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Beliebt Rechnen >

inverselaplace (-40x-80)/(x^2+3x+1)

Lösung

inverse laplace transformation

\frac{- 40 x - 80}{x ^{2} + 3 x + 1}

Lösung

- 40 e^{- \frac{3 t}{2}} cosh (\frac{5 t}{2}) - 85 e^{- \frac{3 t}{2}} sinh (\frac{5 t}{2})

Schritte zur Lösung

L^{- 1} {\frac{- 40 x - 80}{x ^{2} + 3 x + 1}}

Schreibe

\frac{- 40 x - 80}{x ^{2} + 3 x + 1}

um:

- 40 \cdot \frac{x + \frac{3}{2}}{( x + \frac{3}{2} ) ^{2} - \frac{5}{4}} - 20 \cdot \frac{1}{( x + \frac{3}{2} ) ^{2} - \frac{5}{4}}

= L^{- 1} {- 40 \cdot \frac{x + \frac{3}{2}}{( x + \frac{3}{2} ) ^{2} - \frac{5}{4}} - 20 \cdot \frac{1}{( x + \frac{3}{2} ) ^{2} - \frac{5}{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)}

= - 40 L^{- 1} {\frac{x + \frac{3}{2}}{( x + \frac{3}{2} ) ^{2} - \frac{5}{4}}} - 20 L^{- 1} {\frac{1}{( x + \frac{3}{2} ) ^{2} - \frac{5}{4}}}

L^{- 1} {\frac{x + \frac{3}{2}}{( x + \frac{3}{2} ) ^{2} - \frac{5}{4}}} : e^{- \frac{3 t}{2}} cosh (\frac{5 t}{2})

L^{- 1} {\frac{1}{( x + \frac{3}{2} ) ^{2} - \frac{5}{4}}} : e^{- \frac{3 t}{2}} \frac{2}{5} sinh (\frac{5 t}{2})

= - 40 e^{- \frac{3 t}{2}} cosh (\frac{5 t}{2}) - 20 e^{- \frac{3 t}{2}} \frac{2}{5} sinh (\frac{5 t}{2})

Fasse

- 40 e^{- \frac{3 t}{2}} cosh (\frac{5 t}{2}) - 20 e^{- \frac{3 t}{2}} \frac{2}{5} sinh (\frac{5 t}{2})

zusammen:

- 40 e^{- \frac{3 t}{2}} cosh (\frac{5 t}{2}) - 85 e^{- \frac{3 t}{2}} sinh (\frac{5 t}{2})

= - 40 e^{- \frac{3 t}{2}} cosh (\frac{5 t}{2}) - 85 e^{- \frac{3 t}{2}} sinh (\frac{5 t}{2})

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