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Riccati equations

  1. Oct 7, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    So I came across one of these equations. Solve : dy/dx + y2 = 1 + x2 when y(x) = x is a particular solution.

    2. Relevant equations



    3. The attempt at a solution

    So I re-arranged my equation into Riccati form : dy/dx = 1 + x2 - y2

    Now I let : v = [itex]\frac{1}{y-y(x)}[/itex] so that y = x + 1/v

    Thus : dy/dx = -(1/v2)dv/dx + 1

    Subbing these back into my equation yields :

    (-1/v2)dv/dx + 1 = 1 + x2 - (1/v + x)2

    Simplifying everything, I get a seperable linear equation in standard form :

    dv/dx - 2xv = 1 (***)

    So my integrating factor μ must satisfy :

    dμ/dx = -2xμ which implies that μ = e-x2

    So now my equation (***) can be re-written as :

    d/dx [e-x2v] = e-x2

    Integrating both sides and then expressing my integrand in terms of the error function gives me the final answer :

    v = [itex]\frac{\sqrt{\pi} \space erf(x) + d}{2e^{-x^2}}[/itex]

    And finally plugging v back into y yields my solution :

    y = [itex] x + \frac{2e^{-x^2}}{\sqrt{\pi} \space erf(x) + d}[/itex]

    Is this correct? This is my first try at one of these so I'm still in the not sure what I'm doing moment.
     
    Last edited: Oct 7, 2012
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