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

Lösung

inverse laplace transformation

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

H (t) + 4 t + 2 t^{2}

Schritte zur Lösung

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

Schreibe

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

um:

\frac{1}{s} + \frac{4}{s ^{2}} + \frac{4}{s ^{3}}

= L^{- 1} {\frac{1}{s} + \frac{4}{s ^{2}} + \frac{4}{s ^{3}}}

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{1}{s}} + L^{- 1} {\frac{4}{s ^{2}}} + L^{- 1} {\frac{4}{s ^{3}}}

L^{- 1} {\frac{1}{s}} : H (t)

L^{- 1} {\frac{4}{s ^{2}}} : 4 t

L^{- 1} {\frac{4}{s ^{3}}} : 2 t^{2}

= H (t) + 4 t + 2 t^{2}

x \to 3 lim (\frac{x - 3}{x ^{2} + 9})

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

x \to - 8 lim (x + 6)

\frac{d y}{d x} - 2 x y = 10, y (0) = - 8

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