inverselaplace 1/(s[5+s])

Lösung

inverse laplace transformation

\frac{1}{s [ 5 + s ]}

\frac{1}{5} H (t) - \frac{1}{5} e^{- 5 t}

Schritte zur Lösung

L^{- 1} {\frac{1}{s [ s + 5 ]}}

Ermittle den Partialbruch von

\frac{1}{s [ 5 + s ]} : \frac{1}{5 s} - \frac{1}{5 ( s + 5 )}

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

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

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

= \frac{1}{5} H (t) - \frac{1}{5} e^{- 5 t}

y^{^{''}} - 12 y^{^{'}} + 36 y = 0

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

x \to 0 - lim (\frac{18}{tan ( x )})

t a y l or \frac{( x - 1 )}{x}

\frac{d}{d x} (3 x + 7)