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Beliebt Rechnen >

inverselaplace (4s)/((s^2+16))

Lösung

inverse laplace transformation

\frac{4 s}{( s ^{2} + 16 )}

Lösung

cos (4 t) \cdot 4

Schritte zur Lösung

L^{- 1} {\frac{4 s}{( s ^{2} + 16 )}}

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{4 s}{s ^{2} + 16} : F (s) = \frac{4}{s ^{2} + 16}

= \frac{d}{d t} (L^{- 1} {\frac{4}{s ^{2} + 16}}) + L^{- 1} {\frac{4}{s ^{2} + 16}} (0)

L^{- 1} {\frac{4}{s ^{2} + 16}} : sin (4 t)

\frac{d}{d t} (sin (4 t)) = cos (4 t) \cdot 4

Evaluiere

sin (4 t)

am Punkt

0 = 0

= cos (4 t) \cdot 4

Beliebte Beispiele

limit as x approaches 0 of 18-x^2

x \to 0 lim (18 - x^{2})

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

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

limit as x approaches infinity of 8/(x!)

x \to \infty lim (\frac{8}{x !})

derivative y=ln(x+3)

d er i v a t i v e y = ln (x + 3)

integral from 2 to 6 of x^2+1

\int_{2}^{6} x^{2} + 1 d x