PROFESSOR: OK, this lecture,
this day, is differential
equations day.
I just feel even though these
are not on the BC exams, that
we've got everything
we need to actually
see calculus in use.
We've got the derivatives of
the key functions and ready
for a differential equation.
And there it is.
When I look at that equation--
so it's a differential equation
because it has the
derivatives of y as well as
y itself in the equation.
And when I look at it, I see
it's a second order equation
because there's a second
derivative.
It's a linear equation because
second derivative, first
derivative, and y itself are
separate, no multiplying of y
times y prime.
In fact, the only
multiplications are by these
numbers m, r and k, and those
are constant numbers coming
from the application.
So I have a constant
coefficient, linear, second
order, differential equation,
and I'd like to solve it.
And we can do it because it uses
the very functions that
we know how to find
derivatives of.
Let me take two or three special
cases where those
functions appear purely.
So one special case is if
I knock out the second
derivative term, and let
me just choose--
rewrite it in the way it looks
easiest and best. It's just
the derivative of y as
some multiple of y.
It's now first order, and we
know the function that solves
that equation.
When the derivative equals the
function itself, that's the
exponential.
If I want to have an extra
factor a here, then I
need e to the at.
And usually, in fact, we expect
that with a first order
equation, in the solution
there'll be some constant that
we can set later to match
the starting condition.
And that constant
shows up here.
It's just if e to the at solves
that equation, as it
does, because when I take the
derivative, down comes on a,
so does c times e to the at.
That's because the equation
is linear.
So that's a nice one.
Pure exponential.
OK, ready for this one.
So this one, I don't have
this middle term.
I just have the second
derivative and the function.
And let me again change to
letters I like, putting the ky
on the other side with
a minus sign.
So this omega squared
will be k/m when I
reorganize that equation.
My point is we can solve this
equation, the second
derivative equaling minus
the function.
We've met that.
That's this equation
that the sine and
the cosine both solve.
There we get two solutions for
this second order equation.
And just as with the a in this
problem, so with this number
here, I'll just jiggle the sine
and the cosine a little
bit so that we get omega to come
down twice when I take
two derivatives.
So the solution here will be--
one solution will be the cosine
of omega t because two
derivatives of the cosine
is minus the cosine--
that's what this asks for--
with the factor omega
coming out twice.
And another solution will
be sine of omega t
for the same reason.
And again, now with a second
order equation, I'm expecting
a couple of constants to be
able to choose later.
And here they are: c cosine
omega t and d sine omega t.
That's the general solution
to that equation.
So we know that.
And these are the two
important ones.
There's another less important
one and an
extremely simple one.
Suppose all that went away, and
I just had as a third very
special case d second y
dt squared equals 0.
We sure know the solution
to that.
What functions have second
derivative equals 0?
Well, a constant function
does, certainly.
A constant, even its first
derivative is 0
much less its second.
And then t does.
Its first derivative is
1 and then the second
derivative is 0.
So there show up the
powers of t.
Well, the first two powers, t to
the 0 and t to the 1, show
up in that very special case.
Sine and cosine show up here.
e to the at shows up here.
And now let me tell you the
good part of this lecture.
The solution to this equation
and in fact to equations of
third, fourth, all orders, are
products of these ones that we
know: exponentials times sines
and cosines times powers of t.
That's all we need to solve
constant coefficient
differential equations.
So I plan now to go ahead
and solve the--
and move toward this equation.
I should have said that's a
fundamental equation of
engineering.
m stands for some mass.
Oh yeah, let me draw a picture,
and you'll see why I
chose to choose t rather than
x, because things are
happening in time.
And what is happening?
Let me show you.
What's happening in time is
typically this would be a
problem with us some kind of a
spring hanging down, and on
that spring is a mass m.
OK, so what happens if I--
I've pulled that mass down, so
I've stretched the spring, and
then I let go.
Then what does the spring do?
Well, the spring will pull the
mass back upwards, and it will
pull it back up to the point
where it squeezes, compresses
the spring.
The spring will be--
like instead of being
stretched out,
it'll be the opposite.
It'll be compressed in.
And then when compressed in,
it'll push the mass.
Being compressed, the
spring will push.
It'll push the mass down again
and up again, and I get
oscillation.
Oscillation is what
I'm seeing here.
