inverselaplace 1/((s+3)^4)

Lösung

inverse laplace transformation

\frac{1}{( s + 3 ) ^{4}}

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

Schritte zur Lösung

L^{- 1} {\frac{1}{( s + 3 ) ^{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 + 3 ) ^{4}} : a = - 3, F (s) = \frac{1}{s ^{4}}

= e^{- 3 t} L^{- 1} {\frac{1}{s ^{4}}}

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

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

\int 36 x (2 x^{2} - 5)^{8} d x

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

\frac{d x}{d t} = \frac{( e ^{t} - e ^{- t} )}{3 + x}, x (0) = 0

d er i v a t i v e y = x e^{2 x}

\frac{\partial}{\partial x} (ln (x^{2} + y^{2} + 6))