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!

A differential equation : x^2 * u'' = k*u

  1. Sep 7, 2008 #1
    Other than guessing what is the formal way of solving following types of DE :

    x^2 * u'' = k*u

    or more generally :

    x^n * u'' + x^m * u' = k*u

    u is a function of x.
    thanks.
     
  2. jcsd
  3. Sep 8, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The first is an "Euler type" or "equi-potential" equation. The change of variable t= ln(x) changes it into an equation with constant coefficients:
    [tex]\frac{du}{dx}= \frac{du}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{du}{dt}[/tex]
    [tex]\frac{d^2u}{dx^2}= \frac{d }{dx}(\frac{1}{x}\frac{du}{dt})[/tex]
    [tex]= -\frac{1}{x^2}\frac{du}{dt}+ \frac{1}{x}\frac{d }{dt}\frac{du}{dt}[/tex]
    and changing that last d/dt to d/ds introduces another 1/x
    [tex]= -\frac{1}{x^2}\frac{du}{dt}+ \frac{1}{x^2}\frac{d^2u}{dt^2}[/tex]
    Thus
    [tex]x^2\frac{d^2u}{dx^2}= x^2(\frac{1}{x^2}\frac{d^2u}{dt^2}- \frac{1}{x^2}\frac{du}{dt}[/tex]
    [tex]= \frac{d^2u}{dt^2}- \frac{du}{dt}[/tex]

    So the first equation is just
    [tex]\frac{d^2u}{dt^2}- \frac{du}{dt}= ku[/tex]
    a linear equation with constant coefficients which has characteristic equation
    [tex]r^2- r- k= 0[/tex]
    That has roots
    [tex]\frac{1\pm\sqrt{1+4k}}{2}[/itex]
    so the general solution of the equation in terms of t is
    [tex]u(t)= e^t\left(C_1e^{\sqrt{1+4k}t}+ C_2e^{-\sqrt{1+4k}t}\right)[/tex]

    In terms of x,
    [tex]u(x)= e^{ln x}\left(C_1e^{\sqrt{1+4k}(ln x)}+ C_2e^{-\sqrt{1+4k}(ln x)}\right)[/tex]
    [tex]= x\left(C_1x^{\sqrt{1+4k}}+ C_2x^{-\sqrt{1+4k}}\right)[/tex]

    There is no general method for the general equation.
     
  4. Sep 8, 2008 #3
    That was nice. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A differential equation : x^2 * u'' = k*u
Loading...