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Beliebt Rechnen >

inverselaplace s/((s^2-4)(s+2))

Lösung

inverse laplace transformation

\frac{s}{( s ^{2} - 4 ) ( s + 2 )}

Lösung

- \frac{1}{8} e^{- 2 t} + \frac{e ^{- 2 t} t}{2} + \frac{1}{8} e^{2 t}

Schritte zur Lösung

L^{- 1} {\frac{s}{( s ^{2} - 4 ) ( s + 2 )}}

Ermittle den Partialbruch von

\frac{s}{( s ^{2} - 4 ) ( s + 2 )} : - \frac{1}{8 ( s + 2 )} + \frac{1}{2 ( s + 2 ) ^{2}} + \frac{1}{8 ( s - 2 )}

= L^{- 1} {- \frac{1}{8 ( s + 2 )} + \frac{1}{2 ( s + 2 ) ^{2}} + \frac{1}{8 ( s - 2 )}}

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}{8 ( s + 2 )}} + L^{- 1} {\frac{1}{2 ( s + 2 ) ^{2}}} + L^{- 1} {\frac{1}{8 ( s - 2 )}}

L^{- 1} {\frac{1}{8 ( s + 2 )}} : \frac{1}{8} e^{- 2 t}

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

L^{- 1} {\frac{1}{8 ( s - 2 )}} : \frac{1}{8} e^{2 t}

= - \frac{1}{8} e^{- 2 t} + \frac{e ^{- 2 t} t}{2} + \frac{1}{8} e^{2 t}

Beliebte Beispiele

derivative of 2/3 ln(1+3cos^2(x))

\frac{d}{d x} (\frac{2}{3} ln (1 + 3 cos^{2} (x)))

tangent f(x)=x^2-2,\at x=1

t an g e n t f (x) = x^{2} - 2, at x = 1

derivative of 6x^3e^x

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

integral from 0 to pi/4 of 2sec^2(t)

\int_{0}^{\frac{π}{4}} 2 sec^{2} (t) d t

derivative f(x)=(3x)/(1-x)

d er i v a t i v e f (x) = \frac{3 x}{1 - x}