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Math Help - Riccati Equation

  1. #1
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    Riccati Equation

    Consider the initial value problem for a Riccati equation:

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

    Find the true solution y(x). How many solutions have you found?

    This is actually part of a project involving us researching the riccati equation. So, I haven't learned anything about this topic in class, so I'm really not sure how to begin.

    I've found guide on the website http://www.sosmath.com/diffeq/first/...i/riccati.html but I'm not sure how reliable it is.
    Last edited by tn2k7; April 3rd 2009 at 07:29 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Only for clarification purpose: the 'initial value problem' to be 'attacked' is this...

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

    ... or some different?...

    Kind regards

    \chi \sigma
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  3. #3
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    Sorry, I copy and pasted and the subscripts and superscripts came out wrong. It should be:

    y' + y^{2} + 1 = 0, y(0) = 1
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  4. #4
    MHF Contributor chisigma's Avatar
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    All right!... the Riccati equation is written as...

    y^{'}= p(x) + q(x)\cdot y + r(x)\cdot y^{2} (1)

    In general the solution of (1) is possible only if you know one particular solution and that is in most cases an hard problem. The equation you have proposed however is...

    y^{'}= -(1+y^{2}) , y(0)=1 (2)

    ... and, because the independent variable x doesn't partecipate, we can use a different approach. Let's suppose in (2) to swap the x and y, so that the equation becomes...

    x^{'}= -\frac{1}{1+y^{2}} , x(1)=0 (3)

    In this case you can separate the variables, integrate immediately and obtain...

    x= - \tan^{-1} y + c (4)

    The constant c can be obained with the condition x(1)=0 and that gives c= - \frac {\pi}{4}. The (4) is easily invertible, so that the 'true solution' of (2) is...

    y= -\tan (x - \frac{\pi}{4}) (5)

    Kind regards

    \chi \sigma
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