We're going to start. We are going to start studying
today, and for quite a while, the linear second-order
differential equation with constant coefficients.
In standard form, it looks like,
there are various possible choices for the variable,
unfortunately, so I hope it won't disturb you
much if I use one rather than another.
I'm going to write it this way in standard form.
I'll use y as the dependent variable.
Your book uses little p and little q.
I'll probably switch to that by next time.
But, for today, I'd like to use the most
neutral letters I can find that won't interfere with anything
else. So, of course call the constant
coefficients, respectively,
capital A and capital B. I'm going to assume for today
that the right-hand side is zero.
So, that means it's what we call homogeneous.
The left-hand side must be in this form for it to be linear,
it's second order because it involves a second derivative.
These coefficients, A and B, are understood to be
constant because, as I said, it has constant
coefficients. Of course, that's not the most
general linear equation there could be.
In general, it would be more general by making this a
function of the dependent variable, x or t,
whatever it's called. Similarly, this could be a
function of the dependent variable.
Above all, the right-hand side can be a function of a variable
rather than simply zero. In that case the equation is
called inhomogeneous. But it has a different physical
meaning, and therefore it's customary to study that after
this. You start with this.
This is the case we start with, and then by the middle of next
week we will be studying more general cases.
But, it's a good idea to start here.
Your book starts with, in general, some theory of a
general linear equation of second-order,
and even higher order. I'm asking you to skip that for
the time being. We'll come back to it next
Wednesday, it two lectures, in other words.
I think it's much better and essential for your problems at
for you to get some experience with a simple type of equation.
And then, you'll understand the general theory,
how it applies, a lot better,
I think. So, let's get experience here.
The downside of that is that I'm going to have to assume a
couple of things about the solution to this equation,
how it looks; I don't think that will upset
you too much. So, what I'm going to assume,
and we will justify it in a couple lectures,
that the general solution, that is, the solution involving
arbitrary constants, looks like this.
y is equal-- The arbitrary constants occur in a certain
special way. There is c one y one plus c two
y two. So, these are two arbitrary
constants corresponding to the fact that we are solving a
second-order equation. In general, the number of
arbitrary constants in the solution is the same as the
order of the equation because if it's a second-order equation
because if it's a second-order equation, that means somehow or
other, it may be concealed. But you're going to have to
integrate something twice to get the answer.
And therefore, there should be two arbitrary
constants. That's very rough,
but it sort of gives you the idea.
Now, what are the y1 and y2? Well, as you can see,
if these are arbitrary constants, if I take c2 to be
zero and c1 to be one, that means that y1 must be a
solution to the equation, and similarly y2.
So, where y1 and y2 are solutions.
Now, what that shows you is that the task of solving this
equation is reduced, in some sense,
to finding just two solutions of it, somehow.
All we have to do is find two solutions, and then we will have
solved the equation because the general solution is made up in
this way by multiplying those two solutions by arbitrary
constants and adding them. So, the problem is,
where do we get that solutions from?
But, first of all, or rather, second or third of
all, the initial conditions enter into the,
I haven't given you any initial conditions here,
but if you have them, and I will illustrate them when
I work problems, the initial conditions,
well, the initial values are satisfied by choosing c1 and c2,
are satisfied by choosing c1 and c2 properly.
So, in other words, if you have an initial value
problem to solve, that will be taken care of by
the way those constants, c, enter into the solution.
Okay, without further ado, there is a standard example,
which I wish I had looked up in the physics syllabus for the
first semester. Did you study the
spring-mass-dashpot system in 8.01?
I'm embarrassed having to ask you.
You did? Raise your hands if you did.
Okay, that means you all did. Well, just let me draw an
instant picture to remind you. So, this is a two second
review. I don't know how they draw the
picture. Probably they don't draw
picture at all. They have some elaborate system
here of the thing running back and forth.
Well, in the math, we do that all virtually.
So, here's my system. That's a fixed thing.
Here's a little spring. And, there's a little car on
the track here, I guess.
So, there's the mass, some mass in the little car,
and motion is damped by what's called a dashpot.
