inverselaplace (4s)/(s^2+9)

Lösung

inverse laplace transformation

\frac{4 s}{s ^{2} + 9}

4 cos (3 t)

Schritte zur Lösung

L^{- 1} {\frac{4 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{4 s}{s ^{2} + 9} : F (s) = \frac{4}{s ^{2} + 9}

= \frac{d}{d t} (L^{- 1} {\frac{4}{s ^{2} + 9}}) + L^{- 1} {\frac{4}{s ^{2} + 9}} (0)

L^{- 1} {\frac{4}{s ^{2} + 9}} : \frac{4}{3} sin (3 t)

\frac{d}{d t} (\frac{4}{3} sin (3 t)) = 4 cos (3 t)

Evaluiere

\frac{4}{3} sin (3 t)

am Punkt

0 = 0

= 4 cos (3 t)

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 7 y^{3}))

\frac{\partial}{\partial x} (ln (x e^{- 6 x}))

x \to + 0 lim ((\frac{1}{x ^{2}})^{2})

n \to - \infty lim (e^{n})

n = 0 \sum \infty e^{2 n}