inverselaplace (ce^{-as})/(s^2)

Lösung

inverse laplace transformation

\frac{c e ^{- a s}}{s ^{2}}

H (t - a) c (t - a)

Schritte zur Lösung

L^{- 1} {\frac{c e ^{- a s}}{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 - a) L^{- 1} {\frac{c}{s ^{2}}} (t - a)

L^{- 1} {\frac{c}{s ^{2}}} : c t

= H (t - a) c (t - a)

n = 2 \sum \infty e^{5 - 6 n}

\frac{d ^{3}}{d x ^{3}} (2 x)

\int_{1}^{3} 9 arctan (1 x) d x

d er i v a t i v e f (x) = x^{3} 1 - x^{2}

a re a y = 1 + x, y = \frac{( x + 3 )}{3}