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# Using the Laplace Transform to solve a nonhomogenous eq : Solving a non-homogeneous differential equation using the Laplace Transform

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- It's been over a year since I last did a video with the
- differential equations playlist, and I thought I
- would start kicking up, making a couple of videos.
- And I think where I left, I said that I would do a
- non-homogenous linear equation using the Laplace Transform.
- So let's do one, as a bit of a warm-up, now that we've had
- a-- or at least I've had a one-year hiatus.
- Maybe you're watching these continuously, so you're
- probably more warmed up than I am.
- So if we have the equation the second derivative of y plus y
- is equal to sine of 2t.
- And we're given some initial conditions here.
- The initial conditions are y of 0 is equal to 2, and y
- prime of 0 is equal to 1.
- And where we left off-- and now you
- probably remember this.
- You probably recently watched the last video.
- To solve these, we just take the Laplace
- Transforms of all the sides.
- We solve for the Laplace Transform of the function.
- Then we take the inverse Laplace Transform.
- If that doesn't make sense, then let's just do it in this
- video, and hopefully the example
- will clarify all confusion.
- So in the last video-- it was either the last one or the
- previous one-- I showed you that the Laplace Transform of
- the second derivative of y is equal to s squared times the
- Laplace Transform of y-- and we keep lowering the degree on
- s-- so minus s times y of 0.
- You can kind of think of it as taking the derivative.
- This is an integral.
- It's not exactly the anti-derivative of this.
- But the Laplace Transform it is an integral.
- The Transform is an integral.
- So y of 0 is kind of one derivative away from that.
- And then minus y prime of 0.
- And then we could also rewrite this.
- And this is just a purely notational issue.
- I could write this, instead of writing the Laplace Transform
- of y all the time, I could write this as s squared times
- capital Y of s-- because this is going to be a function of
- s, not a function of y-- minus s times y of 0 minus
- y prime y of 0.
- These are going to be numbers, right?
- These aren't functions.
- These are the function evaluated at 0, or the
- derivative of the function evaluated at 0.
- And we know what these values are.
- y of 0, right here, is 2, and y prime of 0 is 1.
- It was given to us.
- So if we take the Laplace Transforms of both sides of
- this equation, first we're going to want to take the
- Laplace Transform of this term right there, which we've
- really just done.
- The Laplace Transform of the second derivative is s squared
- times the Laplace Transform of the function, which we write
- as capital Y of s, minus this, minus 2s-- they gave us that
- initial condition-- and then minus 1.
- Right?
- This term right here is just 1, so minus 1.
- So that's term right there.
- Then we want to take the Laplace
- Transform of y by itself.
- So this is just plus Y of s, right, the Laplace
- Transform of y.
- So I'll just rewrite Laplace Transform of y.
- I'm just rewriting it in this notation.
- Y of S.
- It's good to get used to either one.
- This is going to be equal to the Laplace
- Transform of sine of 2t.
- And I showed you in a video last year that we showed what
- the Laplace Transform of sine of at is, but I'll write it
- down here just so you remember it.
- Laplace Transform of the sine of at is equal to a over s
- squared plus a squared.
- So the Laplace Transform of sine of 2t.
- Here, a is 2.
- This is going to be 2 over s squared plus 4.
- So if we take the Laplace Transform of both sides of
- this, the right-hand side is going to be 2 over s
- squared plus 4.
- Now what we can do is we can separate out all the Y of s
- terms. And so we can factor, well I guess we could say,
- factor out their coefficients, so that's a Y of s term,
- that's a Y of s term.
- And so we could write the left-hand side here as s
- squared-- that's that term-- plus 1-- the coefficient on
- that term-- s squared plus 1, times Y of s.
- Let me do it in green.
- So this is Y of s, and this Y of s, times Y of s, and then
- we have the non-Y of s terms. These two right here.
- So minus 2s, minus 1, is equal to 2 over s squared plus 4.
- We can add 2s plus 1 to both sides, to essentially move
- this to the right-hand side, and we're left with s squared
- plus 1, times Y of s, is equal to 2 over s squared plus 4,
- plus 2s, plus 1.
- Now we can divide both sides of this equation by s squared
- plus 1, and we get the Laplace Transform of Y.
- Y of s is equal to-- let me switch colors-- it's equal to
- 2 over s squared plus 4 times this thing right here.
- I'm dividing both sides of this equation by this term
- right there.
- So times s squared plus 1-- it's in the denominator so I'm
- dividing by it-- plus 2s plus 1-- I have to divide both of
- those terms by the s squared plus 1-- divided by s squared
- plus 1, divided by s squared plus 1.
