inverselaplace {(10s)/(s^2+16)}

Lösung

inverse laplace transformation

{\frac{10 s}{s ^{2} + 16}}

10 cos (4 t)

Schritte zur Lösung

L^{- 1} {\frac{10 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{10 s}{s ^{2} + 16} : F (s) = \frac{10}{s ^{2} + 16}

= \frac{d}{d t} (L^{- 1} {\frac{10}{s ^{2} + 16}}) + L^{- 1} {\frac{10}{s ^{2} + 16}} (0)

L^{- 1} {\frac{10}{s ^{2} + 16}} : \frac{5}{2} sin (4 t)

\frac{d}{d t} (\frac{5}{2} sin (4 t)) = 10 cos (4 t)

Evaluiere

\frac{5}{2} sin (4 t)

am Punkt

0 = 0

= 10 cos (4 t)

\frac{d}{d x} (\frac{4 x}{5})

f^{^{'}} (x) = 12 x^{3} - 63 x^{2} - 72 x + 4

d er i v a t i v e y = \frac{f ( x )}{x ^{6}}

\int_{0}^{1} x - x^{k} d x

d er i v a t i v e 3