inverselaplace 1/((s^4-1))

Lösung

inverse laplace transformation

\frac{1}{( s ^{4} - 1 )}

- \frac{1}{2} sin (t) - \frac{1}{4} e^{- t} + \frac{1}{4} e^{t}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

= - \frac{1}{2} sin (t) - \frac{1}{4} e^{- t} + \frac{1}{4} e^{t}

\int \frac{1}{7 + 5 x} d x

t an g e n t f (x) = \frac{x - 1}{x + 1}, at x = 0

t an g e n t x^{2} + x y - y^{2} = - 4, (2, 4)

y^{^{'}} = \frac{5 x ^{5} e ^{\frac{y}{x}} + x ^{3} y ^{2}}{x ^{4} y}

\int (\frac{x + 5}{x ^{2} + 5 x + 6}) d x