laplacetransform {6e^{-3t}-t^2+2t-8}

Lösung

laplace transformation

{6 e^{- 3 t} - t^{2} + 2 t - 8}

\frac{6}{s + 3} - \frac{2}{s ^{3}} + \frac{2}{s ^{2}} - \frac{8}{s}

Schritte zur Lösung

L {6 e^{- 3 t} - t^{2} + 2 t - 8}

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)}

= 6 L {e^{- 3 t}} - L {t^{2}} + 2 L {t} - L {8}

L {e^{- 3 t}} : \frac{1}{s + 3}

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

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

L {8} : \frac{8}{s}

= 6 \cdot \frac{1}{s + 3} - \frac{2}{s ^{3}} + 2 \cdot \frac{1}{s ^{2}} - \frac{8}{s}

Fasse

6 \frac{1}{s + 3} - \frac{2}{s ^{3}} + 2 \frac{1}{s ^{2}} - \frac{8}{s}

zusammen:

\frac{6}{s + 3} - \frac{2}{s ^{3}} + \frac{2}{s ^{2}} - \frac{8}{s}

= \frac{6}{s + 3} - \frac{2}{s ^{3}} + \frac{2}{s ^{2}} - \frac{8}{s}

(y - x) e^{x} + (1 + e^{x}) \frac{d y}{d x} = 0

\int_{0}^{1} 2 π x (x - x) d x

\int_{1}^{7} (6 x^{2} + 2 x - 3) d x

n = 10 \sum \infty \frac{2}{3} (\frac{3}{4})^{n}

\int (1 + x)^{8} d x