inverselaplace 3/(s(s^2+6s))

Lösung

inverse laplace transformation

\frac{3}{s ( s ^{2} + 6 s )}

- \frac{1}{12} H (t) + \frac{t}{2} + \frac{1}{12} e^{- 6 t}

Schritte zur Lösung

L^{- 1} {\frac{3}{s ( s ^{2} + 6 s )}}

Ermittle den Partialbruch von

\frac{3}{s ( s ^{2} + 6 s )} : - \frac{1}{12 s} + \frac{1}{2 s ^{2}} + \frac{1}{12 ( s + 6 )}

= L^{- 1} {- \frac{1}{12 s} + \frac{1}{2 s ^{2}} + \frac{1}{12 ( s + 6 )}}

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}{12 s}} + L^{- 1} {\frac{1}{2 s ^{2}}} + L^{- 1} {\frac{1}{12 ( s + 6 )}}

L^{- 1} {\frac{1}{12 s}} : \frac{1}{12} H (t)

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

L^{- 1} {\frac{1}{12 ( s + 6 )}} : \frac{1}{12} e^{- 6 t}

= - \frac{1}{12} H (t) + \frac{t}{2} + \frac{1}{12} e^{- 6 t}

\int_{0}^{10} 400 - 20 t d t

d er i v a t i v e 16 e^{x}

s l o p e (3 - 1) (3 - 4)

\frac{d y}{d x} = 7 x + y - 7

\frac{d y}{d t} = - y^{2}, y (0) = \frac{1}{2}