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Beliebt Rechnen >

inverselaplace (2s)/(s^2+9)

Lösung

inverse laplace transformation

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

Lösung

2 cos (3 t)

Schritte zur Lösung

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

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

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

\frac{d}{d t} (\frac{2}{3} sin (3 t)) = 2 cos (3 t)

Evaluiere

\frac{2}{3} sin (3 t)

am Punkt

0 = 0

= 2 cos (3 t)

Beliebte Beispiele

fläche y=2sqrt(x),y= x/2

limit as x approaches-6 of (x^2+8)/(x+6)

derivative of (cos(ln(x))/x)

limit as x approaches 3 of 3x-3ax+2a