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Thread: a first order separable differential equations solution

  1. #1

    a first order separable differential equations solution

    Hi to all,

    I have this beautifull diff. eq. : y'=(1-y^2)/(1-x^2)
    the solution is not a problem. (also with Derive6 or Ti89) y=(c(x+1)+x-1)/(c(x+1)-x+1)
    but the book says : y=(x+c)/(cx+1) and verifying is correct too.
    axes traslstions ? or something else ?

    Thanks.

  2. #2
    Elite Member
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    Quote Originally Posted by kotongo View Post
    Hi to all,

    I have this beautifull diff. eq. : y'=(1-y^2)/(1-x^2)
    the solution is not a problem. (also with Derive6 or Ti89) y=(c(x+1)+x-1)/(c(x+1)-x+1)
    but the book says : y=(x+c)/(cx+1) and verifying is correct too.
    axes traslstions ? or something else ?

    Thanks.
    Combine constant. For example, c+1 simply becomes C.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  3. #3
    Elite Member
    Join Date
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    Rochester, NY
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    2,608
    Quote Originally Posted by kotongo View Post
    Hi to all,

    I have this beautifull diff. eq. : y'=(1-y^2)/(1-x^2)
    the solution is not a problem. (also with Derive6 or Ti89) y=(c(x+1)+x-1)/(c(x+1)-x+1)
    but the book says : y=(x+c)/(cx+1) and verifying is correct too.
    axes traslstions ? or something else ?

    Thanks.
    They are just using a different C, perhaps after simplifying. (They probably used a different method to get it.)

    If we rewrite yours to make it look a little more like theirs, we get [tex]\frac{(c+1)x + (c-1)}{(c-1)x + (c+1)}[/tex].

    Try dividing numerator and denominator by (c+1), and see if you can make it look just like theirs, once you define a new constant C.

  4. #4
    Perfect Thanks a lot !



    (%i1) solve((c*(x+1)+x-1)/(c*(x+1)-x+1)=(x+k)/(k*x+1),k);
    c - 1
    (%o1) [k = -----]
    c + 1
    Aslo with maxima online

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