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

Lesson 6: Finding general solutions using separation of variables

## Want to join the conversation?

• What is an example of a situation where treating dy,dx,dt,.. algebraically would cause a problem?
• Is there anywhere I can read the "mathematically rigorous" explanation on why this works?
• 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).
• 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?
• 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.
• Is there any other way to find particular solution with more rigorous method?
• 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??
• Can you not say that:

(1/y) dy = (1) dx

and then integrate both sides?
• Sure.
(1 ∕ 𝑦)𝑑𝑦 = 𝑑𝑥 ⇔ ∫(1 ∕ 𝑦)𝑑𝑦 = ∫𝑑𝑥 ⇔ ln(𝑦) + 𝐶 = 𝑥 + 𝐶 ⇔ 𝑦 = 𝑒ˣ
• 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 at "the rigour you need to show that this is ok in this situation is not an easy thing to show" ?
• 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.
• 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.