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# Complex roots of the characteristic equations 3: Lets do an example with initial conditions!

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- Let's do a couple of problems where the roots of the
- characteristic equation are complex.
- And just as a little bit of a review, and we'll put this up
- here in the corner so that it's useful for us.
- We learned that if the roots of our characteristic equation
- are r is equal to lambda plus or minus mu i, that the
- general solution for our differential equation is y is
- equal to e to the lambda x times c1 with some constant
- cosine of mu x, plus c2 times sine of mu x.
- And with that said, let's do some problems.
- So let's see, this first one that our differential
- equation-- I'll do this one in blue-- our differential
- equation is the second derivative, y prime prime plus
- 4y prime plus 5y is equal to 0.
- And they actually give us some initial conditions.
- They say y of 0 is equal to 1, and y prime of
- 0 is equal to 0.
- So now we'll actually be able to figure out a particular
- solution, or the particular solution, for this
- differential equation.
- So let's write down the characteristic equation.
- So it's r squared plus 4r plus 5 is equal to 0.
- Let's break out our quadratic formula.
- The roots of this are going to be negative b.
- So minus 4 plus or minus the square root of b squared.
- So that's 16.
- Minus 4 times a-- well that's 1-- times 1, times c times 5.
- All of that over 2 times a.
- a is 1 so all of that over 2.
- And see this simplifies.
- This equals minus 4 plus or minus 16-- this is 20, right?
- 4 times 1 times 5 is 20.
- So 16 minus 20 is minus 4.
- Minus 4 over 2.
- And that equals minus 4 plus or minus 2i, right?
- Square root of minus 4 is 2i.
- All that over 2.
- And so our roots to the characteristic equation are
- minus 2-- just dividing both by 2-- minus 2 plus or minus,
- we could say i or 1i, right?
- So if we wanted to do some pattern match, if we just
- wanted to do it really fast, what's our lambda?
- Our lambda is just minus 1.
- Let me write that down.
- That's our lambda.
- What's our mu?
- Well mu is the coefficient on the i, so mu is 1.
- mu is equal to 1.
- And now we're ready to write down our general solution.
- So the general solution to this differential equation is
- y is equal to e to the lambda x-- well lambda is minus 2--
- minus 2x times c1 cosine of mu x-- but mu is just 1-- so c1
- cosine of x, plus c2 sine of mu x, when mu is
- 1, so sine of x.
- Fair enough.
- Now let's use our initial conditions to find the
- particular solution, or a particular solution.
- So, when x is 0, y is equal to 1.
- So y is equal to 1 when x is 0.
- So 1 is equal to-- let's substitute x is 0 here.
- So e to the minus 2 times 0, that's just 1.
- So this whole thing becomes 1, so we could just ignore it.
- It's just 1 times this thing.
- So I'll write that down.
- e to the 0 is 1 times c1 times cosine of 0, plus c2
- times sine of 0.
- Now what's sine of 0?
- Sine of 0 is 0.
- So this whole term is going to be 0.
- Cosine of 0 is 1.
- So there, we already solved for c1.
- We get this, this is 1.
- So 1 times c1 times 1 is equal to 1.
- So we get our first coefficient.
- c1 is equal to 1.
- Now let's take the derivative of our general solution.
- And we could even substitute c1 in here, just so that we
- have to stop writing c1 all the time.
- And we can solve for c2.
- So right now we know that our general solution is y-- we
- could call this our pseudo-general solution,
- because we already solved for c1-- y is equal to e to the
- minus 2x times c1, but we know that c1 is 1, so I'll write
- times cosine of x, plus c2 times sine of x.
- Now let's take the derivative of this, so that we can use
- the second initial condition.
- So y prime is equal to-- we're going to have to do a little
- bit of product rule here.
- So what's the derivative of the first expression?
- It is minus 2e to the minus 2x.
- And we multiply that times the second expression.
- Cosine of x plus c2 sine of x.
- And then we add that to just the regular first expression.
- So plus e to the minus 2x times the derivative of the
- second expression.
- So what's the derivative of cosine of x?
- It's minus sine of x.
- And then what's the derivative of c2 sine of x?
- Well , it's plus c2 cosine of x.
- And let's see if we can do any kind of simplification here.
- Well actually, the easiest way, instead of trying to
- simplify it algebraically and everything, let's just use our
- initial condition.
- Our initial condition is y prime of 0 is equal to 0.
- Let me write that down.
- Second initial condition was y prime of 0 is equal to 0.
- So y prime, when x is equal to 0, is equal to 0.
- And let's substitute x is equal to 0 into this thing.
- We could have simplified this more, but let's not worry
- about that right now.
- So if x is 0, this is going to be 1, right?
- E to the 0.
- e to the 0 is 1, so we're left with just minus 2, right?
- Minus 2 times e to the 0, times cosine of 0, that's 1,
- plus c2 times sine of 0.
- Sine of 0 is 0.
- So that's just 1 plus 0, plus e to the minus 2 times 0.
- That's just 1.
- Times minus sine of 0.
- Sine of 0 is just 0.
- Plus c2 times cosine of 0.
- Cosine of 0 is 1.
- So plus c2.
- That simplified things, didn't it?
- So let's see, we get 0 is equal to-- this is just 1--
- minus 2 plus c2.
- Or we get c2 is equal to 2.
- Add 2 to both sides. c2 is equal to 2.
- And then we have our particular solution.
- I know it's c2 is equal to 2, c1 is equal to 1.
- Actually, let me erase some of this, just so that we can go
- from our general solution to our particular solution.
- So we had figured out, you can remember, c1 is 1 and c2 is 2.
- That's easy to memorize.
- So I'll just delete all of this.
- I'll write it nice and big.
- So our particular solution, given these initial
- conditions, were, or are, or is y of x is equal to-- this
- was a general solution-- e to the minus 2x times-- we solved
- for c1, it's equal to 1.
- So we can just write cosine of x.
- And then we solved for c2.
- We figured out that that was 2.
- Plus 2 sine of x.
- And there you go.
- We have our particular solution to this-- sorry where
- did I write it-- to this differential equation with
- these initial conditions.
- And what's neat is, when we originally kind of proved this
- formula-- when we originally showed this formula-- we had
- all of these i's and we simplified.
- We said c2, it was a combination of some other
- constants times some i's.
- And we said, oh we don't know whether they're imaginary or
- not, so let's just merge them into some constant.
- But what's interesting is this particular solution has no i's
- anywhere in it.
- And so, that tells us a couple of neat things.
- One that, if we had kept this c2 in terms of some multiple
- of i's, and our constants actually would have had i's
- and they would have canceled out, et cetera.
- And it also tells us that this formula is useful beyond
- formulas that just involve imaginary numbers.
- For example, this differential equation, I don't see an i
- anywhere here.
- I don't see an i anywhere here.
- And I don't see an i anywhere here.
- But given this differential equation, to get to the
- solution, we had to use imaginary numbers in between.
- And I think this is the first time-- if I'm remembering all
- my playlists correctly-- this is the first time that we used
- imaginary numbers for something useful.
- We used it as an intermediary tool where we got a real, a
- non-imaginary solution, to a real problem, a
- non-imaginary problem.
- But we used imaginary numbers as a tool to solve it.
- So, hopefully, you found that slightly interesting.
- And I'll see you in the next video.

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