Cauchy-Euler ODEs question

  • Thread starter Andreol263
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  • #1
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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...
 

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  • #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.
 
  • #3
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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?
 
  • #4
jambaugh
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[itex]B x^{-k}=\frac{B}{x^k} = \frac{B}{0}[/itex] when [itex]x=0[/itex].
 
  • #5
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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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