Can You Pull Out 1/dx in Differential Equations?

  • Context: Undergrad 
  • Thread starter Thread starter RandomGuy88
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Discussion Overview

The discussion centers around the validity of manipulating differential equations by pulling out a term of 1/dx. Participants explore the implications and meanings of such an operation within the context of differential calculus.

Discussion Character

  • Debate/contested, Conceptual clarification, Mathematical reasoning

Main Points Raised

  • One participant questions the validity of pulling out 1/dx, asking if it is a meaningful operation.
  • Another participant suggests that it is possible to pull out 1/dx, but notes that certain requirements must be met regarding the functions involved.
  • A different participant expresses skepticism about the meaning of 1/dx, indicating that they believe it may not have any significance.
  • One participant provides an example involving df/dt, suggesting that pulling out 1/dt is straightforward and does not involve any special considerations.

Areas of Agreement / Disagreement

Participants express differing views on the validity and meaning of pulling out 1/dx, indicating that there is no consensus on the topic.

Contextual Notes

Some participants mention the need for specific conditions or requirements for the functions involved, but these conditions are not fully articulated or agreed upon.

Who May Find This Useful

Individuals interested in differential equations, calculus, and mathematical manipulation of expressions may find this discussion relevant.

RandomGuy88
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I am wondering if this is valid.

(de/dx) + (1/p)(dP/dx) = (1/dx)(de +(1/p)dP)

Basically are you allowed to pull a 1/dx out of the equation?
 
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What would 1/dx even mean?
 
Yes it is possible, although a mathematician would not like it :biggrin:. There are certain requirements the function must meet before you are allowed to treat the differentials like parts of a fraction which I don't know off the top of my head.
 
Thanks for the replies. I am not sure what 1/dx would mean. In fact that is why I am asking this question, because I didn't think it would mean anything and therefore is wrong.
 
Let's say I have [tex]{df \over dt} = 3t^2[/tex] and you just pulled out the 1/dt to get [tex]{1 \over {dt}}(df) = 3t^2[/tex]. There is nothing special going on.
 

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