GILBERT STRANG: OK, this is
my second video on the Laplace transform, and this one will
be about solving second order equations. So let me remember the plan. So there's our second
order equation. And I'm taking
this example first, with the delta function
on the right-hand side. You remember, that's
a key example. And actually, we have a special
letter for the solution. This is an impulse. And the solution is
the impulse response. And I use a little g. So I should have
turned this y into a g. So that g is the
solution, starting from 0 initial
conditions, with a delta. So what's the plan? We want to take the
transform of every term. We have to check, what is
the transform, the Laplace transform of the delta function? You remember the definition
of the transform. It's this integral. You take your
function-- whatever it is, here delta-- multiply
by e to the minus st, and integrate. s equals 0 to infinity. Well, easy to integrate
with a delta function there. It's 0, except where the
impulse is, at t equals 0. And at t equals 0, this is 1. So the answer is 1. That's the nice Laplace
transform of the impulse. And now I want the transform
of the impulse response. So the impulse response
comes from this equation. So now, transform every term. So the transform of g
will be called capital G. And the transform of
the delta function is 1. And the derivative,
you remember, has an extra factor, s. The transform of the derivative
from the first Laplace transform video
was s times g of s. And the second
derivative, another s. So s squared, times g of s. We're not surprised to see the
very familiar quadratic, whose roots are the two exponents
s1 and s2, showing up here. We've seen this every
time, because we have constant coefficients. We always see this quantity. And now we see that, look,
the transform, capital G. I divide by that. It's exactly the
transfer function. So that's connecting in the
idea of the transfer function to the Laplace transform
of the impulse response, because I have this
in the denominator. OK, so I want now-- I've taken
the transform of the equation. I've got the transform of
little g, the impulse response. So now I'm ready to find G,
the impulse response by invert Laplace transform. How do I find the function
with that Laplace transform? Right now, it's 1
over a quadratic. The whole idea of
partial fractions is, split this so this
G of s, final step. It's G of s. Well, you remember that
this is the polynomial that has the two roots, s1 and s2. I'm going to write that
as 1 over s minus s1, and s minus s2. Those are the two roots
from the quadratic formula-- the two poles, we could
say, poles of G of s. And now I want to use
partial fractions. So I want to separate
this into two fractions. And it turns out that they
are 1 over s minus s1, minus 1 over s minus s2. And it turns out
there's a factor there to make it correct. You could check. When you put these, this
over a common denominator, you get this. And when you put that
over a common denominator, you'll get a numerator,
which you have to cancel. OK. So then, now I have
two simple poles. And I could write here what
I know, what g of t is. Remember, the function
with that transform is just e to the s1 t. The one most important
of all transforms is the transform
of the exponential, is that simple pole. So now I invert the
Laplace transform. So this gives me
an e to the s1 t, minus the function with that
transform, is e to the s2 t. And I still have this
constant, s1 minus s2. I've re-discovered
the impulse response. It's a solution to my
equation with impulse force. And it's that
particular function that plays such
an important part in the whole subject of constant
coefficient differential equations. Because you see it's
the critical thing here. Here's the critical
transfer function, and here is the inverse
Laplace transform. The Laplace transform
of that is that. Good. That's that example. And now I just want to take
another function than delta. OK. I could go all the way to
take f of t, any f of t. But let me stay with examples. So now I'm going to do y double
prime, plus b y prime, plus cy. They're all the most
important example. The best one we could do
would be cosine of omega t. An oscillating problem,
with an oscillating force, at a frequency different
from the natural frequency, and we also have damping there. So this is the standard,
spring mass dashpot problem, or RLC circuit
problem, certainly RLC circuits, highly important. And you might remember the
solution gets a little messy. A solution gets a little messy. It's highly important,
but-- so I'll carry it through
to the last step. But I probably won't
take that final step. Just, what I want you to
see is another example of the inverse
