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Beliebt Rechnen >

inverselaplace (3*s)/(5s^2+125)

Lösung

inverse laplace transformation

\frac{3 \cdot s}{5 s ^{2} + 125}

Lösung

\frac{3}{5} cos (5 t)

Schritte zur Lösung

L^{- 1} {\frac{3 s}{5 s ^{2} + 125}}

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{3 s}{5 s ^{2} + 125} : F (s) = \frac{3}{5 s ^{2} + 125}

= \frac{d}{d t} (L^{- 1} {\frac{3}{5 s ^{2} + 125}}) + L^{- 1} {\frac{3}{5 s ^{2} + 125}} (0)

L^{- 1} {\frac{3}{5 s ^{2} + 125}} : \frac{3}{25} sin (5 t)

\frac{d}{d t} (\frac{3}{25} sin (5 t)) = \frac{3}{5} cos (5 t)

Evaluiere

\frac{3}{25} sin (5 t)

am Punkt

0 = 0

= \frac{3}{5} cos (5 t)

Beliebte Beispiele

y^{^{'}} + 2 x y = 10 x

derivative of ((x+8/(x-8))^5)

\frac{d}{d x} ((\frac{x + 8}{x - 8})^{5})

zusammenfügen 1+(a/(x^2))

j o in 1 + (\frac{a}{x ^{2}})

limit as x approaches 0 of (f(x))/(x^2)

x \to 0 lim (\frac{f ( x )}{x ^{2}})

laplacetransform e^{-2t}t^2

l a pl a ce t r an s f or m e^{- 2 t} t^{2}