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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 7

Lesson 1: Modeling situations with differential equations

# Differential equations introduction

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.A (LO)
,
FUN‑7.A.1 (EK)
Differential equations are equations that relate a function with one or more of its derivatives. This means their solution is a function! Learn more in this video.

## Want to join the conversation?

• At we see d^2 y / d x^2. Where is that x^2 coming from?
• As Sal mentioned at , that is the Leibniz notation for second derivatives, with the use of d's being in Leibniz notation (they're not fractions), while the accents such as y', y", and f'(x) and f"(x) being the Lagrange notation for first and second derivatives, and there's also a Newton notation with dots over the y. They are equivalent notations, but sometimes Lagrange and Newton notation are shorter to write, while Leibniz notation is clearer or more explicit regarding the variables which are being used like y and x. The Leibniz notation is named after Gottfried Leibniz, one of the creators of calculus. See:
http://en.wikipedia.org/wiki/Leibniz%27s_notation.
• Don't have a question about this video, but I don't really have a good category to put it... I'm not exactly good at math so forgive me if this question seems really silly.

About a week ago, I was wondering in my math class what the graph x^x looks like, and since we were using the desmos calculator in class, I tried it out. I noticed the function abruptly stopped at the y-axis and did not extend into the second or third quatrains at all. I recognized that some values are complex and cannot be graphed, yet some can, such as (-1, -1) or (-2, 0.25). This lead me to wonder whether there are points in between the integers, and there are! Such as when x = -1/3, yet not all are real numbers. Then I started to find as many points I could in the negative section to see if there was a pattern. I was using pencil and paper for most of my problems to determine whether the y values (given x) were real or not. I was finding y when x = -7/3 when I noticed something strange...

(-7/3)^(-7/3)
= ((-7/3)^(-1))^(7/3)
= (-3/7)^(7/3)
= ((-3/7)^2) * ((-3/7)^(1/3))
= (9/49) * -0.75394744112
= -0.13848014224

Which is completely real... But when you type in (-7/3)^(-7/3) onto most calculators, google included:

You will get this value: 0.0692400711 - 0.119927321i
Which is complex... I asked my math teacher and he said it should be real...

If you have any insight on the weirdness, please give me some insight below! Thanks for your time! :D

P.S.
I know this is also the case for -5/3, try that on paper too...
The 1 calculator that I found that showed a real solution was Symbolab... You can check that out too.
• The reason comes from the Fundamental Theorem of Algebra. In essence, the takeaway from it in the context of this problem is as follows:
When you take the nth root of a number, there are always n different roots

So, if you're interested in cube roots of a number (as you are in this example), you will always have three roots. Some may be imaginary, some may be real. The theorem does not specify. All it specifies is that there will always be three different roots.

So, in the case here, what the calculators are giving you called the "principal root". This is "the most obvious" root. While you might not think it is, when dealing with complex numbers, this is the principal root and it's a little too difficult to explain why without drawings.

The roots of your number are:
* 0.0692400711 - 0.119927321i
* -0.13848
* 0.0692400711 + 0.119927321i

I'm upvoting your question because it's a great question, and very well observed. And let this be a lesson: always be careful what you plug into your calculator, because your calculator will just blindly calculate numbers. It is up to you to make decisions to determine what the answer is at the end of the day.

Edit: This question was so good, I decided to write a more in-depth answer to it on my blog (where I have the freedom to use mathematical notation and add pictures. You can see it here: http://aommaster.com/blog/?p=1667)
• I have been told that differential equations are very useful in physics. Since I don't have the needed knowledge in physics, could you tell me how true the above statement is?
• It's true, I know they come up frequently in Heat Transfer type applications and they are very useful for electrical engineering as well.
• At , how did he come to the conclusion that y1(x) = e^-3x?
• The differential equation `y'' + ay' + by = 0` is a known differential equation called "second-order constant coefficient linear differential equation". Since the derivatives are only multiplied by a constant, the solution must be a function that remains almost the same under differentiation, and `eˣ` is a prime example of such a function.

If you were to solve this equation, you would start with a general solution and from there get a more specific solution, in this case a good starting point would be `y(x) = Ce^(Ax)` , where `A` and `C` would be constants that you try to limit by inserting this general solution on the differential equation.

After doing that you will find that `A` can be either `1` or `-3` for this particular differential equation (`y'' + 2y' - 3y = 0`); so the general solution becomes `y(x) = C₁eˣ + C₂e⁻³ˣ`, where `C₁` and `C₂` are only limited by initial conditions (not given in this problem, so you can choose whatever values you want and it will work).

Sal choose `C₁ = 0` and `C₂ = 1`, so that is why he gave as a specific solution `y(x) = e⁻³ˣ`
• What is the necessity of using differential equations over algebric equations?
• Algebraic equations has numerical solutions whereas differential equations have functions as their solutions. This is significant because if you have a relationship that evolves over time, a numerical solution is a constant which doesn't really model how things are changing; whereas functions can model changes over time. A good example of this is inflation rate of a balloon, if you wanted to see how much the volume is changing per instant then you'd have to write down a differential equation - and the solution would give you an answer. You can use this, then, to approximate things - like how much a balloon can stretch? How long will it take for the balloon to burst? And other similar questions.
• What does the second derivative mean?
• It is the derivative of the first derivative. For example, let's find the second derivative of x^2.
First we find the first derivative of x^2. By power rule, it is 2x.
Then the second derivative of x^2 is the derivative of 2x, which by power rule, is 2.
Thus the second derivative of x^2 is 2.
• The differentials I'm learning about in math class include things like dy=f'(x)(dx) and (delta)x and (delta)y, but these topics don't show up in this Differential video. Is the thing that I'm learning actually differentials, or are they under a separate name?
• Differentials, like `dx`, `dy`, represent a infinitesimal change in the variable, and are first introduced as part of basic calculus (or even precalculus, but without explaining what they are).

Differential equations are much more advanced, and should be studied once you have a firm knowledge of both differential calculus and integral calculus.
• When doing differentiation for trigonometric ratios .I saw a question that d/dx (tan(2x))
and its answer is given as 2sec(2x)^2.Here 2x is an angle then why there is a need of differentiating the angle 2x .Can't we simply write it as sec(2x)^2
• I have a pdf called differential equations demystified has anyone read it? I'm not sure whether to read that or do it on here then read that pdf?
• Differential Equations Demystified covers all of the D.E. material that Khan Academy does, plus a lot more. Khan Academy covers only some very basic parts of differential equations.

Neither is a prerequisite to the other. Both will assume you already have a solid grasp of calculus, but should fully explain the actual material.
• At what is the name of the third notation?

## Video transcript

- [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.