inverselaplace (2(s-1)e^{-2s})/(s^2-2s+2)

Lösung

inverse laplace transformation

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

H (t - 2) \cdot 2 e^{t - 2} cos (t - 2)

Schritte zur Lösung

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

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

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

= H (t - 2) \cdot 2 e^{t - 2} cos (t - 2)

t an g e n t y = x^{2} + 2, (- 4, 18)

a re a y = 7 x, y = \frac{7}{4} x, y = 78 - x^{2}

\int - \frac{1}{50} d x

\int_{0}^{1} \frac{5 x ^{4}}{1 + x ^{5}} d x

\int (6 x - 2)^{- 4} d x