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### Course: Differential equations > Unit 1

Lesson 4: Separable equations- Separable equations introduction
- Addressing treating differentials algebraically
- Worked example: identifying separable equations
- Worked example: finding a specific solution to a separable equation
- Worked example: separable equation with an implicit solution
- Particular solutions to separable differential equations
- Separable equations (old)
- Separable equations example (old)

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# Addressing treating differentials algebraically

Addressing treating differentials algebraically.

## Want to join the conversation?

- What is an example of a situation where treating dy,dx,dt,.. algebraically would cause a problem?(39 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(17 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).(14 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?(5 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.(18 votes)

- Is there any other way to find particular solution with more rigorous method?(8 votes)
- Can you not say that:

(1/y) dy = (1) dx

and then integrate both sides?(4 votes)- Sure.

(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" ?(4 votes)- I know it is very late but here is the thing.. dy/dx is the rate of change of some function y(say).Now cancelling/transposing the rate equation doesn't seem mathematically correct right?(I mean dx is infinitesimal change of something how can you just multiply with a variable it kinda looks weird). And regarding the y-substitution you are actually not changing anything. You are putting 'y' as u

Nolan :)(1 vote)

- 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.(3 votes)- I am not an expert in nonstandard analysis, so take this answer with a grain of salt; parts could be inaccurate.

Introductory calculus courses typically study the real numbers, whereas nonstandard analysis studies the hyperreal numbers. Although many of the concepts are the same, you're ultimately studying a different object with different properties.

Many (all?) of the theorems we care about in Calculus BC carry over by the Transfer Principle, but in higher-level math, they might not. It's not rigorous in that we don't understand, or ask about, when the transfer principle applies; we just hope it'll work.

Compare, for example, the study of integers vs. rationals. The arithmetical operations (addition, subtraction, multiplication) obey the same rules, but more complicated questions like "what polynomials have zeroes" do not.

Likewise, if we're just studying introductory calculus, the operations obey the same rules and we don't care whether we're working with the reals or hyperreals, but at a higher level, they're not the same. Additionally, the conventional definitions are different (e.g. are we taking delta-epsilon limits, or dealing with infinitesimals?) which affects how the subject is taught.

(Also compare how we write divergent limits as`lim [x->a] f(x) = infinity`

whereas in the extended reals, infinity is an actual number, and the limit actually does equal infinity.)(1 vote)

- This is how I find it mathematically valid though:

Say that dy/dx = y, divide both sides by y: (1/y)(dy/dx)=1

Then integrate both sides with respect to x:

int[(1/y)(dy/dx)dx] = int(dx)

And you have y in the integral and its derivative dy/dx. That's simple u substitution.(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)

## Video transcript

- [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.