inverselaplace 1/((s^2-1)^2)

Lösung

inverse laplace transformation

\frac{1}{( s ^{2} - 1 ) ^{2}}

\frac{1}{4} e^{- t} + \frac{e ^{- t} t}{4} - \frac{1}{4} e^{t} + \frac{e ^{t} t}{4}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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}{4 ( s + 1 )}} + L^{- 1} {\frac{1}{4 ( s + 1 ) ^{2}}} - L^{- 1} {\frac{1}{4 ( s - 1 )}} + L^{- 1} {\frac{1}{4 ( s - 1 ) ^{2}}}

L^{- 1} {\frac{1}{4 ( s + 1 )}} : \frac{1}{4} e^{- t}

L^{- 1} {\frac{1}{4 ( s + 1 ) ^{2}}} : \frac{e ^{- t} t}{4}

L^{- 1} {\frac{1}{4 ( s - 1 )}} : \frac{1}{4} e^{t}

L^{- 1} {\frac{1}{4 ( s - 1 ) ^{2}}} : \frac{e ^{t} t}{4}

= \frac{1}{4} e^{- t} + \frac{e ^{- t} t}{4} - \frac{1}{4} e^{t} + \frac{e ^{t} t}{4}

\int_{1}^{3} \frac{2 t ^{2} + 1}{t} d t

\int \frac{3 sin ( x )}{cos ^{3} ( x )} d x

\int \frac{2 x + 3}{x + 7} d x

\int (x + 8)^{2} d x

\int_{- 2}^{3} \frac{10}{x ^{4}} d x