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Studying for a test? Prepare with these 3 lessons on Second order linear equations.

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# 2nd order linear homogeneous differential equations 1

Video transcript

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.