laplacetransform (t-5)^3

Lösung

laplace transformation

(t - 5)^{3}

\frac{6}{s ^{4}} - \frac{30}{s ^{3}} + \frac{75}{s ^{2}} - \frac{125}{s}

Schritte zur Lösung

L {(t - 5)^{3}}

Schreibe

(t - 5)^{3}

um:

t^{3} - 15 t^{2} + 75 t - 125

= L {t^{3} - 15 t^{2} + 75 t - 125}

Verwende die lineare Eigenschaft der Laplace-Transformation:

Für Funktionen

f (t), g (t)

und Konstanten

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {t^{3}} - 15 L {t^{2}} + 75 L {t} - L {125}

L {t^{3}} : \frac{6}{s ^{4}}

L {t^{2}} : \frac{2}{s ^{3}}

L {t} : \frac{1}{s ^{2}}

L {125} : \frac{125}{s}

= \frac{6}{s ^{4}} - 15 \cdot \frac{2}{s ^{3}} + 75 \cdot \frac{1}{s ^{2}} - \frac{125}{s}

Fasse

\frac{6}{s ^{4}} - 15 \frac{2}{s ^{3}} + 75 \frac{1}{s ^{2}} - \frac{125}{s}

zusammen:

\frac{6}{s ^{4}} - \frac{30}{s ^{3}} + \frac{75}{s ^{2}} - \frac{125}{s}

= \frac{6}{s ^{4}} - \frac{30}{s ^{3}} + \frac{75}{s ^{2}} - \frac{125}{s}

(y) d x - 4 (x + y^{6}) d y = 0

t an g e n t f (x) = 3 - 4 x^{2}, (4, - 61)

d er i v a t i v e 5 cos (t)

\frac{d}{d x} (3 e^{x} sin (x))

y^{^{''}} + 16 y = - 16 sec^{2} (4 x)