– [Voiceover] Let’s

now introduce ourselves to the idea of a differential equation. And as we’ll see, differential equations are super useful for

modeling and simulating phenomena and understanding

how they operate. But we’ll get into that later. For now let’s just think

about or at least look at what a differential equation actually is. So if I were to write, so

let’s see here is an example of differential equation,

if I were to write that the second derivative of y plus two times the first derivative of y

is equal to three times y, this right over here is

a differential equation. Another way we could write it if we said that y is a function of

x, we could write this in function notation. We could write the second

derivative of our function with respect to x plus two

times the first derivative of our function is equal to

three times our function. Or if we wanted to use

the Leibniz notation, we could also write, the

second derivative of y with respect to x plus two

times the first derivative of y with respect to x is

equal to three times y. All three of these equations

are really representing the same thing, they’re saying

OK, can I find functions where the second

derivative of the function plus two times the first

derivative of the function is equal to three times

the function itself. So just to be clear,

these are all essentially saying the same thing. And you might have just

caught from how I described it that the solution to a

differential equation is a function, or a class of functions. It’s not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions,

or a class of functions. It’s important to contrast this relative to a traditional equation. So let me write that down. So a traditional equation,

maybe I shouldn’t say traditional equation,

differential equations have been around for a while. So let me write this as

maybe an algebraic equation that you’re familiar with. An algebraic equation

might look something like, and I’ll just write up a simple quadratic. Say x squared plus three x

plus two is equal to zero. The solutions to this algebraic equation are going to be numbers,

or a set of numbers. We can solve this, it’s

going to be x plus two times x plus one is equal to zero. So x could be equal to

negative two or x could be equal to negative one. The solutions here are

numbers, or a set of values that satisfy the equation. Here it’s a relationship

between a function and its derivatives. And so the solutions, or the solution, is going to be a function

or a set of functions. Now let’s make that a

little more tangible. What would a solution to

something like any of these three, which really represent the same thing, what would a solution actually look like? Actually let me move

this over a little bit. Move this over a little bit. So we can take a look at what some of these solutions could look like. Let me erase this a little. This little stuff that

I have right over here. So I’m just gonna give you

examples of solutions here. We’ll verify that these

indeed are solutions for I guess this is really

just one differential equation represented in different ways. But you’ll hopefully

appreciate what a solution to a differential equation looks like. And that there is often

more than one solution. There’s a whole class of functions

that could be a solution. So one solution to this

differential equation, and I’ll just write it as our first one. So one solution, I’ll call it y one. And I could even write it as y one of x to make it explicit that

it is a function of x. One solutions is y one of x is equal to e to the negative three x. And I encourage you to

pause this video right now and find the first derivative of y one, and the second derivative of

y one, and verify that it does indeed satisfy this differential equation. So I’m assuming you’ve had a go at it. So let’s work through this together. So that’s y one. So the first derivative of y one, so we just have to do the chain rule here, the derivative of negative

three x with respect to x is just negative three. And the derivative of e

to the negative three x with respect to negative three x is just e to the negative three x. And if we take the second

derivative of y one, this is equal to the same exact idea, the derivative of this is

three times negative three is going to be nine e

to the negative three x. And now we could just

substitute these values into the differential

equation, or these expressions into the differential

equation to verify that this is indeed going to be

true for this function. So let’s verify that. So we first have the

second derivative of y. So that’s that term right over there. So we have nine e to the negative three x plus two times the first derivative. So that’s going to be two

times this right over here. So it’s going to be minus

six, I’ll just write plus negative six e to

the negative three x. Notice I just took this two

times the first derivative. Two times the first derivative

is going to be equal to, or needs to be equal to, if

this indeed does satisfy, if y one does indeed satisfy

the differential equation, this needs to be equal to three times y. Well three times y is three

times e to the negative three x. Three e to the negative three x. Let’s see if that indeed is true. So these two terms right over here, nine e to the negative three x, essentially minus six e

to the negative three x, that’s gonna be three e

to the negative three x. Which is indeed equal to three

e to the negative three x. So y one is indeed a solution

to this differential equation. But as we’ll see, it is

not the only solution to this differential equation. For example, let’s say y

two is equal to e to the x is also a solution to this

differential equation. And I encourage you to

pause the video again and verify that it’s a solution. So assuming you’ve had a go at it. So the first derivative of this

is pretty straight-forward, is e to the x. Second derivative, one

of the profound things of the exponential function,

the second derivative here is also e to the x. So the second derivative,

let me just do it in those same colors. So the second derivative

is going to be e to the x plus two times e to the x is indeed going to be equal

to three times e to the x. This is absolutely going to be true. E to the x plus two e to the

x is three e to the x. So y two is also a solution

to this differential equation. So that’s a start. In the next few videos,

we’ll explore this more. We’ll start to see what

the solutions look like, what classes of solutions are, techniques for solving them, visualizing solutions to

differential equations, and a whole toolkit for

kind of digging in deeper.