inverselaplace (20)/((s^2+4)(s-4))

Lösung

inverse laplace transformation

\frac{20}{( s ^{2} + 4 ) ( s - 4 )}

- cos (2 t) - 2 sin (2 t) + e^{4 t}

Schritte zur Lösung

L^{- 1} {\frac{20}{( s ^{2} + 4 ) ( s - 4 )}}

Ermittle den Partialbruch von

\frac{20}{( s ^{2} + 4 ) ( s - 4 )} : \frac{- s - 4}{s ^{2} + 4} + \frac{1}{s - 4}

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

Schreibe um

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

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

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

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

= - cos (2 t) - 4 \cdot \frac{1}{2} sin (2 t) + e^{4 t}

Fasse

- cos (2 t) - 4 \frac{1}{2} sin (2 t) + e^{4 t}

zusammen:

- cos (2 t) - 2 sin (2 t) + e^{4 t}

= - cos (2 t) - 2 sin (2 t) + e^{4 t}

x \to 2 lim (tan (\frac{π x}{4}))

\frac{\partial}{\partial x} (ln (\frac{2 x y}{7 z}))

t an g e n t f (x) = 1 3 x, at x = 1

\int (x + 2) sin (x) d x

\int_{- 1}^{2} (20 - x^{4}) d x