GILBERT STRANG: OK. So this is about the
world's fastest way to solve differential equations. And you'll like that method. First we have to
see what equations will we be able to solve. Well, linear,
constant coefficients. I made all the coefficients
1, but no problem to change those to A, B, C.
So the nice left-hand side. And on the right-hand side,
we also need something nice. We want a nice function. And I'll tell you which
are the nice functions. So I can say right away
that e to the exponentials are nice functions, of course. They're are always at the
center of this course. So for example,
equal e to the st. That would be a nice function. OK. And the key is, we're looking
for a particular solution, because we know how to
find null solutions. We're looking for particular
solution for this equation. One function, some
function that solves this equation with
right-hand side e to the st. And the point is, we
know what to look for. We just have some
coefficient to find. And we'll find that by
substituting in the equation. Now, do you remember
what we look for when the right-hand
side is e to the st? Then look for y equals some
constant times e to the st, right? When f of t-- maybe I'll put
the equal sign down there. If f of t is e to the st, then I
just look for a multiple of it. That's one coefficient to be
determined by substitute this into the equation. Do you remember the results? So this is our best example. When I put this in the equation,
I'll get the derivative brings an s. Second derivative
brings another s. So I get s squared and an s
and a 1 times y e to the st is equal to e to the st.
We've done that before. Here we see it as a case with
undetermined coefficient y. But by plugging it
in, I've discovered that y is 1 over that. So that's a nice function then.
e to the st is a nice function. What are the other
nice functions? So now, let me move to the other
board, next board, and ask, what other right-hand
sides could we solve? So I'll keep this
left-hand side equal to. So e to the st was 1. What about t? What about t? A polynomial. Well, that only has one term. So what would be a particular
solution to that equation? So I really have to say, what
is the-- try y particular equals-- now, if
I see a t there, then I'm going to
look for a t in y. And I'll also look
for a constant. So a plus bt would be the
correct form to look for. Let me just show
you how that works. So this now has two
undetermined coefficients. And we determine them by
putting that into the equation and making it right. So try yp is a plus
bt in this equation. OK, the second derivative
of a plus bt is 0. The first derivative
of that is b. So I get a b from that. And y itself is a plus bt. And that's supposed to give t. You see, I plugged it in. I got to this equation. Now I can determine a
and b by matching t. So then b has to be 1. We get b equal to 1. So the t equals t. But if b is 1, I need a to
b minus 1 to cancel that. So a is minus 1. And my answer is
minus 1 plus 1t. t minus 1. And if I put that into the
equation, it will be correct. So I have found a
particular solution, and that's my goal,
because I know how to find null solutions. And then together, that's
the complete solution. So we've learned what
to try with polynomials. With a power of t,
we want to include that power and all lower
powers, all the way down through the constants. OK. With exponentials, we just have
to include the exponential. What next? How about sine t or cosine t? Say sine t. So that case works. Now we want to try y double
prime plus y prime plus y equals, say, sine t. OK. What form do we assume for that? Well, I can tell you quickly. We assume a sine t in it. And we also need to
assume a cosine t. The rule is that the things
we try-- so I'll try y. y particular is what
we're always finding. Some c1 cos t, and
some c2 sine of t. That will do it. In fact, if I plug that in,
and I match the two sides, I determine c1 and
c2, I'm golden. Let me just comment
on that, rather than doing out every step. Again, the steps are
just substitute that in and make the equation correct
by choosing a good c1 and c2. I just noticed
that, you remember from Euler's great
formula that the cosine is a combination of e to the
it and e to the minus it. So in a way, we're really
using the original example. We're using this example,
e to the st, with two s's, e to the it, and
e to the minus it. So we have two
exponentials in a cosine. So I'm not surprised that there
are two constants to find. And now, finally, I
have to say, is this the end of nice functions? So nice functions include
exponentials, polynomials. These are really exponentials,
complex exponentials. And no, there's one
more possibility that we can deal with
in this simple way. And that possibility
is a product of-- so now I'll show you what
to do if it was t times sine t. Suppose we have the
right-hand side, the f of t, the forcing term,
is t times sine t. What is the form to assume? That's really all you have to
