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Math Help - Ricatti equation

  1. #1
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    Ricatti equation

    Im not sure how to to this question, I already did the first part (verify that it is a solution of the equation).

    Im stuck in the part in red. I dont understand the u = y - (3/x) bit, what does it mean?

    I thought I should just use

    y(x) = y + (1/w), with y=3/x?

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  2. #2
    MHF Contributor chisigma's Avatar
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    The DE that was studied three centuries ago by the Italian mathematician 'Conte' Jacopo Riccati has the form...

    y^{'} = a(x) y^{2} + b(x) y + d(x) (1)

    ... and in general can be solved only if we know a particular solution of it. In that case if g(x) is a particular solution of (1) and we set...

    y(x)= g(x) + \frac{1}{z(x)} (2)

    ... after some algebraic steps we obtain from (1)...

    z^{'} = -\{2 a(x) g(x) + b(x)\} z - a(x) (3)

    ... that is linear in z. In our case the DE is...

    y^{'} = - y^{2} + \frac{6}{x^{2}} (4)

    ... and it is easy to verify that \frac {3}{x} is a solution. At this point if You set in (2) u(x)= \frac{1}{z(x)} and g(x) = \frac{3}{x} You obtain the substitution...

    u(x)= y(x) - \frac{3}{x} (5)

    ... exactly as in your book...

    Kind regards

    \chi \sigma
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  3. #3
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    It is a special case , we don't need to use Riccati's method .

     y' = - y^2 + \frac{6}{x^2}

     y' = \frac{6-(xy)^2}{x^2}

    Sub.  xy = u ,  y = \frac{u}{x}

     y' = \frac{ u'x - u }{x^2} so we have

     u'x - u = 6-u^2

     u'x = 6+u-u^2

     \frac{du}{dx} x = (2+u)(3-u)

     \int \frac{du}{(2+u)(3-u)} = \int \frac{dx}{x}

     \frac{1}{5} \int \left[ \frac{1}{2+u} + \frac{1}{3-u} \right]~du =\ln(kx)

     \frac{1}{5} \ln{\big{|} \frac{2+u}{3-u}\big{|}} = \ln(kx)

     (kx)^5 = \big{|}\frac{2+u}{3-u}\big{|}

    Let  a = k^5

     ax^5 = \big{|}\frac{2+xy}{3-xy}\big{|}

    If we neglect the sign of absolute value , we will find that the solution is :


     y = \frac{3}{x} - \frac{5}{x(ax^5+1)}
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  4. #4
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    btw, what does it mean by "find an ordinary differential equation for u = y - 3/x?

    Quote Originally Posted by chisigma View Post
    The DE that was studied three centuries ago by the Italian mathematician 'Conte' Jacopo Riccati has the form...

    y^{'} = a(x) y^{2} + b(x) y + d(x) (1)

    ... and in general can be solved only if we know a particular solution of it. In that case if g(x) is a particular solution of (1) and we set...

    y(x)= g(x) + \frac{1}{z(x)} (2)

    ... after some algebraic steps we obtain from (1)...

    z^{'} = -\{2 a(x) g(x) + b(x)\} z - a(x) (3)

    ... that is linear in z. In our case the DE is...

    y^{'} = - y^{2} + \frac{6}{x^{2}} (4)

    ... and it is easy to verify that \frac {3}{x} is a solution. At this point if You set in (2) u(x)= \frac{1}{z(x)} and g(x) = \frac{3}{x} You obtain the substitution...

    u(x)= y(x) - \frac{3}{x} (5)

    ... exactly as in your book...

    Kind regards

    \chi \sigma
    Thanks!

    This is what I did so far, continuing what you did (is it correct?) how would i proceed from there?



    Quote Originally Posted by simplependulum View Post
     ax^5 = \big{|}\frac{2+xy}{3-xy}\big{|}

    If we neglect the sign of absolute value , we will find that the solution is :


     y = \frac{3}{x} - \frac{5}{x(ax^5+1)}
    Thanks! Learning another method for this question is helpful also!

    erm, im not sure how you got from the 2nd last line to the last line, could you show me? Thanks!
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  5. #5
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    Quote Originally Posted by gomes View Post
    Thanks! Learning another method for this question is helpful also!

    erm, im not sure how you got from the 2nd last line to the last line, could you show me? Thanks!

    Make  y as the subject :


      +/- ax^5 = \frac{2+xy}{3-xy}

    Use another constant  k to represent  +/- a

     kx^5 (3-xy) = 2+xy

     3kx^5 - kx^5 (xy) = 2+ xy

     3kx^5 -2 = xy(1 + kx^5)

     xy = \frac{3kx^5 - 2 }{1 + kx^5}

     xy = \frac{3(kx^5+1)- 2-3 }{1 + kx^5}

     xy = 3 - \frac{5}{1+kx^5}

     y = \frac{3}{x} - \frac{5}{x(1+kx^5)}


    From  \frac{du}{dx} = -u^2 - \frac{6u}{x}

    sub.  u = 1/z

     \frac{du}{dx} = - \frac{1}{z^2} ~\frac{dz}{dx}

     - \frac{1}{z^2} ~\frac{dz}{dx} = -( \frac{1}{z^2} + \frac{6}{zx} )

     \frac{dz}{dx} = 1 + \frac{6z}{x}

    \frac{dz}{dx} - \frac{6}{x} ~z = 1


    The correct Riccati's method should be substituting

     y = f(x) + \frac{1}{u} where  f(x) is one of the solution ... this makes the procedures clearer .
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  6. #6
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    thanks, got it now
    Last edited by gomes; June 2nd 2010 at 03:53 AM.
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