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Transforming an ODE

  1. Sep 17, 2008 #1
    Hi, I have a problem and I am wondering if anyone can help... There is
    this ODE, where P(x) and Q(x) are known functions:

    y''(x)+P(x) y'(x)+Q(x) y(x)=0 (1)

    This ODE cannot be solved analytically in general. However I can solve the following one (for the specific P(x) and Q(x) I have only):

    f''(x)+P(x) f'(x)+Q(x)/x f(x)=0 (2)

    The difference is in the third term :Q(x) => Q(x)/x. Does anyone know
    of a transformation y(x)=>f(x) such that eq. (1) can be transformed
    into eq. (2), which is solvable (as I said with the specific P & Q I
    have, not in general)?

    Cheers
     
  2. jcsd
  3. Sep 17, 2008 #2

    HallsofIvy

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    That will surely depend upon the "specific" P and Q!
     
  4. Sep 17, 2008 #3
    I bet it does! I was just wondering if it can be done for arbitrary functions... However, here you go:

    [tex]P(x)=\frac{3}{x}+\frac{H'[x]}{H[x]}[/tex]

    [tex]Q(x)=\frac{c}{x^5 H(x)^2}[/tex]

    Again, H(x) is arbitrary... sort of... Nevertheless, it is easy to see that eq (2) with these definitions is solvable (for arbitrary H(x)) while eq. (1) is not!

    Any ideas? Thanks in advance
     
  5. Sep 17, 2008 #4
    Your 2nd equation doesn't have any derivatives whereas your first one does, was that a typo?
     
  6. Sep 17, 2008 #5
    Hmmm, when I said eqs. (1) & (2) I obviously meant form my first post... The point is to transform:

    y''(x)+P(x) y'(x)+Q(x) y(x)=0............Equation (1)

    into

    f''(x)+P(x) f'(x)+Q(x)/x f(x)=0...........Equation (2)

    given the P(x) & Q(x), where:
    [tex]P(x)=\frac{3}{x}+\frac{H'(x)}{H(x)}[/tex]
    and
    [tex]Q(x)=\frac{c}{x^5 H(x)^2}[/tex]

    As I mentioned earlier, the reason I would like to find such a transformation (if at all possible) is because with these definitions of P(x) and Q(x), equation (2) for f(x) is solvable, while (1) is not. So knowing the solution for f(x) and the transformation allows in principle to find y(x) which is the desiteratum. Any ideas?
     
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