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# Laplace/Step Function Differential Equation: Hairy differential equation involving a step function that we use the Laplace Transform to solve.

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- Let's apply everything we've learned to an actual
- differential equation.
- Instead of just taking Laplace transforms and taking their
- inverse, let's actually solve a problem.
- So let's say that I have the second derivative of my
- function y plus 4 times my function y is equal to sine of
- t minus the unit step function 0 up until 2 pi of t times
- sine of t minus 2 pi.
- Let's solve this differential equation, an
- interpretation of it.
- And I actually do a whole playlist on interpretations of
- differential equations and how you model it, but you know,
- you can kind of view this is a forcing function.
- That it's a weird forcing function of this being applied
- to some weight with, you know, this is the
- acceleration term, right?
- The second derivative with respect to time is the
- acceleration.
- So the mass would be 1 whatever units, and then as a
- function of its position, this is probably some type of
- spring constant.
- Anyway, I won't go there.
- I don't want to waste your time with the interpretation
- of it, but let's solve it.
- We can do more about interpretations later.
- So we're going take the Laplace transform of both
- sides of this equation.
- So what's the Laplace transform of
- the left-hand side?
- So the Laplace transform of the second derivative of y is
- just s squared, so now I'm taking the Laplace transform
- of just that.
- The Laplace transform of s squared times the Laplace
- transform of y minus-- lower the degree there once-- minus
- s times y of 0 minus y prime of 0.
- So clearly, I must have to give you some initial
- conditions in order to do this properly.
- And then plus 4 times the Laplace transform of y is
- equal to-- what's the Laplace transform of sine of t?
- That should be second nature by now.
- It's just 1 over s squared plus 1.
- And then we have minus the Laplace
- transform of this thing.
- And I'll do a little side note here to figure out the Laplace
- transform of this thing right here.
- And we know, I showed it to you a couple of videos ago, we
- showed that the Laplace transform-- actually I could
- just write it out here.
- This is going to be the same thing as the Laplace transform
- of sine of t, but we're going to have to multiply it by e to
- the minus-- if you remember that last formula-- e to the
- minus cs, where c is 2 pi.
- Actually, let me write that down.
- I decided to write it down, then I decided, oh, no, I
- don't want to do this.
- But let me write that.
- So the Laplace transform of the unit step function that
- goes up to c times some function shifted by c is equal
- to e to the minus cs times the Laplace transform of just the
- original function times the Laplace transform of f of t.
- So if we're taking the Laplace transform of this
- thing, our c is 2 pi.
- Our f of t is just sine of t, right?
- So then this is just going to be equal to-- if we just do
- this piece right here-- it's going to be equal to e to the
- minus cs-- our c is 2 pi-- e to the minus 2 pi s times the
- Laplace transform of f of t. f of t is just sine of t before
- we shifted.
- This is f of t minus 2 pi.
- So f of t is just going to be sine of t.
- So it's going to be times 1 over s squared plus 1.
- This is the Laplace transform of sine of t.
- So let's go back to where we had left off.
- So we've taken the Laplace transform of both sides of
- this equation.
- And clearly, I have some initial conditions here, so
- the problem must have given me some and I just forgot to
- write them down.
- So let's see, the initial conditions I'm given, and they
- are written kind of in the margin here, they tell us--
- I'll do it in orange, they tell us that y of 0 is equal
- to 0, and y prime of 0 is equal to 0.
- That makes the math easy.
- That's 0 and that's 0.
- So let's see if I can simplify my equation.
- So the left-hand side, let's factor
- out the Laplace transform.
- So let's factor out this term and that term.
- So we get the Laplace transform of y times this plus
- this times s squared plus 4 is equal to the right-hand side.
- And what's the right-hand side?
- We could simplify this.
- Well, I'll just write it out.
- I don't want to do too many steps at once.
- It's 1 over s squared plus 1 and then plus-- or minus
- actually, this is a minus-- minus the Laplace transfer of
- this thing, which was e to the minus 2 pi s over s
- squared plus 1.
- So if we divide both sides of this equation by the s squared
- plus 4, then we get the Laplace transform of y is
- equal to-- and actually, I can just merge these two.
- They're the same denominator.
- So before I even divide by s squared plus 4, that
- right-hand side will look like this.
- It will look like with a denominator of s squared plus
- 1 and you have a numerator of 1 minus e to the minus 2 pi s.
