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Beliebt Rechnen >

inverselaplace 1/((s+6)(s-4))

Lösung

inverse laplace transformation

\frac{1}{( s + 6 ) ( s - 4 )}

Lösung

- \frac{1}{10} e^{- 6 t} + \frac{1}{10} e^{4 t}

Schritte zur Lösung

L^{- 1} {\frac{1}{( s + 6 ) ( s - 4 )}}

Ermittle den Partialbruch von

\frac{1}{( s + 6 ) ( s - 4 )} : - \frac{1}{10 ( s + 6 )} + \frac{1}{10 ( s - 4 )}

= L^{- 1} {- \frac{1}{10 ( s + 6 )} + \frac{1}{10 ( s - 4 )}}

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}{10 ( s + 6 )}} + L^{- 1} {\frac{1}{10 ( s - 4 )}}

L^{- 1} {\frac{1}{10 ( s + 6 )}} : \frac{1}{10} e^{- 6 t}

L^{- 1} {\frac{1}{10 ( s - 4 )}} : \frac{1}{10} e^{4 t}

= - \frac{1}{10} e^{- 6 t} + \frac{1}{10} e^{4 t}

Beliebte Beispiele

derivative of 4^{(8x^4+cos(x)^3})

\frac{d}{d x} (4^{(8 x^{4} + c o s (x))^{3}})

integral of 4x^3-3x^2+4x-3

\int 4 x^{3} - 3 x^{2} + 4 x - 3 d x

t^2-t*y*y^'+y^2=0

t^{2} - t \cdot y \cdot y^{^{'}} + y^{2} = 0

integral of 6tan^5(x)sec^2(x)

\int 6 tan^{5} (x) sec^{2} (x) d x

limit as x approaches 0-of 8/(5x^{1/5)}

x \to 0 - lim (\frac{8}{5 x ^{\frac{1}{5}}})