inverselaplace 1/(s^2(s^2+4))

Lösung

inverse laplace transformation

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

\frac{t}{4} - \frac{1}{8} sin (2 t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

L^{- 1} {\frac{1}{4 ( s ^{2} + 4 )}} : \frac{1}{8} sin (2 t)

= \frac{t}{4} - \frac{1}{8} sin (2 t)

a re a y = 3 x - 6, (2, 4)

\frac{d}{d x} (\frac{2 x ^{3} - 5 x ^{2} + 4 x}{( x - 1 ) ^{2}})

\frac{d}{d x} (\frac{2}{5 x ^{2}})

\int (- 2) x d x

\int_{0}^{1} x^{2} 6 x + 8 d x