inverselaplace 1/(s^3+s^4)

Lösung

inverse laplace transformation

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

H (t) - t + \frac{t ^{2}}{2} - e^{- t}

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{3} + s ^{4}}}

Ermittle den Partialbruch von

\frac{1}{s ^{3} + s ^{4}} : \frac{1}{s} - \frac{1}{s ^{2}} + \frac{1}{s ^{3}} - \frac{1}{s + 1}

= L^{- 1} {\frac{1}{s} - \frac{1}{s ^{2}} + \frac{1}{s ^{3}} - \frac{1}{s + 1}}

Wende die lineare Eigenschaftder inversen Laplace-Transformation an:

Für Funktionen

f (s), g (s)

und Konstanten

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{s}} - L^{- 1} {\frac{1}{s ^{2}}} + L^{- 1} {\frac{1}{s ^{3}}} - L^{- 1} {\frac{1}{s + 1}}

L^{- 1} {\frac{1}{s}} : H (t)

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

L^{- 1} {\frac{1}{s ^{3}}} : \frac{t ^{2}}{2}

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

= H (t) - t + \frac{t ^{2}}{2} - e^{- t}

\frac{d}{d x} (5 x tan (x))

y^{^{'}} = t + t y^{2}, y (2) = 0

\frac{d}{d x} (\frac{12}{x ^{3}})

\frac{d}{d x} (\frac{x - 9 x x}{x})

d er i v a t i v e f (x) = \frac{x}{5}