laplacetransform te^{-t}cos(5t)

Lösung

laplace transformation

t e^{- t} cos (5 t)

- \frac{- s ^{2} - 2 s + 24}{( s ^{2} + 2 s + 26 ) ^{2}}

Schritte zur Lösung

L {e^{- t} t cos (5 t)}

Wende Laplace-Transformationstabelle an:

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

Für

t e^{- t} cos (5 t) : f (t) = e^{- t} cos (5 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- t} cos (5 t)})

L {e^{- t} cos (5 t)} : \frac{s + 1}{( s + 1 ) ^{2} + 25}

\frac{d}{d s} (\frac{s + 1}{( s + 1 ) ^{2} + 25}) = \frac{- s ^{2} - 2 s + 24}{( s ^{2} + 2 s + 26 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 2 s + 24}{( s ^{2} + 2 s + 26 ) ^{2}}

= - \frac{- s ^{2} - 2 s + 24}{( s ^{2} + 2 s + 26 ) ^{2}}

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

x \to 0 lim (sech^{2} (x))

\frac{\partial}{\partial r} (r sin (t))

\int cot^{5} (x) csc^{3} (x) d x

x \to 2 lim (3 x^{4})