inverselaplace (-s-1)/(s^2+s+10)

Lösung

inverse laplace transformation

\frac{- s - 1}{s ^{2} + s + 10}

- e^{- \frac{t}{2}} cos (\frac{39 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{39 t}{2} )}{39}

Schritte zur Lösung

L^{- 1} {\frac{- s - 1}{s ^{2} + s + 10}}

Schreibe

\frac{- s - 1}{s ^{2} + s + 10}

um:

- \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}

= L^{- 1} {- \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{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 + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}} - \frac{1}{2} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}}

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}} : e^{- \frac{t}{2}} cos (\frac{39 t}{2})

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{39}{4}}} : e^{- \frac{t}{2}} \frac{2}{39} sin (\frac{39 t}{2})

= - e^{- \frac{t}{2}} cos (\frac{39 t}{2}) - \frac{1}{2} e^{- \frac{t}{2}} \frac{2}{39} sin (\frac{39 t}{2})

Fasse

- e^{- \frac{t}{2}} cos (\frac{39 t}{2}) - \frac{1}{2} e^{- \frac{t}{2}} \frac{2}{39} sin (\frac{39 t}{2})

zusammen:

- e^{- \frac{t}{2}} cos (\frac{39 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{39 t}{2} )}{39}

= - e^{- \frac{t}{2}} cos (\frac{39 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{39 t}{2} )}{39}

\frac{d}{d x} (tan (x) cot (x))

\frac{d}{d x} (5 x^{2} + 4 x^{3})

x \to - 6 lim (x - 6 - x)

t an g e n t cos (5 + x^{2})

in v erse l a pl a ce \frac{3 ^{2}}{s ( s + 3 ) ^{2}}