Can someone please explain me what is up...

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Can someone please explain me what is up with this kind of use of dx?

Why do they write Differential Equations this way? And how do I interpret an equation like this one?

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>>7776836

infinitesimals

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>>7776836

Its just a bunch of bullshit, op. Don't listen to the teachers.

Basically it means that everything after the [math]d[/math] is zero.

For example, in your equation you have [math]x^2ydx-(x^3+y^3)dy=0[/math] well because you have a dx on the first term and a dy on the second term on the lhs those terms are both zero and you get [math]0=0[/math] which is what we mathematicians call taughtology because you know it ;)

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It's a fair notation. For this one just divide everything by dx and then 0/dx is just 0, dx/dx is 1, and dy/dx is your differential

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>>7776836

Writing a DE in this form is useful because it tells you that there is a function M such that

[math]\frac{dM}{dx}=0[/math]

and that

[math]\frac{\partial M}{\partial x}=x^2 y[/math]

[math]\frac{\partial M}{\partial y}=-(x^3+y^3)[/math]

This is enough to solve for M. Then, forcing the derivative in x to be zero, you acquire an expression just with x and y. Solve for y if possible. That's your solution.

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Its a vector field, specifically a gradient field, written in differential form. The function you find after a few steps is the potential of said vector field.

Equivalently its the solution to a non linear system of equations about a locally linear neighborhood.

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>>7777202

As a general rule saying an ODE is exact is the same as saying a vector field is conservative. Most students (engineers) take calc 3 and ODEs and never realize theyre doing a lot of the same shit.

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>>7776836

pretend everything is a function of some other variable, call it t. Then dx means dx/dt, dy means dy/dt, etc., and you can define "dx/dy" to mean the ratio (dx/dt)/(dy/dt).

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>>7777209

I wish I didn't have have an easy calc 3 teacher that didn't require me to learn anything at all. I don't remember much of anything from that class.

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>>7777022

>doesn't understand differential forms

Yeah, sure, a "mathematician"

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>>7778013

Don't worry, undergrad, you'll learn this stuff eventually. Just keep trying and you'll get there.

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>>7777022

>we mathematicians

riiight

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>>7778595

Are you a mathematician? If not then you simply won't understand

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>>7778017

Kill yourself you inbred mongrel. What kind of a crappy school do you have to go to that doesn't teach differential forms and vector analysis in calc 3 or sooner?

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So what is a differential form?

I understand the definition here:

https://proofwiki.org/wiki/Definition:Differential_Form

But how is dx defined through them? (let's stay at functions from R to R)

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>>7778712

Imagine you have a vector U = (3,-5) in R^2.

If you write

e1=(1,0)

and

e2=(0,1),

then since

(a,0) = a·(1,0)

and

(x, y)+(u, v) = (x+u, y+v),

you have

U = 3·e1 + (-5)·e2

At the very base, the 1-form

V = (x^2 y)·dx + (-x^3-y^3)·dy

is a vector in this sense, except e1 is called dx and so on.

What's more, you define an operator d, which acts on functions f(x,y) as follows

(df)(x,y) := (df/dx)·dx + (df/dy)·dy

This enables you to map functions on a space to the vector - it's a formalization of the gradient,

which however extends to a function of rings (of such and more complicated forms).

What's more: d is actually a natural transfomation w.r.t. the functor that assigns the rings to the manifold.

With the right product structure (wedge product of forms), you can give d the structure of a (co-)boundary operator and this is known (see Langs algebra) to interesting groups and you can now investigate the smooth surfaces algebraically

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known to induce*

(i.e. cohomology groups, but that's just a fancy name for saying you can define simple algebraic structures using the vector/tensor fields on the manifold)

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>>7778712

Simply put, a differential form is something you integrate. A 0-form is just a function f(x), a 1-form is f(x)dx, a 2-form is f(x,y)dxdy, a 3-form is f(x,y,z)dxdydz, etc.

The theory of differential forms elaborates on this structure and addresses things like how to treat dxdydz vs dydxdz. You also learn about the exterior derivative which takes an n-form to an n+1 - form. Thus if you have a 0-form f(x) and you take d(f(x)) you will get f'(x)dx which is a 1-form.

There's a lot more to it if you study it from the perspective of differential geometry/topology which is what you should do if you care a lot about this subject but to just have an idea use the above explanation.

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>>7778640

No school teaches differential form or vector analysis in calc 3. None. These are topics for a first/second (vector analysis) and second/third (differential forms) class in analysis. Show me a syllabus that does what you claim or fuck right off.

>>7778712

If you're not a math major it's not something you need to worry about (you need plenty of prerequisites in analysis). Consider it shorthand for the chain rule and you're good to go.

This is what I mean. Consider the diff eq:

x*dx = dy

This is simply a shorthand for x = (d/dx)y, which you can solve by integration in x, giving y = x^2 + c

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>>7778826

If you're a physics student and don't know what a differential form is, then you're an experimentalist.

>muh

d F=0

d*F=j

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dx=0

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>>7778832

The demographic of /sci/ consists strongly of people in their first / second year, and almost all ODE discussion is invariably about the first, introductory class on applied differential equations for engineers. When they come and ask about what they're doing means, and people here start talking about advanced topics in functional analysis, it's not doing them any good. Of course they're going to learn it eventually, but you shouldn't be pushing it into them prematurely.

Bonus: If they aren't in the path to learning it, pushing something like this onto them makes them think they understand it and they start spamming about it everywhere like a meme.

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>>7776836

If we take the left-hand-side of the equation you have written to be a 1-form

[math] \omega = x^2 y dx - (x^3 - y^3)dy [/math]

since [math] \omega =0 [/math] we know it is a closed form and, assuming its domain is contactable, also an exact form.

By the Poincare Lemma we have,

[math] d \alpha = \omega [/math]

where alpha is a 0-form

now, from the definition of the exterior derivative,

[math] d \alpha = \frac{\partial \alpha}{\partial x}dx+\frac{\partial \alpha}{\partial y}dy = \omega [/math]

we now want to solve for [math] \alpha [/math] .

When we equate the two 1-forms above, we get two equations for alpha,

[math] \frac{\partial \alpha}{\partial x}=x^2 y [/math]

and

[math] \frac{\partial \alpha}{\partial y}=-(x^3+y^3) [/math]

This is what you learn in an elementary differential equations class but this is the (brief) explanation of why it works.

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OP look up differential forms. It's a perfectly normal notation.

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>>7778826

My class did in Calc 2. But I went to UPenn. Course name was Math260 but I don't think they teach the same thing every semester at all since the professor can kind of teach whatever he/she wants.

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>>7778981

A little knowledge is a dangerous thing... it's obviously not exact, and there is no such alpha.

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>>7779039

Oh shit ahahaha, now I look retarded.

I didn't even look at the equation because I saw someone else said it was exact and thought I'd throw in a little explanation.

I swear I'm not that stupid lol....

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