inverselaplace (s-1)/(2s^2+s+6)

Lösung

inverse laplace transformation

\frac{s - 1}{2 s ^{2} + s + 6}

\frac{1}{2} e^{- \frac{t}{4}} cos (\frac{47 t}{4}) - \frac{5 e ^{- \frac{t}{4}} sin ( \frac{47 t}{4} )}{2 47}

Schritte zur Lösung

L^{- 1} {\frac{s - 1}{2 s ^{2} + s + 6}}

Schreibe

\frac{s - 1}{2 s ^{2} + s + 6}

um:

\frac{1}{2} \cdot \frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{47}{16}} - \frac{5}{8} \cdot \frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{47}{16}}

= L^{- 1} {\frac{1}{2} \cdot \frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{47}{16}} - \frac{5}{8} \cdot \frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{47}{16}}}

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}{2} L^{- 1} {\frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{47}{16}}} - \frac{5}{8} L^{- 1} {\frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{47}{16}}}

L^{- 1} {\frac{s + \frac{1}{4}}{( s + \frac{1}{4} ) ^{2} + \frac{47}{16}}} : e^{- \frac{t}{4}} cos (\frac{47 t}{4})

L^{- 1} {\frac{1}{( s + \frac{1}{4} ) ^{2} + \frac{47}{16}}} : e^{- \frac{t}{4}} \frac{4}{47} sin (\frac{47 t}{4})

= \frac{1}{2} e^{- \frac{t}{4}} cos (\frac{47 t}{4}) - \frac{5}{8} e^{- \frac{t}{4}} \frac{4}{47} sin (\frac{47 t}{4})

Fasse

\frac{1}{2} e^{- \frac{t}{4}} cos (\frac{47 t}{4}) - \frac{5}{8} e^{- \frac{t}{4}} \frac{4}{47} sin (\frac{47 t}{4})

zusammen:

\frac{1}{2} e^{- \frac{t}{4}} cos (\frac{47 t}{4}) - \frac{5 e ^{- \frac{t}{4}} sin ( \frac{47 t}{4} )}{2 47}

= \frac{1}{2} e^{- \frac{t}{4}} cos (\frac{47 t}{4}) - \frac{5 e ^{- \frac{t}{4}} sin ( \frac{47 t}{4} )}{2 47}

\int_{2}^{4} (2 x^{2} - 3 x + 1) d x

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = \frac{82}{5}

\int 1 - 81 x^{2} d x

\int 7 cos^{2} (x) d x