inverselaplace 2/((4s^2-8s))

Lösung

inverse laplace transformation

\frac{2}{( 4 s ^{2} - 8 s )}

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

Schritte zur Lösung

L^{- 1} {\frac{2}{( 4 s ^{2} - 8 s )}}

Ermittle den Partialbruch von

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

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

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

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

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

\frac{d}{d x} (sin (x) + arcsin (x))

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

12 x^{3} - 3 y^{2} x^{4} + 1 (\frac{d y}{d x}) = 0

\int_{0}^{\frac{1}{2}} (2 x - 1)^{- 3} d x

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