inverselaplace x/(x^2+0.0184)

Lösung

inverse laplace transformation

\frac{x}{x ^{2} + 0.0184}

cos (0.13564 \dots t)

Schritte zur Lösung

L^{- 1} {\frac{x}{x ^{2} + 0.0184}}

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{x}{x ^{2} + 0.0184} : F (x) = \frac{1}{x ^{2} + 0.0184}

= \frac{d}{d t} (L^{- 1} {\frac{1}{x ^{2} + 0.0184}}) + L^{- 1} {\frac{1}{x ^{2} + 0.0184}} (0)

L^{- 1} {\frac{1}{x ^{2} + 0.0184}} : \frac{1}{0.13564 \dots} sin (0.13564 \dots t)

\frac{d}{d t} (\frac{1}{0.13564 \dots} sin (0.13564 \dots t)) = cos (0.13564 \dots t)

Evaluiere

\frac{1}{0.13564 \dots} sin (0.13564 \dots t)

am Punkt

0 = 0

= cos (0.13564 \dots t)

x \to - 4 lim (8 x + 1)

\int x^{4} e^{8 x^{5}} d x

d er i v a t i v e f (x) = e^{3}

t an g e n t f (x) = x^{5}, (1, 1)

\frac{d}{d x} (sec (x) tan (y))