inverselaplace ((-2s+24))/((s^2-13))

Lösung

inverse laplace transformation

\frac{( - 2 s + 24 )}{( s ^{2} - 13 )}

- \frac{13 + 12}{13} e^{- 13 t} + \frac{- 13 + 12}{13} e^{13 t}

Schritte zur Lösung

L^{- 1} {\frac{( - 2 s + 24 )}{( s ^{2} - 13 )}}

Ermittle den Partialbruch von

\frac{- 2 s + 24}{s ^{2} - 13} : - \frac{13 + 12}{13 ( s + 13 )} + \frac{- 13 + 12}{13 ( s - 13 )}

= L^{- 1} {- \frac{13 + 12}{13 ( s + 13 )} + \frac{- 13 + 12}{13 ( s - 13 )}}

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{13 + 12}{13 ( s + 13 )}} + L^{- 1} {\frac{- 13 + 12}{13 ( s - 13 )}}

L^{- 1} {\frac{13 + 12}{13 ( s + 13 )}} : \frac{13 + 12}{13} e^{- 13 t}

L^{- 1} {\frac{- 13 + 12}{13 ( s - 13 )}} : \frac{- 13 + 12}{13} e^{13 t}

= - \frac{13 + 12}{13} e^{- 13 t} + \frac{- 13 + 12}{13} e^{13 t}

\int 9 tan (x) sec^{3} (x) d x

\int sin (2 x) (e^{1 + c o s^{2} (x)}) + 3 d x

\int cos^{4} (θ) d θ

\frac{d}{d x} (6 x^{3})

2 x y \frac{d y}{d x} = x^{2} + 2 y^{2}