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A 2nd order DE

  1. Jan 5, 2007 #1
    help me solve
    ( 1 / x^2 ) d / dx [ x^2 ( df / dx ) ] = - f^( 3 / 2 )
    I think it needs a series sol'n but it's tough
     
  2. jcsd
  3. Jan 5, 2007 #2

    dextercioby

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    A series solution doesn't work. The equation is nonlinear in "f".

    Daniel.
     
  4. Jan 5, 2007 #3

    HallsofIvy

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    This is the same as
    [tex]\frac{df^2}{dx^2}+ \frac{2}{x}\frac{df}{dx}= f^{\frac{3}{2}[/itex]
    As dextercioby said, it is non-linear. There are no general methods for solving non-linear differential equations- not even series solutions.
     
    Last edited: Jan 5, 2007
  5. Jan 5, 2007 #4

    dextercioby

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    I think Halls meant

    [tex] \frac{d^{2}f}{dx^{2}}+\frac{2}{x}\frac{df}{dx}=f^{3/2} [/tex]

    but the exact form is less relevant. The nonlinearity is.

    Daniel.
     
  6. Jan 5, 2007 #5

    HallsofIvy

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    Actually, I had that but wrote "\fra {2}{x}" instead of "\frac{2}{x}"!
     
  7. Jan 6, 2007 #6
    There is a simple solution of the form [tex]f (x) = \frac{A}{x^4}[/tex], if that's any help. (I haven't worked out what the value of A is yet).

    edit: by my first estimate, A = 144.
     
    Last edited: Jan 6, 2007
  8. Jan 6, 2007 #7
    you can substitute in:
    [tex]g(x^n)=f(x)[/tex]
    then you'll get:
    [tex]nx^{n-1}g''(x^n)+(n^2-n)x^{n-2}g'(x^n)+2nx^{n-2}g'(x^n)=g^{\frac{3}{2}}(x^n)[/tex]

    let n be -1 to eliminate the first derivative term. Then maybe you can do some tricks and simplify...
     
    Last edited: Jan 6, 2007
  9. Jan 10, 2007 #8
    Thanks Matt, but..

    Thanks but a minus sign is being omitted here! Look at the original. It seems that 144 i / r^4 is a sol'n but it's imaginary. But it might be a clue. My intuition tells me that the real sol'n will have a nice form.
     
  10. Jan 12, 2007 #9
    Yeah - sorry 'bout the sign. I wasn't using the equation in the initial post.
     
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