inverselaplace (18)/((s+1)^2(s^2+4))

Lösung

inverse laplace transformation

\frac{18}{( s + 1 ) ^{2} ( s ^{2} + 4 )}

\frac{36}{25} e^{- t} + \frac{18}{5} e^{- t} t - \frac{36}{25} cos (2 t) - \frac{27 sin ( 2 t )}{25}

Schritte zur Lösung

L^{- 1} {\frac{18}{( s + 1 ) ^{2} ( s ^{2} + 4 )}}

Ermittle den Partialbruch von

\frac{18}{( s + 1 ) ^{2} ( s ^{2} + 4 )} : \frac{36}{25 ( s + 1 )} + \frac{18}{5 ( s + 1 ) ^{2}} + \frac{- 36 s - 54}{25 ( s ^{2} + 4 )}

= L^{- 1} {\frac{36}{25 ( s + 1 )} + \frac{18}{5 ( s + 1 ) ^{2}} + \frac{- 36 s - 54}{25 ( s ^{2} + 4 )}}

Schreibe um

= L^{- 1} {- \frac{36 s}{25 ( s ^{2} + 4 )} + \frac{36}{25 ( s + 1 )} + \frac{18}{5 ( s + 1 ) ^{2}} - \frac{54}{25 ( 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)}

= \frac{36}{25} L^{- 1} {\frac{1}{s + 1}} + \frac{18}{5} L^{- 1} {\frac{1}{( s + 1 ) ^{2}}} - \frac{36}{25} L^{- 1} {\frac{s}{s ^{2} + 4}} - \frac{54}{25} L^{- 1} {\frac{1}{s ^{2} + 4}}

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

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

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

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

= \frac{36}{25} e^{- t} + \frac{18}{5} e^{- t} t - \frac{36}{25} cos (2 t) - \frac{54}{25} \cdot \frac{1}{2} sin (2 t)

Fasse

\frac{36}{25} e^{- t} + \frac{18}{5} e^{- t} t - \frac{36}{25} cos (2 t) - \frac{54}{25} \frac{1}{2} sin (2 t)

zusammen:

\frac{36}{25} e^{- t} + \frac{18}{5} e^{- t} t - \frac{36}{25} cos (2 t) - \frac{27 sin ( 2 t )}{25}

= \frac{36}{25} e^{- t} + \frac{18}{5} e^{- t} t - \frac{36}{25} cos (2 t) - \frac{27 sin ( 2 t )}{25}

\frac{\partial}{\partial z} (y - z)

x \to 5 lim (\frac{1}{x ^{2} + 1})

d er i v a t i v e g (x) = x^{2} + 6 x^{3}

\frac{\partial}{\partial x} (7 x^{2} y + 6 x y^{3})

\frac{d y}{d x} = \frac{sin ( x )}{y}