inverselaplace 1/((s-1)^2*(s+5))

Lösung

inverse laplace transformation

\frac{1}{( s - 1 ) ^{2} \cdot ( s + 5 )}

- \frac{1}{36} e^{t} + \frac{e ^{t} t}{6} + \frac{1}{36} e^{- 5 t}

Schritte zur Lösung

L^{- 1} {\frac{1}{( s - 1 ) ^{2} ( s + 5 )}}

Ermittle den Partialbruch von

\frac{1}{( s - 1 ) ^{2} ( s + 5 )} : - \frac{1}{36 ( s - 1 )} + \frac{1}{6 ( s - 1 ) ^{2}} + \frac{1}{36 ( s + 5 )}

= L^{- 1} {- \frac{1}{36 ( s - 1 )} + \frac{1}{6 ( s - 1 ) ^{2}} + \frac{1}{36 ( s + 5 )}}

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}{36 ( s - 1 )}} + L^{- 1} {\frac{1}{6 ( s - 1 ) ^{2}}} + L^{- 1} {\frac{1}{36 ( s + 5 )}}

L^{- 1} {\frac{1}{36 ( s - 1 )}} : \frac{1}{36} e^{t}

L^{- 1} {\frac{1}{6 ( s - 1 ) ^{2}}} : \frac{e ^{t} t}{6}

L^{- 1} {\frac{1}{36 ( s + 5 )}} : \frac{1}{36} e^{- 5 t}

= - \frac{1}{36} e^{t} + \frac{e ^{t} t}{6} + \frac{1}{36} e^{- 5 t}

y^{^{''}} + 100 y = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 cos (x)

t y^{^{'}} + (t + 1) y = t, y (ln (2)) = 1

\int (6 sin (2 x) + sec^{2} (x)) d x

\int (125 x^{9}) d x