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Math Help - Ricatti equation linearization

  1. #1
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    Ricatti equation linearization

    if the quadratic term in a Ricatti equation q2 is nonzero, y'=q0+q1+q2y^2 then you can make a substitution v=yq2, S=q2q0, R=q1+(q2'/q2) which satisfies a Riccati, v'=v^2+R(x)v+S(x)=(yq2)'=q0q2+(q1+q2'/q2)v+v^2
    with double substitution v=-u'/u, u now satisfies a linear 2nd ODE:
    u''-R(x)u'+S(x)u=0, v'=-(u'/u)'=-(u''/u)+v^2, u''/u=v^2-v'=-S+Ru'/u, therefore we have reduced the Ricatti to an equation
    u''+Su-Ru'=0, solving this equation will lead to a solution y=-u'/(q2u).....what was the original motivation behind the invention of the substitutions? my eyes are so bad I'm refonting
    Last edited by mathlover10; September 12th 2013 at 02:08 AM.
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  2. #2
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    Re: Ricatti equation linearization

    Hey mathlover.

    All substitutions in mathematics are typically used to take one representation to another that is more useful or easier to use in the context of the problem.

    In terms of DE's, the goal is to find the unique set of functions that satisfy the DE. It may be a family functions or it may be a unique one.

    At the end of the day, you are trying to take a DE and find a function representation satisfying that. What the DE corresponds to and why its useful is usually answered by physicists and other scientists, or applied mathematicians. The theoretical/pure side is interested in getting a solution, knowing a proper solution exists, and knowing some of its other properties (like convergence, differentiation, etc).
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