inverselaplace s/(2s^2+1)

Lösung

inverse laplace transformation

\frac{s}{2 s ^{2} + 1}

\frac{1}{2} cos (\frac{1}{2} t)

Schritte zur Lösung

L^{- 1} {\frac{s}{2 s ^{2} + 1}}

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

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

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

\frac{d}{d t} (\frac{1}{2} sin (\frac{1}{2} t)) = \frac{1}{2} cos (\frac{1}{2} t)

Evaluiere

\frac{1}{2} sin (\frac{1}{2} t)

am Punkt

0 = 0

= \frac{1}{2} cos (\frac{1}{2} t)

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

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

x \to 1 lim (\frac{x ^{3}}{x ^{2} - 1})

d er i v a t i v e f (x) = 6^{7 x^{2}}

\int - \frac{1}{x + 1} d x