Simple differential equation

  • #1
John O' Meara
330
0
Two concentric spherical conductors radii a and b are at potentials V1 and 0 respectively. The potential V at a distance x from their common centre is given by [tex] \frac{d[x^2\frac{dV}{dx}]}{dx}=0 \\ [/tex]. Find V in terms of x, a, b and V1 ( Note V=V1 when x=a and V=0 when x=b).
I just find it difficult to get started on this, I seem not able to separate out the variables dV and x. Thanks for the help.
 

Answers and Replies

  • #2
daveb
548
1
Well, if (as you've written it) the derivative is 0, then x^2V' would be constant. That is separable.
 
  • #3
John O' Meara
330
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Do I write down x^2V' =0, then [tex] x^2 \int dV =0 [/tex]
 
  • #4
arunbg
594
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No, you write
[tex]x^2\frac{dV}{dx} = k(constant)[/tex]
 
  • #5
Office_Shredder
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EDIT: Bah, I screwed up somewhere in the latex. arunbg wrote the essential part anyway
 
  • #6
John O' Meara
330
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Is that constant k not =0? I mean the [tex] d[x^2\frac{dV}{dx}}] = dx \times 0 = 0 \\ [/tex] In other words where did you get the k. Anyway I have the following [tex] V=k\int \frac{1}{x^2}dx \\ [/tex].
 
  • #7
cristo
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You can't separate the original equation like that. You have [tex]\frac{d}{dx}\left(x^2V'\right)=0[/tex]. Integrating both sides wrt x gives [tex]\int \frac{d}{dx}\left(x^2V'\right)dx=k \Rightarrow x^2V'=k[/tex]

Your next part is correct. Now integrate that.
 
Last edited:
  • #8
John O' Meara
330
0
Thanks cristo for your help.
 

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