inverselaplace (s+1)/((s^2-4s)(s+5))

Lösung

inverse laplace transformation

\frac{s + 1}{( s ^{2} - 4 s ) ( s + 5 )}

- \frac{1}{20} H (t) + \frac{5}{36} e^{4 t} - \frac{4}{45} e^{- 5 t}

Schritte zur Lösung

L^{- 1} {\frac{s + 1}{( s ^{2} - 4 s ) ( s + 5 )}}

Ermittle den Partialbruch von

\frac{s + 1}{( s ^{2} - 4 s ) ( s + 5 )} : - \frac{1}{20 s} + \frac{5}{36 ( s - 4 )} - \frac{4}{45 ( s + 5 )}

= L^{- 1} {- \frac{1}{20 s} + \frac{5}{36 ( s - 4 )} - \frac{4}{45 ( s + 5 )}}

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}{20 s}} + L^{- 1} {\frac{5}{36 ( s - 4 )}} - L^{- 1} {\frac{4}{45 ( s + 5 )}}

L^{- 1} {\frac{1}{20 s}} : \frac{1}{20} H (t)

L^{- 1} {\frac{5}{36 ( s - 4 )}} : \frac{5}{36} e^{4 t}

L^{- 1} {\frac{4}{45 ( s + 5 )}} : \frac{4}{45} e^{- 5 t}

= - \frac{1}{20} H (t) + \frac{5}{36} e^{4 t} - \frac{4}{45} e^{- 5 t}

x \to 2 lim (ln (x^{2} - 4))

\frac{d}{d x} (\frac{e ^{x}}{x \cdot cos ( x )})

f^{^{''}} + 16 f = sin (2 x)

\int_{6}^{12} 144 - x^{2} d x

x^{2} y^{^{''}} - 6 y = 0, y (1) = 6, y^{^{'}} (1) = - 2