inverselaplace (e^{-4s})/((s+2)^3)

Lösung

inverse laplace transformation

\frac{e ^{- 4 s}}{( s + 2 ) ^{3}}

\frac{e ^{- 2 (t - 4)} H ( t - 4 ) ( t - 4 ) ^{2}}{2}

Schritte zur Lösung

L^{- 1} {\frac{e ^{- 4 s}}{( s + 2 ) ^{3}}}

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{1}{( s + 2 ) ^{3}}} (t - 4)

L^{- 1} {\frac{1}{( s + 2 ) ^{3}}} : \frac{e ^{- 2 t} t ^{2}}{2}

= H (t - 4) \frac{e ^{- 2 (t - 4)} ( t - 4 ) ^{2}}{2}

= \frac{e ^{- 2 (t - 4)} H ( t - 4 ) ( t - 4 ) ^{2}}{2}

x \to 0 + lim (\frac{2 x}{x ^{2}})

\frac{d}{d x} (\frac{5 - 4 x ^{2} + x ^{5}}{x ^{3}})

\int \frac{36}{36 + e ^{x}} d x

x \to 4 lim (\frac{3 x}{x - 4})

\frac{\partial}{\partial x} (e^{x y^{2}})