And what are examples of
oscillation in real life?
The spring or a clock,
especially a grandfather clock
that's swinging back and forth,
back and forth, so I'll
just put a clock.
Music, a violin string
is oscillating.
That's where the beautiful
sound comes from.
Our heart is in and out,
in and out, a regular
oscillation.
I could add molecules.
They oscillate extremely
quickly.
So this equation that I'm
aiming for comes up in
biology, in chemistry--
for molecules, it's a very
important equation--
in physics and mechanics and
engineering for springs.
It comes up in economics.
It's everywhere.
And by choosing constant
coefficients, I have the basic
model and the simplest model.
OK, so I'll talk in this
language of springs, but it's
all these oscillations that lead
to equations like that.
Actually, this model often
would have r equals 0.
So let me take that case
r equals 0 again.
So where does the equation
come from?
Can I do two cents worth of
physics and then go back to
the math, solving
the equation.
The physics is just remembering
Newton's
Law: f equals ma.
So there is the m, the mass.
The a is the acceleration.
That's the second derivative,
right?
You don't mind if I write that
as second derivative.
Acceleration, the mass
is constant.
We're not going at the speed of
light here so we can assume
that mass is not being
converted to energy.
It's mass.
And then what's the force?
Well, for this spring
force, for this
spring, what did we say?
So y is going this way.
It's the disposition, the
displacement of the spring.
When it's down like that, when
y is positive, the spring is
pulling back.
The force from the spring
is pulling back.
And the force from the spring
is proportional to y.
And the proportionality constant
is my number k.
That fact that I
just said is--
and I guess it's pulling
opposite to a positive y.
When y is positive, and this
spring is way down, the force
is pulling it up,
pulling in the
negative direction, upwards.
So I need that minus sign.
So that k is the stiffness
of the spring.
This is Hooke's Law, that the
force coming from a spring is
proportional to the stretch.
And the constant in there is
Hooke's constant, the spring
constant k.
So now that if I put the minus
ky over there as plus ky, you
see my equation again.
But this is the one that we were
able to solve right here.
Do you see that this is the
case where r is 0 in this
first model?
r is 0 because r involves
resistance: r for resistance,
air resistance, damping.
And right now, I don't
have that.
I just have a little spring
that'll oscillate forever.
It'll oscillate forever
following sine and cosine.
That's exactly what the
spring will do.
It will just go on forever,
and the c and the d, their
constants, will depend
on how it started.
Did it start from rest?
If it started from rest, there
would be no sine term.
It's easy to find
c and d later.
The real problem is to solve the
equation, and we've done
it for this equation.
OK, and let's just
remember this.
Let me repeat that when I put
that onto the other side and
divide by k, then I see the--
oh, divided by m, sorry--
then I see the omega squared.
So omega squared is k/m.
Right.
Oh, it is k/m, right.
OK, that's the simple case,
pure sines and cosines.
Now I'm letting in
some resistance.
So now I'm coming back
to my equation.
Let me write it again.
m y double prime--
second derivative--
plus 2r y prime--
first derivative--
plus ky--
0-th derivative equals 0.
I want to solve that equation
now for any
numbers m and r and k.
OK, well, the nice thing is that
the exponential function
takes us right to the answer,
the best plan here.
So this is the most important
equation you would see in a
differential equations course.
And then the course kind of goes
past, and then you easily
forget this is the most
important and the simplest.
Why is it simple?
Because if the key idea--
this is the key idea: try
y equals e to the--
an exponential e to the
something times t.
Let me call that something
lambda.
You might have preferred c.
I'm happy with c or any
other number there.
Yeah, OK, I'll call that lambda,
just because it gives
it a little Greek importance.
All right, so I try this.
I substitute that into the
equation, and I'm going to
choose lambda to make
things work.
OK, but here's the key.
The key idea is so easy.
Now, put it into the equation.
So what happens when I put--
when this is y, take
two derivatives.
Well, let me start with
taking no derivatives.
So I have down here the k
e to the lambda t, and
over here is a 0.
And now let me back up to
the first derivative.
So that's 2r times
the derivative
of this guy y prime.
I'm just substituting this
into the equation.
So what's the derivative?
We know that the derivative of
this brings down the lambda.
We already did it.
lambda e to the lambda
t, right?
That's the derivative.
