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

Lösung

inverse laplace transformation

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

cos (2 t) \cdot 2

Schritte zur Lösung

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

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

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

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

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

Evaluiere

sin (2 t)

am Punkt

0 = 0

= cos (2 t) \cdot 2

d er i v a t i v e y = ln (8 - 3 x)

\int e^{- u} u d u

d er i v a t i v e x sin (\frac{8}{x})

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

\int \frac{2 x + 5}{( x - 4 ) ( x + 1 )} d x