Bsp 24
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  1. #1
    dj_m.o.h.t.'s Avatar
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    Bsp 24

    Man zeige: u=ln Wurzel (x-a)^2+(y-b)^2 genügt der Differentialgleichung uxx+uyy=0!

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

    ux = 1 / Wurzel (x-a)^2+(y-b)^2 * 1/2 * [(x-a)^2 + (y-b)^2)^-1/2 * [2 * (x-a)] = x-a / (x-a)^2 + (y-b)^2

    uxx = (x-a)^2+(y-b)^2-(x-a)*2*(x-a) / [(x-a)^2+(y-b)^2]^2 = (y-b)^2 - (x-a)^2 / [(x-a)^2 + (y-b)^2]

    uy = y-b / (x-a)^2 + (y-b)^2

    uyy = (x-a)^2 - (y-b)^2 / [(x-a)^2 + (y-b)]^2

    uxx + uyy = 0

    (y-b)^2 - (x-a)^2 + (x-a)^2 - (y-b)^2 / [(x-a)^2 + (y-b)^2]^2 = 0 / [(x-a)^2 + (y-b)^2]^2 = 0 => w. A.

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