inverselaplace 1/(s^2(s^2+2^2))

Lösung

inverse laplace transformation

\frac{1}{s ^{2} ( s ^{2} + 2 ^{2} )}

\frac{t}{4} - \frac{1}{8} sin (2 t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{s ^{2} ( s ^{2} + 2 ^{2} )} : \frac{1}{4 s ^{2}} - \frac{1}{4 ( s ^{2} + 4 )}

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

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

L^{- 1} {\frac{1}{4 ( s ^{2} + 4 )}} : \frac{1}{8} sin (2 t)

= \frac{t}{4} - \frac{1}{8} sin (2 t)

y^{^{'}} + 6 x y = 0

l a pl a ce t r an s f or m 2 t - e^{- 3 t}

\int (x^{2} e^{3 x}) d x

\frac{\partial}{\partial v} (\frac{1}{3} (u + v))

\frac{d}{d z} (4 - z)