AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 7Lesson 5: Finding general solutions using separation of variables
- Separable equations introduction
- Addressing treating differentials algebraically
- Separable differential equations
- Separable differential equations: find the error
- Worked example: separable differential equations
- Separable differential equations
- Worked example: identifying separable equations
- Identifying separable equations
- Identify separable equations
Addressing treating differentials algebraically
Addressing treating differentials algebraically.
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- What is an example of a situation where treating dy,dx,dt,.. algebraically would cause a problem?(35 votes)
- I would say when taking the second derivative of a parametric function, if you use differentials algebraically you will most likely falter. It's too difficult to type up, but this is a great page, scroll down to see what I mean: http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx
If you try taking (d/dx) (dy/dx), you will surely not get anywhere close to the correct answer, which is [(d^2y/dt^2) (dx/dt) - (dy/dt) (d^2x/dt^2)]/(dx/dt)^3
Here are other opinions: https://math.stackexchange.com/questions/726950/how-is-it-that-treating-leibniz-notation-as-a-fraction-is-fundamentally-incorrec(16 votes)
- Is there anywhere I can read the "mathematically rigorous" explanation on why this works?(21 votes)
- I recommend Keislers 'elementary infinitesimal calculus', it provides a lot of visual intuition and covers single variable, multivariate, and vector calculus as well as some differential equations. It is available for free I believe as well as his follow up 'foundations of infinitesimal calculus' which provides more rigour (which I am not personally interested in).(12 votes)
- Is there any other way to find particular solution with more rigorous method?(8 votes)
- MIT OCW's course on differential equations is very long - ~40 lectures on the topic whereas KhanAcademy has nowhere near that amount of material. Is KhanAcademy's material enough or should we treat just as a introductory rather than complete course on the topic?(4 votes)
- Khan Academy's differential equations content is introductory, enough to cover the aspects of differential equations covered in AP Calculus. It is not a full undergraduate course in differential equations.(8 votes)
- Can you not say that:
(1/y) dy = (1) dx
and then integrate both sides?(4 votes)
(1 ∕ 𝑦)𝑑𝑦 = 𝑑𝑥 ⇔ ∫(1 ∕ 𝑦)𝑑𝑦 = ∫𝑑𝑥 ⇔ ln(𝑦) + 𝐶 = 𝑥 + 𝐶 ⇔ 𝑦 = 𝑒ˣ(6 votes)
- I am still not feeling good about this type of treating of dx notation. just symbolic manipulation isn't enough :((
I want to have a powerful intuition. how can I get it??(6 votes)
- Can't we just do the u-substitution here ?
∫(1/y)*(dy/dx)dx, u = y => du/dx = dy/dx, then ∫(1/u)du ?
I mean, it's the same logic, isn't it ? We can call this not u-substitution but y-substitution, same drill.
Basically, if we need to find an antiderivative of equation in form y(x) * dy/dx, we just need to throw away the derivative, cause it's a consequence of the chain rule on y(x), and find an antiderivative of y(x) only.
What's so difficult here that Sal says at2:16"the rigour you need to show that this is ok in this situation is not an easy thing to show" ?(2 votes)
- It would be great if someone could explain what, exactly, is not mathematically rigorous with this treating of differentials (or, if it is mathematically rigorous, then what did Sal mean in the video), especially given that:
(1) it apparently works flawlessly, and
(2) in the 1960's there's been the non-standard analysis approach, which, if I understand correctly, sought to give infinitesimals such a solid foundation that this treating of derivatives as the ratio of infinitesimals could now be considered legitimate.(2 votes)
- If you compare dy/dx and dx/dy of an equation, are they called "inverses?" Is there a mathematical term for this?(2 votes)
- How/Where do i find the rigorous proof of that I can manipulate dx/dy algebraicaly?(2 votes)
- I think the two main things to look at would proof of u-substition and chain rule.I believe covers chain rule proof not sure about u-substitution.(1 vote)
- [Instructor] So when you first learn calculus, you learn that the derivative of some function f, could be written as f prime of x is equal to the limit as, then there's multiple ways of doing this, the change in x approaches zero of f of x plus our change in x, minus f of x, over our change in x. And you learn multiple notations for this. For example, if you that y is equal f of x, you might write this as y prime. You might write this as d-y, d-x, which you'll often hear me say is the derivative of y, with respect to x, and that you can do the derivative of f with respect to x because y is equal to our function. But then later on when you, especially when you start getting into differential equations, you see people start to treat this notation as an actual algebraic expression. For example, you will learn or you might have already seen. If you're trying to solve the differential equation, the derivative of y with respect to x, is equal to y. So the rate of change of y with respect to x is equal to the value of y itself. This is one of, the most basic differential equations you might see. You'll see this technique, where people just say, "Well, let's just multiply both sides by d-x." Just treating d-x like as if it's some algebraic expression. So you multiply both sides by d-x and then you have, so that would cancel out algebraically, and so you see people treat it like that. So you have d-y is equal to y times d-x, and then they'll say, "Okay let's divide both sides by y." Which is a reasonable thing to do. Y is an algebraic expression. So if you divide both sides by y, you get one over y, d-y is equal to d-x. And then folks will integrate both sides to find a general solution to this differential equation. But my point on this video isn't to think about how do you solve a differential equation here, but to think about this notion of using, what we call differentials. So a d-x, or a d-y, and treating them algebraically like this. Treating them as algebraic expressions, where I can just multiply both sides by just d-x or d-y, or divide both sides by d-x or d-y. And I don't normally say this, but the rigor you need to show that this is okay in this situation, is not an easy thing to say. And so to just feel reasonably okay about doing this, this is a little bit hand wavy, it's not super mathematically rigorous. But it has proven to be a useful tool for us, to find these solutions. And conceptually the way that I think about a d-y, or a d-x, is this is the super small change in y, in response to a super small change in x. And that's essentially what this definition of the limit is telling us. Especially as Delta x approaches zero, we're going to have a super small change in x, as Delta x approaches zero. And then we're going to have a resulting super small change in y. So that's one way that you can feel a little bit better, of... And this is actually one of the justifications for this type of notation. As you can view this, what's the resulting super small, or what's the super small change in y, for a given super small change in x? Which is giving us the sense of what's the limiting value of the slope? As we go from the slope of the secant line to a tangent line? And if you view it that way you might feel a little bit better about using the differentials, or creating them algebraically. Let me just multiply both sides by that super small change in x. So the big picture is, this is a technique that you'll often see, in introductory differential equations classes. Introductory multi-variable classes and introductory calculus classes. But it's not very mathematically rigorous, to just treat differentials like algebraic expressions. But even thought it's not very mathematically rigorous to do it willy nilly like that. It has proven to be very useful. Now as you get more sophisticated in your mathematics there are rigorous definitions of a differential. Where you can get a better sense of where it is mathematically rigorous to use it and where it isn't. But the whole point here is, if you felt a little weird feeling about multiplying both sides by d-x or dividing both sides by d-x or d-y. Your feeling was mathematically justified, because it's not a very rigorous thing to do. At least until you have more rigor behind it, but I will tell you that if you're an introductory student it is a reasonable thing to do as you explore and manipulate some of these basic differential equations.