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Cauchy-Euler ODEs question

  1. Dec 25, 2015 #1
    Why these ODEs when applied some boundary conditions, like x = 0, their solution of the form Ax^k + Bx^(-k), B WILL have to go to zero?Like some problems which involve spherical harmonics...
     
  2. jcsd
  3. Dec 25, 2015 #2

    jambaugh

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    The negative exponent of the B coefficient term implies divergent solution at x=0. If you are considering solutions with physical interpretation such as electron wave functions then the physical application is contradicted by the divergence at infinity. Those solutions don't apply and you set B = 0.
     
  4. Dec 25, 2015 #3
    Yeah i understand why when x->infinity the solution is inconsistent, cause it's blow up, but i simply don't get it WHY x=0 implies in a divergent solution and B has to be 0, there's a more deep explanation?
     
  5. Dec 25, 2015 #4

    jambaugh

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    [itex]B x^{-k}=\frac{B}{x^k} = \frac{B}{0}[/itex] when [itex]x=0[/itex].
     
  6. Dec 25, 2015 #5
    Oh damn, i'm stupid , i was thinking in that, but it appears so simple that i forget this option because some texts are very confusing , and thank you so much for killing that existential question for me :D...
     
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