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Solving second order linear homogeneous differential equation

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the set of functions from [itex](-1,1)→ℝ[/itex] which are solutions of:

    [itex](x^{2}-1)y''+xy'-4y = 0[/itex]

    2. Relevant equations

    3. The attempt at a solution

    OK, I'm not really sure how to go about solving this equation, I have only previously attempted problems where the functions in [itex]x[/itex] are constant.

    There is a hint which says to use the change of variable:
    [itex]x=cos(θ)[/itex]

    doing this I get:

    (1): [itex](cos^{2}(θ)-1)y''+cos(θ)y'-4y = 0[/itex]

    which can be rearranged to give:

    (2): [itex]sin^{2}(θ)y''-cos(θ)y'+4y = 0[/itex]
    or
    (3): [itex](\frac{cos(2θ)-1}{2})y''+cos(θ)y'-4y = 0[/itex]

    No idea what to do next!
    Any pointers would be great, thank you!
     
  2. jcsd
  3. Apr 29, 2012 #2
    Looks like Heun's Differential Equation to me.
     
  4. Apr 29, 2012 #3
    I've never come across that before :/

    How can I use that to solve it?

    rearrange to give:

    [itex]y''+\frac{x}{(x+1)(x-1)}y'-\frac{4}{(x+1)(x-1)} = 0[/itex] ?

    Then I can't see why the change of variable hint has been given.
     
  5. Apr 29, 2012 #4

    HallsofIvy

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    This is completely wrong. You have replaced x with [itex]cos(\theta)[/itex] but you have left y' and y'' as derivatives with respect to x.
    If [itex]x= cos(\theta)[/itex] then [itex]\theta= cos^{-1}(x)[/itex] so that
    [tex]\frac{d\theta}{dx}= \frac{1}{\sqrt{1- x^2}}[/tex]
    and
    [tex]\frac{dy}{dx}= \frac{dy}{d\theta}\frac{d\theta}{dx}= \frac{1}{\sqrt{1- x^2}}\frac{dy}{d\theta}[/tex]

     
  6. Apr 29, 2012 #5

    HallsofIvy

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    This "rearrangement" is wrong.

    The correct arrangement is
    [tex]y''+\frac{x}{(x+1)(x-1)}y'-\frac{4y}{(x+1)(x-1)} = 0[/tex]

    Now, how does that help?

    The substitution, [itex]x= cos(\theta)[/itex], if done correctly, simplifies this a great deal.
     
    Last edited: Apr 29, 2012
  7. Apr 29, 2012 #6
    OK, thanks a lot!

    Right, am I correct in substituting the following:

    [itex]x = cos(θ)[/itex]

    [itex]\frac{dy}{dx} = (\frac{-1}{sin(θ)})\frac{dy}{dθ}[/itex]

    [itex]\frac{d^{2}y}{dx^{2}} = (\frac{-cos(θ)}{sin^{3}(θ)})\frac{d^{2}y}{dθ^{2}}[/itex]

    ???


    If so, I get:

    [itex]cos(θ)\frac{d^{2}y}{dθ^{2}}-cos(θ)\frac{dy}{dθ}-4sin(θ)y = 0[/itex]

    What next?
    Or have I made some stupid error?
     
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