inverselaplace (10)/((s-2)(s^3+s))

Lösung

inverse laplace transformation

\frac{10}{( s - 2 ) ( s ^{3} + s )}

e^{2 t} - 5 H (t) + 4 cos (t) - 2 sin (t)

Schritte zur Lösung

L^{- 1} {\frac{10}{( s - 2 ) ( s ^{3} + s )}}

Ermittle den Partialbruch von

\frac{10}{( s - 2 ) ( s ^{3} + s )} : \frac{1}{s - 2} - \frac{5}{s} + \frac{4 s - 2}{s ^{2} + 1}

= L^{- 1} {\frac{1}{s - 2} - \frac{5}{s} + \frac{4 s - 2}{s ^{2} + 1}}

Schreibe um

= L^{- 1} {\frac{1}{s - 2} + \frac{4 s}{s ^{2} + 1} - \frac{5}{s} - \frac{2}{s ^{2} + 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 - 2}} - 5 L^{- 1} {\frac{1}{s}} + 4 L^{- 1} {\frac{s}{s ^{2} + 1}} - 2 L^{- 1} {\frac{1}{s ^{2} + 1}}

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

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

L^{- 1} {\frac{s}{s ^{2} + 1}} : cos (t)

L^{- 1} {\frac{1}{s ^{2} + 1}} : sin (t)

= e^{2 t} - 5 H (t) + 4 cos (t) - 2 sin (t)

2 t y^{^{'}} = 4 y

\frac{d P}{d t} = 0.08 P (1 - \frac{P}{1000})

\int \frac{x ^{2}}{( 21 + 4 x - x ^{2} )} d x

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\int 2 x (8 x^{2} - 20)^{2} d x