inverselaplace {1/(s^4-9)}

Lösung

inverse laplace transformation

{\frac{1}{s ^{4} - 9}}

- \frac{1}{6 3} sin (3 t) - \frac{1}{12 3} e^{- 3 t} + \frac{1}{12 3} e^{3 t}

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{4} - 9}}

Ermittle den Partialbruch von

\frac{1}{s ^{4} - 9} : - \frac{1}{6 ( s ^{2} + 3 )} - \frac{1}{12 3 ( s + 3 )} + \frac{1}{12 3 ( s - 3 )}

= L^{- 1} {- \frac{1}{6 ( s ^{2} + 3 )} - \frac{1}{12 3 ( s + 3 )} + \frac{1}{12 3 ( s - 3 )}}

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}{6 ( s ^{2} + 3 )}} - L^{- 1} {\frac{1}{12 3 ( s + 3 )}} + L^{- 1} {\frac{1}{12 3 ( s - 3 )}}

L^{- 1} {\frac{1}{6 ( s ^{2} + 3 )}} : \frac{1}{6 3} sin (3 t)

L^{- 1} {\frac{1}{12 3 ( s + 3 )}} : \frac{1}{12 3} e^{- 3 t}

L^{- 1} {\frac{1}{12 3 ( s - 3 )}} : \frac{1}{12 3} e^{3 t}

= - \frac{1}{6 3} sin (3 t) - \frac{1}{12 3} e^{- 3 t} + \frac{1}{12 3} e^{3 t}

d er i v a t i v e y = 9 e^{x} + \frac{4}{3 x}

d er i v a t i v e - 5 t + 3 + t^{7}

\int_{- 2}^{2} t^{3} (3 + t^{4})^{3} d t

\int (0.3 y^{4} - 8 y^{- 3} + 2) d y

x \to - 1 + lim (x)