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Beliebt Rechnen >

inverselaplace (1+5s^2)/(s^2(s-9))

Lösung

inverse laplace transformation

\frac{1 + 5 s ^{2}}{s ^{2} ( s - 9 )}

Lösung

- \frac{1}{81} H (t) - \frac{t}{9} + \frac{406}{81} e^{9 t}

Schritte zur Lösung

L^{- 1} {\frac{1 + 5 s ^{2}}{s ^{2} ( s - 9 )}}

Ermittle den Partialbruch von

\frac{1 + 5 s ^{2}}{s ^{2} ( s - 9 )} : - \frac{1}{81 s} - \frac{1}{9 s ^{2}} + \frac{406}{81 ( s - 9 )}

= L^{- 1} {- \frac{1}{81 s} - \frac{1}{9 s ^{2}} + \frac{406}{81 ( s - 9 )}}

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}{81 s}} - L^{- 1} {\frac{1}{9 s ^{2}}} + L^{- 1} {\frac{406}{81 ( s - 9 )}}

L^{- 1} {\frac{1}{81 s}} : \frac{1}{81} H (t)

L^{- 1} {\frac{1}{9 s ^{2}}} : \frac{t}{9}

L^{- 1} {\frac{406}{81 ( s - 9 )}} : \frac{406}{81} e^{9 t}

= - \frac{1}{81} H (t) - \frac{t}{9} + \frac{406}{81} e^{9 t}

Beliebte Beispiele

integral of 8e^{5x}

(2x+ye^{y/x})dx-xe^{y/x}dy=0,y(1)=4

tangent y=x^2-5x+5

integral of (x^2)/(sqrt(361-x^2))

integral of (4x+2)/(x^2+x+5)