A dashpot is the sort of thing, you see them in everyday life
as door closers. They're the thing up above that
you never notice that prevent the door slamming shut.
So, if you take one apart, it looks something like this.
So, that's the dash pot. It's a chamber with a piston.
This is a piston moving in and out, and compressing the air,
releasing it, is what damps the motion of the
thing. So, this is a dashpot,
it's usually called. And, here's our mass in that
little truck. And, here's the spring.
And then, the equation which governs it is,
let's call this x. I'm already changing,
going to change the dependent variable from y to x,
but that's just for the sake of example, and because the track
is horizontal, it seems more natural to call
it x. There's some equilibrium
position somewhere, let's say, here.
That's the position at which the mass wants to be,
if the spring is not pulling on it or pushing on it,
and the dashpot is happy. I guess we'd better have a
longer dashpot here. So, this is the equilibrium
position where nothing is happening.
When you depart from that position, then the spring,
if you go that way, the spring tries to pull the
mass back. If it goes on the site,
the spring tries to push the mass away.
The dashpot, meanwhile, is doing its thing.
And so, the force on the, m x double prime.
That's by Newton's law, the force, comes from where?
Well, there's the spring pushing and pulling on it.
That force is opposed. If x gets to be beyond zero,
then the spring tries to pull it back.
If it gets to the left of zero, if x gets to be negative,
that that spring force is pushing it this way,
wants to get rid of the mass. So, it should be minus kx,
and this is from the spring, the fact that is proportional
to the amount by which x varies. So, that's called Hooke's Law.
Never mind that. This is a law.
That's a law, Newton's law,
okay, Newton, Hooke with an E,
and the dashpot damping is proportional to the velocity.
It's not doing anything if the mass is not moving,
even if it's stretched way out of its equilibrium position.
So, it resists the velocity. If the thing is trying to go
that way, the dashpot resists it.
It's trying to go this way, the dashpot resists that,
too. It's always opposed to the
velocity. And so, this is a dash pot
damping. I don't know whose law this is.
So, it's the force coming from the dashpot.
And, when you write this out, the final result,
therefore, is it's m x double prime plus c x prime,
it's important to see where the various terms are,
plus kx equals zero. And now, that's still not in standard form.
To put it in standard form, you must divide through by the
mass. And, it will now read like
this, plus k divided by m times x equals zero.
And, that's the equation governing the motion of the
spring. I'm doing this because your
problem set, problems three and four, ask you to look at a
little computer visual which illustrates a lot of things.
And, I didn't see how it would make, you can do it without this
interpretation of spring-mass-dashpot,
-- -- but, I think thinking of it
of these constants as, this is the damping constant,
and this is the spring, the constant which represents
the force being exerted by the spring, the spring constant,
as it's called, makes it much more vivid.
So, you will note is that those problems are labeled Friday or
Monday. Make it Friday.
You can do them after today if you have the vaguest idea of
what I'm talking about. If not, go back and repeat
8.01. So, all this was just an
example, a typical model. But, by far,
the most important simple model.
Okay, now what I'd like to talk about is the solution.
What is it I have to do to solve the equation?
So, to solve the equation that I outlined in orange on the
board, the ODE, our task is to find two
solutions. Now, don't make it too trivial.
There is a condition. The solution should be
independent. All that means is that y2
should not be a constant multiple of y1.
I mean, if you got y1, then two times y1 is not an
acceptable value for this because, as you can see,
you really only got one there. You're not going to be able to
make up a two parameter family. So, the solutions have to be
independent, which means, to repeat, that neither should
be a constant multiple of the other.
They should look different. That's an adequate explanation.
Okay, now, what's the basic method to finding those
solutions? Well, that's what we're going
to use all term long, essentially,
studying equations of this type, even systems of this type,
with constant coefficients. The basic method is to try y
equals an exponential. Now, the only way you can
fiddle with an exponential is in the constant that you put up on
top. So, I'm going to try y equals e
to the rt. Notice you can't tell from that
what I'm using as the independent variable.
But, this tells you I'm using t.
And, I'm switching back to using t as the dependent
variable. So, T is the independent
variable. Why do I do that?
