inverselaplace ((s+1)^2)/(s^2-s-6)

Lösung

inverse laplace transformation

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

δ (t) - \frac{1}{5} e^{- 2 t} + \frac{16}{5} e^{3 t}

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{( s + 1 ) ^{2}}{s ^{2} - s - 6} : 1 - \frac{1}{5 ( s + 2 )} + \frac{16}{5 ( s - 3 )}

= L^{- 1} {1 - \frac{1}{5 ( s + 2 )} + \frac{16}{5 ( s - 3 )}}

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} {1} - L^{- 1} {\frac{1}{5 ( s + 2 )}} + L^{- 1} {\frac{16}{5 ( s - 3 )}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{1}{5 ( s + 2 )}} : \frac{1}{5} e^{- 2 t}

L^{- 1} {\frac{16}{5 ( s - 3 )}} : \frac{16}{5} e^{3 t}

= δ (t) - \frac{1}{5} e^{- 2 t} + \frac{16}{5} e^{3 t}

\frac{d}{d x} (\frac{7}{x ^{8} + 5 x + 1})

\frac{d}{d x} (sin (14 x))

y^{^{'}} = 1 + y^{2} + x + x y^{2}

\int \frac{( 1 + x )}{x} d x

x \to 0 lim (2 x + 12)