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Beliebt Rechnen >

inverselaplace (2s)/((s^2+25)(s-1))

Lösung

inverse laplace transformation

\frac{2 s}{( s ^{2} + 25 ) ( s - 1 )}

Lösung

- \frac{1}{13} cos (5 t) + \frac{5 sin ( 5 t )}{13} + \frac{1}{13} e^{t}

Schritte zur Lösung

L^{- 1} {\frac{2 s}{( s ^{2} + 25 ) ( s - 1 )}}

Ermittle den Partialbruch von

\frac{2 s}{( s ^{2} + 25 ) ( s - 1 )} : \frac{- s + 25}{13 ( s ^{2} + 25 )} + \frac{1}{13 ( s - 1 )}

= L^{- 1} {\frac{- s + 25}{13 ( s ^{2} + 25 )} + \frac{1}{13 ( s - 1 )}}

Schreibe um

= L^{- 1} {- \frac{s}{13 ( s ^{2} + 25 )} + \frac{25}{13 ( s ^{2} + 25 )} + \frac{1}{13 ( s - 1 )}}

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{1}{13} L^{- 1} {\frac{s}{s ^{2} + 25}} + \frac{25}{13} L^{- 1} {\frac{1}{s ^{2} + 25}} + \frac{1}{13} L^{- 1} {\frac{1}{s - 1}}

L^{- 1} {\frac{s}{s ^{2} + 25}} : cos (5 t)

L^{- 1} {\frac{1}{s ^{2} + 25}} : \frac{1}{5} sin (5 t)

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

= - \frac{1}{13} cos (5 t) + \frac{25}{13} \cdot \frac{1}{5} sin (5 t) + \frac{1}{13} e^{t}

Fasse

- \frac{1}{13} cos (5 t) + \frac{25}{13} \frac{1}{5} sin (5 t) + \frac{1}{13} e^{t}

zusammen:

- \frac{1}{13} cos (5 t) + \frac{5 sin ( 5 t )}{13} + \frac{1}{13} e^{t}

= - \frac{1}{13} cos (5 t) + \frac{5 sin ( 5 t )}{13} + \frac{1}{13} e^{t}

Beliebte Beispiele

derivative y=(2x)/(x^2-1)

d er i v a t i v e y = \frac{2 x}{x ^{2} - 1}

fläche y=2-x^2,y=x

a re a y = 2 - x^{2}, y = x

limit as x approaches-2 of (-5x-10)/(x^3+2x^2)

x \to - 2 lim (\frac{- 5 x - 10}{x ^{3} + 2 x ^{2}})

derivative (6x)/(sqrt(x^2+2))

d er i v a t i v e \frac{6 x}{x ^{2} + 2}

(dy)/(dx)= 1/(xsqrt(16x^2-1))

\frac{d y}{d x} = \frac{1}{x 16 x ^{2} - 1}