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Separation of variables , but for 2nd order

  1. Jul 13, 2012 #1
    "separation of variables", but for 2nd order

    Ok, I know how to separate variables in solving an ODE. I am unable to understand a solution I have for a problem which was the result of reduction of order- we end up with u''*sinx-2u'*cosx=0
    so turn this into u''/u'=-2cosx/sinx
    At this point I don't understand the process. I've tried looking at it as making the left side a "u substitution", or rewriting it d(du/dt)/dt / du/dt and integrating both sides,
    also, tried making a substition w=u', but this seems hard to get u back in the end.
    Can someone explain thoroughly what the process looks like in solving this?
     
  2. jcsd
  3. Jul 13, 2012 #2

    tiny-tim

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    hi ericm1234! :wink:
    well, then that's w'/w = -2cosx/sinx,

    so ln(w) = … ? :smile:
     
  4. Jul 13, 2012 #3
    Re: "separation of variables", but for 2nd order

    Thanks, got it now, not as complicated as I was making it.
     
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