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

Lösung

inverse laplace transformation

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

- \frac{3}{16} H (t) + \frac{3 t}{8} + \frac{3}{16} e^{- 2 t} - \frac{3 e ^{- 2 t} t ^{2}}{8}

Schritte zur Lösung

L^{- 1} ⎩ ⎨ ⎧ \frac{( 3 s + 3 )}{( s ^{2} ( s + 2 ) ^{3} )} ⎭ ⎬ ⎫

Ermittle den Partialbruch von

\frac{3 s + 3}{s ^{2} ( s + 2 ) ^{3}} : - \frac{3}{16 s} + \frac{3}{8 s ^{2}} + \frac{3}{16 ( s + 2 )} - \frac{3}{4 ( s + 2 ) ^{3}}

= L^{- 1} {- \frac{3}{16 s} + \frac{3}{8 s ^{2}} + \frac{3}{16 ( s + 2 )} - \frac{3}{4 ( s + 2 ) ^{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{3}{16 s}} + L^{- 1} {\frac{3}{8 s ^{2}}} + L^{- 1} {\frac{3}{16 ( s + 2 )}} - L^{- 1} {\frac{3}{4 ( s + 2 ) ^{3}}}

L^{- 1} {\frac{3}{16 s}} : \frac{3}{16} H (t)

L^{- 1} {\frac{3}{8 s ^{2}}} : \frac{3 t}{8}

L^{- 1} {\frac{3}{16 ( s + 2 )}} : \frac{3}{16} e^{- 2 t}

L^{- 1} {\frac{3}{4 ( s + 2 ) ^{3}}} : \frac{3 e ^{- 2 t} t ^{2}}{8}

= - \frac{3}{16} H (t) + \frac{3 t}{8} + \frac{3}{16} e^{- 2 t} - \frac{3 e ^{- 2 t} t ^{2}}{8}

\frac{d P}{d t} + P = 10

d er i v a t i v e f (x) = 8 - x^{3}

d er i v a t i v e cos (x^{- 7})

y^{^{'}} = \frac{3 x + 2 y}{3 x + 2 y + 2}

y d y = \frac{2 x}{( 1 + x ^{2} )} d x, y (0) = 2