 2 times s plus 3.
- Notice in all differential equations, the hairiest part's
- always the algebra.
- So now what we do is we match up.
- We say, well, let's add the s terms here.
- And we could say that the numerators have to equal each
- other, because the denominators are equal.
- So we have A plus Bs plus 3A plus 2B is equal to 2s plus B.
- So the coefficient on s, on the right-hand side, is 2.
- The coefficient on the left-hand side is A plus B, so
- we know that A plus B is equal to 2.
- And then on the right-hand side, we see 3A plus 2B must
- be equal to-- oh, this is a 13.
- Did I say B?
- This is a 13.
- That's a 13.
- It looks just like a B, right?
- That was 2s plus 13.
- Anyway, so on the right-hand side I get, it was 3A plus 2B
- is equal to 13.
- Now we have two equations with two unknowns,
- and what do we get?
- I know this is very tiresome, but it'll be
- satisfying in the end.
- Because you'll actually solve something
- with the Laplace Transform.
- So let's multiply the top equation by 2, or let's just
- say minus 2.
- So we get minus 2A minus 2B equals minus 4.
- And then we get-- add the two equations-- you get A is equal
- to-- these cancel out-- A is equal to 9.
- Great.
- If A is equal to 9, what is B equal to?
- B is equal to 9 plus what is equal to 2?
- Or 2 minus 9 is minus 7.
- And we have done some serious simplification.
- Because now we can rewrite this whole expression as the
- Laplace Transform of y is equal to A over s plus 2, is
- equal to 9 over s plus 2, minus 7 over s plus 3.
- Or another way of writing it, we could write it as equal to
- 9 times 1 over s plus 2, minus 7 times 1 over s plus 3.
- Why did I take the trouble to do this?
- Well hopefully, you'll recognize this was actually
- the second Laplace Transform we figured out.
- What was that?
- I'll write it down here just so you remember it.
- It was the Laplace Transform of e to the at, was equal to 1
- over s minus a.
- That was the second Laplace Transform we figured out.
- So this is interesting.
- This is the Laplace Transform of what?
- So if we were to take the inverse Laplace Transform--
- actually let me just stay consistent.
- So that means that this is the Laplace Transform of y, is
- equal to 9 times the Laplace Transform of what?
- If we just do pattern matching, if this is s minus
- a, then a is minus 2.
- So 9 times the Laplace Transform of e
- to the minus 2t.
- Does that make sense?
- Take this, put it in this one, which we figured out, and you
- get 1 over s plus 2.
- And let me clean this up a little bit, because I'm going
- to need that real estate.
- I'll write this.
- I'll leave that there, because we'll still use that.
- And then we have minus 7 times-- this is the Laplace
- Transform of what?
- This is the Laplace Transform of e to the minus 3t.
- This pattern matching, you're like, wow, if you saw this,
- you would go to your Laplace Transform table, if you didn't
- remember it, you'd see this.
- You're like, wow, that looks a lot like that.
- I just have to figure out what a is.
- I have s plus 3.
- I have s minus a.
- So in this case, a is equal to minus 3.
- So if a is equal to minus 3, this is the Laplace Transform
- of e to the minus 3t.
- So now we can take the inverse Laplace-- actually,
- before we do that.
- We know that because the Laplace Transform is a linear
- operator-- and actually now I can delete this down here-- we
- know that the Laplace Transform is a linear
- operator, so we can write this.
- And you normally wouldn't go through all of these steps.
- I just really want to make you understand what we're doing.
- So we could say that this is the same thing as the Laplace
- Transform of 9e to the minus 2t, minus 7e to the minus 3t.
- Now we have something interesting.
- The Laplace Transform of y is equal to the Laplace
- Transform of this.
- Well if that's the case, then y must be equal to 9e to the
- minus 2t, minus 7e to the minus 3t.
- And I never proved to you, but the Laplace Transform is
- actually a 1:1 Transformation.
- That if a function's Laplace Transform, if I take a
- function against the Laplace Transform, and then if I were
- take the inverse Laplace Transform, the only function
- whose Laplace Transform that that is, is
- that original function.
- It's not like two different functions can have the same
- Laplace Transform.
- Anyway, a couple of things to think about here.
- Notice, we had that thing that kind of looked like a
- characteristic equation pop up here and there.
- And we still have to solve a system of two equations with
- two unknowns.
- Those are both things that we had to do when we solve an
- initial value problem, when we use just traditional, the
- characteristic equation.
- But here it happened all at once.
- And frankly it was a little bit hairier because we had to
- do all this partial fraction expansion.
- But it's pretty neat.
- The Laplace Transform got us something useful.
- In the next video I'll actually do a non-homogeneous
- equation, and show you that the Laplace Transform applies
- equally well there.
- So it's kind of a more consistent theory of solving
- differential equations, instead of kind of guessing
- solutions, and solving for coefficients and all of that.
- See you in the next video.