laplacetransform t^3+3t^2+3t+1

Lösung

laplace transformation

t^{3} + 3 t^{2} + 3 t + 1

\frac{6}{s ^{4}} + \frac{6}{s ^{3}} + \frac{3}{s ^{2}} + \frac{1}{s}

Schritte zur Lösung

L {t^{3} + 3 t^{2} + 3 t + 1}

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}} + 3 L {t^{2}} + 3 L {t} + L {1}

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

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

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

L {1} : \frac{1}{s}

= \frac{6}{s ^{4}} + 3 \cdot \frac{2}{s ^{3}} + 3 \cdot \frac{1}{s ^{2}} + \frac{1}{s}

Fasse

\frac{6}{s ^{4}} + 3 \frac{2}{s ^{3}} + 3 \frac{1}{s ^{2}} + \frac{1}{s}

zusammen:

\frac{6}{s ^{4}} + \frac{6}{s ^{3}} + \frac{3}{s ^{2}} + \frac{1}{s}

= \frac{6}{s ^{4}} + \frac{6}{s ^{3}} + \frac{3}{s ^{2}} + \frac{1}{s}

\frac{d}{d t} (t^{- 2} ln (t))

\frac{\partial}{\partial x} (1 - x^{2})

\frac{\partial}{\partial y} (x^{4} sin (x y^{3}))

x \to \infty lim (\frac{lo g _{2} ( x )}{x})

f (x) = 2 x^{4}