inverselaplace 1/(s(s+4)^2)

Lösung

inverse laplace transformation

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

\frac{1}{16} H (t) - \frac{1}{16} e^{- 4 t} - \frac{e ^{- 4 t} t}{4}

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{s ( s + 4 ) ^{2}} : \frac{1}{16 s} - \frac{1}{16 ( s + 4 )} - \frac{1}{4 ( s + 4 ) ^{2}}

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

L^{- 1} {\frac{1}{16 s}} : \frac{1}{16} H (t)

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

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

= \frac{1}{16} H (t) - \frac{1}{16} e^{- 4 t} - \frac{e ^{- 4 t} t}{4}

\frac{d}{d x} ((3 x^{2} + 5)^{3} (3 x - 1)^{2})

\int (1 + x^{2})^{- 1} d x

t an g e n t f (x) = (e^{5 x} - 4)^{2}, (0, 9)

\frac{d}{d x} (sin (x) - x \cdot cos (x))

\frac{d y}{d x} + y cos (x) = sin (x) cos (x)