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# Complex roots of the characteristic equations 3

Video transcript

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.