WEBVTT
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GILBERT STRANG: OK.
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So can I begin with a few
words about the big picture
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of solving
differential equations?
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So if that was a
nonlinear equation,
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we would go to
computer solutions.
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And Cleve Moler is making
a parallel video series
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about the Matlab suite
of codes for solving
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differential equations.
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Then when that
equation is linear,
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as it is here, with constant
coefficients, as like the 1,
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minus 3, and the
2, we can always
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get a formula for the answer.
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Involves an integral.
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There's still one integral to do
involving the impulse response.
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And you'll see that.
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But there are few, the most
beautiful, the most simple
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equations when the right-hand
side has a special form.
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And that's one
possibility-- t, or t
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squared we could deal
with, or e to the t.
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We'll see all those.
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Then we know what the
solution looks like.
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For example-- first of all,
we know the null solutions,
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of course.
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That's when this is 0.
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And then I just wrote for e to
the st, and I find s could be 1
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or s could be 2.
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Those are two solutions
with right side 0.
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So we can match
initial conditions.
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But we need a
particular solution.
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And that's where this
one is especially simple.
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So the idea is to try the form
we know the solution will have.
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So I'm going to try a
particular solution.
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When I see a t there, I'll want
a t in the particular solution.
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But I also need
the constant term.
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So I'll try a plus bt.
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What I mean by try is put
that into the equation,
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match the left side
and the right side,
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and we have a solution.
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And I'll just go
ahead and do that.
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So there's several--
this video is mostly
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about the list of
possible nice functions.
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And that's one of them.
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OK.
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So if I put that
into the equation,
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the second derivative
is 0 for that.
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So I have minus 3.
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The first derivative is just b.
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Then I have plus 2 times y
itself, which is a plus bt.
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So the left side of the
equation is just this much.
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And it has to equal 4t.
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And that we can make happen.
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I see 2bt and 4t here,
so I want b equal to 2.
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And so when b is 2, then
I have 4t matching 4t.
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And then I have minus
3b plus 2a should be 0.
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Minus 3b plus 2a, the constant
part there we don't want,
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so that should be 0.
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We already know b is 2,
so that's minus 6 plus 2a.
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a is 3.
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a is 3.
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Now, that's perfectly satisfied
by b equal 2a equal 3,
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and that's the answer.
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Correct answer is
b was 2, a was 3,
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and we don't have
to say try anymore.
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We got it.
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Done.
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One other right-hand side.
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Once you get a nice one like
this, you look for more.
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If it was t squared,
we would assume--
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what would we assume if
it was a 4t squared there?
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We would assume a plus
bt plus ct squared.
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We want to match
the right-hand side.
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Now, a different type
of right-hand side
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we know already.
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What if this was e to
the-- say, e to the 3t?
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Or e to the st. Let me put any
exponent there for the moment.
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So now we have a
different right-hand side.
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A very nice function.
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The best function of
differential equations.
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And now what we will try
with this right-hand side,
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e to the st. We've
seen it before.
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The particular solution
is just y e to the st.
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The undetermined coefficients
were a and b in the first time.
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Here the undetermined
coefficient
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is capital Y. I'm just going
to plug that into the equation
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and match the left
side and right side.
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And I'll determine this
coefficient, capital Y.
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So what happens when I put
that into the equation?
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I get second
derivative brings down
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an s squared, and first
derivative, we have a minus 3.
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The first derivative brings
down an s, and I have plus 2.
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All that is Y e
to the st, right?
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I put in Y e to the
st. The derivatives
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brought down this
familiar polynomial.
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And it's all supposed
to match e to the st.
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We can do that.
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That's a perfect match.
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The left side has the same
form as the right side.
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I can cancel the e
to the st's, and I
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discover that Y is 1 over
s squared minus 3s plus 2.
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Good.
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That's the coefficient,
1 over that.
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Let's see.
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I have to make two comments.
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One comment is that s-- we're
totally golden if that s
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is imaginary.
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If s is i omega,
well, minus i omega,
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or both, both
possibilities, those
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will give us sine and cosine.
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So we'll add those
to the nice function.
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So the nice functions
are t, polynomials.
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E to the st, exponentials.
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Sines and cosines.
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And you see this
worked perfectly.
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Well, perfectly except
we have to be sure
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that we don't have 1/0.
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When would we have 1/0?
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If s is 1, this would be 1 minus
3 plus 2, that would be 1/0.
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I can't deal with s equal
1 with that assumption.
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Doesn't work.
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Also, s equal 2.
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What's special about s
equal 1 and s equal 2?
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Those make that 0.
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They give the null solutions.
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And I know that if this
happened to be e to the t,
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and a null solution
was also e to the t,
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I have to fix the particular
solution by giving it
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an extra factor t.
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Let me do that at the
end of the lecture.
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That's the resonance idea,
where my null solution
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is the same as what I hope
for the particular solution.
