inverselaplace (s^2)/((s-3)(s^2+5))

Lösung

inverse laplace transformation

\frac{s ^{2}}{( s - 3 ) ( s ^{2} + 5 )}

\frac{9}{14} e^{3 t} + \frac{5}{14} cos (5 t) + \frac{3 5 sin ( 5 t )}{14}

Schritte zur Lösung

L^{- 1} {\frac{s ^{2}}{( s - 3 ) ( s ^{2} + 5 )}}

Ermittle den Partialbruch von

\frac{s ^{2}}{( s - 3 ) ( s ^{2} + 5 )} : \frac{9}{14 ( s - 3 )} + \frac{5 s + 15}{14 ( s ^{2} + 5 )}

= L^{- 1} {\frac{9}{14 ( s - 3 )} + \frac{5 s + 15}{14 ( s ^{2} + 5 )}}

Schreibe um

= L^{- 1} {\frac{5 s}{14 ( s ^{2} + 5 )} + \frac{9}{14 ( s - 3 )} + \frac{15}{14 ( s ^{2} + 5 )}}

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)}

= \frac{9}{14} L^{- 1} {\frac{1}{s - 3}} + \frac{5}{14} L^{- 1} {\frac{s}{s ^{2} + 5}} + \frac{15}{14} L^{- 1} {\frac{1}{s ^{2} + 5}}

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

L^{- 1} {\frac{s}{s ^{2} + 5}} : cos (5 t)

L^{- 1} {\frac{1}{s ^{2} + 5}} : \frac{1}{5} sin (5 t)

= \frac{9}{14} e^{3 t} + \frac{5}{14} cos (5 t) + \frac{15}{14} \cdot \frac{1}{5} sin (5 t)

Fasse

\frac{9}{14} e^{3 t} + \frac{5}{14} cos (5 t) + \frac{15}{14} \frac{1}{5} sin (5 t)

zusammen:

\frac{9}{14} e^{3 t} + \frac{5}{14} cos (5 t) + \frac{3 5 sin ( 5 t )}{14}

= \frac{9}{14} e^{3 t} + \frac{5}{14} cos (5 t) + \frac{3 5 sin ( 5 t )}{14}

n = 0 \sum \infty \frac{e ^{n}}{n ^{2}}

\frac{\partial}{\partial z} (2 x z^{3} e^{y^{2}})

\int_{5}^{9} x^{2} d x

a re a 5 - x^{2}, 17 - 7 x

\int c t d t