I Solving for constants in a differential equation

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?
 
Physics news on Phys.org
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##.
 
  • Like
Likes Tio Barnabe
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.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top