And what about this one?
This one is going to be
an m y double prime.
What happens with
two derivatives?
Bring down lambda twice, two
times, so I have lambda
squared e to the lambda t.
And then in a minute, you know
what I'm going to do.
I'm going to cancel that common
factor e to the lambda
t, which is never 0 so I can
safely divide it out, and then
write the equation we get.
OK, let me write that out
more with more space.
m lambda squared plus
2r lambda--
taking that--
plus k is 0.
This is the equation.
It's just an ordinary
quadratic equation.
It's a high school
algebra equation.
Lambda appears squared because
we had two derivatives.
We had a second derivative, and
so I need the quadratic
formula to know--
I expect two answers,
two lambdas.
And that's normal for a second
order equation, and I'll get
two solutions: e to the lambda
1t and e to the lambda 2t.
Two different exponentials will
both solve the problem.
All right, what's the lambda?
Well, can I just recall
the quadratic formula?
Well, it's a little messy, but
it's not too bad here, just to
show that I remember it.
And the 2 there is kind of
handy with the quadratic
formula because then I just get
minus an r plus or minus
the square root of r squared.
And it's not minus 4km, but
because of the 2 there, it's
just minus km.
And then I divide by m.
OK, well, big deal.
I get two roots.
Let me use numbers.
So what you see there
is like solving the
differential equation.
Not too bad.
Now let me put in numbers to
show what's typical, and, of
course, as those numbers
change, we'll
see different lambdas.
And actually, as the numbers
change, that will take us
between the exponential stuff
and the oscillating stuff.
All right, let me take one
where I think it start--
I think this will be-- so
this is example one.
I'll choose m equal 1 y
double prime plus--
let me take r to be 3, so
then I have 6y prime.
And let me choose k
to be 8y equals 0.
All right, now we've
got numbers.
So the numbers are m equals
1, r equals 3, k equals 8.
And I claim we can write the
solution to that equation, or
the two solutions, because there
will be two lambdas.
OK, so when I try e to the
lambda t, plug it in, I'll get
lambda twice, and then I'll get
6 lambda once, and then
I'll get 8 without a lambda
coming down, all multiplied by
e to the lambda t, which I'm
canceling, equals 0.
So I solve that equation either
directly by recognizing
that it factors into lambda
plus 2 times lambda plus 4
equals 0 or by plugging
in r and k and m in
the quadratic formula.
Either way, I'm learning that
lambda is minus 2 or minus 4.
Those are the two solutions.
The two decay rates, you could
call them, because they're up
in the exponent.
So what's the solution?
y of t, the solution to that
equation, the general solution
with a constant c and a constant
d is an e to the
minus 2t and an e
to the minus 4t.
The two lambdas are in the
exponent, and we've solved it.
So that's the point.
We have the ability to solve
differential equations based
on the three most important
derivatives we know:
exponential, sines-cosines,
powers of t.
OK, ready for example two?
Example two, I'm just going to
change that 8 to a 10, so
you're going to see
a 10 show up here.
All right, but that will
make a difference.
It won't just be some
new numbers.
There'll be a definite
difference here.
OK, let me go over across
here to the one with 10.
OK, so now my equation is 1y
double prime, 6y prime still,
and now 10y is equals to 0,
remembering prime means
derivative.
OK, so again, I try y is
e to the lambda t.
I plug it in.
When I have two derivatives,
bring down lambda squared.
One derivative brings
down lambda.
No derivatives leaves
the 10, equals 0.
That's my equation for lambda.
Ha!
I don't know how to
factor that one.
And in fact, I better use the
quadratic formula just to show
what happens here.
So the quadratic formula will
be the two roots lambda.
Can I remember that
dumb formula?
Minus r plus or minus the square
root of r squared minus
km, all divided by m.
I got the 2's and the
4's out of it by
taking r to be 3 here.
OK, so it's minus 3 plus or
minus the square root of r
squared is 9 minus 1 times 10.
k and m is 10 divided by 1.
Ha!
You see something different's
going on here.
I have the square root
of a negative number.
Over there, if I wrote out that
square root, you would
have seen the square
root of plus 1.
That gave me minus 3 plus
1 or minus 3 minus 1.
That was the minus
2 and minus 4.
Now I'm different.
Now seeing the square root of
minus 1, so this is minus 3
plus or minus i.