The answer is because somebody thought of doing it,
probably Euler, and it's been a tradition
that's handed down for the last 300 or 400 years.
Some things we just know. All right, so if I do that,
as you learned from the exam, it's very easy to differentiate
exponentials. That's why people love them.
It's also very easy to integrate exponentials.
And, half of you integrated instead of differentiating.
So, we will try this and see if we can pick r so that it's a
solution. Okay, well, I will plug in,
then. Substitute, in other words,
and what do we get? Well, for y double prime,
I get r squared e to the rt. That's y double prime because each time you
differentiate it, you put an extra power of r out
in front. But otherwise,
do nothing. The next term will be r times,
sorry, I forgot the constant. Capital A times r e to the rt, and then there's the last term,
B times y itself, which is B e to the rt. And, that's supposed to be
equal to zero. So, I have to choose r so that
this becomes equal to zero. Now, you see,
the e to the rt occurs as a factor in every
term, and the e to the rt is never zero.
And therefore, you can divide it out because
it's always a positive number, regardless of the value of t.
So, I can cancel out from each term.
And, what I'm left with is the equation r squared plus ar plus
b equals zero. We are trying to find values of r that satisfy that equation.
And that, dear hearts, is why you learn to solve
quadratic equations in high school, in order that in this
moment, you would be now ready to find how spring-mass systems
behave when they are damped. This is called the
characteristic equation. The characteristic equation of
the ODE, or of the system of the spring mass system,
which it's modeling, the characteristic equation of
the system, okay? Okay, now, we solve it,
but now, from high school you know there are several cases.
And, each of those cases corresponds to a different
behavior. And, the cases depend upon what
the roots look like. The possibilities are the roots
could be real, and distinct.
That's the easiest case to handle.
The roots might be a pair of complex conjugate numbers.
That's harder to handle, but we are ready to do it.
And, the third case, which is the one most in your
problem set is the most puzzling: when the roots are
real, and equal. And, I'm going to talk about
those three cases in that order. So, the first case is the roots
are real and unequal. If I tell you they are unequal,
and I will put down real to make that clear.
Well, that is by far the simplest case because
immediately, one sees we have two roots.
They are different, and therefore,
we get our two solutions immediately.
So, the solutions are, the general solution to the
equation, I write down without further ado as y equals c1 e to
the r1 t plus c2 e to the r2 t. There's our solution.
Now, because that was so easy, and we didn't have to do any
work, I'd like to extend this case a little bit by using it as
an example of how you put in the initial conditions,
how to put in the c. So, let me work a specific
numerical example, since we are not going to try
to do this theoretically until next Wednesday.
Let's just do a numerical example.
So, suppose I take the constants to be the damping
constant to be a four, and the spring constant,
I'll take the mass to be one, and the spring constant to be
three. So, there's more damping here,
damping force here. You can't really talk that way
since the units are different. But, this number is bigger than
that one. That seems clear,
at any rate. Okay, now, what was the
characteristic equation? Look, now watch.
Please do what I do. I've found in the past,
even by the middle of the term, there are still students who
feel that they must substitute y equals e to the rt,
and go through that whole little derivation to find that
you don't do that. It's a waste of time.
I did it that you might not ever have to do it again.
Immediately write down the characteristic equation.
That's not very hard. r squared plus 4r plus three
equals zero. And, if you can write down its roots immediately,
splendid. But, let's not assume that
level of competence. So, it's r plus three times r
plus one equals zero. Okay, you factor it. This being 18.03,
a lot of the times the roots will be integers when they are
not, God forbid, you will have to use the
quadratic formula. But here, the roots were
integers. It is, after all,
only the first example. So, the solution,
the general solution is y equals c1 e to the negative,
notice the root is minus three and minus one,
minus 3t plus c2 e to the negative t. Now, suppose it's an initial
value problem. So, I gave you an initial
condition. Suppose the initial conditions
were that y of zero were one. So, at the start, the mass has been moved over to
the position, one, here.
Well, we expected it, then, to start doing that.
But, this is fairly heavily damped.
