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Beliebt Rechnen >

inverselaplace s/(s^2+9)

Lösung

inverse laplace transformation

\frac{s}{s ^{2} + 9}

Lösung

cos (3 t)

Schritte zur Lösung

L^{- 1} {\frac{s}{s ^{2} + 9}}

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} + 9} : F (s) = \frac{1}{s ^{2} + 9}

= \frac{d}{d t} (L^{- 1} {\frac{1}{s ^{2} + 9}}) + L^{- 1} {\frac{1}{s ^{2} + 9}} (0)

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

\frac{d}{d t} (\frac{1}{3} sin (3 t)) = cos (3 t)

Evaluiere

\frac{1}{3} sin (3 t)

am Punkt

0 = 0

= cos (3 t)

Beliebte Beispiele

derivative of (2x-9/(sqrt(x)))

\frac{d}{d x} (\frac{2 x - 9}{x})

integral of 3x^3e^{-x^2}

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

derivative y= 1/(7t+3)

d er i v a t i v e y = \frac{1}{7 t + 3}

integral from 0 to 15 of x/(sqrt(x+1))

\int_{0}^{15} \frac{x}{x + 1} d x

fläche-x^3+3,y=x,x=-1,x=1

a re a - x^{3} + 3, y = x, x = - 1, x = 1