inverselaplace (-5s)/((s^2+1)^2)

Lösung

inverse laplace transformation

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

- \frac{5 t sin ( t )}{2}

Schritte zur Lösung

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

Wende inverse Transformationsregel an: if

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

dann

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

Für

\frac{- 5 s}{( s ^{2} + 1 ) ^{2}} : F (s) = \frac{- 5}{( s ^{2} + 1 ) ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{- 5}{( s ^{2} + 1 ) ^{2}}}) + L^{- 1} {\frac{- 5}{( s ^{2} + 1 ) ^{2}}} (0)

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

\frac{d}{d t} (- \frac{5}{2} (sin (t) - t cos (t))) = - \frac{5 t sin ( t )}{2}

Evaluiere

- \frac{5}{2} (sin (t) - t cos (t))

am Punkt

0 = 0

= - \frac{5 t sin ( t )}{2}

x \to 1 lim (\frac{3}{x - 1})

\frac{\partial}{\partial x} (8 - x^{2})

\int_{1}^{100} \frac{21}{x ^{3}} d x

d er i v a t i v e ln (x + x^{2} - 7)

2 y^{^{''}} - 6 y^{^{'}} + 5 y = 0