inverselaplace 1/(s^2+4s+8)

Lösung

inverse laplace transformation

\frac{1}{s ^{2} + 4 s + 8}

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

Schritte zur Lösung

L^{- 1} {\frac{1}{s ^{2} + 4 s + 8}}

\frac{1}{s ^{2} + 4 s + 8} = \frac{1}{( s + 2 ) ^{2} + 4}

= L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 4}}

Wende inverse Transformationsregel an: if

L^{- 1} {F (s)} = f (t)

dann

L^{- 1} {F (s - a)} = e^{a t} f (t)

Für

\frac{1}{( s + 2 ) ^{2} + 4} : a = - 2, F (s) = \frac{1}{s ^{2} + 4}

= e^{- 2 t} L^{- 1} {\frac{1}{s ^{2} + 4}}

L^{- 1} {\frac{1}{s ^{2} + 4}} : \frac{1}{2} sin (2 t)

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

d er i v a t i v e g (x) = \frac{3}{6 x + 8}

\int x^{2} - 2 x - 3 d x

(1 - 2 y + x) \frac{d y}{d x} + y = 2 x - 1

\int sin (1 - 2 x) cos (1 + 2 x) d x

\frac{d}{d x} ((8 x^{2} - 5 x) (13 x^{2} + 4))