1. Not finding help here? Sign up for a free 30min 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!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Science Advisor
    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!!!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: 2nd order O.D.E.
  1. 2nd order O.D.E (Replies: 3)

  2. Second order O.D.E (Replies: 1)