inverselaplace (5s)/((s+7)(s^2+4))

Lösung

inverse laplace transformation

\frac{5 s}{( s + 7 ) ( s ^{2} + 4 )}

- \frac{35}{53} e^{- 7 t} + \frac{35}{53} cos (2 t) + \frac{10 sin ( 2 t )}{53}

Schritte zur Lösung

L^{- 1} {\frac{5 s}{( s + 7 ) ( s ^{2} + 4 )}}

Ermittle den Partialbruch von

\frac{5 s}{( s + 7 ) ( s ^{2} + 4 )} : - \frac{35}{53 ( s + 7 )} + \frac{35 s + 20}{53 ( s ^{2} + 4 )}

= L^{- 1} {- \frac{35}{53 ( s + 7 )} + \frac{35 s + 20}{53 ( s ^{2} + 4 )}}

Schreibe um

= L^{- 1} {\frac{35 s}{53 ( s ^{2} + 4 )} + \frac{20}{53 ( s ^{2} + 4 )} - \frac{35}{53 ( s + 7 )}}

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{35}{53} L^{- 1} {\frac{1}{s + 7}} + \frac{35}{53} L^{- 1} {\frac{s}{s ^{2} + 4}} + \frac{20}{53} L^{- 1} {\frac{1}{s ^{2} + 4}}

L^{- 1} {\frac{1}{s + 7}} : e^{- 7 t}

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

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

= - \frac{35}{53} e^{- 7 t} + \frac{35}{53} cos (2 t) + \frac{20}{53} \cdot \frac{1}{2} sin (2 t)

Fasse

- \frac{35}{53} e^{- 7 t} + \frac{35}{53} cos (2 t) + \frac{20}{53} \frac{1}{2} sin (2 t)

zusammen:

- \frac{35}{53} e^{- 7 t} + \frac{35}{53} cos (2 t) + \frac{10 sin ( 2 t )}{53}

= - \frac{35}{53} e^{- 7 t} + \frac{35}{53} cos (2 t) + \frac{10 sin ( 2 t )}{53}

\int_{0}^{2} \frac{30 t ^{2}}{4 + t ^{3}} d t

\int \frac{x ^{3} + x}{x + 1} d x

\int sin (5) (t) cos (4) (t) d t

x \to \infty lim (6 x^{3})

d er i v a t i v e 2 π r - π r^{2}