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We'll now move from the world of first order differential

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equations to the world of second order

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differential equations.

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And what does that mean?

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That means that we're now going to start involving the

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second derivative.

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And the first class that I'm going to show you-- and this

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is probably the most useful class when you're studying

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classical physics-- are linear second order

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differential equations.

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So what is a linear second order differential equation?

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I think I touched on it a little bit in our very first

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intro video.

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But it's something that looks like this.

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If I have a of x-- so some function only of x-- times the

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second derivative of y, with respect to x, plus b of x,

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times the first derivative of y, with respect to x, plus c

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of x, times y is equal to some function that's only a

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function of x.

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So just to review our terminology, y is the second

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order because the highest derivative here is the second

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derivative, so that makes it second order.

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And what makes it linear?

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Well all of the coefficients on-- and I want to be careful

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with the term coefficients, because traditionally we view

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coefficients as always being constants-- but here we have

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functions of x as coefficients.

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So in order for this to be a linear differential equation,

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a of x, b of x, c of x and d of x, they all have to be

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functions only of x, as I've drawn it here.

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And now, before we start trying to solve this

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generally, we'll do a special case of this, where a, b, c

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are constants and d is 0.

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So what will that look like?

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So I can just rewrite that as A-- so now A is not a function

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anymore, it's just a number-- A times the second derivative

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of y, with respect to x, plus B times the first derivative,

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plus C times y.

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And instead of having just a fourth constant, instead of d

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of x, I'm just going to set that equal to 0.

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And by setting this equal to 0, I have now introduced you

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to the other form of homogeneous

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differential equation.

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And this one is called homogeneous.

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And I haven't made the connection yet on how these

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second order differential equations are related to the

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first order ones that I just introduced-- to these other

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homogeneous differential equations I introduced you to.

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I think they just happen to have the same name, even

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though they're not that related.

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So the reason why this one is called homogeneous is because

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you have it equal to 0.

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So this is what makes it homogeneous.

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And actually, I do see more of a connection between this type

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of equation and milk where all the fat is spread out, because

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if you think about it, the solution for all homogeneous

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equations, when you kind of solve the equation, they

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always equal 0.

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So they're homogenized, I guess is the best way that I

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can draw any kind of parallel.

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So we could call this a second order linear because A, B, and

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C definitely are functions just of-- well, they're not

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even functions of x or y, they're just constants.

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So second order linear homogeneous-- because they

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equal 0-- differential equations.

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And I think you'll see that these, in some ways, are the

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most fun differential equations to solve.

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And actually, often the most useful because in a lot of the

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applications of classical mechanics, this is all you

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need to solve.

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But they're the most fun to solve because they all boil

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down to Algebra II problems. And I'll touch

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on that in a second.

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But let's just think about this a little bit.

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Think about what the properties of these

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solutions might be.

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Let me just throw out something.

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Let's say that g of x is a solution.

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So that means that A times g prime prime, plus B times g

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prime, plus C times g is equal to 0.

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Right?

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These mean the same thing.

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Now, my question to you is, what if I have some

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constant times g?

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Is that still a solution?

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So my question is, let's say some constant c1 gx-- c1 times

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g-- is this a solution?

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Well, let's try it out.

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Let's substitute this into our original equation.

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So A times the second derivative of this would just

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be-- and I'll switch colors here; let me switch to brown--

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so A times the second derivative of this would be--

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the constant, every time you take a derivative, the

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constant just carries over-- so that'll just be A times c1

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g prime prime, plus-- the same thing for the first

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derivative-- B times c1 g prime, plus C-- and this C is

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different than the c1 c-- times g.

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And let's see whether this is equal to 0.

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So we could factor out that c1 constant, and we get c1 times

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Ag prime prime, plus Bg prime, plus Cg.

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And lo and behold, we already know.

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Because we know that g of x is a solution, we know

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that this is true.

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So this is going to be equal to 0.

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Because g is a solution.

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So if this is 0, c1 times 0 is going to be equal to 0.

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So this expression up here is also equal to 0.

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Or another way to view it is that if g is a solution to

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this second order linear homogeneous differential

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equation, then some constant times g is also a solution.

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So this is also a solution to the differential equation.

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And then the next property I want to show you-- and this is

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all going someplace, don't worry.

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The next question I want to ask you is, OK, we know that g

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of x is a solution to the differential equation.

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What if I were to also tell you that h

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of x is also a solution?

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So my question to you is, is g of x plus h of x a solution?

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If you add these two functions that are both solutions, if

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you add them together, is that still a solution of our

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original differential equation?

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Well, let's substitute this whole thing into our original

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differential equation, right?

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So we'll have A times the second

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derivative of this thing.

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Well, that's straightforward enough.

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That's just g prime prime, plus h prime prime, plus B

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times-- the first derivative of this thing-- g prime plus h

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prime, plus C times-- this function-- g plus h.

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And now what can we do?

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Let's distribute all of these constants.

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We get A times g prime prime, plus A times h prime prime,

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plus B times the first derivative of g, plus B times

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the first derivative of h, plus C times g,

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plus C times h.

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And now we can rearrange them.

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And we get A-- let's take this one; let's take all the g

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terms-- A times the second derivative of g, plus B times

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the first derivative, plus C times g-- that's these three

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terms-- plus A times the second derivative of h, plus B

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times the first derivative, plus C times h.

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And now we know that both g and h are solutions of the

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original differential equation.

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So by definition, if g is a solution of the original

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differential equation, and this was the left-hand side of

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that differential equation, this is going to be equal to

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0, and so is this going to be equal to 0.

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So we've shown that this whole expression is equal to 0.

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So if g is a solution of the differential equation-- of

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this second order linear homogeneous differential

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equation-- and h is also a solution, then if you were to

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add them together, the sum of them is also a solution.

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So in general, if we show that g is a solution and h is a

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solution, you can add them.

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And we showed before that any constant times

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them is also a solution.

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So you could also say that some constant times g of x

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plus some constant times h of x is also a solution.

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And maybe the constant in one of the

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cases is 0 or something.

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I don't know.

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But anyway, these are useful properties to maybe

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internalize for second order homogeneous linear

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differential equations.

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And in the next video, we're actually going to apply these

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properties to figure out the solutions for these.

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And you'll see that they're actually straightforward.

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I would say a lot easier than what we did in the previous

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first order homogeneous difference equations, or the

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exact equations.

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This is much, much easier.

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I'll see you in the next video.

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