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

Lösung

inverse laplace transformation

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

H (t) - t - cos (t) + sin (t)

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

Schreibe um

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

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}} - L^{- 1} {\frac{1}{s ^{2}}} - L^{- 1} {\frac{s}{s ^{2} + 1}} + L^{- 1} {\frac{1}{s ^{2} + 1}}

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

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

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

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

= H (t) - t - cos (t) + sin (t)

\frac{d}{d x} (\frac{4 x ^{2} + 6 x + 4}{x})

x \to 7 lim (\frac{∣ x - 7 ∣}{x - 7})

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

\int 2 x (x^{2} + 1)^{2} d x

\frac{\partial}{\partial x} (y ln (x^{4} + y^{3} + 1))