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Sugestions to solve this equation

  1. Nov 13, 2012 #1
    Any sugestions on how to find the solutions to this equation?


    [itex]y'' +\frac{b'}{b} y' - \frac{a^2}{b^2}y=0[/itex]

    where [itex]a[/itex] is a constant
     
  2. jcsd
  3. Nov 13, 2012 #2
    I assume that b is a constant as well. That's a pretty standard homogeneous linear equation; just use the standard methods (can you write out its characteristic equation?).
     
  4. Nov 13, 2012 #3
    No, [itex]b[/itex] is a function, if it were a constant its derivative would be zero and the term with [itex]y'[/itex] would disappear
     
  5. Nov 13, 2012 #4

    haruspex

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    You can obtain a lot of solutions by setting b(x) = Axn and solving for y. That won't give you a general solution though.
     
  6. Nov 14, 2012 #5
    rewrite

    [itex]by''+b'y'-\frac{a^2}{b}y=0[/itex]

    to the standard form:

    [itex]z''=-c^2(t)z[/itex],

    with

    [itex]c(t)=-\frac{2a + b'(t)}{2b}[/itex]

    the general solution of the transformed equation is then

    [itex]z=Asin(cx)+Bcos(cx)[/itex]

    Then get the solution of y by transforming back:
    [itex]y=z e^{\int{\frac{b'}{2b}dx}} = z\frac{b'}{2b}[/itex]

    [itex]y=(A\sin(cx)+B\cos(cx))\frac{b'}{2b}[/itex]

    That's of course, assuming a,b are such that all steps are valid
     
  7. Nov 14, 2012 #6

    haruspex

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    Looks to me that if you substitute that in the original equation there'll be an unbalanced b''' term.
     
  8. Nov 15, 2012 #7
    Oops, you're right, I was thinking too simple! The transformation to standard form does not lead to an expression that can be easily solved by a ricatti equation.

    Actually, maple gave the following:

    [itex]y=A\sinh(\int -\frac{a}{b(x)}dx) + B\cosh(\int -\frac{a}{b(x)}dx)[/itex]

    hope this helps a bit...
     
  9. Nov 15, 2012 #8
    That suggests the substitution

    [itex] t =∫(a/b(x)) dx [/itex]

    which works wonders :).
     
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