Strange way of solving a linear 2nd order DE

  • Thread starter Alexrey
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Homework Statement


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]

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.
 

Answers and Replies

  • #2
haruspex
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[tex]x^{'}=-(6/\eta)x[/tex]
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?
 
  • #3
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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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