inverselaplace 6/((s-2)^4*(s+2)^2)

Lösung

inverse laplace transformation

\frac{6}{( s - 2 ) ^{4} \cdot ( s + 2 ) ^{2}}

- \frac{3}{128} e^{2 t} + \frac{9 e ^{2 t} t}{128} - \frac{3 e ^{2 t} t ^{2}}{32} + \frac{e ^{2 t} t ^{3}}{16} + \frac{3}{128} e^{- 2 t} + \frac{3 e ^{- 2 t} t}{128}

Schritte zur Lösung

L^{- 1} {\frac{6}{( s - 2 ) ^{4} ( s + 2 ) ^{2}}}

Ermittle den Partialbruch von

\frac{6}{( s - 2 ) ^{4} ( s + 2 ) ^{2}} : - \frac{3}{128 ( s - 2 )} + \frac{9}{128 ( s - 2 ) ^{2}} - \frac{3}{16 ( s - 2 ) ^{3}} + \frac{3}{8 ( s - 2 ) ^{4}} + \frac{3}{128 ( s + 2 )} + \frac{3}{128 ( s + 2 ) ^{2}}

= L^{- 1} {- \frac{3}{128 ( s - 2 )} + \frac{9}{128 ( s - 2 ) ^{2}} - \frac{3}{16 ( s - 2 ) ^{3}} + \frac{3}{8 ( s - 2 ) ^{4}} + \frac{3}{128 ( s + 2 )} + \frac{3}{128 ( s + 2 ) ^{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{3}{128 ( s - 2 )}} + L^{- 1} {\frac{9}{128 ( s - 2 ) ^{2}}} - L^{- 1} {\frac{3}{16 ( s - 2 ) ^{3}}} + L^{- 1} {\frac{3}{8 ( s - 2 ) ^{4}}} + L^{- 1} {\frac{3}{128 ( s + 2 )}} + L^{- 1} {\frac{3}{128 ( s + 2 ) ^{2}}}

L^{- 1} {\frac{3}{128 ( s - 2 )}} : \frac{3}{128} e^{2 t}

L^{- 1} {\frac{9}{128 ( s - 2 ) ^{2}}} : \frac{9 e ^{2 t} t}{128}

L^{- 1} {\frac{3}{16 ( s - 2 ) ^{3}}} : \frac{3 e ^{2 t} t ^{2}}{32}

L^{- 1} {\frac{3}{8 ( s - 2 ) ^{4}}} : \frac{e ^{2 t} t ^{3}}{16}

L^{- 1} {\frac{3}{128 ( s + 2 )}} : \frac{3}{128} e^{- 2 t}

L^{- 1} {\frac{3}{128 ( s + 2 ) ^{2}}} : \frac{3 e ^{- 2 t} t}{128}

= - \frac{3}{128} e^{2 t} + \frac{9 e ^{2 t} t}{128} - \frac{3 e ^{2 t} t ^{2}}{32} + \frac{e ^{2 t} t ^{3}}{16} + \frac{3}{128} e^{- 2 t} + \frac{3 e ^{- 2 t} t}{128}

\int x^{7} - 3 x + 2 d x

f (x) = cos (sin (x))

\int_{0}^{1} \frac{3 x}{e ^{2 x}} d x

a re a y = 4 x, y = x^{2}

a re a x^{3}, 0, 2