- Now, in order to be able to take the inverse Laplace
- Transform of this, I need to get it in some type of simple
- fraction form.
- These are actually easier to do, but this was one's a
- little bit difficult.
- I want to do some partial fraction decomposition to
- break this up into maybe simpler fractions.
- So what I want to do, I'm going to do a little bit of an
- aside here.
- And this really is the hardest part of these problems. The
- algebra, breaking this thing up.
- So I'm going to break this up.
- So let me write this this way.
- 2 over s squared plus 4 times s squared plus 1.
- I'm going to break this up into two fractions.
- This is the partial fraction decomposition.
- One fraction is s squared plus 4.
- And the other fraction is s squared plus 1.
- And since both of these denominators are of degree 2,
- the numerators are going to be of degree 1.
- So they're going to be some-- let me write it this way--
- this one will be As plus B.
- And then this one will be Cs plus D.
- This is just pure algebra.
- This is just partial fraction decomposition.
- I've made a couple of videos on it.
- And I'm saying that I'm assuming that this expression
- right here can be broken up into two
- expressions of this form.
- And I need to solve for A, B, C, and D.
- So let's see how we can do that.
- So if I were to start with these two and add them up,
- what do I get?
- I would have to multiply these times-- so my denominator, my
- common denominator, would be this thing again-- it would be
- s squared plus 4 times s squared plus 1.
- And now I'm going to have to multiply the As plus B times
- this s squared plus 1.
- This, as it is right now, these two terms
- would cancel out.
- You'll just get this term, but I need to add it
- to this right here.
- So you get plus Cs plus D times s squared plus 4.
- And now let's see what we could do to match up the terms
- here with this number 2 right here.
- So let's multiply all of this out.
- So As times s squared is As to the 3rd.
- As times 1 is plus As.
- B times s squared, so plus Bs squared, and then you have B
- times 1 is plus B.
- And then you have Cs times s squared, that's Cs to the 3rd.
- And then Cs times 4, so it's plus 4Cs.
- These problems are tiring.
- And I also have a cold, so this is especially tiring, but
- I'll soldier forward.
- Where was I?
- So I multiplied the C's times each of these, now I have to
- multiply the D's.
- So plus Ds squared-- that's D times that one-- plus D times
- 4, so plus 4D.
- So that's all of them.
- And I just wrote it this way so I have the common degree
- terms under each other.
- So if I were to add the entire numerator, I get-- and I'll
- just switch colors, somewhat arbitrarily-- I get A plus C
- times s to the 3rd plus-- let me write the s squared term
- next-- plus B plus D times s squared-- now I'll write this
- s term-- plus A plus 4C times s plus B plus 4D.
- This is just the numerator.
- This is when I just added these two things up.
- This whole thing up here simplifies to this.
- I don't know if the word simplify is appropriate.
- But it becomes this expression right here.
- And that's just the numerator.
- The denominator is still what we had written before.
- The denominator is still the s squared plus 4, times the s
- squared plus 1.
- Of course, I have to show that this is a fraction.
- And this is going to be equal to this thing over here.
- 2 over s squared plus 4 times s squared plus 1.
- Now, why did I go through this whole mess right here?
- Well, the reason why I went through it is because we
- should be able to solve for A, B, C, and D.
- So let's see, A plus C.
- This is the coefficient on the s cubed term.
- Do we see any s cubed terms here?
- No, we see no s cubed terms here.
- So A plus C-- let me write this down-- A plus C must be
- equal to 0, because we see nothing here that
- has an s to the third.
- B plus D is a coefficient on the s squared term.
- Do we see any s squared terms here?
- No, so B plus D must be equal to 0.
- A plus 4C are the coefficient of the s term.
- I see no s term over here.
- So A plus 4C must also be equal to 0.
- And then finally, we look at just the constant terms. And
- we do have a constant term on the left-hand
- side of this equation.
- We have 2.
- so B plus 4D-- I didn't want to make it that thick-- B plus
- 4D must be equal to 2.
- This just seems like these linear equations are pretty
- easy to solve for.
- Let's subtract this from this.
- So A-- or let me subtract the bottom one from the top one--
- so A minus A, that's 0A.
- And then C minus 4C minus 3C is equal to 0.
- And so you get C is equal to 0.
- If C is equal to 0, A plus C is equals to 0, A must be
- equal to 0.
- And let's do the same thing here.
- Let's subtract this from that.
- So you get B minus B is 0, and then minus 3D-- that's just D
- minus 4D-- and then 0 minus 2 is equal to minus 2.
