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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.
  2. jcsd
  3. Sep 8, 2008 #2


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    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]
    [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
    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.
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