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Beliebt Rechnen >

inverselaplace s/(s^2+2)

Lösung

inverse laplace transformation

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

Lösung

cos (2 t)

Schritte zur Lösung

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

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

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

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

Evaluiere

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

am Punkt

0 = 0

= cos (2 t)

Beliebte Beispiele

tangent f(x)=4x-32sqrt(x),\at x=4

t an g e n t f (x) = 4 x - 32 x, at x = 4

integral of sqrt(1+t^2)

\int 1 + t^{2} d t

limit as x approaches-1 of 1/(x^2)

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

tangent-x^3(-2.8)

t an g e n t - x^{3} (- 2.8)

integral of (x^2)/(e^{2x)}

\int \frac{x ^{2}}{e ^{2 x}} d x