laplacetransform te^{-0.003t}cos(10t)

Lösung

laplace transformation

t e^{- 0.003 t} cos (10 t)

- \frac{- s ^{2} - 0.006 s + 99.999991}{( s ^{2} + 0.006 s + 100.000009 ) ^{2}}

Schritte zur Lösung

L {e^{- 0.003 t} t cos (10 t)}

Wende Laplace-Transformationstabelle an:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

Für

t e^{- 0.003 t} cos (10 t) : f (t) = e^{- 0.003 t} cos (10 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 0.003 t} cos (10 t)})

L {e^{- 0.003 t} cos (10 t)} : \frac{s + 0.003}{( s + 0.003 ) ^{2} + 100}

\frac{d}{d s} (\frac{s + 0.003}{( s + 0.003 ) ^{2} + 100}) = \frac{- s ^{2} - 0.006 s + 99.999991}{( s ^{2} + 0.006 s + 100.000009 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 0.006 s + 99.999991}{( s ^{2} + 0.006 s + 100.000009 ) ^{2}}

= - \frac{- s ^{2} - 0.006 s + 99.999991}{( s ^{2} + 0.006 s + 100.000009 ) ^{2}}

\int_{0}^{1} \frac{1}{( x + 1 ) ( x ^{2} + 1 )} d x

t an g e n t f (x) = x^{4} + 8 x^{3} + 2 x + 2, at x = 1

\frac{\partial}{\partial x} (\frac{12 x}{x + y})

d er i v a t i v e f (x) = (- x^{3} + x^{2} + x) cos (x)

\int x^{2} - 2 x + 1 d x