inverselaplace 5/(2(s-e)^2+50)

Lösung

inverse laplace transformation

\frac{5}{2 ( s - e ) ^{2} + 50}

e^{e t} \frac{1}{2} sin (5 t)

Schritte zur Lösung

L^{- 1} {\frac{5}{2 ( s - e ) ^{2} + 50}}

Wende inverse Transformationsregel an: if

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

dann

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

Für

\frac{5}{2 ( s - e ) ^{2} + 50} : a = e, F (s) = \frac{5}{2 s ^{2} + 50}

= e^{e t} L^{- 1} {\frac{5}{2 s ^{2} + 50}}

L^{- 1} {\frac{5}{2 s ^{2} + 50}} : \frac{1}{2} sin (5 t)

= e^{e t} \frac{1}{2} sin (5 t)

in v erse l a pl a ce \frac{25}{s ^{3} ( s ^{2} + 4 s + 5 )}

\int 3 x^{2} - x^{3} d x

y^{^{'}} = (4 x + y)^{2}

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

\frac{d}{d x} (sin (x + \frac{π}{2}))