inverselaplace (s^2+12)/(s^4-16)

Lösung

inverse laplace transformation

\frac{s ^{2} + 12}{s ^{4} - 16}

- \frac{1}{2} sin (2 t) - \frac{1}{2} e^{- 2 t} + \frac{1}{2} e^{2 t}

Schritte zur Lösung

L^{- 1} {\frac{s ^{2} + 12}{s ^{4} - 16}}

Ermittle den Partialbruch von

\frac{s ^{2} + 12}{s ^{4} - 16} : - \frac{1}{s ^{2} + 4} - \frac{1}{2 ( s + 2 )} + \frac{1}{2 ( s - 2 )}

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

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

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

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

= - \frac{1}{2} sin (2 t) - \frac{1}{2} e^{- 2 t} + \frac{1}{2} e^{2 t}

\int \frac{29 x ^{2}}{x ^{2} - 4} d x

\int \frac{1}{8} t d t

\int \frac{( x )}{( 4 x ^{(2)} - 12 x + 25 )} d x

x \to 4 lim (\frac{x - 5}{( x - 4 ) ^{2}})

x y^{^{'}} + 2 y = - 4 sin (x)