- And, of course, we're dividing both sides of this equation by
- s squared plus 4, so we're going to have to stick that s
- squared plus 4 over here.
- Now, we're at the hard part.
- In order to figure out why, we have to take the inverse
- Laplace transform of this thing.
- So how do we take the inverse Laplace
- transform of this thing?
- That's where the hard part is always, you know, it makes
- solving the differential equation's easy if you know
- the Laplace transforms. So it looks like we're going to have
- to do some partial fraction expansion.
- So let's see if we can do that.
- So we can rewrite this equation right here.
- Actually, let's write it as this, because this'll kind of
- simplify our work.
- Let's factor this whole thing out.
- So we're going to write it as 1 minus e to the minus 2 pi s,
- all of that times-- I'll do it in orange-- all of that times
- 1 over s squared plus 1 times s squared plus 4.
- Now, we need to do some partial fraction expansion to
- simplify this thing right here.
- We're going to do this on the side.
- Maybe I should do this over on the right here.
- This thing-- let me rewrite it-- 1 over s squared plus 1
- times s squared plus 4 should be able to be rewritten as two
- separate fractions, s squared plus 1 and s squared plus 4,
- with the numerators.
- This one would be As plus B.
- It's going to have to have degree 1, because
- this is degree 2.
- Here And then we'd have Cs plus D.
- And so when you add these two things up, you get As plus B
- times s squared plus 4 plus Cs plus D times s squared plus 1,
- all of that over the common denominator.
- We've seen this story before.
- We just have to do some algebra here.
- As you can tell, these differential equations
- problems, they require a lot of stamina.
- You kind of just have to say I will keep moving forward and
- do the algebra that I need to do in order to get the answer.
- And you kind of have to get excited about that notion that
- you have all this algebra to do.
- So let's figure it out.
- So this top can be simplified to As to the third plus Bs
- squared plus 4As plus 4B.
- And then this one, you end up with Cs to the third plus Ds
- squared plus Cs plus D.
- So when you add of these up together, you get-- and this
- is all the algebra that we have to do, for better, for
- worse-- A plus C over s to the third plus B plus D times s
- squared plus 4A plus C times s-- let's scroll over a little
- bit-- plus 4B plus D.
- And now we just have to say, OK, all of this is equal to
- this thing up here.
- This is the numerator.
- We just simplified the numerator.
- This is the numerator.
- That's the numerator right there.
- And all of this is going to be over your original s squared
- plus 1 times your s squared plus 4.
- And we established that this thing should be-- let me just
- write this-- that 1 over s squared plus 1 times s squared
- plus 4 should equal this thing.
- And then you just pattern match on the coefficients.
- This is all just intense partial fraction expansion.
- And you say, look, A plus C is the coefficient of the s cubed
- terms. I don't see any s cubed terms here.
- So A plus C must be equal to 0.
- And then you see, OK, B plus D is the coefficient of the s
- squared terms. I don't see any s squared terms there.
- So B plus D must be equal to 0.
- 4A plus C, the coefficient of the s terms. I don't see any s
- terms over here.
- So 4A plus C must be equal to 0.
- And then we're almost done.
- 4B plus D must be the constant terms. There is a constant
- term there.
- So 4B plus D is equal to 1.
- So let's see if we can do anything here.
- If we subtract this from that, we get minus 3A is equal to 0,
- or A is equal to 0.
- If A is equal to 0, then C is equals to 0.
- And let's see what we can get here.
- If we subtract this from that, we get minus 3B.
- The D's cancel out.
- It's equal to minus 1, or B is equal to 1/3.
- And then, of course, we have D is equal to minus B, if you
- subtract B from both sides. so D is equal to 1/3.
- So all of that work, and we actually have a
- pretty simple result.
- Our equation, this thing here, can be rewritten as-- the A
- disappeared.
- It's 1/3 over s squared plus 1.
- B was the coefficient on the-- let me make it very clear.
- B was the coefficient on the-- or it was a term on top of the
- s squared plus 1, so that's why I'm using B there.
- And then D is minus B, so D is minus 1.
- So let me make sure I have that.
- B is 1/3 minus-- let me make sure I get that right.
- D is 1/3.
- So, sorry, B as in boy is 1/3, so D is minus 1/3.
- So B, there's a term on top of the s squared plus 1.
- And then you have minus D over the minus 1/3 over s
- squared plus 4.
- This takes a lot of stamina to record this video.
- I hope you appreciate it.