Laplace transform. So what do we need? We need the Laplace
transform of that. I plan to take the Laplace
transform of every term. s squared plus bs plus c. We'll multiply. I'm transforming everything. And here I have to put
the transform of that. OK. So how will I get that? And of course, there might
be a sine omega t in there. I would really like to
get them both at once. So let me put down
them both at once. The cosine and the sine are the
real and imaginary parts of e to the i omega t, right? Euler's great formula. e to the i omega t is cosine
omega t, the real part, plus i sine omega t,
the imaginary part. But now I know the transform of
e to the a t, e to the i omega t. So that transforms
to-- so I want the real and the
imaginary parts of 1 over, you remember what it is. It's just that
simple pole again, s minus the exponent i omega. So I'm going to get the cosine
and the sine at the same time, from one calculation, finding
the real and imaginary parts of this, 1 over s
plus s minus i omega. How do you deal
with a pole, when if you want the real
and imaginary parts, you're happy to get a real
number in the denominator and see the real and
imaginary parts up above. I don't like it when
it's down there. So what I'm going to do is,
real and imaginary parts of, I'll multiply s minus i omega. This is the key trick
with complex numbers. It comes up enough,
so it's good to learn. I multiply that by its
conjugate, s plus i omega over s plus i omega. So I multiplied by 1. But you see, what's
happened now is, real and imaginary parts
of-- I've got what I want, s plus i omega is now up in the
numerator, where I can see it. And what do I have down below? s minus i omega
times s plus i omega, the very important quantity. S squared minus i omega s, plus
i omega s, minus i squared, that's plus 1 omega
squared plus omega squared. OK. We've done it. I've transformed the
cosine and the sine into s over this,
and omega over this. So two at once, two transforms. And of course, as always, we're
able to do, and recognize, and work with the transforms
of a special group of nice functions,
exponentials above all, sines and cosines coming from
exponentials, delta functions. It's a short list. Those are the ones we can
do, and fortunately those are the ones that we need to do. OK, so I'm now ready to
put in the right-- what did that turn out to be? It turned out to be
s from the cosine, I'm not doing the
sine now, the cosine. I'm taking the real part. It's s over that positive s
squared plus omega squared. Are you with me? Left-hand side, all normal. I'm starting from 0
initial conditions, otherwise I would see the
initial values in here. But I don't, because they're 0. And over here, I've
got the transform of the right-hand side. OK? Then I just bring
that down there. So finally, I know y of s
is s over this all-important quadratic, and then I have the
s squared plus omega squared. OK. Well, you see it did get harder. We have s squared from
a quadratic there. We have two quadratics. We have a fourth-degree
polynomial down there. Partial fractions will work. Partial fractions can simplify
this, for any polynomials, but the algebra
gets quickly worse when you get up to degree four. But actually, this can
be, it could be done. And I don't plan to do it. To me, this would
eventually give the solution to this example. But we have other ways
to get that solution. And I believe, for me at least,
the other ways are simpler. I know that the solution is
a combination of cos omega t and sine omega t. And this is the particular
solution I'm talking about. And I can figure out what
those combinations are, because I know the right form. Here, I have to deal with
partial fractions and degree four, down there. I'm going to chicken
out on that one. I won't completely chicken out. I'll say what the
pieces look like. But I won't figure
out all the numbers. OK. So all this thing, I see as
some constant over-- well, this factors into s
minus s1 s minus s2. Those are the two
roots, as we saw above. So I have a linear, a
linear, and a quadratic. And partial fraction says
that I can separate out the first linear,
the second linear, and the third quadratic,
which I could factor too, but-- I could factor
s squared plus omega squared into that
times that, but it brings in imaginary numbers. So I'll put a cs and a d. OK. Four numbers to be determined,
or rather not to be determined. Because I'm going to stop there. What I've discovered
is this part would be the null
solution, would be a null solution because
it's involves e to the s1 t and e to the s2 t. This part, when I find
the inverse transform, it must be the combination of
cos omega t and sine omega t. So that's some
combination, when I find that it's the transform of-- [AUDIO OUT]
