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Second order DE

  1. Dec 3, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Find the general solution to ##A(x)y''+A'(x)y'+\frac{y}{A(x)}=0## where A(x) is a known function and y(x) is the unknown one.
    Hint:Eliminate the term that contains the first derivative.


    2. Relevant equations
    Not sure.


    3. The attempt at a solution
    So I don't really know how to tackle it. I guess that the hint suggests a change of variable that would get rid of the y' term. So I tried z=y'A, z=Ay and z=y/A. All failed to express the ODE as a function of z and its derivative(s) and A and its derivative(s).
    I am therefore stuck. Any other hint will be welcome!
     
  2. jcsd
  3. Dec 3, 2012 #2

    micromass

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    What if you try the following substition [itex]y=Bz[/itex], where B is a known function that is yet to be decided.

    What do you get if you perform this substitution?

    What should we take for B if we want the z' term to vanish?
     
  4. Dec 3, 2012 #3

    fluidistic

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    Good idea! ##x''A'B'+z'(AB+AB'+A'B)+z (A'B'+B/A)=0##.

    ##B=Ce^{-x}/A##. I checked out and indeed the terms in front of ##z'## vanishes.
    I'm left to solve ##z''A'Ce^{-x} \left ( \frac{1}{A}+ \frac{A'}{A^2} \right ) + z \left ( \frac{A'}{A} + \frac{A'^2}{A^2}- \frac{Ce^{-x}}{A^2} \right ) =0##. That really does not look beautiful/easy to me. :eek:
     
  5. Dec 3, 2012 #4

    micromass

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    I get something different here. I get that the term of z' is
    [tex]A^\prime B + 2A B^\prime[/tex]

    The form of my B is also much simpler.
     
  6. Dec 3, 2012 #5

    fluidistic

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    I see, I made a mistake; I don't know why I make so many of them. I reach the same as yours, so ##B=kA^{-1/2}## where k is a constant.

    Edit: I might have rushed through the math but I reach that ##z''+z \left ( 1-\frac{2}{A'} \right ) =0##. Which has different solutions depending on the sign of the term in front of z. I guess I made another mistakes.
     
    Last edited: Dec 3, 2012
  7. Dec 3, 2012 #6

    micromass

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    OK, I get the same B. But there is no real need of the constant. You just need to find one single B that makes z' disappear, so you can take k=1.

    Anyway, with that choice of B, what does your equation become then?
     
  8. Dec 3, 2012 #7

    fluidistic

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    I now get ##B'=-\frac{1}{2}A^{-3/2}##, ##B''=\frac{3}{4}A^{-5/2}A'^2-\frac{A^{-3/2}A''}{2}##.
    The equation becomes ##z''AB+z(AB''+A'B'+B/A)=0##. Now I have to replace the B and its derivatives in that.
    Edit: I've just done it and it's not beautiful so far.
     
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