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Simple differential equation

  1. May 14, 2007 #1
    Two concentric spherical conductors radii a and b are at potentials V1 and 0 respectively. The potential V at a distance x from thier 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.
  2. jcsd
  3. May 14, 2007 #2
    Well, if (as you've written it) the derivative is 0, then x^2V' would be constant. That is separable.
  4. May 14, 2007 #3
    Do I write down x^2V' =0, then [tex] x^2 \int dV =0 [/tex]
  5. May 14, 2007 #4
    No, you write
    [tex]x^2\frac{dV}{dx} = k(constant)[/tex]
  6. May 14, 2007 #5


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    EDIT: Bah, I screwed up somewhere in the latex. arunbg wrote the essential part anyway
  7. May 14, 2007 #6
    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].
  8. May 14, 2007 #7


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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: May 14, 2007
  9. May 14, 2007 #8
    Thanks cristo for your help.
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