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Math Help - unique solution for a homogeneous differential equation

  1. #1
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    unique solution for a homogeneous differential equation

    I'm having trouble getting this question started. Im not sure how to go about solving it nor what method to use. Some of my friends say to divide by x^2, while others tell me other ways.

    x^2 d^2 y/ dx^2 + 2x (x-1) dy/dx - 2(x-1)y=0

    Why doesn't the general theory guarantee a unique solution to the equation satisfying the initial conditions y(0) = 0, y'(0) = 1.

    If someone could explain to me what i need to do, i would much appreciate it

    thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    In a linear DE of the type...

     c_{2} (x) y^{''} + c_{1} (x) y^{'} + c_{0} (x) y=0 (1)

    ... where  x \in [a,b] the condition of existence and unicity of the solution given the 'initial conditions' y(x_{0}) = y_{0}, y'(x_{0}) = y'_{0} requires that c_{2} (x_{0}) \ne 0. In your example is c_{2} (x)= x^{2} = 0_{x=0} so that the requirement isn't satisfied...

    Kind regards

    \chi \sigma
    Last edited by chisigma; August 27th 2009 at 12:29 AM. Reason: ...in (1) fogotten an 'y'... sorry!...
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  3. #3
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    is that all?! thanks man. i appreciate it, i dont know what the others are talking about dividing by x^2
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  4. #4
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    Ok i need more help with a related problem to this question


    <br />
x^2 d^2 y/ dx^2 + 2x (x-1) dy/dx - 2(x-1)y=0<br />

    Using the method of reduction of order, find a second linearly independent
    solution y2(x) of the equation.


    In the previous question i was able to show y1(x) = x is a solution of this equation, by taking the derivatives of y1, and then substituting back into the equation. and finding that x is indeed a solution.

    Though for this question i tried using that guide at the top of the forum for reduction order, and i get an answer that doesnt look right.

    u(x) = |[(e^ (1/3x^22x-3)) / (x^2)]dx cant get the maths coding right :S

    anyway help would be great
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  5. #5
    MHF Contributor chisigma's Avatar
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    The procedure to obain, once you have a solution of this type of equation, another solution independent from the first is illustated here...

    http://www.mathhelpforum.com/math-he...tion-case.html

    The second solution has a singularity in x=0 and that's why you aren't free to impose arbitrary 'initial conditions' in x=0...

    Kind regards

    \chi \sigma
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