Complex and repeated roots of characteristic equation
Complex roots of the characteristic equations 1 What happens when the characteristic equations has complex roots?!
Complex roots of the characteristic equations 1
- We learned in the last several videos, that if I had a linear
- differential equation with constant coefficients in a
- homogeneous one, that had the form A times the second
- derivative plus B times the first derivative plus C
- times-- you could say the function, or the 0
- derivative-- equal to 0.
- If that's our differential equation that the
- characteristic equation of that is Ar squared plus Br
- plus C is equal to 0.
- And if the roots of this characteristic equation are
- real-- let's say we have two real roots.
- Let me write that down.
- So the real scenario where the two solutions are going to be
- r1 and r2, where these are real numbers.
- Then the general solution of this differential equation--
- and watch the previous videos if you don't remember this or
- if you don't feel like it's suitably proven to you-- the
- general solution is y is equal to some constant times e to
- the first root x plus some other constant times e to the
- second root times x.
- And we did that in the last several videos.
- We even did some examples.
- Now my question to you is, what if the characteristic
- equation does not have real roots,
- what if they are complex?
- And just a little bit of review, what
- do I mean by that?
- Well, if I wanted to figure out the roots of this and if I
- was lazy, and I just want to do it without having to think,
- can I factor it, I would just immediately use the quadratic
- equation, because that always works.
- I would say, well the roots of my characteristic equation are
- negative B plus or minus the square root of B
- squared minus 4AC.
- All of that over 2A.
- So what do I mean by non-real roots?
- Well, if this expression right here-- if this B squared minus
- 4AC-- if that's a negative number, then I'm going to have
- to take the square root of a negative number.
- So it will actually be an imaginary number, and so this
- whole term will actually become complex.
- We'll have a real part and an imaginary part.
- And actually, the two roots are going to be conjugates of
- each other, right?
- We could rewrite this in the real and imaginary parts.
- We could rewrite this as the roots are going to be equal to
- minus B over 2A, plus or minus the square root of B squared
- minus 4AC over 2A.
- And if B squared minus 4AC is less than 0, this is going to
- be an imaginary number.
- So in that case, let's just think about what the roots
- look like generally and then we'll actually do some
- problems. So let me go back up here.
- So then the roots aren't going to be two real
- numbers like that.
- The roots, we can write them as two complex numbers that
- are conjugates of each other.
- And I think light blue is a suitable color for that.
- So in that situation, let me write this, the complex
- roots-- this is a complex roots scenario-- then the
- roots of the characteristic equation are going to be, I
- don't know, some number-- Let's call it lambda.
- Let's call it mu, I think that's the convention that
- people use-- actually let me see what they tend to use, it
- really doesn't matter-- let's say it's lambda.
- So this number, some constant called lambda, and then plus
- or minus some imaginary number.
- And so it's going to be some constant mu.
- That's just some constant, I'm not trying to be fancy, but
- this is I think the convention used in most differential
- equations books.
- So it's mu times i.
- So these are the two roots, and these
- are true roots, right?
- Because we have lambda plus mu i, and lambda minus mu i.
- So these would be the two roots, if B squared minus 4AC
- is less than 0.
- So let's see what happens when we take these two roots and we
- put them into our general solution.
- So just like we've learned before, the general solution
- is going to be-- I'll stay in the light blue-- the general
- solution is going to be y is equal to c1 times e to the
- first root-- let's make that the plus version-- so
- lambda plus mu i.
- All of that times x, plus c2 times e to the second root.
- So that's going to be lambda minus mu i times x.
- Let's see if we can do some simplification here, because
- that i there really kind of makes things kind of crazy.
- So let's see if we can do anything to either get rid of
- it or simplify it, et cetera.
- So let's multiply the x out.
- Just doing some algebraic manipulation.
- I'm trying to use as much space as possible.
- So we get y is equal to c1 e to the what?
- Lambda x-- just distributing that x-- plus mu xi, plus c2
- times e to the lambda x minus mu xi.
- Just distributed the x's in both of the terms.
- And let's see what we can do.
- Well, when you add exponents, this is the exact same thing
- as y is equal to c1 e to the lambda x, times e
- to the mu xi, right?
- If you have the same base and you're multiplying, you could
- just add exponents, so this is the same thing as that.
- Plus c2 times e to the lambda x, times e to the minus mu xi.
- Now let's see, we have an e to the lambda x in both of these
- terms, so we can factor it out.
- So we get y is equal to-- let me draw a line here, I don't
- want you to get confused with all this quadratic equation
- stuff-- y is equal to e to the lambda x times c1 e to the mu
- xi-- that's an i-- plus c2 times e to the minus mu xi.
- Now what we can we do?
- And this is where it gets fun.
- If you watched the calculus playlist, especially when I
- talk about approximating functions with series, we came
- up with what I thought was the most amazing result in
- calculus, just from a-- or in mathematics-- just from a
- metaphysical point of view.
- And now we will actually use it for something that you'll
- hopefully see is vaguely useful.
- So here we have two terms that have something times e to the
- something times i.
- And we learned before, Euler's formula.
- And what was Euler's formula?
- I'll write that in purple.
- That e to the i theta, or we could write e to the ix, is
- equal to cosine of x plus i sine of x.
- And what's amazing about that is, if you put negative 1 in
- here, then you get e to the-- oh no, actually if you put pi
- in here-- so e to the i pi is equal to negative 1, right?
- If you substituted this because sine of pi is 0.
- So I thought that was amazing, where you could write e to the
- i 2 pi is equal to 1.
- That's pretty amazing as well.
- And in one equation you have all of the fundamental numbers
- of mathematics.
- That's amazing, but let's get back down to
- earth and get practical.
- So let's see if we can use this to simplify Euler's--
- This is actually a definition, and the definition makes a lot
- of sense, because when you do the power series
- approximation, or the Maclaurin series
- approximation, of e to the x, it really looks like cosine of
- x plus i times the power series approximation of x.
- But anyway, we won't go into that now.
- I have like six or seven videos on it.
- But let's use this to simplify this up here.
- So we can rewrite that as y is equal to e to the lambda x
- times-- let's do the first one-- c1.
- It's e to the mu xi, so instead of an x
- we have a mu x.
- That will be equal to cosine of whatever is in front of the
- i, so cosine of mu x plus i sine of mu x.
- And then plus c2 times what?
- Times cosine of minus mu x plus i sine of minus mu x.
- And let's see if we can simplify this further.
- So one thing that you might-- So let's distribute the c's.
- So now we get-- I'll do it in a different color-- actually
- I'm running out of time, so I'll continue
- this in the next video.
- See you soon.
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This is great, I finally understand quadratic functions!
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