1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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


    User Avatar
    Science Advisor
    Gold Member
    2017 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:

    y''y' = -2y^{-3}y' \implies y''y' = \frac{d(y^-2)}{dt} \implies \int{y''y'}dt = y^{-2} + c1
    \implies \frac{y'^{2}}{2} = y^{-2} + c1 + c2 \implies y' = \pm \sqrt{2(c + y^{-2})}
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted