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}

\int (2 x sin (x^{2})) d x

\int sin (x) x d x

x y^{2} y^{^{'}} = 2 - y^{3}

a re a y = x - 3, y^{2} = 2 x + 18

t an g e n t f (x) = \frac{1}{2 + 3 x}, at x = 2