This is heavily damped. I'm going to assume that the
mass starts at rest. So, the spring is distended.
The masses over here. But, there's no motion at times
zero this way or that way. In other words,
I'm not pushing it. I'm just releasing it and
letting it do its thing after that.
Okay, so y prime of zero, I'll assume, is zero. So, it starts at rest,
but in the extended position, one unit to the right of the
equilibrium position. Now, all you have to do is use
these two conditions. Notice I have to have two
conditions because there are two constants I have to find the
value of. All right, so,
let's substitute, well, we're going to have to
calculate the derivative. So, why don't we do that right
away? So, this is minus three c1 e to
the minus 3t minus c2 e to the negative t. And now, if I substitute in at
zero, when t equals zero, what do I get?
Well, the first equation, the left says that y of zero
should be one. And, the right says this is
one. So, it's c1 plus c2. That's the result of
substituting t equals zero. How about substituting? What should I substitute in the
second equation? Well, y prime of zero is zero. So, if the second equation,
when I put in t equals zero, the left side is zero according
to that initial value, and the right side is negative
three c1 minus c2. You see what you end up with,
therefore, is a pair of simultaneous linear equations.
And, this is why you learn to study linear set of pairs of
simultaneous linear equations in high school.
These are among the most important.
Solving problems of this type are among the most important
applications of that kind of algebra, and this kind of
algebra. All right, what's the answer
finally? Well, if I add the two of them,
I get minus 2c1 equals one. So, c1 is equal to minus one
half. And, if c1 is minus a half, then c2 is minus 3c1. So, c2 is three halves. The final question is,
what does that look like as a solution?
Well, in general, these combinations of two
exponentials aren't very easy to plot by yourself.
That's one of the reasons you are being given this little
visual which plots them for you. All you have to do is,
as you'll see, set the damping constant,
set the constants, set the initial conditions,
and by magic, the curve appears on the
screen. And, if you change either of
the constants, the curve will change nicely
right along with it. So, the solution is y equals
minus one half e to the minus 3t plus three halves e to the
negative t. How does it look? Well, I don't expect you to be
able to plot that by yourself, but you can at least get
started. It does have to satisfy the
initial conditions. That means it should start at
one, and its starting slope is zero.
So, it starts like that. These are both declining
exponentials. This declines very rapidly,
this somewhat more slowly. It does something like that.
If this term were a lot, lot more negative,
I mean, that's the way that particular solution looks.
How might other solutions look? I'll draw a few other
possibilities. If the initial term,
if, for example, the initial slope were quite
negative, well, that would have start like
this. Now, just your experience of
physics, or of the real world suggests that if I give,
if I start the thing at one, but give it a strongly negative
push, it's going to go beyond the equilibrium position,
and then come back again. But, because the damping is
big, it's not going to be able to get through that.
The equilibrium position, a second time,
is going to look something like that.
Or, if I push it in that direction, the positive
direction, that it starts off with a positive slope.
But it loses its energy because the spring is pulling it.
It comes and does something like that.
So, in other words, it might go down.
Cut across the equilibrium position, come back again,
it do that? No, that it cannot do.
I was considering giving you a problem to prove that,
but I got tired of making out the problems set,
and decided I tortured you enough already,
as you will see. So, anyway, these are different
possibilities for the way that can look.
This case, where it just returns in the long run is
called the over-damped case, over-damped.
Now, there is another case where the thing oscillates back
and forth. We would expect to get that
case if the damping is very little or nonexistent.
Then, there's very little preventing the mass from doing
that, although we do expect if there's any damping at all,
we expect it ultimately to get nearer and nearer to the
equilibrium position. Mathematically,
what does that correspond to? Well, that's going to
correspond to case two, where the roots are complex.
The roots are complex, and this is why,
let's call the roots, in that case we know that the
roots are of the form a plus or minus bi.
There are two roots, and they are a complex
conjugate. All right, let's take one of
them. What does a correspond to in
terms of the exponential? Well, remember,
the function of the r was, it's this r when we tried our
exponential solution. So, what we formally,
this means we get a complex solution.
