inverselaplace (4se^{-4s})/(2s^2+50)

Lösung

inverse laplace transformation

\frac{4 s e ^{- 4 s}}{2 s ^{2} + 50}

H (t - 4) \cdot 2 cos (5 (t - 4))

Schritte zur Lösung

L^{- 1} {\frac{4 e ^{- 4 s} s}{2 s ^{2} + 50}}

Wende inverse Transformationsregel an: if

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

dann

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

H (t)

Heaviside (Schritt) Funktion ist

= H (t - 4) L^{- 1} {\frac{4 s}{2 s ^{2} + 50}} (t - 4)

L^{- 1} {\frac{4 s}{2 s ^{2} + 50}} : 2 cos (5 t)

= H (t - 4) \cdot 2 cos (5 (t - 4))

in v erse l a pl a ce \frac{s + 1}{s ^{2} + s + 1}

y^{^{'}} = y (2 x - 5)

\int 24 (1 - x)^{5} d x

\frac{d}{d x} (A (2 x e^{2 x} + e^{2 x} \cdot 2 x^{2}))

d er i v a t i v e f (x) = x (3 x^{4} - x)