Solving for constants in a differential equation

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Discussion Overview

The discussion revolves around solving for the constant ##B## in the differential equation $$\frac{dU( \theta)}{d \theta} + 2Br = 0$$. Participants explore the implications of treating ##B## as a constant versus a function of ##r##, and the resulting contradictions that arise from different approaches to differentiation.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that differentiating the equation with respect to ##r## leads to a result of ##B = 0##, while not differentiating gives ##B = -\frac{1}{2r} \frac{dU( \theta)}{d \theta}##, highlighting a contradiction.
  • Another participant suggests that treating ##B## as a function of ##r## leads to a different set of equations, which can be reconciled, indicating that the treatment of ##B## is crucial to the outcome.
  • A further comment emphasizes that the equation's structure implies that ##U## is a dependent variable and ##\theta## is independent, questioning the treatment of ##r## as a variable in this context.

Areas of Agreement / Disagreement

Participants express differing views on whether ##B## should be treated as a constant or a function of ##r##, leading to unresolved contradictions in the results derived from the differential equation.

Contextual Notes

There are limitations regarding the assumptions made about the nature of ##B## and the roles of the variables involved, which remain unresolved in the discussion.

Tio Barnabe
I feel so sorry when I found myself trapped in a basic problem like this one, but let's go ahead...

Suppose we have the following equation, knowing that ##B## is a constant, $$\frac{dU( \theta)}{d \theta} + 2Br = 0$$ where we want to solve for ##B##. If we differentiate the above equation with respect to ##r## we get that ##B = 0##. But if we don't, we find $$B = -\frac{1}{2r} \frac{dU( \theta)}{d \theta}$$ That is, two seemgly valid but contradicting results. What am I missing here?
 
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You treat ##B## simultaneously as a constant (first case) and as a function of ##r## (second case).

If ##B=B(r)## then differentiation gets you ##B'r+B=0## for the first equation, and ##B'=\dfrac{1}{2r^2} \dfrac{dU(\theta)}{d\theta}=-\dfrac{B}{r}## for the second, which are equal.

If ##B## is a constant, then differentiation gives ##B=0## and thus ##\dfrac{dU(\theta)}{d\theta} =0## and the second equation is also true: ##0=0##.
 
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Thanks!
 
Tio Barnabe said:
Suppose we have the following equation, knowing that ##B## is a constant, ##\frac{dU( \theta)}{d \theta} + 2Br = 0##
As written, it doesn't make much sense to treat r as a variable. The differential equation indicates that U is a dependent variable, and ##\theta## is the independent variable.
 

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