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Homework Help: 2nd order O.D.E.

  1. Oct 17, 2005 #1
    for the following question:

    my problem:
    i don't have a clue how to get a hand on this one! any suggestions?
  2. jcsd
  3. Oct 17, 2005 #2
    [tex]yy'' = 2\left( {y'} \right)^2 [/tex]. The independent variable doesn't seem to be there. So perhaps [tex]p\left( y \right) = y' \Rightarrow y'' = p\frac{{dp}}{{dy}}[/tex] so that [tex]yp\frac{{dp}}{{dy}} = 2p^2 [/tex]. It would also be a good idea to not 'cancel' a p from both sides.
  4. Oct 17, 2005 #3
    may i ask:
    how'd you think of that?
  5. Oct 17, 2005 #4


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    That's a fairly standard "reduction of order" method.

    If y"= f(y, y') so that there is no x explicitely in the equation, then
    Letting u= y' gives, by the chain rule, y"= u'= (du/dy)(dy/dx)= u(du/dy) resulting in the first order equation u(du/dy)= f(y,u) with y as the independent variable and u as the dependent variable.
  6. Oct 18, 2005 #5


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    Homework Helper

    You can divide by p if p is not zero. If p=0 then that means y=constant. Note that this solution satisfies the ODE.
  7. Oct 18, 2005 #6
    thank you!!!
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