inverselaplace (50s)/(s^2+25)

Lösung

inverse laplace transformation

\frac{50 s}{s ^{2} + 25}

50 cos (5 t)

Schritte zur Lösung

L^{- 1} {\frac{50 s}{s ^{2} + 25}}

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{50 s}{s ^{2} + 25} : F (s) = \frac{50}{s ^{2} + 25}

= \frac{d}{d t} (L^{- 1} {\frac{50}{s ^{2} + 25}}) + L^{- 1} {\frac{50}{s ^{2} + 25}} (0)

L^{- 1} {\frac{50}{s ^{2} + 25}} : 10 sin (5 t)

\frac{d}{d t} (10 sin (5 t)) = 50 cos (5 t)

Evaluiere

10 sin (5 t)

am Punkt

0 = 0

= 50 cos (5 t)

\frac{\partial}{\partial x} (e^{- x} y)

\int_{- 2}^{5} - x^{3} + 9 x d x

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

d er i v a t i v e f (x) = - 5 e^{- 2.5 x^{2}} x

x \to 0 lim (x^{3} + 2^{2} - x)