know is what form to assume? OK. Now, that t-- so we have here
a product, a polynomial times a sine or cosine or exponential. I could have done t
e to the st there. But what do I have to do
when the t shows up there? Then I have to try something
more with that t in there. So now I have a product
of polynomial times sine, cosine, or exponential. So what I try is at plus--
or rather, a plus bt. I try a product. Times cos t and c
plus dt times sine t. That's about as bad a case
as we're going to see. But it's still quite pleasant. So what do I see there? Because of the t here, I
needed to assume polynomials up to that same degree 1. So a plus bt. Had to do that, just
the way I did up there when there was a t. But now it multiplies sine t. So I have to allow sine
t and also cosine t. The pattern is, really,
we've sort of completed the list of nice functions. Exponentials, polynomials, and
polynomial times exponential. That's really what
a nice function is. A polynomial times
an exponential. Or we could have a
sum of those guys. We could have two or three
polynomial times exponential, like there and another one. And that's still
a nice function. And what's the real
key to nice functions? The key point is, why is this
such a good bunch of functions? Because, if I take
its derivative, I get a function
of the same form. If I take the derivative
of that right-hand side, and I use the
product rule, you see I'll get this times
the derivative of that. So I'll have something
looking with a sine in there. And I get this
times the derivative of that, which is just a b. So again, it fits the same
form, polynomial times cosine, polynomial times sine. So here I have a case
where I have actually four coefficients. But they'll all fall
out when you plug that into the equation. You just match terms
and your golden. So it really is a
straightforward method. Straightforward. So the key about
nice functions is-- and they're nice for
Laplace transforms, they're nice at every step. But it's the same good functions
that we keep discovering as our best examples. The key about nice functions
is that the-- that's a form of a nice function
because its derivative has the same form. The derivative of that function
fits that pattern again. And then the second
derivative fits. All the derivatives fit. So when we put them in the
equation, everything fits. And always in the last
minute of a lecture, there's a special case. There's a special case. And let's remember what that is. So special case when we
have to change the form. And why would we
have to do that? Let me do y double prime
minus y, say, is e to the t. What is special about that? What's special is that
this right-hand side, this f function,
solves this equation. If I try e to the
t, it will fail. Try y equals some Y e to the t. Do you see how
that's going to fail? If I put that into the
equation, the second derivative will cancel the y and I'll
have 0 on the left side. Failure, because that's
the case called resonance. This is a case of
resonance, when the form of the right-hand
side is a null solution at the same time. It can't be a
particular solution. It won't work because
it's also a null solution. And do you remember how
to escape resonance? How to deal with resonance? What happens with resonance? The solution is a
little more complicated, but it fits everything here. We have to assume to allow a t. We have to allow a t. So instead of this multiple,
the and in this thing, we have to allow-- so I'm
going to assume-- I have to have-- I need a t in there. Oh, no. Actually, I don't. I just need a t. That would do it. When there's resonance,
take the form you would normally
assume and multiply by that extra factor t. Then, when I substitute that
into the differential equation, I'll find Y's quite safely. I'll find Y entirely safely. So I do that. So that's the resonant case,
the sort of special situation when e to the t solved this. So we need something new. And the way we get the right new
thing is to have a t in there. So when I plug that in, I take
the second derivative of that, subtract off that
itself, match e to the t. And that will tell
me the number Y. Perhaps it's 1/2 or 1. I won't do it. Maybe I'll leave
that as an exercise. Put that into the equation and
determine the number capital Y. OK. Let me pull it together. So we have certain
nice functions, which we're going to see
again, because they're nice. Every method works well
for these functions. And these functions are
exponentials, polynomials, or polynomials
times exponentials. And within exponentials,
I include sine and cosine. And for those functions,
we know the form. We plug it into the equation. We make it match. We choose these
undetermined coefficients. We determine them so that
they solve the equation. And then we've got a
particular solution. So this is the best
equations to solve to find particular solutions. Just by knowing the right form
and finding the constants, it did come out of the
particular equation. OK. All good, thanks.
