inverselaplace (2s+2)/(s+10)

Lösung

inverse laplace transformation

\frac{2 s + 2}{s + 10}

2 δ (t) - 18 e^{- 10 t}

Schritte zur Lösung

L^{- 1} {\frac{2 s + 2}{s + 10}}

Ermittle den Partialbruch von

\frac{2 s + 2}{s + 10} : 2 - \frac{18}{s + 10}

= L^{- 1} {2 - \frac{18}{s + 10}}

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} {2} - L^{- 1} {\frac{18}{s + 10}}

L^{- 1} {2} : 2 δ (t)

L^{- 1} {\frac{18}{s + 10}} : 18 e^{- 10 t}

= 2 δ (t) - 18 e^{- 10 t}

\frac{\partial}{\partial y} (3 x^{2} - 2 y^{3} + 6 x y^{2} - 8)

t an g e n t 2 cos (x) + cos^{2} (x)

x \to - 2 - lim (\frac{1}{x + 2})

f (x) = 9 ln (x)

\frac{\partial}{\partial x} (\frac{2 x y ^{2}}{1 + x ^{2} y ^{2}})