We'll now move from the world of first order differential
equations to the world of second order
differential equations.
And what does that mean?
That means that we're now going to start involving the
second derivative.
And the first class that I'm going to show you-- and this
is probably the most useful class when you're studying
classical physics-- are linear second order
differential equations.
So what is a linear second order differential equation?
I think I touched on it a little bit in our very first
intro video.
But it's something that looks like this.
If I have a of x-- so some function only of x-- times the
second derivative of y, with respect to x, plus b of x,
times the first derivative of y, with respect to x, plus c
of x, times y is equal to some function that's only a
function of x.
So just to review our terminology, y is the second
order because the highest derivative here is the second
derivative, so that makes it second order.
And what makes it linear?
Well all of the coefficients on-- and I want to be careful
with the term coefficients, because traditionally we view
coefficients as always being constants-- but here we have
functions of x as coefficients.
So in order for this to be a linear differential equation,
a of x, b of x, c of x and d of x, they all have to be
functions only of x, as I've drawn it here.
And now, before we start trying to solve this
generally, we'll do a special case of this, where a, b, c
are constants and d is 0.
So what will that look like?
So I can just rewrite that as A-- so now A is not a function
anymore, it's just a number-- A times the second derivative
of y, with respect to x, plus B times the first derivative,
plus C times y.
And instead of having just a fourth constant, instead of d
of x, I'm just going to set that equal to 0.
And by setting this equal to 0, I have now introduced you
to the other form of homogeneous
differential equation.
And this one is called homogeneous.
And I haven't made the connection yet on how these
second order differential equations are related to the
first order ones that I just introduced-- to these other
homogeneous differential equations I introduced you to.
I think they just happen to have the same name, even
though they're not that related.
So the reason why this one is called homogeneous is because
you have it equal to 0.
So this is what makes it homogeneous.
And actually, I do see more of a connection between this type
of equation and milk where all the fat is spread out, because
if you think about it, the solution for all homogeneous
equations, when you kind of solve the equation, they
always equal 0.
So they're homogenized, I guess is the best way that I
can draw any kind of parallel.
So we could call this a second order linear because A, B, and
C definitely are functions just of-- well, they're not
even functions of x or y, they're just constants.
So second order linear homogeneous-- because they
equal 0-- differential equations.
And I think you'll see that these, in some ways, are the
most fun differential equations to solve.
And actually, often the most useful because in a lot of the
applications of classical mechanics, this is all you
need to solve.
But they're the most fun to solve because they all boil
down to Algebra II problems. And I'll touch
on that in a second.
But let's just think about this a little bit.
Think about what the properties of these
solutions might be.
Let me just throw out something.
Let's say that g of x is a solution.
So that means that A times g prime prime, plus B times g
prime, plus C times g is equal to 0.
Right?
These mean the same thing.
Now, my question to you is, what if I have some
constant times g?
Is that still a solution?
So my question is, let's say some constant c1 gx-- c1 times
g-- is this a solution?
Well, let's try it out.
Let's substitute this into our original equation.
So A times the second derivative of this would just
be-- and I'll switch colors here; let me switch to brown--
so A times the second derivative of this would be--
the constant, every time you take a derivative, the
constant just carries over-- so that'll just be A times c1
g prime prime, plus-- the same thing for the first
derivative-- B times c1 g prime, plus C-- and this C is
different than the c1 c-- times g.
And let's see whether this is equal to 0.
So we could factor out that c1 constant, and we get c1 times
Ag prime prime, plus Bg prime, plus Cg.
And lo and behold, we already know.
Because we know that g of x is a solution, we know
that this is true.
So this is going to be equal to 0.
Because g is a solution.
So if this is 0, c1 times 0 is going to be equal to 0.
So this expression up here is also equal to 0.
Or another way to view it is that if g is a solution to
this second order linear homogeneous differential
equation, then some constant times g is also a solution.
So this is also a solution to the differential equation.
And then the next property I want to show you-- and this is
all going someplace, don't worry.
The next question I want to ask you is, OK, we know that g
of x is a solution to the differential equation.
What if I were to also tell you that h
of x is also a solution?
So my question to you is, is g of x plus h of x a solution?
If you add these two functions that are both solutions, if
you add them together, is that still a solution of our
original differential equation?
Well, let's substitute this whole thing into our original
differential equation, right?
So we'll have A times the second
derivative of this thing.
Well, that's straightforward enough.
That's just g prime prime, plus h prime prime, plus B
times-- the first derivative of this thing-- g prime plus h
prime, plus C times-- this function-- g plus h.
And now what can we do?
Let's distribute all of these constants.
We get A times g prime prime, plus A times h prime prime,
plus B times the first derivative of g, plus B times
the first derivative of h, plus C times g,
plus C times h.
And now we can rearrange them.
And we get A-- let's take this one; let's take all the g
terms-- A times the second derivative of g, plus B times
the first derivative, plus C times g-- that's these three
terms-- plus A times the second derivative of h, plus B
times the first derivative, plus C times h.
And now we know that both g and h are solutions of the
original differential equation.
So by definition, if g is a solution of the original
differential equation, and this was the left-hand side of
that differential equation, this is going to be equal to
0, and so is this going to be equal to 0.
So we've shown that this whole expression is equal to 0.
So if g is a solution of the differential equation-- of
this second order linear homogeneous differential
equation-- and h is also a solution, then if you were to
add them together, the sum of them is also a solution.
So in general, if we show that g is a solution and h is a
solution, you can add them.
And we showed before that any constant times
them is also a solution.
So you could also say that some constant times g of x
plus some constant times h of x is also a solution.
And maybe the constant in one of the
cases is 0 or something.
I don't know.
But anyway, these are useful properties to maybe
internalize for second order homogeneous linear
differential equations.
And in the next video, we're actually going to apply these
properties to figure out the solutions for these.
And you'll see that they're actually straightforward.
I would say a lot easier than what we did in the previous
first order homogeneous difference equations, or the
exact equations.
This is much, much easier.
I'll see you in the next video.