Bsp 28
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  1. #1
    dj_m.o.h.t.'s Avatar
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    Bsp 28

    Es sei F(x,y)=x^3-3xy+y^3-1=0! Man berechne y' und y''!

  2. #2
    dj_m.o.h.t.'s Avatar
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    Meine Lösung!

    3x^2 - 3y - 3x * y' + 3y^2 * y' = 0

    y' * (3y^2 - 3x) = 3y - 3x^2

    y' = y - x^2 / y^2 - x

    y'' * (y^2 - x) + y' * (2y * y' - 1) = y' - 2x

    y'' * (y^2 - x) + (y')^2 * 2y - y' = y' - 2x

    y'' * (y^2 - x) + (y - x^2 / y^2 - x)^2 * 2y - 2 * (y - x^2 / y^2 - x) = -2x

    y'' * (y^2 - x) = 2 * (y - x^2 / y^2 - x) - (y - x^2 / y^2 - x) * 2y - 2x

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