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So I have to change
that particular solution
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with an extra factor t.
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Let me keep going
here to be sure we
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get all the possibilities.
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Again, we're getting a very
small set of nice functions.
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But fortunately, they appear
quite often in applications.
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Constants, linear like
4t, exponentials like e
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to the 5t, oscillation
like e to the i omega t.
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So let me make a list.
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So polynomials, like the
forcing function could be t,
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could be 1.
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Constant, that would
be like a ramp.
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That would be like
a step function.
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They could be 3 to
the st, but not s
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equals 1 and 2 in this problem.
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They could be e to
the i omega t, and e
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to the minus i omega t, which
leads me directly to cosine
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of omega t and sine of omega t.
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I'm creating here a list
of the nice functions--
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polynomials, exponentials,
cosine and sine,
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because those come
from exponentials,
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and finally, I can
multiply these.
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I could have, for example, t e
to the st. I could allow that.
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And now, all I have
to do is tell you,
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what form do I try to plug-in?
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The form will have
undetermined coefficients.
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I'll substitute in
the equation, and I'll
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determine the coefficients.
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So here we saw a plus bt.
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That was good.
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That was the Y. Here we saw
Y e to the st. That was good.
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Here, what will I have?
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If I have cosine, I need
the sine there also.
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So I'll have to allow
a combination of those.
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Say M on this plus N of that.
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If I tried to do
cosines alone, I
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would be in danger of taking
the derivative, getting a sine,
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and having nothing to match.
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So I'll take that.
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Now, final question,
or next question
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is, if I multiply two of these,
the product rule is still
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going to tell me that
derivatives of that
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have the same form as that.
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Derivatives of
this t e to the st
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have the same-- they also
involve t or e to the st,
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with factors s, from
the product rule.
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So what do I assume here?
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Well, when I see t
there, I have to include,
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as I did up there,
also constants.
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And if I saw a t squared,
I would go up to t squared.
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I'd have three coefficients.
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Now, that e to the
st, I can keep.
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So what I've put on
the right-hand side
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is the right form to assume.
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It's just like good advice.
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Put that into the
differential equation when
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this is the right-hand side.
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Match left side
with the right side.
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That will tell you a and b.
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And you have the answer.
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You have the
particular solution.
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You have a particular solution,
and that's what you wanted.
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OK.
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And if I had t
cosine omega t, do
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you want to see in
the whole business?
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If I had t cosine
omega t-- oh dear,
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it's getting a little messy.
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I'd need an a plus bt
to deal with the t.
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And I'd need a cosine
to deal with the cosine.
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And then, just to make the
problem a little messier, can't
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be helped, I'd need the sine.
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So I need a sine omega t.
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And I'm going to need
a c plus dt there.
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OK.
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I'm up to four coefficients
to determine by plugging in.
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It uses more ink, but it
doesn't use more thinking.
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You just put it in,
match all terms,
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and you discover A, B, C, and D.
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Finally, I have to say
that word about resonance
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that I mentioned earlier.
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The possibility that if
s happened to be 1 or 2
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on the right-hand side,
that's also a null solution.
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This method would give me
1/0, infinite answer, no good.
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And we know what to do.
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We know how to deal
with resonance.
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So with resonance-- so could
I just finally complete
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the whole story.
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Resonance.
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In this example, s equal 1 or 2.
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What would I do with f
of t equals e to the t?
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If that was the right-hand side,
that would give me resonance.
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It's got the exponent
s equal 1, which
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is also in the null
solution, e to the t.
00:14:35.730 --> 00:14:37.420
So what do I try?
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So I try-- everybody knows what
happens when there's resonance.
00:14:42.960 --> 00:14:45.110
When you have this doubling up.
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You need an extra
factor t to rescue.
00:14:48.470 --> 00:14:51.810
So you would try y of t.
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y, this is the particular
solution I'm looking for.
00:14:55.100 --> 00:15:02.790
You would try t e
to the t, with a Y.
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And you would plug that in.
00:15:06.650 --> 00:15:10.170
You would find the right
number for capital Y.
00:15:10.170 --> 00:15:12.940
And you'd have the
particular solution.
00:15:12.940 --> 00:15:16.220
Only, I think, doing a
few examples of this,
00:15:16.220 --> 00:15:19.970
you get the knack of
assume the right form,
00:15:19.970 --> 00:15:24.130
put it into the equation, match
left side with right side,
00:15:24.130 --> 00:15:27.660
and that reveals the
undetermined coefficients.
00:15:27.660 --> 00:15:32.590
It tells you what a and b
and capital Y and c and d,
00:15:32.590 --> 00:15:34.740
tells you what they
all have to be.
00:15:34.740 --> 00:15:37.420
So this is a really
good method that
00:15:37.420 --> 00:15:42.560
applies to really
nice right-hand sides.
00:15:42.560 --> 00:15:43.060
Good.
00:15:43.060 --> 00:15:44.610
Thanks.