Name: Second order homogeneous linear differential equations with constant coefficients
Uploaded: 2024-08-17T09:01:18.000Z
Duration: 11 min 44 s
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Future simple

In this video, I'll show you guys how to approach to solve a second-order linear differential equation with constant coefficients, and in this case, we want to focus on the right-hand side to be 0, and when the right-hand side is equal to 0, this is called a homogeneous situation.

To-infinitive

And once again, a, b, and c, they are constants, and in this case, we don't want a to be 0, otherwise this term will be gone, right?

All right, as we can see, we have y, y' which is the first derivative, and this is y'', the second derivative.

And it seems like these are just the constant multiples of the original first and the second derivative, right?

Do we know a function so that its derivative is just a constant of its original?

If I start with y equals to, well, if I have, let's say, e to some power times x, well, in this case, we use t, by the way, for second-order, use t.

Well, if you differentiate that, you get y' and you know this is going to give you e to the 3t, the function part stays exactly the same, but the chain rule says I will have to multiply by 3, right?

And you see, this 3 stays, and this is just the original, so 3 times y, right?

And likewise, I can do it again. y'' is going to be, we will keep this 3 times e to the 3t, but then we multiply this 3, right, for the derivative by the chain rule. 3 times 3 is 9, and e to the 3t is the original.

And you see, whenever you have an e to some power, the first derivative, the second derivative, they are just going to be the constant multiple of the original, right?

So what this is telling me is that this will suggest, let me just put this down, this suggests that for the function, we should have the form y equals to e to some power times t.

I don't know what this number should be, earlier I just used 3, right?

So in general, let me put down r times t.

And once again, for the second-order situation, we usually use t because there are a lot of applications that impose time and things like that.

First of all, we begin by saying y is equal to e to the rt power, and the idea is that I'm just going to go ahead, differentiate this twice, and then plug in.

Hopefully, we can squeeze out some conditions that will help us to solve for this kind of differential equation.

All right, y' is going to be e to the rt times r, let me just put r in the front, and then y'' is going to be, this right here repeats, so r e rt, but then we multiply by another r, which will be r times r, which is r².

And now, let me put all this into their corresponding.

Here we have a, y'' is r² e to the rt, and then we add it with b, so we put on plus b, y'' is r e to the rt, and then we continue, plus c, y is e to the rt, and this right here is equal to 0.

And now, can we kind of squeeze out a condition?

Every term has e to the rt, so we can factor it out, of course, e to the rt, and that will give us ar² plus br plus c equals to 0.

All right, this is an exponential part, right?

This right here, we know, is never 0, isn't it?

Never 0, let me just spell it out much better.

So, when we have this quantity times that quantity, since e to the rt is never 0, so you can either divide it out, or you can just forget about it, because we just want to focus on this.

So that means we must have this part, we must have, let me just write it down, we must have the situation that ar² plus br plus c equal to 0.

In fact, this right here is the condition that we need, because from here, this is pretty much just a quadratic equation, isn't it?

From here, we can solve for r, and we can just plug into the r here, and we can generate the building blocks of the solution.

And I'll show you guys what I mean by that.

And before I show you guys an example, let me tell you guys that this equation here has a name.

This is called the characteristic equation, and this is also called the auxiliary equation.

And now, let's go ahead and solve this for y'' minus 5y' minus 6y is equal to 0.

The first step is we have to take this and change it to its corresponding characteristic equation.

Earlier, when we start with a y'', we will end up ar², right?

That means right here, the first term is going to give me 4r².

Minus 5, and the y' corresponds with r, we change that to r right here, and then minus 6.

If you have y, we didn't have any r, right?

And now, we just have to solve this quadratic equation and do it whichever way that you would like, and I will factor this out for you guys.

And I'll show you guys with the tic-tac-toe factoring, and to do so, I want to ask myself, what times 4 gives me 4r²?

And let me tell you guys the correct combination, which is 4r times r.

And once again, let me tell you guys the correct combination, which is negative 2 plus 3 in these boxes like this, in this order.

Well, I will convince you guys this is correct.

To do so, you cross multiply 4r times negative 2, which is negative 8r, and then you take 3 times r, which is 3r.

It is correct because negative 8r plus 3r is negative 5r, so this is correct, isn't it?

And now we have to read the answers correctly from the boxes here.

So the first factor is going to be 4r plus 3, and the second one is r minus 2, and we still have equal to 0, of course.

All right, for the first one, we know r is equal to negative 3 over 4, and for the second one, we know r is equal to 2.

And you see, we end up with two different r values.

In this case, we will end up with two different building blocks for the solution, e to the first r, which is this, and times t.

The second one is going to be e to the 2t, right?

So let me just write down the building blocks for you guys first.

This right here is going to give me e to the negative 3 over 4t, and the second one is going to give me e to the 2t.

And I'm going to explain to you guys what do I mean by building blocks.

First of all, let me just show you guys that both of them will satisfy the original differential equation, and just for simplicity purpose, let me just check this one for you guys only, okay?

Let me just say this is the second one I get, so let me put down y2. y2 equals to e to the 2t, and then what I have to do is differentiate this, which is going to give me 2e2t, right?

And then I'll do it again. y2 double prime, which is 4e2t, right?

And now I will plug in all this into the original, and you see it will give me 4, right?

And y double prime, which is that, so we multiply by 4e2t, and then minus 5y prime, which is that, which is times 2e2t, and then minus 6y, which is that, e to the 2t, like this.

This is 16 minus 10, which is 6, and then minus 6, of course, everything ends up to be 0, so 0 is equal to 0, so it checks.

This right here definitely works, isn't it?

And you can do exactly the same thing for this, but you can also take my word for it, it will work out nicely as well.

All right, now, these are just the function parts.

What we'll be doing is that, well, remember, when we're solving differential equations, we have that constant, right?

And keep in mind, whenever we have the second-order differential equation, we will have two different constants.

Well, do I put on plus C, or divide by C?

The truth is, we multiply the function part by C1 and C2, so this right here is C1, and this right here is C2.

So, let me just demonstrate this right here on the side for you guys real quick, because now, you see, when I multiply this function by C, so let me just put on C in black like this, guess what?

I will just have to multiply everything by C, isn't it?

So this right here will end up with C, C, C, C, C, C, C.

And then, pretty much this term, we'll now have C right here, and this function here, this part here, has a C here as well.

This right here, just multiply by C, right?

We're just plugging everything accordingly.

When I have this C here, I will still end up with 0 on the left-hand side, of course still equal to 0.

The point of that is to show you, these are the places where C should be.

C1 and C2 are just the multiple of the function part.