inverselaplace 2/(s^3(s^2+9))

Lösung

inverse laplace transformation

\frac{2}{s ^{3} ( s ^{2} + 9 )}

- \frac{2}{81} H (t) + \frac{t ^{2}}{9} + \frac{2}{81} cos (3 t)

Schritte zur Lösung

L^{- 1} {\frac{2}{s ^{3} ( s ^{2} + 9 )}}

Ermittle den Partialbruch von

\frac{2}{s ^{3} ( s ^{2} + 9 )} : - \frac{2}{81 s} + \frac{2}{9 s ^{3}} + \frac{2 s}{81 ( s ^{2} + 9 )}

= L^{- 1} {- \frac{2}{81 s} + \frac{2}{9 s ^{3}} + \frac{2 s}{81 ( s ^{2} + 9 )}}

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{2}{81 s}} + L^{- 1} {\frac{2}{9 s ^{3}}} + L^{- 1} {\frac{2 s}{81 ( s ^{2} + 9 )}}

L^{- 1} {\frac{2}{81 s}} : \frac{2}{81} H (t)

L^{- 1} {\frac{2}{9 s ^{3}}} : \frac{t ^{2}}{9}

L^{- 1} {\frac{2 s}{81 ( s ^{2} + 9 )}} : \frac{2}{81} cos (3 t)

= - \frac{2}{81} H (t) + \frac{t ^{2}}{9} + \frac{2}{81} cos (3 t)

\frac{d x}{d t} + 2 t x = t

d er i v a t i v e 640 (1.06)^{t}

d er i v a t i v e f (x) = 2 x^{2} - 32

\frac{d}{d x} (ln (1 + sin (x)))

\frac{d}{d x} ((f (x) (6 x))^{2} - 2)