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

Lösung

inverse laplace transformation

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

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

Schritte zur Lösung

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

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

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

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

\int e^{l n (2 x)} d x

x \to 0 lim (\frac{5 x cot ( 3 x )}{2 sec ( x )})

d er i v a t i v e f (t) = 2 t sin (π t)

\int \frac{4 x ^{2} + 21 x + 54}{x ^{2} + 6 x + 13} d x

x \to 8 lim (6 x^{2} - 9 x + 1)