1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook