inverselaplace 8/((2s+1)^3)

Lösung

inverse laplace transformation

\frac{8}{( 2 s + 1 ) ^{3}}

\frac{e ^{- \frac{t}{2}} t ^{2}}{2}

Schritte zur Lösung

L^{- 1} {\frac{8}{( 2 s + 1 ) ^{3}}}

Wende inverse Transformationsregel an: if

L^{- 1} {F (s)} = f (t)

dann

L^{- 1} {F (s - a)} = e^{a t} f (t)

Für

\frac{8}{( 2 s + 1 ) ^{3}} : a = - \frac{1}{2}, F (s) = \frac{8}{( 2 s ) ^{3}}

= e^{- \frac{1}{2} t} L^{- 1} {\frac{8}{( 2 s ) ^{3}}}

L^{- 1} {\frac{8}{( 2 s ) ^{3}}} : \frac{t ^{2}}{2}

= \frac{e ^{- \frac{t}{2}} t ^{2}}{2}

\frac{\partial}{\partial x} (2 x + 3 y - 4)

d er i v a t i v e f (x) \frac{2}{x}

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

144 y^{^{'''}} + 168 y^{^{''}} + 49 y^{^{'}} = 0

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