inverselaplace s/(s^2+s+3)

Lösung

inverse laplace transformation

\frac{s}{s ^{2} + s + 3}

e^{- \frac{t}{2}} cos (\frac{11 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{11 t}{2} )}{11}

Schritte zur Lösung

L^{- 1} {\frac{s}{s ^{2} + s + 3}}

Schreibe

\frac{s}{s ^{2} + s + 3}

um:

\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}}

= L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}} - \frac{1}{2} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{11}{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{11}{4}}} - \frac{1}{2} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}}}

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}}} : e^{- \frac{t}{2}} cos (\frac{11 t}{2})

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{11}{4}}} : e^{- \frac{t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

= e^{- \frac{t}{2}} cos (\frac{11 t}{2}) - \frac{1}{2} e^{- \frac{t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

Fasse

e^{- \frac{t}{2}} cos (\frac{11 t}{2}) - \frac{1}{2} e^{- \frac{t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

zusammen:

e^{- \frac{t}{2}} cos (\frac{11 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{11 t}{2} )}{11}

= e^{- \frac{t}{2}} cos (\frac{11 t}{2}) - \frac{e ^{- \frac{t}{2}} sin ( \frac{11 t}{2} )}{11}

\frac{d}{d x} (\frac{4}{3} π r (x) (x)^{3})

\frac{d}{d x} (\frac{- x ^{2} - 8}{( x ^{2} - 8 ) ^{2}})

\frac{d x}{d t} = - \frac{6 x}{20 + 2 t}

\frac{d}{d x} ((3 x^{2} + 1)^{3})

a re a y = x^{3} - 3 x^{2} - x + 3, - 1 < x \leq 3