The complex solution y equals e to this, let's use one of them,
let's say, (a plus bi) times t. The question is, what do we do that?
We are not really interested, I don't know what a complex
solution to that thing means. It doesn't have any meaning.
What I want to know is how y behaves or how x behaves in that
picture. And, that better be a real
function because otherwise I don't know what to do with it.
So, we are looking for two real functions, the y1 and the y2.
But, in fact, what we've got is one complex
function. All right, now,
a theorem to the rescue: this, I'm not going to save for
Wednesday because it's so simple.
So, the theorem is that if you have a complex solution,
u plus iv, so each of these is a function of time,
u plus iv is the complex solution to a real differential
equation with constant coefficients.
Well, it doesn't have to have constant coefficients.
It has to be linear. Let me just write it out to y
double prime plus A y prime plus B y equals zero. Suppose you got a complex
solution to that equation. These are understood to be real
numbers. They are the damping constant
and the spring constant. Then, the conclusion is that u
and v are real solutions. In other words,
having found a complex solution, all you have to do is
take its real and imaginary parts, and voila,
you've got your two solutions you were looking for for the
original equation. Now, that might seem like
magic, but it's easy. It's so easy it's the sort of
theorem I could spend one minute proving for you now.
What's the reason for it? Well, the main thing I want you
to get out of this argument is to see that it absolutely
depends upon these coefficients being real.
You have to have a real differential equation for this
to be true. Otherwise, it's certainly not.
So, the proof is, what does it mean to be a
solution? It means when you plug in A (u
plus iv) plus, prime,
plus B times u plus iv, what am I supposed to get? Zero.
Well, now, separate these into the real and imaginary parts.
What does it say? It says u double prime plus A u
prime plus B u, that's the real part of
this expression when I expand it out.
And, I've got an imaginary part, too, which all have the
coefficient i. So, from here,
I get v double prime plus i times A v prime plus
i times B v. So, this is the imaginary part.
Now, here I have something with a real part plus the imaginary
part, i, times the imaginary part is zero.
Well, the only way that can happen is if the real part is
zero, and the imaginary part is separately zero.
So, the conclusion is that therefore this part must be
zero, and therefore this part must be zero because the two of
them together make the complex number zero plus zero i.
Now, what does it mean for the real part to be zero?
It means that u is a solution. This, the imaginary part zero
means v is a solution, and therefore,
just what I said. u and v are solutions to the
real equation. Where did I use the fact that A
and B were real numbers and not complex numbers?
In knowing that this is the real part, I had to know that A
was a real number. If A were something like one
plus i, I'd be screwed,
I mean, because then I couldn't say that this was the real part
anymore. So, saying that's the real
part, and this is the imaginary part, I was using the fact that
these two numbers, constants, were real constants:
very important. So, what is the case two
solution? Well, what are the real and
imaginary parts of (a plus b i) t?
Well, y equals e to the at + ibt.
Okay, you've had experience. You know how to do this now.
That's e to the at times, well, the real part is,
well, let's write it this way. The real part is e to the at
times cosine b t. Notice how the a and b enter into the expression.
That's the real part. And, the imaginary part is e to
the at times the sine of bt. And therefore, the solution,
both of these must, therefore, be solutions to the
equation. And therefore,
the general solution to the ODE is y equals, now,
you've got to put in the arbitrary constants.
It's a nice thing to do to factor out the e to the at. It makes it look a little
better. And so, the constants are c1
cosine bt and c2 sine bt. Yeah, but what does that look
like? Well, you know that too.
This is an exponential, which controls the amplitude.
But this guy, which is a combination of two
sinusoidal oscillations with different amplitudes,
but with the same frequency, the b's are the same in both of
them, and therefore, this is, itself,
a purely sinusoidal oscillation.
So, in other words, I don't have room to write it,
but it's equal to, you know.
It's a good example of where you'd use that trigonometric
identity I spent time on before the exam.
Okay, let's work a quick example just to see how this
works out. Well, let's get rid of this.
