inverselaplace (4s+1)/(2s^2-5s+6)

Lösung

inverse laplace transformation

\frac{4 s + 1}{2 s ^{2} - 5 s + 6}

2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + \frac{12}{23} e^{\frac{5 t}{4}} sin (\frac{23 t}{4})

Schritte zur Lösung

L^{- 1} {\frac{4 s + 1}{2 s ^{2} - 5 s + 6}}

Schreibe

\frac{4 s + 1}{2 s ^{2} - 5 s + 6}

um:

2 \cdot \frac{s - \frac{5}{4}}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}} + 3 \cdot \frac{1}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}

= L^{- 1} {2 \cdot \frac{s - \frac{5}{4}}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}} + 3 \cdot \frac{1}{( s - \frac{5}{4} ) ^{2} + \frac{23}{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)}

= 2 L^{- 1} {\frac{s - \frac{5}{4}}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}} + 3 L^{- 1} {\frac{1}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}}

L^{- 1} {\frac{s - \frac{5}{4}}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}} : e^{\frac{5 t}{4}} cos (\frac{23 t}{4})

L^{- 1} {\frac{1}{( s - \frac{5}{4} ) ^{2} + \frac{23}{16}}} : e^{\frac{5 t}{4}} \frac{4}{23} sin (\frac{23 t}{4})

= 2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + 3 e^{\frac{5 t}{4}} \frac{4}{23} sin (\frac{23 t}{4})

Fasse

2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + 3 e^{\frac{5 t}{4}} \frac{4}{23} sin (\frac{23 t}{4})

zusammen:

2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + \frac{12}{23} e^{\frac{5 t}{4}} sin (\frac{23 t}{4})

= 2 e^{\frac{5 t}{4}} cos (\frac{23 t}{4}) + \frac{12}{23} e^{\frac{5 t}{4}} sin (\frac{23 t}{4})

d er i v a t i v e f (x) = π^{x^{2}}

y^{^{'}} = 2

\int 8 x^{- 5} d x

\int \frac{1}{x ^{2} + 100} d x

t y^{^{'}} = 2 y + 2 t^{3} sin (3 t), y (\frac{π}{3}) = 0