- OK, so let me rewrite everything, just so we can get
- back to the problem because when you take the partial
- fraction detour, you forget-- not even to speak of the
- problem, you forget what day it is.
- Let's see, so you get the Laplace transform of y is
- equal to 1 minus e to the minus 2 pi s times what that
- mess that we just solved for, times-- and I'll
- write it like this.
- 1/3 times 1 over s squared plus 1 minus 1/3 times--
- actually, let me write it this way.
- Because I have this s squared plus 4, so I really want to
- have a 2 there.
- So I want to have a 2 in the numerator, so you want to have
- a 2 over s squared plus 4.
- So if I put a 2 in the numerator there, I have to
- divide this by 2 as well.
- So let me change this to a 6.
- Minus 1/6 times 2 is minus 1/3.
- So I did that just so I get this in the form of the
- Laplace transform of sine of t.
- Now, let's see if there's anything that
- I can do from here.
- This is an epic problem.
- I'll be amazed if I don't make a careless
- mistake while I do this.
- So we can rewrite everything.
- Let's see if we can simplify this.
- And by simplifying it, I'm just going to make it longer.
- We can write the Laplace transform of y is equal to--
- I'm just going to multiply the 1 out, and then I'm going to
- multiply the e to the minus 2 pi s out.
- So if you multiply the 1 out, you get 1/3 times 1 over s
- squared plus 1-- I'm just multiplying the 1 out-- minus
- 1/6-- these are all the 1's times the 1-- times 2 over s
- squared plus 4.
- And then I'm going to multiply the minus e.
- Let me just switch colors, do the minus e.
- So then you get minus e to the minus 2 pi s over 3 times 1
- over s squared plus 1.
- And then the minus and the minus cancel out, so you get
- plus e to the minus 2 pi s over 6 times 2 over s
- squared plus 4.
- Now, taking the inverse Laplace transform of these
- things are pretty straightforward.
- So let's do that.
- Let's take the inverse Laplace transform of the whole thing.
- And we get y is equal to the inverse Laplace transform of
- this guy right here, is just 1/3 sine of t-- I don't have
- to write a parentheses there-- sine of t, and then this is
- minus 1/6 times-- this is the Laplace
- transform of sine of 2t.
- That's that term right there.
- Now, these are almost the same, but we have this little
- pesky character over here.
- We have this e to the minus 2 pi s.
- And there, we just have to remind ourselves-- I'll write
- it here in the bottom.
- We just have to remind ourselves that the Laplace
- transform of the unit step function-- I'll put the pi
- there, just 2 pi times f of t minus 2 pi-- I should put as
- the step function of t-- is equal to e to the minus 2 pi s
- times the Laplace transform of just-- or let me just write it
- this way-- times the Laplace transform of f of t.
- So if we view f of t as just sine of t or sine of 2t, then
- we can kind of backwards pattern match.
- And we'll have to shift it and multiply it by
- the unit step function.
- So I want to make that clear.
- If you didn't have this guy here, the inverse Laplace
- transform of this guy would be the same thing as this guy.
- It'd just be sine of t.
- The inverse Laplace transform of this guy
- would be sine of 2t.
- But we have this pesky character here, which
- essentially, instead of having the inverse Laplace transform
- just being our f of t, it's going to be our f of t shifted
- by 2 pi times the unit step function, where
- it steps up at 2pi.
- So this is going to be minus 1/3 times the unit step
- function, where c is 2 pi of t times-- instead of sine of t--
- sine of t minus 2pi.
- And then we're almost done.
- I'll do it in magenta to celebrate it.
- Plus this very last term, which is 1/6 times the unit
- step function 2 pi of t, the unit step function that steps
- up at 2 pi times sine of-- and we have to be careful here.
- Wherever we had a t before, we're going to replace it with
- a t minus 2 pi.
- So sine of, instead of 2t, is going to be 2
- times t minus 2 pi.
- And there you have it.
- We finally have solved our very hairy problem.
- We could take some time if we want to simplify
- this a little bit.
- In fact, we might as well.
- At the risk of making a careless mistake at the last
- moment, let me see if I can make any simplifications here.
- Well, we could factor out this guy right here, but other than
- that, that seems about as simple as we can get.
- So this is our function of t that satisfies our otherwise
- simple-looking differential equation that we had up here.
- This looked fairly straightforward, but we got
- this big mess to actually satisfy that equation, given
- those initial conditions that we had initially.

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