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Thread: second order x function 14

  1. #1
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    second order x function 14

    x^2y''+xy'-0.25=0
    how to solve it?
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  2. #2
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    \displaystyle x^2\,\frac{d^2y}{dx^2}+ x\,\frac{dy}{dx} - \frac{1}{4} = 0

    \displaystyle \frac{d^2y}{dx^2} + \frac{1}{x}\,\frac{dy}{dx} - \frac{1}{4x^2} = 0

    \displaystyle \frac{d^2y}{dx^2} + \frac{1}{x}\,\frac{dy}{dx} = \frac{1}{4x^2}.


    Now let \displaystyle u = \frac{dy}{dx} so that \displaystyle \frac{du}{dx} = \frac{d^2y}{dx^2} and the DE becomes

    \displaystyle \frac{du}{dx} + \frac{1}{x}\,u = \frac{1}{4x^2}.

    This is first order linear, so use the Integrating Factor method to find \displaystyle u, which you can then use to find \displaystyle y.
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