inverselaplace (s+5)/(s^2+5s+9)

Lösung

inverse laplace transformation

\frac{s + 5}{s ^{2} + 5 s + 9}

e^{- \frac{5 t}{2}} cos (\frac{11 t}{2}) + \frac{5 e ^{- \frac{5 t}{2}} sin ( \frac{11 t}{2} )}{11}

Schritte zur Lösung

L^{- 1} {\frac{s + 5}{s ^{2} + 5 s + 9}}

Schreibe

\frac{s + 5}{s ^{2} + 5 s + 9}

um:

\frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{11}{4}} + \frac{5}{2} \cdot \frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{11}{4}}

= L^{- 1} {\frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{11}{4}} + \frac{5}{2} \cdot \frac{1}{( s + \frac{5}{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{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{11}{4}}} + \frac{5}{2} L^{- 1} {\frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{11}{4}}}

L^{- 1} {\frac{s + \frac{5}{2}}{( s + \frac{5}{2} ) ^{2} + \frac{11}{4}}} : e^{- \frac{5 t}{2}} cos (\frac{11 t}{2})

L^{- 1} {\frac{1}{( s + \frac{5}{2} ) ^{2} + \frac{11}{4}}} : e^{- \frac{5 t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

= e^{- \frac{5 t}{2}} cos (\frac{11 t}{2}) + \frac{5}{2} e^{- \frac{5 t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

Fasse

e^{- \frac{5 t}{2}} cos (\frac{11 t}{2}) + \frac{5}{2} e^{- \frac{5 t}{2}} \frac{2}{11} sin (\frac{11 t}{2})

zusammen:

e^{- \frac{5 t}{2}} cos (\frac{11 t}{2}) + \frac{5 e ^{- \frac{5 t}{2}} sin ( \frac{11 t}{2} )}{11}

= e^{- \frac{5 t}{2}} cos (\frac{11 t}{2}) + \frac{5 e ^{- \frac{5 t}{2}} sin ( \frac{11 t}{2} )}{11}

y^{^{''}} + 4 y = t^{2} + 3 e^{t}, y (0) = 0, y^{^{'}} (0) = 2

\int y^{3} d y

t an g e n t y = (\frac{( x - 1 )}{( x + 1 )})^{2}, at x = 0

2 \frac{d y}{d x} + 10 y = 1

\int_{0}^{3} 9 - x^{2} d x