Okay, let's now make the damping, since this is showing
oscillations, it must correspond to the case
where the damping is less strong compared with the spring
constant. So, the theorem is that if you
have a complex solution, u plus iv, so each of these is
a function of time, u plus iv is the
complex solution to a real differential equation with
constant coefficients. A stiff spring,
one that pulls with hard force is going to make that thing go
back and forth, particularly at the dipping is
weak. So, let's use almost the same
equation as I've just concealed. But, do you remember a used
four here? Okay, before we used three and
we got the solution to look like that.
Now, we will give it a little more energy by putting some
moxie in the springs. So now, the spring is pulling a
little harder, bigger force,
a stiffer spring. Okay, the characteristic
equation is now going to be r squared plus 4r plus five is
equal to zero. And therefore,
if I solve for r, I'm not going to bother trying
to factor this because I prepared for this lecture,
and I know, quadratic formula time, minus four plus or minus
the square root of b squared, 16, minus four times five,
16 minus 20 is negative four all over two.
And therefore, that makes negative two plus or
minus, this makes, simply, i.
2i divided by two, which is i.
So, the exponential solution is e to the negative two,
you don't have to write this in.
You can get the thing directly. t, let's use the one with the
plus sign, and that's going to give, as the real solutions,
e to the negative two t times cosine t,
and e to the negative 2t times the sine of t. And therefore,
the solution is going to be y equals e to the negative 2t
times (c1 cosine t plus c2 sine t). If you want to put initial
conditions, you can put them in the same way I did them before.
Suppose we use the same initial conditions: y of zero
equals one, and y prime of zero equals one,
equals zero, let's say, wait,
blah, blah, blah, zero, yeah.
Okay, I'd like to take time to actually do the calculation,
but there's nothing new in it. I'd have to take,
calculate the derivative here, and then I would substitute in,
solve equations, and when you do all that,
just as before, the answer that you get is y
equals e to the negative 2t, so,
choose the constants c1 and c2 by solving linear equations,
and the answer is cosine t, so, c1 turns out to be one,
and c2 turns out to be two, I hope.
Okay, I want to know, but what does that look like?
Well, use that trigonometric identity.
The e to the negative 2t is just a real
factor which is going to reproduce itself.
The question is, what is cosine t plus 2 sine t
look like? What's its amplitude as a pure
oscillation? It's the square root of one
plus two squared. Remember, it depends on looking at that little triangle,
which is one, two, this is a different scale
than that. And here is the square root of
five, right? And, here is phi,
the phase lag. So, it's equal to the square
root of five. So, it's the square root of
five times e to the negative 2t,
and the stuff inside is the cosine of, the frequency is one.
Circular frequency is one, so it's t minus phi,
where phi is this angle. How big is that,
one and two? Well, if this were the square
root of three, which is a little less than
two, it would be 60 infinity. So, this must be 70 infinity.
So, phi is 70 infinity plus or minus five, let's say.
So, it looks like a slightly delayed cosine curve,
but the amplitude is falling. So, it has to start.
So, if I draw it, here's one, here is,
let's say, the square root of five up about here.
Then, the square root of five times e to the negative 2t
looks maybe something like this.
So, that's square root of five e to the negative 2t. This is cosine t,
but shoved over by not quite pi over two.
It starts at one, and with the slope zero.
So, the solution starts like this.
It has to be guided in its amplitude by this function out
there, and in between it's the cosine curve.
But it's moved over. So, if this is pi over two,
the first time it crosses, it's 72 infinity to the right
of that. So, if this is pi over two, it's pi over two plus
70 infinity where it crosses. So, it must be doing something
like this. And now, on the other side,
it's got to stay within the same amplitude.
So, it must be doing something like this.
Okay, that gets us to, if this is the under-damped
case, because if you're trying to do this with a swinging door,
it means the door's going to be swinging back and forth.
Or, our little mass now hidden, but you could see it behind
that board, is going to be doing this.
But, it never stops. It never stops.
It doesn't realize, but not in theoretical life.
So, this is the under-damped. All right, so it's like
Goldilocks and the Three Bears. That's too hot,
and this is too cold. What is the thing which is just
right? Well, that's the thing you're
going to study on the problem set.
