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 differential equation

  1. Sep 20, 2003 #1

    When I was younger, a teacher of mine gave me
    this problem for training:


    This was a test for the admission to a well known
    french Engineering School. What was interesting was that
    the test contains a question that was unsolved at the
    time this examination was given. If someone has the solution,
    I'm very interested to know it.

    Summary: the goal is to find a solution of
    xy''+2y' +x/y=0 defined on [0,1] such that y(1)=e where e is a
    given real number. For that purpose one sets g_0=e and
    g_{n+1} = e+T(1/g_n) where
    T(f)(x) = (1/x-1)\int_0^x t^2f(t)dt + \int_x^1 (t-t^2)f(t)dt

    g_{2p} converges to g and g_{2p+1} converges to G. The open
    question is to show that g=G ....

    Any idea is welcome
  2. jcsd
  3. May 31, 2004 #2
    I don't pretend to understand what this is all about, but make the following query/observations:

    1)Is it correct to note that this is a non-linear DE in that the power (-1) of 1/y disqualifies it as linear.

    2)It appears to be a second order DE in that y'' is the highest order derivative present. As such, is it not correct that 2 initial conditions are required for solution? The author provides y(1) = e which is one initial condition.

    The rest makes little sense to me, no disrespect intended.

    Last edited: May 31, 2004
  4. Jun 1, 2004 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    You are correct that this is a non-linear second order differential equation. In general it is not possible to find a closed form solutionto such a problem and, in these questions, we are not asked to solve the equation but to find some properties of the equation. The last part asks for a graph of [all] solutions that satisfy y(1)= e. There are, of course, an infinite number of such solutions.
  5. Jun 1, 2004 #4
    [tex] xy'' + 2y'+ \frac{x}{y}= 0\quad[/tex] can be rewritten as [tex]\quad y'' +2 \frac{y'}{x}+ y^{-1}=0[/tex]
    Here is how I think the query reads:
    [tex]xy''+2y'+\frac{x}{y}=0\mbox{ defined on } [0,1] \mbox{ such that } y(1)=e[/tex]
    where e is a given real number. For that purpose one sets
    [tex]g_0=e \mbox{ and }g_{n+1}=e+T(\frac{1}{g_n})\mbox{ where }T[f(x)]=(\frac{1}{x-1})\int_0^xt^2f(t)\;\mathrm{dt}[/tex]
    [tex]+\int _x^1(t-t^2)f(t)\;\mathrm{ dt}[/tex]

    [tex]g_{2p}[/tex] converges to g and [tex] g_{2p+1} [/tex] converges to G. The open question is to show that g=G...

    So, if one agrees with the way I have written the orginal query and recognizes the elaborate T[ransform?] function for [tex]g_{n+1}[/tex] ,
    why not illuminate the rest of us as the the thrust of this problem?
    Last edited: Jun 11, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: 2nd order differential equation