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Derivation step in ODE solution in textbook

  1. Sep 5, 2014 #1
    This is not homework but is part of the solution process of an ODE and I cannot understand how the author made a derivation step. After a change of variable in the original ODE, the ODE in the new independent variable has a standard method of solution. But instead of using this method, the author takes a shortcut but the exact steps are skipped and I cannot reproduce the transition on my own.

    1. The problem statement, all variables and given/known data

    How do we go from: [itex]y'' + 2y^{-3} = 0[/itex] to: [itex]y' = \sqrt{2(c_1 + 1/y^2)}[/itex] ?

    The only hint given by the author is: "...but in this case we can multiply through by y' and integrate directly to obtain..."

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I tried myltiplying through by y' and integrating both sides by parts but that gave me nothing I could use to go further.. I cannot even get back the first eq. by differentiating the second..
     
  2. jcsd
  3. Sep 5, 2014 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Think about the chain and product rule of integration. If I say more, it's more or less the solution ;-)).

    BTW: The textbook is incomplete! In front of the square root should be a [itex]\pm[/itex] sign!
     
  4. Sep 5, 2014 #3
    Thanks vanhees71! Your comment helped
    So the intermediate steps are:

    [tex]
    y''y' = -2y^{-3}y' \implies y''y' = \frac{d(y^-2)}{dt} \implies \int{y''y'}dt = y^{-2} + c1
    [/tex]
    [tex]
    \implies \frac{y'^{2}}{2} = y^{-2} + c1 + c2 \implies y' = \pm \sqrt{2(c + y^{-2})}
    [/tex]
     
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