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Strange way of solving a linear 2nd order DE

  1. Nov 16, 2012 #1
    1. The problem statement, all variables and given/known data
    I was given a DE of the form: [tex]\Phi^{''}+(6/\eta)\Phi^{'}=0[/tex] where the next step was given as [tex]\Phi^{'} \propto \eta^{-6}[/tex] where the answer came out to be [tex]\Phi \propto \eta^{-5} + constant[/tex]

    3. The attempt at a solution
    My attempt was to set [tex]\Phi^{'}=x[/tex] where I would then get [tex]x^{'}=-(6/\eta)x[/tex] and then solve for a seperable DE, but my answer was incorrect. Any help would be appreciated, thanks guys.

    Any help would be appreciated, thanks guys.
     
  2. jcsd
  3. Nov 16, 2012 #2

    haruspex

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    To clarify, η is the independent variable, and differentiation is wrt that? If so [tex]dx/x=-6d\eta/\eta[/tex] yes? Isn't it straightforward from there?
     
  4. Nov 16, 2012 #3
    About 5 minutes after posting this I figured it out. :/ Thanks for replying though, I appreciate it, that's exactly what I got. Cheers!
     
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