inverselaplace (2s)/(s^2+6)

Lösung

inverse laplace transformation

\frac{2 s}{s ^{2} + 6}

2 cos (6 t)

Schritte zur Lösung

L^{- 1} {\frac{2 s}{s ^{2} + 6}}

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} + 6} : F (s) = \frac{2}{s ^{2} + 6}

= \frac{d}{d t} (L^{- 1} {\frac{2}{s ^{2} + 6}}) + L^{- 1} {\frac{2}{s ^{2} + 6}} (0)

L^{- 1} {\frac{2}{s ^{2} + 6}} : \frac{2}{3} sin (6 t)

\frac{d}{d t} (\frac{2}{3} sin (6 t)) = 2 cos (6 t)

Evaluiere

\frac{2}{3} sin (6 t)

am Punkt

0 = 0

= 2 cos (6 t)

\int (\frac{5}{x} - \frac{4}{x ^{5}}) d x

\frac{d y}{d t} = \frac{8}{7} y

\frac{\partial}{\partial x} ((2 x + 3) (y - 2))

t an g e n t y = x^{3} - 3 x + 1

\frac{d}{d x} (lo g_{3} (sin (x^{2})))