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

Lösung

inverse laplace transformation

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

e^{t} + e^{t} t

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

= e^{t} + e^{t} t

\int x^{2} sin (2 x) d x

\frac{d}{d x} ((a x^{2} + b x + c) e^{\frac{- x}{6}})

x \to 7 lim (\frac{5 x}{x + 2})

t \to - 2 lim (t^{2} - 6 t - 16)

y^{^{''}} - 2 y^{^{'}} + y = e^{t} arctan (t)