inverselaplace 1/(s^2)(1-e^{-5s})

Lösung

inverse laplace transformation

\frac{1}{s ^{2}} (1 - e^{- 5 s})

t - H (t - 5) (t - 5)

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2}} (- e^{- 5 s} + 1)}

Schreibe

\frac{1}{s ^{2}} (1 - e^{- 5 s})

um:

\frac{1}{s ^{2}} - \frac{e ^{- 5 s}}{s ^{2}}

= L^{- 1} {\frac{1}{s ^{2}} - \frac{e ^{- 5 s}}{s ^{2}}}

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 ^{2}}} - L^{- 1} {\frac{e ^{- 5 s}}{s ^{2}}}

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

L^{- 1} {\frac{e ^{- 5 s}}{s ^{2}}} : H (t - 5) (t - 5)

= t - H (t - 5) (t - 5)

d er i v a t i v e 12 x^{3} + 24 x^{2}

\frac{\partial}{\partial y} (x^{2} - 4 x y + 4 y^{2} + 2)

n = 1 \sum \infty cos (\frac{2}{n})

x \to 1 + lim (\frac{7}{x ^{3} - 1})

x \frac{d y}{d x} + 3 y = 2 x^{5}, y (2) = 7