What is the name of this equation?

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The equation discussed is y = 3 + 2x³, which is classified as an ordinary equation. The conversation highlights the differentiation of this equation, resulting in dy/dx = 6x². The terms "dy" and "dx" represent differentials, which are essential concepts in calculus. Participants clarified that there is no special name for the equation beyond its standard form.

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Goethe10
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Hello everybody, What is this equation called? I'm from non-English speaking country and I need to look up materials about how to solve it, Thank you.
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Welcome to PF!

Hello Goethe10! Welcome to PF! :smile:
Goethe10 said:
y = 3 + 2x3

(why have you written a "d" there? :confused:)

This is an ordinary equation,

and the four possible answers are dervatives of it,

except that instead of writing dy/dx = f(x), they've written them dy = f(x)dx :wink:
 
tiny-tim said:
Hello Goethe10! Welcome to PF! :smile:


(why have you written a "d" there? :confused:)

This is an ordinary equation,

and the four possible answers are dervatives of it,

except that instead of writing dy/dx = f(x), they've written them dy = f(x)dx :wink:

Haha, sorry.
It was just smudge done by paint tool, not an actual D, sorry about confusion :P
What do you mean "ordinary equation" Can you please give me the full name? Thanks.
 
Goethe10 said:
What do you mean "ordinary equation" Can you please give me the full name? Thanks.

i mean it's just an equation, there's nothing special about, there's nothing to look up! :smile:

(just differentiate it!)
 
Ah, I see! :D Thank you
So
3+2x^3 would be
6x^2
Can you please tell me what does "dy" means?
 
Goethe10 said:
So
3+2x^3 would be
6x^2

correct! :smile:
Can you please tell me what does "dy" means?

if you'd written the equations in full, you would have written:

y = 3+2x3

So dy/dx = 6x2

The "dy" and "dx" in "dy = 6x2dx" are called differentials (or infinitesimal differentials).

You can always rewrite any dy/dx equation so that the dy and dx are on opposite sides …

look that up in almost any book on calculus :wink:
 
Thank you! :)
 

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