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Help me please

  1. May 23, 2004 #1
    Help me please!!!!!!!

    Can you help me to solve this:

    (d² e(r)/dr²)+(1/r)*(d e(r)/dr)=0

    There is no initials conditions, please use general form
     
  2. jcsd
  3. May 23, 2004 #2

    arildno

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    Hint:
    Convince yourself of the following equality:
    [tex]\frac{d^{2}e}{dr^{2}}+\frac{1}{r}\frac{de}{dr}=\frac{1}{r}\frac{d}{dr}(r\frac{de}{dr})[/tex]
     
  4. May 23, 2004 #3
    Yes, but....

    And, if i want to find the general form of e(r)?
     
  5. May 23, 2004 #4

    arildno

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    You get:
    [tex]\frac{d}{dr}(r\frac{de}{dr})=0[/tex]
    This differential equation can be directly integrated to find the general solution.
     
    Last edited: May 23, 2004
  6. May 23, 2004 #5
    Maybe this?

    If I understand, you've made:
    1/r*d/dr*(r*de/dr)=0
    so:
    d/dr*(r*de/dr)=0

    If i integrate, i find:
    e(r)=A*ln(r)+B

    Wrong or not ?
     
  7. May 23, 2004 #6
    Thanks

    THANKS!!!!! you save me!!!
     
  8. May 23, 2004 #7
    I would also do it this way:

    Rewriting:

    [tex]e''+\frac{1}{r}e'=0[/tex]

    I would then multiply through by r^2:

    [tex]r^2e''+re'=0[/tex]

    I would recognize this as a d.e. of the Euler-Cauchy form:

    [tex]x^2y''+axy' + by=0[/tex]

    In the case of the given equation, a=1 and b=0. The characteristic equation for the Euler-Cauchy is:

    [tex]m^2+(a-1)m+b=0[/tex]

    In our case:

    [tex]
    \begin{align*}
    m^2&=0\\
    m&=0
    \end{align*}
    [/tex]

    For the case of a real double root in the characteristic equation, the general solution for the Euler-Caucy is given as:

    [tex]y=(A + B\ln x)x^m[/tex]

    So in our case:

    [tex]
    \begin{align*}
    e(r)&=(A + B\ln r)x^0\\
    e(r)&=A + B\ln r
    \end{align*}
    [/tex]

    I guess this solution depends on having the Euler-Cauchy form available to you in your course, which may not be the case.
     
  9. May 23, 2004 #8

    HallsofIvy

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    Another way to do this problem is let u= e' so that u'= e" and the equation reduces to the separable first order equation u'+ (1/r)u= 0. Then du/u= -dr/r and so
    ln(u)= -ln(r)+ C1 or u= e'= C1/r. Integrating again, e= C1ln|r|+ C2.
     
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