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Beliebt Rechnen >

inverselaplace (5se^{-s})/(2s^2+32)

Lösung

inverse laplace transformation

\frac{5 s e ^{- s}}{2 s ^{2} + 32}

Lösung

H (t - 1) \frac{5}{2} cos (4 (t - 1))

Schritte zur Lösung

L^{- 1} {\frac{5 e ^{- s} s}{2 s ^{2} + 32}}

Wende inverse Transformationsregel an: if

L^{- 1} {F (s)} = f (t)

dann

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Wo

H (t)

Heaviside (Schritt) Funktion ist

= H (t - 1) L^{- 1} {\frac{5 s}{2 s ^{2} + 32}} (t - 1)

L^{- 1} {\frac{5 s}{2 s ^{2} + 32}} : \frac{5}{2} cos (4 t)

= H (t - 1) \frac{5}{2} cos (4 (t - 1))

Beliebte Beispiele

integral of (x+1)/((x+2)^3)

\int \frac{x + 1}{( x + 2 ) ^{3}} d x

9y^{''}-12y^'+4y=e^{-3x}

9 y^{^{''}} - 12 y^{^{'}} + 4 y = e^{- 3 x}

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

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

derivative of x^2+8

\frac{d}{d x} (x^{2} + 8)

(\partial)/(\partial y)(sqrt(x)e^{xy})

\frac{\partial}{\partial y} (x e^{x y})