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Math Help - Riccati's Equation

  1. #1
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    Post Riccati's Equation

    Greetings,
    I came across the folllowing question on Ross, Differential Equations and was unable to solve it:
    y'=-y^2+xy+1 with one solution given as f(x)=x, you are asked to find the general solution, my solution is as follows:
    y=f(x)+\frac{1}{v}
    \frac{d}{dx}(x+\frac{1}{v})=-(x+\frac{1}{v})^2+x(x+\frac{1}{v})+1
    -\frac{1}{v^2}\frac{dv}{dx}=-\frac{x}{v}-\frac{1}{v^2}
    Writing it in the standard form leads:
    \frac{dv}{dx}-vx=1
    I found the integrating factor:
    e^{\int{-xdx}}=e^\frac{-x^2}{2}
    My final result is:
    v=e^\frac{x^2}{2}\int{e^\frac{-x^2}{2}dx}+ce^\frac{x^2}{2}
    Afterwards I could not evaluate the indefinite integral and this seems to be an introductory question, do I have a mistake here ?

    Any help is appreciated.
    Thanks
    Last edited by JohnDoe; November 20th 2011 at 02:15 AM. Reason: Mistake in Standard form
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  2. #2
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    Re: Riccati's Equation

    Looks good to me!
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  3. #3
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    Re: Riccati's Equation

    I did a quick search on Wolfram Alpha and I think that integral cannot be evaluated with the use of elementary functions and it is called the error function if I am not mistaken?
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