inverselaplace (1-e^{-2s})/(6s^2)

Lösung

inverse laplace transformation

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

\frac{t}{6} - \frac{1}{6} H (t - 2) (t - 2)

Schritte zur Lösung

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

Schreibe um

= L^{- 1} {\frac{1}{6 s ^{2}} - \frac{e ^{- 2 s}}{6 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)}

= \frac{1}{6} L^{- 1} {\frac{1}{s ^{2}}} - \frac{1}{6} L^{- 1} {\frac{e ^{- 2 s}}{s ^{2}}}

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

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

= \frac{1}{6} t - \frac{1}{6} H (t - 2) (t - 2)

Fasse

\frac{1}{6} t - \frac{1}{6} H (t - 2) (t - 2)

zusammen:

\frac{t}{6} - \frac{1}{6} H (t - 2) (t - 2)

= \frac{t}{6} - \frac{1}{6} H (t - 2) (t - 2)

x \to 0 lim (- cos (x) + 1)

y = x (1 - x)^{6}

in v erse l a pl a ce f (s) = s - 3 (s^{2} + 4 s + 13)

a re a y^{2} = 4 x, y^{2} = \frac{4}{x}

n = 1 \sum \infty \frac{( 2 n ) !}{5 ^{n} ( n ! ) ^{2}}