- And then you get D is equal to-- what do we get?-- D is
- equal to 2/3.
- Minus 2 divided by minus 3 is 2/3, and then-- this isn't a
- minus here, I wrote that there later-- we said B plus D is
- equal to 0.
- So B must be the opposite of D, right?
- We could write B is equal to minus D, or B is
- equal to minus 2/3.
- Let's remember all of this and go back to our original
- problem, because we've kind of-- actually
- let me just be clear.
- We can rewrite 2 over s squared plus 4 times s
- squared plus 1.
- We can rewrite this as, well, A is 0, B is minus 2/3.
- So this is equal to minus 2/3 over s squared plus 4.
- And then C is 0, we figured that out.
- And then D is 2/3.
- So plus 2/3 over s squared plus 1.
- So all of that work that I just did, that was just to
- break up this piece right here.
- That was just to break up that piece right there.
- And then, of course, we have these other two pieces here
- that we can't forget about.
- So after all of this work, what do we have?
- And I'm going to make sure I don't make a
- careless mistake here.
- We get the Laplace Transform of Y-- as you can see, the
- algebra is the hardest part here-- is equal to this first
- term-- I'm just going back-- this first term, which I've
- now decomposed into this.
- So it's minus-- let me write it this way-- minus 1/3-- and
- I think you're going to see in a second why I'm writing this
- way-- minus 1/3 times 2 over s squared plus 4, and then plus
- 2/3 times 1 over s squared plus 1.
- And you're probably saying, Sal, why are you
- writing it this way?
- Well you can already immediately see that this is
- the Laplace Transform of sine of 2t.
- This is the Laplace Transform of sine of t.
- So I wanted to write this 2 here, because this is 2, this
- is 2 squared.
- This is 1, this is 1 squared.
- So I wanted to write it in this form.
- This was just the first term.
- We have two more terms to worry about.
- I don't want to make a careless mistake.
- I have 2s over s squared plus 1.
- So let me write that down.
- So plus 2 times s over s squared plus 1, plus-- last
- one-- plus 1 over s squared plus 1.
- Now we just take the inverse Laplace Transform of the whole
- thing, and then we'll know what Y is.
- So, you know, just to remember the Laplace Transform.
- So this is going to be a little inverse.
- This is going to be sine of 2t.
- Let me just write, just so we have it here, so you know I'm
- not doing some type of voodoo.
- The Laplace Transform of sine of at is equal to a over s
- squared plus a squared.
- And the Laplace Transform of cosine of at is equal to s
- over s squared plus a squared.
- Let's just remember those two things when we take the
- inverse Laplace Transform of both sides of this equation.
- The inverse Laplace Transform of the Laplace Transform of y,
- well that's just y.
- y-- maybe I'll write it as a function of t-- is equal to--
- well this is the Laplace Transform of sine of 2t.
- You can just do some pattern matching right here.
- If a is equal to 2, then this would be the Laplace Transform
- of sine of 2t.
- So it's minus 1/3 times sine of 2t plus 2/3 times-- this is
- the Laplace Transform of sine of t.
- If you just make a is equal to 1, sine of t's Laplace
- Transform is 1 over s squared plus 1.
- So plus 2/3 times the sine of t-- let me do the next one in
- blue, just because it was already written in blue-- plus
- 2 times-- this is the Laplace Transform of cosine of t.
- If you make a is equal to 1, then the cosine t Laplace
- Transform is s over s squared plus 1.
- So 2 times cosine of t-- and then one last term-- plus--
- this is just like this one over here, this is just the
- Laplace Transform of sine of t-- plus sine of t.
- And we're almost done.
- We're essentially done, but there's a little bit more
- simplification we can do.
- I have 2/3 times the sign of t here, and then I have another
- 1 sine of t here, so I can add the 2/3 to the 1.
- What's 2/3 plus 1, or 3/3?
- It's 5/3.
- So I can write y of t is equal to minus 1/3 sine of 2t plus--
- these two terms I'm just going to add up--
- plus 5/3 sine of t.
- And then I have this last term here, plus 2cosine of t.
- So this was a hairy problem, a lot of work.
- And we saw that the hardest part really was just the
- partial fraction decomposition that we did up here, not
- making any careless mistakes.
- But at the end, we got a pretty neat answer that's not
- too complicated, that satisfies this non-homogenous
- differential equation.
- We were able to incorporate the boundary
- conditions as we did it.
- Anyway, hopefully you found that vaguely satisfying.
- This is a good warm-up after a year of no
- differential equations.

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