Beispiel 61
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Thread: Beispiel 61

  1. #1
    dj_m.o.h.t.'s Avatar
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    Beispiel 61

    Angabe:

    Man löse die Differentialgleichungen!

    y'' + y = 1/cosx (mittels Variation der Konstanten)

    Lösung:

    yh: L...Lamda

    L^2 + 1 = 0 => L^2 = -1 => L1,2 = +-i

    => yh = c1 * cosx - c2 * sinx

    Warum diese Lösung:

    Siehe folgendes: e^Alpha+i*Beta = e^Alpha * (cosBeta - i * sinBeta)
    Darum!

    Variation der Konstanten:

    Statt c1 und c2 => z1 und z2

    y = z1 * cosx - z2 * sinx
    y' = z1' * cosx - z1 * sinx - z2' * sinx - z2 * cosx
    Bedingung: z1' * cosx - z2' * sinx = 0
    y'' = -z1' * sinx - z1 * cosx - z2' * cosx + z2 * sinx

    Einsetzen:

    -z1' * sinx - z1 * cosx - z2' * cosx + z2 * sinx + z1 * cosx - z2 * sinx = 1/cosx
    -z1' * sinx - z2' * cosx = 1/cosx

    z1' * cosx - z2' * sinx = 0 => z1' = z2' * sinx/cosx

    -z2' * sin^2x/cosx - z2' * cosx = 1/cosx
    -z2' * (sin^2x + cos^2x) =1

    sin^2x + cos^2x = 1

    => z2' = -1 => z2 = -x

    z1' = - sinx/cosx => z1' = -tanx => z1 = -ln|cosx|

    y= yh + yp => y = c1 * cosx - c2 * sinx - ln|cosx| * cosx + x * sinx

  2. #2

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    Gilt nicht folgendes:
    Nullstellen:
    +i => cosx
    und für die konjugierte
    -i => sinx

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