So, just right is called critically damped.
It's what people aim for in trying to damp motion that they
don't want. Now, what's critically damped?
It must be the case just in between these two.
Neither complex, nor the roots different.
It's the case of two equal roots.
So, r squared plus Ar plus B equals zero
has two equal roots. Now, that's a very special
equation. Suppose we call the root,
since all of these, notice these roots in this
physical case. The roots always turn out to be
negative numbers, or have a negative real part.
I'm going to call the root a. So, r equals negative a,
the root. a is understood to be a
positive number. I want that root to be really
negative. Then, the equation looks like,
the characteristic equation is going to be r plus a,
right, if the root is negative a, squared because it's a double
root. And, that means the equation is
of the form r squared plus two times a r plus a squared equals
zero. In other words,
the ODE looked like this. The ODE looked like y double
prime plus 2a y prime plus, in other words, the damping and the spring
constant were related in this special wake,
that for a given value of the spring constant,
there was exactly one value of the damping which produced this
in between case. Now, what's the problem
connected with it? Well, the problem,
unfortunately, is staring us in the face when
we want to solve it. The problem is that we have a
solution, but it is y equals e to the minus at.
I don't have another root to get another solution with.
And, the question is, where do I get that other
solution from? Now, there are three ways to
get it. Well, there are four ways to
get it. You look it up in Euler.
That's the fourth way. That's the real way to do it.
But, I've given you one way as problem number one on the
problem set. I've given you another way as
problem number two on the problem set.
And, the third way you will have to wait for about a week
and a half. And, I will give you a third
way, too. By that time,
you won't want to see any more ways.
But, I'd like to introduce you to the way on the problem set.
And, it is this, that if you know one solution
to an equation, which looks like a linear
equation, in fact, the piece can be functions of
t. They don't have to be constant,
so I'll use the books notation with p's and q's.
y prime plus q y equals zero. If you know one solution,
there's an absolute, ironclad guarantee,
if you'll know that it's true because I'm asking you to prove
it for yourself. There's another of the form,
having this as a factor, one solution y one,
let's call it, y equals y1 u is
another solution. And, you will be able to find
u, I swear. Now, let's, in the remaining
couple of minutes carry that out just for this case because I
want you to see how to arrange the work nicely.
And, I want you to arrange your work when you do the problem
sets in the same way. So, the way to do it is,
the solution we know is e to the minus at.
So, we are going to look for a solution to this differential
equation. That's the differential
equation. And, the solution we are going
to look for is of the form e to the negative at times u. Now, you're going to have to
make calculation like this several times in the course of
the term. Do it this way.
y prime equals, differentiate,
minus a e to the minus a t u plus e to the minus a
t u prime. And then, differentiate again. The answer will be a squared. You differentiate this:
a squared e to the negative at u.
I'll have to do this a little fast, but the next term will be,
okay, minus, so this times u prime,
and from this are you going to get another minus.
So, combining what you get from here, and here,
you're going to get minus 2a e to the minus a t u prime. And then, there is a final
term, which comes from this, e to the minus a t u double
prime. Two of these,
because of a piece here and a piece here combine to make that.
And now, to plug into the equation, you multiply this by
one. In other words,
you don't do anything to it. You multiply this line by 2a,
and you multiply that line by a squared, and you add them. On the left-hand side,
I get zero. What do I get on the right?
Notice how I've arrange the work so it adds nicely.
This has a squared times this, plus 2a times that,
plus one times that makes zero. 2a times this plus one times
this makes zero. All that survives is e to the a
t u double prime, and therefore,
e to the minus a t u double prime is equal to zero. So, please tell me,
what's u double prime? It's zero.
So, please tell me, what's u?
It's c1 t plus c2. Now, that gives me a whole family of solutions.
Just t would be enough because all I am doing is looking for
one solution that's different from e to the minus a t. And, that solution,
therefore, is y equals e to the minus a t times t. And, there's my second
solution. So, this is a solution of the
critically damped case. And, you are going to use it in
three or four of the different problems on the problem set.
But, I think you can deal with virtually the whole problem set,
except for the last problem, now.
