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

Lösung

inverse laplace transformation

\frac{1}{( s + 1 ) ^{4}}

\frac{e ^{- t} t ^{3}}{6}

Schritte zur Lösung

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

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

L^{- 1} {\frac{1}{s ^{4}}} : \frac{t ^{3}}{6}

= \frac{e ^{- t} t ^{3}}{6}

\frac{\partial}{\partial x} (y e^{6 x})

d er i v a t i v e f (x) = (3) e^{- 5 x}

\int_{0}^{2} 4 t^{7} d t

\int (\frac{x ^{2} - 1}{x ^{\frac{3}{2}}}) d x

x \to 0 lim (\frac{5 ^{x} - 1 0 ^{x}}{x})