So I see the solution y of t.
You see, this is i here, the
square root of minus 1.
We can deal with that.
It's a complex number,
an imaginary number.
And the combination minus 3 plus
i is a complex number,
and we have to accept that
that's our lambda.
So I have any multiple of c,
and the lambda here is
minus 3 plus i t.
And the second solution is
e minus 3 minus i t.
I found the general solution.
We could say done, except you
might feel, well, how did
imaginary--
what are we going to do with
these imaginary numbers here?
How did they get in this
perfectly real
differential equation?
Well, they slipped in because
the solutions
were not real numbers.
The solutions were minus
3 plus or minus i.
But we can get real again.
So this is one way to write the
answer, but I just want to
show you using the earlier
lecture, using the beautiful
fact that Euler discovered.
So now let me complete this
example by remembering Euler's
great formula for e to the it.
Because you see we have an e
to the minus 3t, perfectly
real, decaying.
The spring is slowing down
because of air resistance.
But we also have
an e to the it.
That's what Euler's formulas
about and Euler's formula says
that e to the it is the
cosine of t plus i
times the sine of t.
So it's through Euler's
formula that these
oscillations are coming in.
The direct method led
to an e to the it.
But the next day, Euler
realized that e
to the minus it--
or probably being Euler, it
didn't take till next day--
will be minus i sine t.
So both e to the it and e to
the minus it, they both can
get replaced by sines
and cosines.
So in place of e to the
it, I'll put that.
In place of e to the minus
it, I put that one.
The final result is--
can I just jump to that?
The final result is that with
some different constants, we
have the cosine--
oh!
let me not forget e
to the minus 3t.
That's part of this answer.
I'm damping this out by e to the
minus 3t, this resistance
r, times cosine of t and the e
to the minus 3t sine of t.
OK, that's good.
General solution, back
to a real numbers.
It describes a damped
oscillation.
It's damped out.
It's slowing down.
Rather, it's decaying.
The amplitude is-- the
spring is like--
it's like having a shock
absorber or something.
It's settling down to the center
point pretty fast. But
as it settles, it's goes back
across that center point,
oscillates across.
OK, now you might finally ask,
the last step of this lecture,
where do powers of t come in?
Where does t come in?
So far we've seen exponentials
come in.
We've seen sines and
cosines come in.
Can I do a last example just
here in the corner, which will
be y double prime, 6 y prime,
and now this time instead of 8
or 10, I'm going to
take 9y equals 0.
OK, can we use this as example
3, which we can solve?
You know that I'm going to
try y equals to lambda t.
Let me substitute that.
I'll get lambda squared coming
from two derivatives, lambda
coming from one derivative, 9
coming from no derivatives.
I've got my quadratic equation
that's supposed
to give me two lambdas.
Little problem here.
When I factor this, it factors
into lambda plus 3
squared equals 0.
So the answer, the lambda,
is minus 3 twice.
Twice!
The two lambdas happen to hit
the same value: minus 3.
OK, we don't have any
complex stuff here.
It's two real values that
happen to coincide.
And when that happens, well,
minus 3 tells us that a
solution e to the
minus 3t works.
Where do we get the
other solution?
We want two solutions to a
second order equation.
I can't just use e to
the minus 3t again.
I need a second solution, and
it shows up at this--
it's typical of math
that it shows--
something special happens at
this special situation of a
double root, and the solutions
will be e to the minus 3t and
t e to the minus 3t.
That's the last step, and
maybe I just put it--
well, where am I going
to put it?
I'll bring this down and
just put it here.
So now I'm solving this
particular problem y is a
multiple of e to the minus 3t.
Good, but I can't just repeat
it for the second one.
So the second one, it just turns
out that then is when a
little factor t appears.
You might like to substitute
that for practice with the
product rule.
If you substitute that in the
differential equation, you'll
find everything cancels,
and it works.
So the conclusion is linear
constant coefficient
differential equations are
completely solved by trying e
to the lambda t and finding
that number lambda.
If it's a real number,
we have exponentials.
If it's an imaginary number,
we have sines and cosines.
If it's a repeated number,
we have an extra
factor t showing up.
That's the exceptional event
in that particular case.
So there you go.
Differential equations
with constant
coefficients we can handle.
You can handle.
Thank you.
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