inverselaplace (x^2+1)/(x^2-x-6)

Lösung

inverse laplace transformation

\frac{x ^{2} + 1}{x ^{2} - x - 6}

δ (t) - e^{- 2 t} + 2 e^{3 t}

Schritte zur Lösung

L^{- 1} {\frac{x ^{2} + 1}{x ^{2} - x - 6}}

Ermittle den Partialbruch von

\frac{x ^{2} + 1}{x ^{2} - x - 6} : 1 - \frac{1}{x + 2} + \frac{2}{x - 3}

= L^{- 1} {1 - \frac{1}{x + 2} + \frac{2}{x - 3}}

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} {1} - L^{- 1} {\frac{1}{x + 2}} + L^{- 1} {\frac{2}{x - 3}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{1}{x + 2}} : e^{- 2 t}

L^{- 1} {\frac{2}{x - 3}} : 2 e^{3 t}

= δ (t) - e^{- 2 t} + 2 e^{3 t}

d er i v a t i v e f (x) = 8 sin^{6} (x)

x \to 4 + lim (\frac{1}{∣ 4 - x ∣})

s l o p e y = x^{3}, at x = 5

\frac{d}{d x} (3 x^{4} - 2 x^{3} - 12 x^{2} + 18 x + 15)

\frac{d}{d x} (\frac{1}{ln ( x ^{2} )})