Linear homogeneous equations
2nd Order Linear Homogeneous Differential Equations 1 Introduction to 2nd order, linear, homogeneous differential equations with constant coefficients.
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- 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.
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