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

Lösung

inverse laplace transformation

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

\frac{3}{4} e^{2 t} + \frac{3 e ^{2 t} t}{8} - \frac{3}{4} e^{4 t} + \frac{9 e ^{4 t} t}{8} - \frac{3 e ^{4 t} t ^{2}}{4} + \frac{e ^{4 t} t ^{3}}{4}

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

L^{- 1} {\frac{9}{8 ( s - 4 ) ^{2}}} : \frac{9 e ^{4 t} t}{8}

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

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

= \frac{3}{4} e^{2 t} + \frac{3 e ^{2 t} t}{8} - \frac{3}{4} e^{4 t} + \frac{9 e ^{4 t} t}{8} - \frac{3 e ^{4 t} t ^{2}}{4} + \frac{e ^{4 t} t ^{3}}{4}

x \frac{d y}{d x} + 1 = x + y

\frac{\partial}{\partial y} (6 - 3 x - 2 y)

\int_{- \infty}^{- 1} e^{- 34 t} d t

n = 1 \sum \infty \frac{2 n}{1 + n ^{4}}

x^{3} y^{^{'}} + 2 x^{2} y = 2 cos (3 x) y^{- \frac{1}{2}}